The recording tableaux in the quantum
Littlewood-Richardson map,
the orthogonal transpose symmetry map, and
the computation of \(\mathfrak{k}\)-highest weight tableaux
March 17, 2026
Recently Watanabe has given an algorithm to compute a bijection, that he calls (quantum) Littlewood-Richardson (LR) map (or quantum LR rule of type AII), between semi-standard Young tableaux of shape a partition with at most \(2n\) parts and pairs of tableaux consisting of a symplectic tableau with shape a partition with at most \(n\) parts, and a recording tableau of skew-shape given by the two previous shapes. The recording tableaux in that algorithm are shown to be equinumerous to Littlewood-Richardson-Sundaram tableaux whose injectivity is shown combinatorially while the surjectivity is concluded via representation theory of a quantum symmetric pair of type AII\(_{2n-1}\). Henceforth, the algorithm to compute the quantum LR map provides a new branching model for the branching multiplicities from \(GL_{2n}( \mathbb{C})\) to \(Sp_{2n}( \mathbb{C})\). Here, as morally suggested by Watanabe, one provides a combinatorial proof for the surjectivity of the quantum LR map which in turn exhibits the restriction of the LR orthogonal transpose symmetry map to LR-Sundaram tableaux. The surjectivity is exhibited via the reverse Schensted insertion on the quantum recording tableaux, ruled by the slack data, followed with the inverse of the reduction map on the bumped entries that we explicitly compute for certain families of symplectic columns. As an application of the explicit surjectivity and henceforth of the inverse of the quantum LR map, we compute and characterize a family of \(\mathfrak{k}\)- highest weight semi-standard tableaux in the recent proof of the Naito-Sagaki conjecture using the Watanabe’s branching rule based on the crystal basis theory for \(\imath\)quantum groups of type AII\(_{2n-1}\). More precisely, that family of tableaux is generated either by quantum record tableaux with one vertical strip or with \(1\)-\(0\)-slack sequences. For a given \(\mathfrak{k}\)-highest weight symplectic tableau, they satisfy certain linear inequalities between multiplicities of columns in that symplectic tableau and multiplicities of the columns obtained by the inversion of the reduce map on the bumped numbers from the given \(\mathfrak{k}\)-highest weight symplectic tableau.
The quantum Littlewood-Richardson map by Watanabe [1] for the pair \((GL_{2n}(\mathbb{C}),Sp_{2n}(\mathbb{C}))\) can be seen as a generalization of the branching rule, known as the Littlewood-Richardson rule, for the pair \((GL_m(\mathbb{C})\times GL_m(\mathbb{C}), GL_m(\mathbb{C}))\). More precisely, a generalization of the G.P. Thomas [2] bijection providing the Littlewood-Richardson rule [3] \[\begin{align} \label{tho0}SST_m(\mu)\times SST_m(\nu) \overset{\sim}\rightarrow\bigsqcup_{\begin{smallmatrix}\lambda\in Par_{\le m}\\ T\in LR(\lambda/\mu,\nu)\end{smallmatrix}}SST_{m}(\lambda) \times \{T\}, \end{align}\tag{1}\] to a bijection providing the quantum Littlewood-Richardson rule [1] \[\begin{align} \label{wat0}\displaystyle SST_{2n}(\lambda) \overset{\sim}\rightarrow\bigsqcup_{\begin{smallmatrix}\mu\in Par_{\le n}\\ \mu\subseteq\lambda\\ Q\in Rec_{2n}(\lambda/\mu)\end{smallmatrix}}SpT_{2n}(\mu) \times \{Q\}, \end{align}\tag{2}\] where \(SpT_{2n}(\mu)\) denotes the set of symplectic semi-standard tableaux of shape \(\mu\), with entries in \([1,2n]\), and the sets of Littlewood-Richardson-Sundaram tableaux \(LRS_{2n}(\lambda/\mu) \subseteq LR_{2n}(\lambda/\mu)\) [4] and of recording tableaux \(Rec_{2n}(\lambda/\mu)\) in 2 satisfy \(LRS_{2n}(\lambda/\mu)\overset{\sim}\longrightarrow Rec_{2n}(\lambda/\mu)\) via a natural bijection.
For the sake of comparison, the G.P. Thomas bijection 1 asserts \[\begin{align} &SST_m(\mu)\times SST_m(\nu)\overset{\sim}\longrightarrow \bigsqcup_{\begin{smallmatrix}\lambda\in Par_{\le m}\\{} \end{smallmatrix}} SST_m(\lambda)\times LR(\lambda/\mu,\nu). \end{align}\] where the recording tableau of the Schensted column insertion of a tableau pair in \(SST_m(\mu)\times SST_m(\nu)\) is stored in the form of an LR tableau in \(LR(\lambda/\mu,\nu)\) for some partition \(\lambda\). Very importantly is that this bijection lifts to a \(\mathfrak{gl}_m\)-crystal isomorphism [5] as in 3 by showing how the tensor product of two \(Gl_m\)-irreducible representations decomposes into irreducible \(GL_m\)-representations, while giving simultaneously the crystal version of the original Littlewood-Richardson (LR) rule [6], the Berenstein-Gelfand-Zelevinsky LR rules [7]–[9] on left and right Gelfand-Tsetlin patterns [10], and notably the integer Knutson-Tao hives [11], [12] as the interlocking of those Gelfand-Tsetlin patterns [13], [14], \[\begin{align} \label{crystalthomas} B(\mu,m)\otimes B(\nu,m)\simeq \bigoplus_{\begin{smallmatrix}T\in LR(\lambda/\mu,\nu)\\ \lambda\in Par_{\le m}\end{smallmatrix}} B(\lambda,m)\times \{T\}\simeq \bigoplus_{\begin{smallmatrix} \lambda\in Par_{\le m}\end{smallmatrix}} B(\lambda,m)^{c_{\mu,\nu}^\lambda}. \end{align}\tag{3}\] Notably, the highest and lowest weights of the connected component \(B(\lambda,m)\times \{T\}\), determined by the recording LR tableau \(T\), exhibit the right respectively left companions of the LR tableau \(T\), and their interlocking the integer hive of boundary \((\lambda,\mu,\nu)\). The multiplicity \(c_{\mu,\nu}^\lambda\) of the crystal \(B(\lambda,m)\) in this decomposition shows that \(LR(\lambda/\mu,\nu)\) is equinumerous to the right respectively left companions in the form of Gelfand-Tsetlin patterns in the Berenstein-Gelfand-Zelevinsky LR rules as well as the number of integer hives with boundary \((\mu,\nu,\lambda)\) (see [15] for details).
It remains to say that beyond the previous information packed in the crystal isomorphism 3 , \(c_{\mu,\nu}^\lambda\) is also the number of integral points of a hive polytope with integral boundary \((\mu,\nu,\lambda)\) (that is, the number of integral hives describing different coupling of three irreducible representations), a Knutson-Tao new description of the Berenstein-Zelevinski polytopes (where integral points are Berenstein-Zelevinsky patterns) [16]. It is then immediate that LR coefficients can be modelled via the Ehrhart (quasi)polynomial [17] of a rational polytope whose polynomiality has been proved by Derksen-Weyman and Rassart in [18], [19]. (For advances in the Ehrhart theory model for \(c_{\mu,\nu}^\lambda\), the reader is referred, for instance, to [20] and [21] and references therein.)
From [22]–[25] and the Henriques-Kamnitzer \(\mathfrak{gl}_m\)-crystal commuter [26], [27], it turns out that the decomposition 3 has a refinement to include LR-Sundaram (LRS)-tableaux [4], [28] and their companion pairs whose interlocking exhibits a flagged hive as in [24]. For a fixed \(n\in \mathbb{N}\), let \(m=2n\), \(\mu\in Par_{\le n}\) a partition with at most \(n\) parts, and \(\nu, \lambda \in Par_{\le 2n}\) partitions with at most \(2n\) parts such that \(\nu\) has even length columns. Let \(LRS_{2n}(\lambda/\mu,\nu)\subseteq LR(\lambda/\mu,\nu)\) be the set of LR-Sundaram tableaux (or LR symplectic tableaux \(LRT^{Sp}_{2n}(\lambda/\mu)\) in [1]) of shape \(\lambda/\mu\) and weight \(\nu\), and let \(SpT_{2n}(\mu)\subseteq SST_{2n}(\mu)\) be the set of symplectic semi-standard tableaux of shape \(\mu\) [1], in natural bijection with the symplectic Geland-Testlin patterns [25]. Then 3 refines as
\[\begin{align} &B(\mu,2n)\otimes B(\nu,2n)\simeq \\ &\bigoplus_{\begin{smallmatrix}T\in LRS(\lambda/\mu,\nu)\\ G^{sp}_\mu \in SpT_{2n}(\mu)\\ \tiny left companion of T\\ \lambda\in Par_{\le 2n}\end{smallmatrix}} B( G^{sp}_\mu\otimes Y(rev\nu) )\times \{T\} \bigsqcup\bigoplus_{\begin{smallmatrix}T\in LR(\lambda/\mu,\nu)\setminus LRS(\lambda/\mu,\nu)\\ G_\mu\in SST_{2n}\\\tiny left companion of T\\ \lambda\in Par_{\le 2n}\end{smallmatrix}} B( G_\mu\otimes Y(\mathrm rev\,\nu) )\times \{T\}\\ &\simeq \bigoplus_{\begin{smallmatrix} G^{Sp}_\mu\otimes Y(rev\nu)\simeq Y(rev\lambda)\\ \lambda\in Par_{\le 2n}\end{smallmatrix}} B(\lambda,2n)^{spc_{\mu,\nu}^\lambda}\bigsqcup\bigoplus_{\begin{smallmatrix} G_\mu\otimes Y(rev\nu)\simeq Y(rev\,\lambda)\\ G_\mu\in SST_{2n}(\lambda)\setminus SpT_{2n}(\mu)\\ \lambda\in Par_{\le 2n}\end{smallmatrix}} B(\lambda,2n)^{c_{\mu,\nu}^\lambda} \end{align}\] where \(spc_{\mu,\nu}^\lambda\) is the cardinality of \(LRS_{2n}(\lambda/\mu,\nu)\) or the number of symplectic semi-standard tableaux \(G^{Sp}_\mu\in SpT_{2n}(\mu)\) with weight \(rev(\lambda-\nu)\) such that \(G^{sp}_\mu\otimes Y(rev\nu)\simeq Y(rev\lambda)\).
Let \(\lambda \in Par_{\le 2n}\) be a partition with at most \(2n\) parts. The Littlewood-Richardson map \(LR^{AII}\) [1] is an one-to-one assignment of a semi-standard tableau \(T\) of shape \(\lambda\in Par_{\le 2n}\) to a pair \((P^{II}(T), Q^{II}(T))\) for some shape \(\mu\in Par_{\le n}\), consisting of a symplectic tableau in \(SpT_{2n}(\mu)\) respectively a recording tableau of skew shape \(\lambda/\mu\) in \(Rec_{2n}(\lambda/\mu)\), \[\begin{align} \label{lrII} LR^{AII}: SST_{2n}(\lambda) \overset{\sim}\longrightarrow \bigsqcup_{\begin{smallmatrix}\mu\in Par_{\le n}\\ \mu\subseteq\lambda\end{smallmatrix}}SpT_{2n}(\mu) \times Rec_{2n}(\lambda/\mu). \end{align}\tag{4}\] Furthermore, for each \(\mu \in Par_{\le n}\) with \(\mu\subset \lambda\), the set of recording tableaux \(Rec_{2n}(\lambda/\mu)\) in the Littlewood-Richardson map \(LR^{AII}\) are in natural bijection with the set of LR-Sundaram tableaux \(LRS_{2n}(\lambda/\mu)\), \[\begin{align} \label{record}Rec_{2n}(\lambda/\mu) \overset{\sim}\longrightarrow LRS_{2n}(\lambda/\mu). \end{align}\tag{5}\] Henceforth, the algorithm defining the map \(LR^{AII}\) gives rise to a bijection between the set of semi-standard tableaux of given shape, say \(\lambda\), and the disjoint union of several copies of the sets of symplectic tableaux for each shape \(\mu\subseteq \lambda\) with at most \(n\) parts, where the multiplicity \(c_\mu^\lambda\) of each shape \(\mu\) equals to the cardinality of \(LRS(\lambda/\mu)\), the number of LR-Sundaram tableaux of shape \(\lambda/\mu\), \[\begin{align} \label{lrsII} SST_{2n}(\lambda) \overset{\sim}\longrightarrow \bigsqcup_{\begin{smallmatrix}\mu\in Par_{\le n}\\ \mu\subseteq\lambda\\ Q\in Rec_{2n}(\lambda/\mu)\end{smallmatrix}}SpT_{2n}(\mu) \times \{Q\}. \end{align}\tag{6}\] Thereby, the algorithm defining the map \(LR^{AII}\) gives a new branching rule for the pair \((Gl_{2n}(\mathbb{C}), Sp_{2n}(\mathbb{C})\). It is then demanding to have a complete combinatorial proof of this new branching rule. As suggested in [1] the bijections 4 and 5 might be proved in the realm of combinatorics.
The injectivity of the Littlewood-Richardson map 4 and the injectivity of the bijection 5 are proved combinatorially in [1] and respectively in [1] by showing, \[Rec_{2n}(\lambda/\mu)\subseteq \widetilde{R}ec_{2n}(\lambda/\mu) \overset{\sim}\hookrightarrow LRS_{2n}(\lambda/\mu).\]
The surjectivity of 4 and 5 is concluded from the representation theory of a quantum symmetric pair of type \(\mathrm{AII}_{2n-1}\) in [1] respectively [1]. More precisely, the quantum symmetric pair of type \(\mathrm{AII}_{2n-1}\) consists of the Drinfeld-Jimbo quantum group \(\mathbf{U}\) and the Letzter \(\imath\)quantum group \(\mathbf{U}\imath\) [29] (a coideal subalgebra of \(\mathbf{U}\)) of the universal enveloping algebras of \(\mathfrak{gl}_{2n}(\mathbb{C})\) respectively \(\mathfrak{sp}_{2n}(\mathbb{C})\). For details on symmetric quantum pairs, we refer the reader to the survey paper [30].
The Littlewood-Richardson map 4 has a \(q\) analogue isomorphism by showing how the irreducible \(\boldsymbol{U}\)-module \(V (\lambda)\) decomposes into irreducible \(\boldsymbol{U}^\imath\)-submodules \(V^\imath(\mu)\) [31] at \(q=\infty\) which allows to conclude the surjectivity of 4 \[\begin{align} LR^{AII}:V(\lambda)\overset{\sim}\rightarrow \bigoplus_{\mu\in Par\le n} V^\imath(\mu)\otimes\mathbb{Q}Rec_{2n}(\lambda/\mu), \end{align}\] where \(\mathbb{Q}Rec_{2n}(\lambda/\mu)\) denotes the \(\mathbb{Q}\)-vector space with basis \(Rec_{2n}(\lambda/\mu)\).
Similarly to the G.P. Thomas bijection 1 whose recording tableaux are Littlewood-Richardson that determine the highest and lowest weight tableaux of a connected component in the tensor product decomposition, the recording tableaux in the quantum Littlewood-Richardson map 2 determine the \(\mathfrak{k}\)-highest and lowest weight semi-standard tableaux in \(SST_{2n}(\lambda)\) as defined in the recent proof of the Naito–Sagaki conjecture [32] via the Watanabe branching rule for \(\imath\)quantum groups. Here [32] \(\mathfrak{k}\) means a certain subalgebra of \(\mathfrak{sl}_{2n}(\mathbb{C})\) that is isomorphic to \(\mathfrak{sp}_{2n}(\mathbb{C})\) and \(\boldsymbol{U}^\imath\) can be seen as a \(q\)-deformation of \(U(\mathfrak{k})\) with a natural embedding in \(\boldsymbol{U}\). It is then useful to have an explicit surjectivity which allows to compute explicitly those highest or lowest \(\mathfrak{k}\)-weight semi-standard tableaux. We note that the first complete proof of the Naito-Sagaki conjecture and that does not not rely on \(\imath\)quantum groups was provided by Schumann–Torres in [33]. Recently another combinatorial proof of the Naito–Sagaki conjecture, which does not rely on either or this paper, appeared in [34].
Our main results assert as follows.
Theorem A (Theorem 1)[The surjectivity of the map in [1] Let \(T\in LRS_{2n}(\lambda/\mu)\) of even weight \(\nu\) and \(\nu^t\) its conjugate partition. Let \(J_1,\dots,J_{\nu_1}\) be the decomposition of \(T\) into vertical strips where each strip is filled with \(1,2,\dots, |J_i|=\nu^t_i\in 2\mathbb{Z}\) for \(i=1,\dots,\nu_1\). Let \(\lozenge\) be the map that relabels each vertical strip \(J_i\) with \(\nu^t_i\) \(i\)’s, for \(i=1,\dots,\nu_1\). Then we get a new tableau \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) of weight \(\nu^t\) and \(\lozenge\) is a bijection between \(LRS_{2n}(\lambda/\mu)\) and \(\widetilde{R}ec_{2n}(\lambda/\mu)\). Equivalently, \(\lozenge\) is the inverse of the map in [1].
Corollary (Section 3.1) The \(LR\) orthogonal transpose symmetry map \(\blacklozenge\) [35] restricts to LR-Sundaram tableaux as \[\begin{align} \blacklozenge: LRS_{2n}(\mu,\nu,\lambda)\overset{\lozenge}\longrightarrow \widetilde{R}ec_{2n}(\mu,\nu^t,\lambda)\underset{\pi\circ t}\hookrightarrow LR({\lambda^t},\nu^t,\mu^t): T\mapsto \lozenge T=Q\mapsto Q^{\pi\circ t}=\blacklozenge T \end{align}\] where \(Q^{\pi\circ t}\) means the \(\pi\)-rotation (transposition) followed with transposition (rotation) of the Young diagram \(D(\lambda)\). In other words, the bijection \(\lozenge\), and therefore the injection in [1], exhibits the restriction of the LR orthogonal transpose symmetry map \(\blacklozenge\), \(c_{\mu,\nu,\lambda}=c_{\lambda^t, \nu^t,\mu^t}\), to LR Sundaram tableaux.
Let \(l\in [0,2n]\). In [1] it is shown that the reduction map on \(SST_{2n}(\varpi_l)\) is the injective assignment ?? \[\begin{align} \mathrm{red}:SST_{2n}(\varpi_l)\rightarrow\bigsqcup_{\begin{smallmatrix} 0\le t\le min\{l,2n-l\}\\ l-t\in 2\mathbb{Z} \end{smallmatrix}}SpT_{2n}(\varpi_t), \;\; \mathbf{a}\mapsto \mathrm{red}( \mathbf{a})=\mathbf{a}\setminus \mathrm{rem}(\mathbf{a}). \end{align}\]
Theorem B [The reduction map, \(\mathrm{red}\), on \(SST_{2n}(\varpi_l)\) [1] is surjective] Let \(l\in [0,2n]\) and \(t\in[0,n]\) such that \(0\le t\le min\{l,2n-l\}\) and \(l-t\in 2 \mathbb{Z}\). Let \(\mathbf{a} = (a_1, \dots , a_t)\) be a column in \(SpT_{2n}(\varpi_t)\). Lemma 1, Theorem 2 and Theorem 3 exhibit a procedure to explicitly compute the inverse of the reduction map in the cases where the symplectic column decomposes into non empty factors \(\boldsymbol{A}_i\) of even length consisting of consecutive integers starting with an odd number, respectively when the symplectic column is such that consecutive integers occur only as an even number followed with an odd number which are within the patterns of the symplectic \(\mathfrak{k}\)-highest (lowest) weight tableaux considered in the last section.
The next theorem is a consequence of the previous one and the reverse Schensted insertion ruled by the slack data [36] (see subsections 4.4 and 4.5). It gives the constructive fundamental step for the unwinding of the sequence of successors [1] in the algorithm to computing the quantum Littlewood-Richardson map. The unwind procedure goes from the last to the first successor, and Theorem C below gives explicitly \({\mathsf{LR}^{AII}}^{-1}(S,Q)\) for the unwind of the last successor where \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) with \(\mu\subset_{vert}\) and \(S\in SpT_{2n}(\mu)\).
Theorem C (Theorem 4)[The inclusion \(Rec_{2n}(\lambda/\mu)\supseteq \widetilde{R}ec_{2n}(\lambda/\mu)\) for [1]] Let \(S\in SpT_{2n}(\mu)\) and \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\), \(\mu\subset_{vert} \lambda\), \(l=\ell(\lambda)\), has slack row index vector \(\mathbf{r}=\{r_1<\dots< r_{t_0}\}\). Then, \(\ell(\mu)\le l\), \(0\le l- t_0\in 2 \mathbb{Z}\), \(l\le 2n-t_0\), and one has the following assertion:
\[\begin{align} \label{inverseint}{\mathsf{LR}^{AII}}^{-1}(S,Q)&= \mathrm{c}\circ (\mathrm{red}_{t_0}^{-1},\mathrm{id})\circ (\underset{{\mathbf{r}}}\leftarrow S)\nonumber\\ &= \mathrm{c}\circ (\mathrm{red}_{t_0}^{-1},\mathrm{id})((a_1,\dots,a_{t_0}),S^1 )\nonumber\\ &= \mathrm{c}\circ (\mathrm{red}_{t_0}^{-1}(a_1,\dots,a_{t_0}), S^1) \nonumber\\ &=\mathrm{red}_{t_0}^{-1}((a_1,\dots,a_{t_0}))\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^1=:S^\mathbf{r}\in SST_{2n}(\lambda). \end{align}\tag{7}\] where \(\mathrm{red}_{t_0}^{-1}((a_1,\dots,a_{t_0}))=T_0T_1\cdots T_{t_0}\in SST_{2n}(\varpi_{\ell(\lambda)})\) in Lemma 1, Theorem 2 and Theorem 3, and \(S^1\in SpT_{2n}({\mu^{(1)}}')\) with \({\mu^{(1)}}'=\mu^{(1)}-\delta_\mathbf{r}\), and \(\lambda=\mu^{(1)}-\delta_\mathbf{r}+\varpi_{\ell(\lambda)}\), \(\mu^{(1)}:=\mu\).
As an application of the explicit surjectivity and consequently the explicit inverse \(\mathsf{LR}^{AII}\) of the quantum Littlewood-Richardson map, we characterize in Corollary 4, Theorem 7, Theorem 8 and their corollaries for \(n=1,3\) respectively \(n=2,4\), the family of \(\mathfrak{k}\)-highest wight tableaux generated either by quantum record tableaux with one vertical strip or with \(1\)-\(0\)-slack sequences. Consider the sequence \(\{u_i\}_{i=1}^n\) of numbers depending on the parity of \(n\) \[\begin{align} \{u_i\}_{i=1}^n=\begin{cases}\{2,3,6,7,10,\dots, 2(n-1)-1,2n\},& n\notin 2 \mathbb{Z},\\ \{2,3,6,7,\dots, 2(n-1),2n-1\},& n\in 2 \mathbb{Z}\end{cases}\label{numbers:uuint} \end{align}\tag{8}\] A symplectic tableau in \(SpT_{2n}(\mu)\) is said to be \(\mathfrak{k}\)-highest weight of shape \(\mu\) if row \(i\) has only numbers \(u_i\) for \(i=1,\dots,n\). For \(n\) odd, one has the following assertion.
Theorem D (Theorem 7)[\(\mathfrak{k}\)-highest weight tableaux produced by \(1\)-\(0\)-slack sequence quantum recording tableaux for \(n\) odd] Let \(S^{H,\mu}\) be the \(\mathfrak{k}\)-highest weight tableau in \(SpT_{2n}(\mu)\), and \(Q\in Rec_{2n}(\lambda/\mu)\) with \(1\)-\(0\)-slack sequence of the form \(\underline \mathbf{t}=(1,\dots,1,0^M)\) and slack vector sequence \(\underline\mathbf{r}\) with slack incidence matrix \(\delta_{\underline\mathbf{r}}\). Then, for some \(0\le k\le n\), \({\mathsf{LR}^{AII}}^{-1}(S^{H,\mu},Q)\) returns the \(\mathfrak{k}\)-highest weight tableau in \(SST_{2n}(\lambda)\), with \({\mathfrak{k}}\)-weight \(\mu\), of the form \[\begin{align} &Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\nonumber}{\m@th.5\bullet}}}\\ &\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle}{\m@th.5\bullet}}}\circledcirc_{j=n}^{k+1}\big(\mathrm{red}_1^{-1}({u_j})\big)^{m_{\mathrm{red}^{-1}(u_j)}}\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}({u_{k}})\big)^{m_{\mathrm{red}^{-1}(u_k)}}\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=k}^{1}\big(\mathrm{red}_1^{-1}({u_j})\setminus\{u_n\}\big)^{m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}}}{\m@th.5\bullet}}}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}, \end{align}\] where \(S^{H,\mu'}\) is the \(\mathfrak{k}\)-highest weight tableau in \(SpT(\mu')\), \(\mu'= \mu-\delta_{\underline \mathbf{r}^+}\), satisfying the identity on the multiplicities \[\displaystyle \sum_{j=n}^{k+1} m_{\mathrm{red}^{-1}(u_j)}+m_{\mathrm{red}^{-1}(u_k)}+\sum_{j=1}^k m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}=|\delta_{\underline\mathbf{r}}|,\] and linear inequalities on the multiplicities of the columns
\[\begin{align} &\mathrm{red}^{-1}(u_i),~~ \mathrm{red}^{-1}(u_i)\setminus \{u_n\},i=2,\dots,n, on the LHS of S^{H,\mu'},\\ & and the symplectic columns u_1u_2\cdots u_i\in SpT_{2n}(\varpi_i) \eqref{noddsymphw-hw}, i=1,\dots,n,\nonumber\\ &u_1u_2\cdots u_n,~~ u_1u_2\cdots u_{n-1},\dots,u_1u_2,~~u_1 \end{align}\] as follows
\[\begin{align} &m_{\mathrm{red}^{-1}(u_i)}+m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}\le m_{u_1\cdots u_{i-1}}, ~i=2,\dots,n,\\ &0\le m_{\mathrm{red}^{-1}(u_n)}-\sum_{i=1}^{n-1}m_{\mathrm{red}^{-1}(u_i)\setminus\{u_n\}}\le m_{u_1\cdots u_{n-1}}. \end{align}\]
With the same setting as above and \(n\) even, one has
Theorem E (Theorem 8)[\(\mathfrak{k}\)-highest weight tableaux produced by \(1\)-\(0\)-slack sequence quantum recording tableaux for \(n\) even] For some \(0\le k\le n\), \({\mathsf{LR}^{AII}}^{-1}(S^{H,\mu},Q)\) returns the \(\mathfrak{k}\)-highest weight tableau in \(SST_{2n}(\lambda)\) with \({\mathfrak{k}}\)-weight \(\mu\), in either form:
\[\begin{align} &Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}(2n-1))^{m_{\mathrm{red}^{-1}(2n-1) }}\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=n-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\big)^{m_{\mathrm{red}^{-1}(u_j) }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}\\ =&Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}((12\cdots (2n-3).(2n-2).(2n-1)\big)^{m_{12\cdots (2n-3).(2n-2).2n-1 }}\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=n-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\big)^{m_{\mathrm{red}^{-1}(u_j) }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}, \end{align}\]
or
\[\begin{align} &Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}(2n-1)\big)^{m_{\mathrm{red}^{-1}(2n-1) }} \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\cdot 2n\big)^{m_{12\cdots 2n-2\cdot 2n}} \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\big)^{m_{(12\cdots (2n-2)) }}\mathpalette\mathbin{\vcenter{\scalebox{\nonumber}{\m@th.5\bullet}}}\\ & \mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=n-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\setminus\{2n\}\big)^{m_{\mathrm{red}^{-1}(u_j)\setminus\{2n\} }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}\\ =&Y(M^{2n}) \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}((12\cdots (2n-3).(2n-2).(2n-1)\big)^{m_{12\cdots (2n-3).(2n-2).2n-1 }}\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\cdot 2n\big)^{m_{12\cdots 2n-2\cdot 2n}}\nonumber\\ &\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\big)^{m_{12\cdots (2n-2) }}\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=n-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\setminus\{2n\}\big)^{m_{\mathrm{red}^{-1}(u_j)\setminus\{2n\} }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'} \end{align}\]
or
\[\begin{align} &Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}(2n-1)\big)^{m_{\mathrm{red}^{-1}(2n-1) }} \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\cdot 2n\big)^{m_{12\cdots 2n-2\cdot 2n}} \mathpalette\mathbin{\vcenter{\scalebox{\nonumber}{\m@th.5\bullet}}}\\ &\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle}{\m@th.5\bullet}}}\circledcirc_{j=n-1}^{k+1}\big(\mathrm{red}_1^{-1}({u_j})\big)^{m_{\mathrm{red}^{-1}(u_j)}} \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}({u_{k}})\big)^{m_{\mathrm{red}^{-1}(u_k)}}\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}(u_k)\setminus\{2n\}\big)^{m_{\mathrm{red}^{-1}(u_k)\setminus\{2n\} }}\nonumber\\ &\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=k-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\setminus\{2n\}\big)^{m_{\mathrm{red}^{-1}(u_j)\setminus\{2n\} }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}, \end{align}\] where the multiplicities satisfy the identity \(\displaystyle \sum_{j=n}^{k+1} m_{\mathrm{red}^{-1}(u_j)}+m_{\mathrm{red}^{-1}(u_k)}+\sum_{j=k}^1 m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}=|\underline\delta_\mathbf{r}|-M\), and linear inequalities on the multiplicities of the columns
\[\begin{align} &\mathrm{red}^{-1}(u_i),~~ \mathrm{red}^{-1}(u_i)\setminus \{2n\},i=1,2,\dots,n, on LHS of S^{H,\mu'},\nonumber\\ & and the symplectic columns u_1u_2\cdots u_i\in SpT_{2n}(\varpi_i) \eqref{nevensymphw-hw}, i=1,\dots,n,\nonumber\\ &u_1u_2\cdots u_n,~ u_1u_2\cdots u_{n-1},\dots,u_1u_2,~~u_1\nonumber \end{align}\] as follows
\[\begin{align} &m_{\mathrm{red}^{-1}(u_i)}+m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}\le m_{u_1\cdots u_{i-1}}, ~i=2,\dots,n\\ & m_{12\cdots (2n-2)2n} =\sum_{i=1}^{n-1} m_{\mathrm{red}^{-1}(u_i)\setminus\{2n\}}. \end{align}\]
We hope that a geometrical object will emerge from the characterization of the \(\mathfrak{k}\)-highest (lowest) weight tableaux by linear inequalities obtained so far to be attached to the quantum LR map as it happens in the Thomas LR map [2]. The coupling of the left and right companions of a recording LR tableau in [2] (determining also the lowest and highest weights of a connected component in the crystal tensor product decomposition 3 ) led to hives [11], [37].
This paper is organized in five sections. Section 2 introduces the relevant notation on semistandard tableaux and recalls the Schensted column insertion and its reverse including the relevant properties on the bumping and reverse bumping routes. Section 3 proves the surjectivity of the map \(\widetilde{R}ec_{2n}(\lambda/\mu) \overset{\sim}\hookrightarrow LRS(\lambda/\mu)\) in Theorem 1 (Theorem A in the Introduction) and how it exhibits the restriction of the LR symmetry \(c_{\mu,\nu,\lambda}=c_{\lambda^t, \nu^t,\mu^t}\), to LR Sundaram tableaux in Subsection 3.1, Corollary 1 in the Introduction. Section 4 recalls the relevant operations of the algorithm that computes the quantum Littlewood-Richardson map and their properties. Lemma 1, Theorem 2 and Theorem 3 (Theorem B in the Introduction) on the surjectivity of the reduction map are ones of the main results. With this on hand we get prepared for the fundamental unwind step of the sequence of successors of that algorithm to prove the surjectivity of the map \[\displaystyle SST_{2n}(\lambda) \overset{\sim}\hookrightarrow\bigsqcup_{\begin{smallmatrix}\mu\in Par_{\le n}\\ \mu\subset\lambda\\ Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\end{smallmatrix}}SpT_{2n}(\mu) \times \{Q\},\] providing we know that \(\widetilde{R}ec_{2n}(\lambda/\mu) \overset{\sim}\rightarrow LRS(\lambda/\mu)\). The unwind steps are determined by \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) with input a symplectic tableau of shape \(\mu\), and consist of reverse column insertion ruled by the slack data introduced in [36] and recalled and expanded in Subsection 4.5 for any quantum recording tableau, followed with reverse removal or the inverse of the reduction map and concatenation in the plactic monoid. The fundamental unwind step is explicitly exhibited in Theorem 4 (Theorem C in the Introduction) and its iteration ruled by the slack data appears in theorems 5 and 6. In the last Section 5, the inverse of the quantum Littlewood-Richardson map is applied to characterize a family of \(\mathfrak{k}\)-highest weight tableaux in the Naito-Sagaki conjecture recent proof [32] based on the Watanabe branching rule for \(\imath\)quantum groups [1]: Corollary 4, Theorem 7, Theorem 8, also in the Introduction above, and their corollaries for \(n=1,3\) respectively \(n=2,4\), provide the family of \(\mathfrak{k}\)-highest wight tableaux generated either by quantum record tableaux with one vertical strip respectively with \(1\)-\(0\)-slack sequences.
The author acknowledges financial support by the Centre for Mathematics of the University of Coimbra (CMUC, https://doi.org/10.54499/UID/00324/2025) under the Portuguese Foundation for Science and Technology (FCT), Grants UID/00324/2025 and UID/PRR/00324/2025.
A partition \(\gamma\) is a weakly decreasing sequence of nonnegative integers \(\gamma_1\ge \gamma_2 \ge \cdots\) such that \(\gamma_k =0\) for some \(k \ge 1\). The maximal \(i\) such that \(\gamma_i> 0\) is called the number of parts or length of \(\gamma\), denoted \(\ell(\gamma)\). For each \(m\ge 0\), the set of partitions of length at most \(m\) is denoted by \(Par_{\le m}\). We assume the inclusion \(Par_{\le m}\subseteq Par_{\le k}\) whenever \(k\ge m\). Thus we often write the partition \(\gamma\) as a vector \(\gamma = (\gamma_1, \gamma_2, \dots,\gamma_k)\) for \(k \ge \ell(\gamma)\). The empty partition is the empty sequence \(()\) and is regarded as the unique partition of length zero. Given \(m\in \mathbb{Z}_{\ge 0}\), \(\varpi_m\) denotes the partition of length \(m\) whose parts are all \(1\), that is, \(\varpi_m =(\underbrace{1,\dots,1}_{m})=:(1^m)\).
The partition \(\gamma\) is said to be even if \(\gamma_{2i-1}=\gamma_{2i}\) for all \(i\ge 1\). In other words, all columns of \(\gamma\) have even length and necessarily the length of \(\gamma\) is even.
A partition \(\gamma\) is identified with its Young diagram \(D(\gamma)\) which is a left and top justified collection of boxes (or cells) with \(\gamma_k\) many boxes in the \(k\)th row for all \(k \in \mathbb{Z}_> 0\). In particular, the empty Young diagram and the partition \(()\) are identified. The number of cells of \(D(\gamma)\) is the sum of the parts of \(\gamma\) and is denoted by \(|\gamma|\). The boxes or cells of the Young diagram of \(\gamma\) are identified by its coordinates \((i,j)\) in the matrix style, that is, \(1\le i\le \ell(\gamma)\) and \(1\le j\le \gamma_i\).
Let \(\gamma\), \(\mu\) partitions with \(\mu \subseteq \gamma\), that is, \(\mu_i\le \gamma_i\) for all \(i\in\mathbb{Z}_>0\), or the Young diagram of \(\mu\) is a subset of the Young diagram of \(\gamma\). The skew-diagram (or skew Young diagram) \(\gamma/\mu\) is defined to be \(D(\gamma)\setminus D(\mu)\). For \(\lambda\subseteq\gamma\) partitions, we write \(\lambda\subseteq_{vert} \gamma\) to mean that \(\gamma/\lambda\) is a vertical strip, that is, the skew-diagram \(\gamma/\lambda\) has at most one box in each row. The number of cells of \(\gamma/\lambda\) is \(|\gamma/\lambda|:=|\gamma|-|\lambda|.\)
A tableau \(T\) of (skew) shape \(\gamma/\mu\) is a map (or a filling of \(D(\gamma)\)) \[T:D(\gamma)\rightarrow \mathbb{Z}_\ge0, \; (i,j)\mapsto T(i,j),\] assigning a positive integer to each box of \(D(\gamma)\setminus D(\mu)\) and \(0\) to the boxes of \(D(\mu)\). Denote by \(\mathrm{Tab}(\gamma/\mu)\) the set of tableaux of shape \(\gamma/\mu\). We say that the tableau \(T\) is semi-standard if an addition the assignment is such that it is weakly increasing as we go from left to right along a row and strictly increasing as we go from top to bottom along a column excluding the boxes in \(\mu\), \[\begin{align} &T(i,j)\le T(i,j+1),\; T(i,j)<T(i+1,j),for all (i,j)\in D(\gamma)\setminus D(\mu),\nonumber\\ & andT(i,j)=0,for (i,j)\in D(\mu), \nonumber \end{align}\] where we set \(T(a,b):=\infty\) if \((a, b) \notin D(\gamma)\). Usually \(T(i,j)\) is just referred as the entry in the box \((i,j)\) and we omit the zeroes in the boxes of \(\mu\). A positive integer \(m\ge \ell(\gamma)\) will be fixed and \([0, m]:=\{0, 1, \dots, m\}\) will be used as a co-domain for the map \(T\). We call \([m]:=\{1, \dots, m\}\) the alphabet of the semi-standard tableau \(T\). In this case, we will denote the set of semi-standard tableaux of shape \(\gamma/\mu\) by \(SST_m(\gamma/\mu)\). When \(\mu=()\), we just write \(SST_m(\gamma)\). The weight or content of \(T\) is the nonnegative vector \(\mathrm{wt}(T)=(T[1],\dots, T[m])\), where \(T[i]:=\#\{(a,b)\in D(\gamma):T(a,b)=i\}\) for \(i\in[m]\), that is, \(T[i]\) is the number of occurrences of \(i\) in the tableau \(T\).
The reverse row word of a semi-standard tableau \(T\), denoted \(w(T)=w_1\cdots w_l\), with \(l\) the number of non zero entries of \(T\), is obtained by reading the entries of its rows (excluding the entry 0) right to left starting from the top row and proceeding downwards. The column word of a semi-standard tableau \(T\) denoted \(w^{col}(T)=w'_1\cdots w_l'\) is the sequence of entries obtained by reading the entries (excluding the entry \(0\)) of its columns from bottom to top and left to right. The reverse column word of \(T\), \(w'_l\cdots w'_1\). is obtained by reading the entries of its columns (excluding the entry 0) right to left starting from the top and proceeding downwards. The weight of the word \(w(T)\) is the weight of \(T\).
A Yamanouchi word is a word \(u_1 \cdots u_l\) such that, for each \(1 \le k \le l\), the weight of the subword \(u_1 \cdots u_k\) is a partition. A Yamanouchi tableau of shape \(\mu\), denoted \(Y(\mu)\), is a semi-standard tableau of shape and weight \(\mu\). That is, its reverse column word is Yamanouchi and \(Y(i,j)=i\) for all \(1\le i\le \ell(\mu)\) and \(1\le j\le\mu_i\).
The Schensted column insertion or column bumping [38], [39], takes a positive integer \(x\) and a tableau \(T\in SST(\gamma)\) and puts \(x\) in a new box at the bottom of the first column (counting from left to right) if it is strictly larger than all the entries of the column. If not, it bumps the smallest entry in the column that is larger than or equal to \(x\). The bumped entry moves to the next column, going to the end if possible, and bumping an element to the next column otherwise. The process terminates when the bumped entry goes to the bottom of the next column, or until it becomes the only entry of a new column. The returned tableau is denoted \(x\mathpalette\mathbin{\vcenter{\scalebox{T}{\m@th.5\bullet}}}:=x \rightarrow T\), that is, \(x\mathpalette\mathbin{\vcenter{\scalebox{T}{\m@th.5\bullet}}}\) is the result of column inserting \(x\) into \(T\), \(x \rightarrow T\). We call to \(x \mathpalette\mathbin{\vcenter{\scalebox{T}{\m@th.5\bullet}}}\) the multiplication of \(x\) and \(T\) in the plactic monoid.
The reverse Schensted column insertion takes a tableau \(U\in SST(\gamma)\) and an entry \(y\) at the bottom of a column of \(U\), whose cell has row coordinate, say \(r_y\), and bumps it while deletes its box. If the column is the first, the new tableau is obtained by deleting the bottom box of the first column of \(U\). If not, the bumped \(y\) moves to the next column on its left and bumps the largest entry in the column that is smaller than or equal to \(y\). In both cases, if the bumped entry belongs to the first column the process ends and returns a pair \((y', ~\underset{ r_y}\leftarrow U)\) consisting of the bumped entry \(y'\le y\) from the first column of \(U\) together with a new tableau denoted \(\underset{r_y}\leftarrow U\), with one box less than \(U\). If not, the process continues until an entry is bumped from the first column of \(U\).
Remark 1. [38] Let us consider two successive column-insertions of \(x < x'\) into a tableau \(T\). First column-inserting \(x\) in \(T\) and then column-inserting \(x'\) in the resulting tableau \(x\rightarrow T\), yielding to two bumping routes \(R\) and \(R'\), and two new boxes \(B\) and \(B'\). Then \(R'\) lies strictly below \(R\), and \(B'\) is Southwest of \(B\).
If \(x\ge x'\), \(R'\) lies weakly above \(R\) and \(B'\) Northeast of \(B\).
Since column insertion is reversible, one has the following reverse bumping route property for reverse column insertion.
Remark 2. Let \(y\) and \(y'\) be two bottom column entries of a tableau \(T\), with \(y'\) strictly Southwest of \(y\) (not in the same row), or let \(y<y'\) be the two bottom entries of a column of \(T\). Let \(r_y\) and \({r_{y'}}\) be the corresponding cell row coordinates.
First consider the reverse column-inserting of \(y'\) and then reverse column-inserting of \(y\) in the resulting tableau \((\underset{r_{y'}}\leftarrow T)\). This yields two reverse bumping routes \(Z'\) respectively \(Z\) such that \(Z'\) lies strictly below \(Z\). In particular, the two new entries \(b'\) respectively \(b\) in the first column of \(T\) are such that \((\underset{r_{y'}}\leftarrow T)(i',1)=b'> (\underset{r_{y}}\leftarrow(\underset{r_{y'}}\leftarrow T))(i,1)=b\) and \(i'>i\).
In case \(y'\) strictly Southwest of \(y\) (not in the same row) and we first consider reverse column-inserting of \(y\) followed with the reverse column-inserting \(y'\) in the resulting tableau \((\underset{r_{y}}\leftarrow T)\) then the corresponding reverse bumping routes \(Z\) and \(Z'\) are such that \(Z'\) lies weakly below \(Z\). The two new entries \(b\) respectively \(b'\) in the first column of \(T\), \((\underset{r_{y'}}\leftarrow(\underset{r_{y}}\leftarrow T))(i',1)=b'>(\underset{r_{y}}\leftarrow T)(i,1)=b\) with \(i'>i\), or \(i=i'\) and \((\underset{r_{y}}\leftarrow T)(i,1)=b\le (\underset{r_{y'}}\leftarrow(\underset{r_{y}}\leftarrow T))(i,1)=b'\).
If two or more columns have the bottom entries in a same row, let \(y'\) be the right most bottom entry in that row. Consider the reverse column-inserting of \(y'\) in \(T\) and then again the reverse column-inserting of \(y'\) in the resulting tableau \(y'\leftarrow T\). This yields two bumping routes \(Z_1\) respectively \(Z_2\) and two new entries \(b_1\) respectively \(b_2\) in the first column of \(T\), such that \((\underset{r_{y'}}\leftarrow(\underset{r_{y'}}\leftarrow T))(i_2,1)=b_2> (\underset{r_{y'}}\leftarrow T))(i_1,1)=b_1\) with \(i_2>i_1\), or \(i_1=i_2\) and \((\underset{r_{y'}}\leftarrow T)(i_1,1)=b_1\le (\underset{r_{y'}}\leftarrow(\underset{r_{y'}}\leftarrow T))(i_1,1)=b_2\).
Let \(\lambda\in Par_{\le m}\) and \(k\in [0,m]\). The following assignment [1] is the combinatorial Pieri’s rule, defined by the column insertion of a column tableau into a tableau is a bijection
\[\begin{align} \mathpalette\mathbin{\vcenter{\scalebox{:}{\m@th.5\bullet}}}SST_m(\varpi_k)\times SST_m(\lambda)&\rightarrow\bigsqcup_{\begin{smallmatrix} \gamma\in Par_{\le m}\\ \lambda\subseteq_{vert} \gamma,~|\gamma/\lambda|=k \end{smallmatrix}} SST_m(\gamma) \nonumber\\ (S,T)&\mapsto S\mathpalette\mathbin{\vcenter{\scalebox{T}{\m@th.5\bullet}}}:=w_k\rightarrow(\cdots\rightarrow( w_2 \rightarrow(w_1\rightarrow T))\cdots) \end{align}\] where \(S=(w_1<w_2<\dots<w_k)\) is a column of length \(k\) with entries in \([1,m]\). In particular, if \(S(i,1)\le T(i,1)\) for every \(1\le i\le m\), then \(S\mathpalette\mathbin{\vcenter{\scalebox{T}{\m@th.5\bullet}}}\) is the concatenation of \(S\) and \(T\).
The process \(S\mathpalette\mathbin{\vcenter{\scalebox{T}{\m@th.5\bullet}}}=w_k\rightarrow(\cdots( w_2 \rightarrow(w_1\rightarrow T))\) is reversible \[\begin{align} \label{reverse}(S, T)=(\underset{\gamma/\lambda}\leftarrow T):=(\underset{r_{w'_{1}}}\leftarrow(\cdots( \underset{r_{w'_{k-1}}} \leftarrow(\underset{r_{w'_k}}\leftarrow S\mathpalette\mathbin{\vcenter{\scalebox{T}{\m@th.5\bullet}}}))\cdots)) \end{align}\tag{9}\] where the entries of the vertical strip \(\gamma/\lambda\) from bottom to top define the word \(w'_k\cdots w'_1\) and \(r_{w'_i}\) indicates the row index of \(w'_i\), \(1\le i\le k\). For \(U\in SST_m(\gamma)\) such that \(\lambda\subseteq_{vert}\gamma\) the inverse map is defined by the reverse column insertion by taking the entries of \(U\) in the vertical strip \(\lambda/\gamma\) from bottom to top, \((\underset{\gamma/\lambda}\leftarrow U)\in SST_m(\varpi_k)\times SST_m(\lambda)\).
Recall a partition \(\nu\) is said to be even if all columns of \(\nu\) have even length and necessarily the length of \(\nu\) is even. In other words, \(\nu_{2i-1}=\nu_{2i}\) for all \(i\ge 1\). In this case, \(\nu^t\) the transpose or conjugate of \(\nu\) has all rows of even length. We start by recalling the definition of Littlewood-Richardson-Sundaram tableau (or symplectic Littlewood-Richardson tableau). Fix \(n\in\mathbb{N}\). Let \(\lambda \in Par_{\le 2n}\) and \(\mu \in Par_{\le n}\).
Definition 1. Let \(\lambda \in Par_{\le 2n}\) and \(\mu \in Par_{\le n }\) be such that \(\mu\subset\lambda\). A semi-standard tableau \(T \in Tab(\lambda/\mu)\) is said to be an \(n\)-symplectic Littlewood-Richardson tableau or Littlewood-Richardson-Sundaram tableau if it satisfies the following.
\(T \in SST_{2n}(\lambda/\mu)\).
The reversed column word of \(T\) is a Yamanouchi word.
The sequence \(\mathrm{wt}(T) = (T[1], T[2],\dots, T[2n])\) is an even partition,
If \(T(i, j) = 2k + 1\) for some \((i, j) \in D(\lambda/\mu)\) and \(k \in\mathbb{Z}_{\ge 0}\) then we have \(i \le n + k\).
For \(\lambda \in Par_{\le 2n}\) and \(\mu \in Par_{\le n}\), the subset of \(SST_{2n}(\lambda/\mu)\) satisfying conditions \((2)\), \((3)\) and \((4)\) in Definition 1 is denoted by \(LRS_{2n}(\lambda/\mu)\).
Remark 3. Condition \((4)\) above can be replaced by \(T(n + i, 1) \ge 2i\) for every \(i\ge 0\).
Remark 4. Let \(T\in LRS_{2n}(\lambda/\mu)\) of content \(\nu\). Let \(N:=\nu_1\) and \(\nu^t=(\nu_1^t,\dots,\nu^t_{N})\) the transpose of \(\nu\). Put \(\mu^{(0)}:=\lambda\) and \(\mu^{(N)}:=\mu\). For \(k=1,\dots,N\), define \(\mu^{(k)}\) the shape obtained by erasing from NE to SW in \(T_{|\mu^{(k-1)}}\), \(T\) restricted to the shape \(\mu^{(k-1)}\), the first rightmost \(\nu^t_k\) cells filled with \(1,2,\dots, \nu^t_k\). Hence
\[\begin{align} \mu^{(0)}=\lambda\supset_{vert} \mu^{(1)}\supset_{vert} \cdots\supset_{vert}\mu^{(N)} =\mu \label{nested1} \end{align}\qquad{(1)}\] This decomposes the shape \(\lambda/\mu\) into \(N\) vertical strips which completely defines \(T\)
\[\begin{align} J_1:=\mu^{(0)}/\mu^{(1)},J_2:= \mu^{(1)}/\mu^{(2)}, J_i=\mu^{(i-1)}/\mu^i,\dots, J_{N}=\mu^{(N-1)}/\mu^{(N)}\label{nested2} \end{align}\qquad{(2)}\]
Then, for \(k=1,\dots,N\), the vertical strip \(J_k=\mu^{(k-1)}/\mu^{(k)}\) has \(\nu^t_k=|\mu^{(k-1)}|-|\mu^{(k)}|\) cells and is filled in \(T\), top to bottom, with \(12\cdots \nu_k^t\). Concatenating those column words as \(12\cdots \nu^t_{N}\mathpalette\mathbin{\vcenter{\scalebox{\cdots}{\m@th.5\bullet}}}\mathpalette\mathbin{\vcenter{\scalebox{1}{\m@th.5\bullet}}}2\mathpalette\mathbin{\vcenter{\scalebox{\cdots}{\m@th.5\bullet}}}\mathpalette\mathbin{\vcenter{\scalebox{\nu}{\m@th.5\bullet}}}^t_2\mathpalette\mathbin{\vcenter{\scalebox{1}{\m@th.5\bullet}}}2\cdots \nu^t_{1}\) gives the reverse-column word of the Yamanouchi tableau of shape \(\nu\) which is also obtained by rectification of the word of \(T\). For an illustration, see Example 1.
We now recall the definition of the set \(\widetilde{R}ec_{2n}(\lambda/\mu)\).
Definition 2. [[1]][] Let \(\lambda \in Par_{\le 2n}\) and \(\mu \in Par_{\le n }\) be such that \(\mu\subset\lambda\). Let \(\widetilde{R}ec_{2n}(\lambda/\mu)\) denote the set of tableaux \(Q\) of shape \(\lambda/\mu\) satisfying the following:
The entries of \(Q\) strictly decrease along the rows from left to right.
The entries of \(Q\) weakly decrease along the columns from top to bottom.
For each \(k > 0\), the number \(Q[k]\) of occurrences of \(k\) is even.
For each \(k > 0\), it holds that
\[Q[k] \ge 2(\ell(\mu^{(k-1)})- n),\] where \(\mu^{(k-1)}\) is the partition such that \[D(\mu^{(k-1)}) = D(\mu) \cup\{(i, j) \in D(\lambda/\mu) | Q(i, j) \ge k\}.\]
For each \(r, k >0\), let \(Q_{\le r}[k]\) denote the number of occurrences of \(k\) in \(Q\) in the \(r\)-th row or above. Then, the following inequality holds: \[Q_{\le r}[k + 1] \le Q_{\le r}[k].\]
Remark 5. Let \(N\) be the largest entry in \(Q\) where we are assuming the blank boxes filled with \(0\). When \(N=0\) one has \(\lambda=\mu \Leftrightarrow\lambda/\lambda=()\Leftrightarrow \mathrm{wt}(Q)=()\). Let \(N\ge 1\). Since \(Q[k]\in 2\mathbb{Z}\) and \(Q[k+1]\le Q[k]\le \ell(\lambda)\le 2n\), for any \(k>0\), the weight of \(Q\) is the partition \(\mathrm{wt}(Q)=(Q[1], \dots, Q[N])\) and its transpose or conjugate is an even partition \(\nu=(\nu_1=N,\dots,\nu_{Q[1]})\). Thus \(Q\) is also defined by the sequence of nested partitions \[\begin{align} \mu^{(0)}=\lambda\supset_{vert} \mu^{(1)}\supset_{vert} \cdots\supset_{vert}\mu^{(N)} =\mu,~~ \end{align}\]
where \(\mu^{(i-1)}/\mu^{(i)}\) is a vertical strip of even length \(Q[i]\ge 2\) satisfying \((R4)\) and \((R5)\) conditions, for \(i=1,\dots,N\).
Proposition 1. Given \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) and \(k>0\), the following conditions hold
\(Q[k]=Q_{\le \ell(\mu^{(k-1)})}[k]\).
If \(1\le i\le \ell(\mu^{(k-1)})\) then \(Q[k]=Q_{\le \ell(\mu^{(k-1)})}[k]\le Q_{\le i}[k]+(\ell(\mu^{(k-1)})-i)\).
\(Q[k] \ge 2(\ell(\mu^{(k-1)})- n)\) (condition \((R4)\)) if and only if \(Q_{\le i}[k] \ge 2(i- n)\), for all \(1\le i\le \ell(\mu^{(k-1)})\).
Proof. \((a)\) Evident because \(J_k:=\mu^{(k-1)}/\mu^{(k)}=\{(i,j)| Q(i,j)=k\}\subseteq \mu^{(k-1)}\) and \(\ell(J_k)=Q[k]\). Therefore \[Q(i,j)=k\Rightarrow 1\le i\le \ell(\mu^{(k-1)}).\]
\((b)\) Let \(1\le i\le \ell(\mu^{(k-1)})\). Then \[\begin{align} &J_k:=\mu^{(k-1)}/\mu^{(k)}=\{(s,j)| Q(s,j)=k, s\le i\}\cup \{(s,j)| Q(s,j)=k, s> i\}\\ &\subseteq \{(s,j)| Q(s,j)=k, s\le i\}\cup \{i+1,\dots, \ell(\mu^{(k-1)})\} \end{align}\] Therefore \(\ell(J_k)=Q[k]\le Q_{\le i}[k]+ (\ell(\mu^{(k-1)}-i).\)
\((c)\) \(1\le i\le \ell(\mu^{(k-1)})\). From \((R4)\), \(Q[k]\ge 2(\mu^{(k-1)}-n)\). By contradiction suppose, \(Q_{\le i}[k] < 2(i- n)\). Then \[\begin{align} Q[k]&\le Q_{\le i}[k]+ (\ell(\mu^{(k-1)}-i)\\ &< 2(i- n)+(\ell(\mu^{(k-1)}-i)\\ &=\ell(\mu^{(k-1)}-2n+i\\ &\le\ell(\mu^{(k-1)}-2n+\ell(\mu^{(k-1)}=2(\ell(\mu^{(k-1)}-n).\\ \end{align}\] Hence \(Q[k]<2(\ell(\mu^{(k-1)}-n)\) a contradiction with \((R4)\). ◻
Corollary 1. \((a)\) For \(\lambda=\mu \in Par_{\le n }\), \(\widetilde{R}ec_{2n}(\lambda/\lambda)=\{D(\lambda)\}\).
\((b)\) \(\label{recolumn}\widetilde{R}ec_{2n}(\varpi_l)\neq \emptyset\) if and only if \(l\) even and \(l\le 2n\). In this case,
\[\begin{align} \widetilde{R}ec_{2n}(\varpi_l)=\{ \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{1}}, {{1}}, {\vdots}, {{1}}, {{1}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \} \end{align}\]
Proof. From \((R3)\), \(l=Q[1]\) is even and from \((R4)\), \(Q[1]=l\ge 2(l-n)\Leftrightarrow l\le 2n\). ◻
Let \(\nu\) be an even partition. Then the \(\nu_1\) columns are of even length and its transpose or conjugate partition \(\nu^t=(\nu^t_1,\dots,\nu^t_{\nu_1})\) is such that \(\nu_i^t\in 2 \mathbb{Z}\), for all \(i\ge 1\).
The following theorem asserts that the conditions on a LR-Sundaram tableau \(T\) translate to conditions on the lengths of their vertical strips \(J_i\) ?? , ?? and gives the right inverse of the map in [1].
Theorem 1. Let \(T\in LRS_{2n}(\lambda/\mu)\) of weight \(\nu\) and \(J_1,\dots,J_{\nu_1}\) its decomposition into vertical strips each strip filled in \(1,2,\dots, |J_i|=\nu^t_i\in 2\mathbb{Z}\) for \(i=1,\dots,\nu_1\). Relabel each vertical strip \(J_i=\mu^{(i-1)}/\mu^{(i)}\) with \(\nu^t_i\) \(i\)’s, for \(i=1,\dots,\nu_1\). We get a new tableau \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) of weight \(\nu^t\). We denote this map by \(\lozenge\).
Proof. We have to prove that \(Q\) satisfies \((R4)\), \(Q[r] \ge 2(\ell(\mu^{(r-1)})- n),\) for each \(r=1,\dots,\nu_1\).
Let \(r>0\), \(n< i\le \ell(\mu^{(r-1)})\) and \((i,j)\) a cell in the vertical strip \(J_r\) such that \(T(i,j)=2k+1\), for some \(k>0\). Since \(T\) is an LR-Sundaram tableau it implies \(n+k\ge i\). One then has \(Q(i,j)=r\) and \(Q_{\le i}[r]=2k+1\) and necessarily \[Q_{\le i}[r]=2k+1> 2(i-n)\Leftrightarrow Q_{\le i}[r]=2k+1\ge 2(i-n)+1\]
Otherwise one obtains \[\begin{align} 2k+1\le 2(i-n)\Leftrightarrow 2k+2n\le 2i-1\Leftrightarrow k+n\le i-1/2\Rightarrow k+n<i \end{align}\] A contradiction with the Sundaram condition.
If \(T(i+1,j')=2(k+1)\) for some \(j'\le j\) and the cell \((i+1,j') \in J_r\) then \(Q(i+1,j')=r\) and \[Q_{\le i+1}[r]=2(k+1)=2k+2\ge 2(i-n)+1+1=2(i+1-n)\]
By induction on \(\nu_1\). If \(\nu_1=1\), then \(\nu=(1^m, 0^{2n-m})\) with \(m\) even. Then \(T\) is the vertical strip \(\lambda/\mu\) filled with \(\{1,\dots,m\) and \(Q[1]=m\). If \(\ell(\lambda)- n\le 0\) there is nothing to prove, \(Q[1]=m>2(\ell(\lambda)-n)\). Otherwise we have to check the \(\ell(\lambda)-n\) entries \(T(n+1,1)=m-(\ell(\lambda)-n)+1,\dots, T(\ell(\lambda),1)=m\).
If \(\ell(\lambda)-n\) is even the entries form a sequence of \(\frac{\ell(\lambda)-n}{2}\) pairs of odd and even pairs in consecutive rows in the same column and this case has been already studied above. In particular \(Q[1]=Q_{\le \ell(\lambda)}[1]=m>2(\ell(\lambda)-n)\)
If \(\ell(\lambda)-n\) is odd we have just to check \(T(n+1,1)=even\le m\) because the remaining \(\ell(\lambda)-n-1\) entries were already studied above. In particular \(Q[1]=Q_{\le \ell(\lambda)}[1]=m>2(\ell(\lambda)-n)\).
Indeed, \(Q_{\le n+1}[1]=T(n+1,1)=even\ge 2(n+1-m)=2\)
For \(\nu_1>1\), let \(T^{(1)}\) be the LRS tableau of shape \(\mu^{(1)}\) obtained from \(T\) by suppressing the string \(J_1\). Then \(T^1\in LRS(\mu^1/\mu)\), of weight the even partition \(\nu^{(1)}=\nu-(1^{\ell(\nu)},0^{2n-\ell(\nu)})\), that is, its columns are still even and its conjugate partition is \((\nu'_2,\dots,\nu'_{\nu_1})\) and with strings \(J_2,\dots J_{\nu_1}\). By induction one has
\[Q[r] \ge 2(\ell(\mu^{(r-1)})- n),for each r=2,\dots,\nu_1\] and for each \(r=2,\dots,\nu_1\) \[Q_{\le i}[r] \ge 2(i- n),for 1\le i \le \ell(\mu^{(r-1)})\]
On the other hand because \(T\) is an LR tableau
\[Q_{\le r}[k + 1] \le Q_{\le r}[k],for r\ge 1, k>0.\]
We want to show that
\[Q[1] \ge 2(\ell(\mu^{(0)})- n)=2(\ell(\lambda)- n)\]
One has \(\ell(\lambda)\ge \ell(\mu^{(1)})\) and \(\mu^{(1)}\subseteq \lambda\), henceforth \[Q_{\le \ell(\mu^{(1)})}[1]\ge Q_{\le \ell(\mu^{(1)})}[2]\ge 2(\ell(\mu^{(1)})- n)\]
If \(\ell(\lambda)= \ell(\mu^{(1)})\), it is done. Otherwise, \(\lambda=(\mu^{(1)},1^{\ell(\lambda)- \ell(\mu^{(1)})})\) (we are omitting the zero entries in \(\lambda\) and \(\mu^{(1)}\)). It remains to check the \(m=\ell(\lambda)-\ell(\mu^{(1)})(\Leftrightarrow \ell(\lambda)=\ell(\mu^{(1)})+m)\) entries of \(T\),
\(T(\ell(\mu^{(1)})+1,1)=Q_{\le \ell(\mu^{(1)})}[1]+1,\dots, T(\ell(\lambda),1)=Q_{\le \ell(\mu^{(1)})}[1]+m=Q[1]\).
As in the case \(\nu_1=1\) it remains to study the case \(m=odd\Leftrightarrow T(\ell(\mu^{(1)})+1,1)=Q_{\le \ell(\mu^{(1)})}[1]+1=even\Leftrightarrow Q_{\le \ell(\mu^{(1)})}[1]=odd\). If \(Q_{\le \ell(\mu^{(1)})}[1]=odd\), then
\[Q_{\le \ell(\mu^{(1)})}[1]=odd\ge Q_{\le \ell(\mu^{(1)})}[2]\ge 2(\ell(\mu^{(1)})- n)=even\Rightarrow Q_{\le \ell(\mu^{(1)})}[1]+1\ge 2(\ell(\mu^{(1)})+1- n)\]
Hence \[Q_{\le \ell(\mu^{(1)})}[1]\ge 2(\ell(\mu^{(1)})- n)+1\Rightarrow Q_{\le \ell(\mu^{(1)})}[1]+1\ge 2(\ell(\mu^{(1)})- n)+2=2((\ell(\mu^{(1)})+1- n)\]
Finally, since the cell \((\ell(\mu^{(1)})+1,1) \in J_1\) it means that \(Q(\ell(\mu^{(1)})+1,1)=1\) and \[Q_{\le \ell(\mu^{(1)})+1}[1]=Q_{\le \ell(\mu^{(1)})}[1]+1\ge 2((\ell(\mu^{(1)})+1- n)\] as desired. (We have assumed that \(n\le \ell(\mu^{(1)})\) which is the worst case.) ◻
Example 1. Let \(n=3\), \(\lambda=(4,3,2,2,1,0)\), \(\mu=(3,1,0)\), \(\nu=(3,3,1,1,0^2)\), \(\nu^t=(4,2,2)\) and \(T\in LRS(\lambda/\mu, \nu)\) and \(Q\in \widetilde{R}ec_{6}(\lambda/\mu,\nu^t)\) \[\begin{align} Q= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{},{},{},{1}}, {{},{2},{1}}, {{3},{2}}, {{3},{1}}, {{1}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in \widetilde{R}ec_{6}(\lambda/\mu,\nu^t)\underset{\sim}{\overset{\lozenge}\longleftarrow} T= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{},{},{},{1}}, {{},{1},{2}}, {{1},{2}}, {{2},{3}}, {{4}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in LRS_6(\lambda/\mu,\nu) \end{align}\] with \(\mathrm{wt}(Q)=(4,2,2,0^3)=\nu^t=(Q[1],Q[2],Q[3])\) and the conjugate partition is the even partition \(\nu=(3,3,1,1,0^2)\). The tableaux \(Q\) and \(T\) are both defined by the sequence of nested partitions where \(N=3=\nu_1\),
\[\mu^{(0)}=\lambda\supset_{vert} \mu^{(1)}=(3,2,2,1,0^2)\supset_{vert} \mu^{(2)}=(3,1,1,1,0^3)\supset_{vert}\mu^{(3)} =\mu\]
This decomposes \(\lambda/\mu\) into vertical strips \[J_1:=\mu^{(0)}/\mu^{(1)},J_2:= \mu^{(1)}/\mu^{(2)}, J_3=\mu^{(2)}/\mu\]
The size of the vertical strip \(J_k=\mu^{(k-1)}/\mu^{(k)}\) is \(Q[k]\), \(|J_k|=|\mu^{(k-1)}/\mu^{(k)}|=|\mu^{(k-1)}|-|\mu^{(k)}|=Q[k]\), \(k=1,2,3\). The vertical strip \(J_i\) in \(T\) is filled with the word \(12\dots \nu_i^t=Q[i]\) and in \(Q\) with \(\nu_i^t=Q[i]\) \(i\)’s. One has \(Q[k]\ge Q[k+1]\), condition \((R5)\) is verified \[Q_{\le r}[k + 1] \le Q_{\le r}[k],for r, k>0.\] as well as condition \((R4)\) is verified. The filling of \(J_3\), \(J_2\) and \(J_1\) in \(T\) are respectively the column words \(12\),\(12\) and \(12345\). Furthermore concatenating the reverse column words of the strings \(J_3,J_2, J_1\) in \(T\) gives the Yamanouchi tableau \[Y(\nu)= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{1},1,1}, {{2},2,2}, {{3}}, {{4}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} }\]
the rectification of the word of \(T\).
In [35] a complete account has been provided on the LR-symmetries under the action of the dihedral group \(\mathbb{Z}_2\times { D}_3\). In particular, the LR-symmetry maps are exhibited on Littlewood-Richardson tableaux as well as on the companion pairs. The LR symmetry maps restricted to LR-Sundaram tableaux do not need to return LR-Sundaram tableaux for the same \(n\) but its characterization play an important role on branching rules from \(Gl_{2n}( \mathbb{C})\) to \(Sp_{2n}( \mathbb{C})\). Such restriction, in the case of the fundamental symmetry map, \(c_{\mu,\nu,\lambda}=c_{\nu,\mu,\lambda}\), has been studied by Kumar-Torres in [24] via hives which in turn reduces to the restriction of right and left companions as discussed in [25], [40]. The restriction of the orthogonal-transpose linear map \(\blacklozenge\) [35], [41] exhibiting \(c_{\mu,\nu,\lambda}=c_{\lambda^t,\nu^t,\mu^t}\) for LR tableaux, to LR-Sundaram tableaux has not been studied before but surfaces in the branching rule defined by the quantum Littlewood-Richardson map. It turns out that the \(Rec_{2n}(\lambda/\mu)=\widetilde{R}ec_{2n}(\lambda/\mu\) characterizes explicitly the restriuction of the LR orthogonal transpose symmetry map on LR-Sundaram tableaux. For convenience let us consider our partitions \(\mu,\nu,\lambda\) inside of a fixed rectangle and define \(\check{\lambda}\) the complement of the Young diagram of \(\lambda\) inside the given rectangle. With this convention \(LRS_{2n}(\mu,\nu,\lambda)\) denotes the set of LR-Sundaram tableaux of shape \({\lambda^\vee}/\mu\) and weight \(\nu\) and similarly for \(\widetilde{R}ec_{2n}(\mu,\nu^t,\lambda)\). See Example 2.
Denote the bijection in Theorem 1 by \(\lozenge\). Then the \(LR\) orthogonal transpose symmetry map \(\blacklozenge\) [35] is defined by \[\begin{align} \blacklozenge:& LRS_{2n}(\mu,\nu,\lambda)&\overset{\lozenge}\longrightarrow &\widetilde{R}ec_{2n}(\mu,\nu^t,\lambda)&\underset{\pi\circ t}\hookrightarrow& LR({\lambda^t},\nu^t,\mu^t)&\\ &T&\mapsto& Q&\mapsto& Q^{\pi\circ t}=\blacklozenge T& \end{align}\] where \(Q^{\pi\circ t}\) means the \(\pi\)-rotation (transposition) followed with transposition (rotation) of \(D(\lambda)\). The extra condition \((R3)\), Definition [def:tilderec], on \(Q\) characterizes the LR tableau \(\blacklozenge T\), the image of the LRS tableau \(T\) by the orthogonal transpose map \(\blacklozenge\). The bijection \(\lozenge\) and its inverse is so natural that we can intertwine LR-Sundaram tableaux with their \(Q\) presentations in \(\widetilde{R}ec_{2n}(\mu,\nu^t,\lambda)\). For notation and further details we refer the reader to [35], [41], [42]. We resume to Example 1.
Example 2. Let \(n=3\), \(\lambda^\vee=(4,3,2,2,1)\), \(\lambda=(3,2,2,1)\) \(\mu=(3,1,0)\), \(\mu^t=(2,1,1,0)\), \(\nu=(3,3,1,1,0^2)\), \(\nu^t=(4,2,2)\) and \(T\in LRS_6(\mu, \nu,\lambda)\) \[\begin{align} T= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{},{},{},{1}}, {{},{1},{2},{}}, {{1},{2},{},{}}, {{2},{3},{},{}}, {{4},{},{},{}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in LRS_6(\mu,\nu,\lambda)\mapsto Q= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{},{},{},{1}}, {{},{2},{1},{}}, {{3},{2},{},{}}, {{3},{1},{},{}}, {{1},{},{},{}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in \widetilde{R}ec_{6}(\mu,\nu^t,\lambda)\mapsto Q^{\pi\circ t}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{},{},{},{},{1}}, {{},{},{},{1},{}}, {{},{1},{2},2,{}}, {{1},{3},{3},{},{}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =\blacklozenge T\nonumber\\ \blacklozenge T \in LR({\lambda^t},\nu^t,\mu^t).\nonumber \end{align}\]
The action of the symmetry map \(\blacklozenge\) restricted to \(LRS_{2n}(\mu, \nu,\lambda)\) returns tableaux in \(LR({\lambda^t},\nu^t,\mu^t)\) satisfying the following conditions on the parts of \(\nu^t\): \(\nu_k^t \in 2 \mathbb{Z}\), and \(\nu_k^t\ge 2(\ell(\mu^{(k-1)})-n)\), for all \(k\ge 1\), where \(\mu^{(k-1)}\) is the partition such that \[D(\mu^{(k-1)}) = D(\mu) \cup\{(i, j) \in D(\lambda/\mu) | Q(i, j) \ge k\}.\]
From now on we fix \(n\in\mathbb{N}\).
Definition 3. [43] A semistandard tableau \(G \in SST_{2n}(\lambda)\) is said to be symplectic if \[G(k, 1) \ge 2k-1,for allk \in [ \ell(\lambda)].\] Let \(SpT_{2n}(\lambda)\) denote the set of all symplectic tableaux of shape \(\gamma\) on the alphabet \([2n]\).
Proposition 2. Let \(G \in SST_{2n}(\lambda)\).
[1] If \(SpT_{2n}(\lambda)\neq\emptyset\) then \(\ell(\lambda)\le n\).
If \(G=(a_1,\dots,a_l)\in SpT_{2n}(\varpi_l)\) then \(l\le n\) and \(l\le \lfloor\frac{a_l+1}{2}\rfloor\).
[1]If \(G\) is not symplectic, then there exists a unique \(i \in[2, 2n]\) such that \[\begin{align} G(i, 1) < 2i -1andG(k, 1) \ge 2k -1for all k \in [1, i -1 ].\label{G1} \end{align}\qquad{(3)}\] Moreover, we have \[\begin{align} G(i - 1, 1) = 2i - 3=2(i-1)-1and G(i, 1) = 2i -2=2(i-1) .\label{G2} \end{align}\qquad{(4)}\]
Proof. (2) follows from \((1)\) and since \(G\) is symplectic, \(a_l\ge 2l-1\Leftrightarrow l\le \frac{a_l+1}{2}\Rightarrow l\le \lfloor\frac{a_l+1}{2}\rfloor\). ◻
Conditions ?? and ?? say that the first symplectic fail needs to be checked in the first column of \(G\) among consecutive integer entries consisting of an odd number followed with an even number, and if the fail occurs it happens for the first time in a such even entry. In other words, if the first column has no consecutive integers consisting of an odd number followed with an even number it is symplectic.
Corollary 2. The following holds:
For \(0\le l\le n\), \(Sp_{2n}(\varpi_l)\neq \emptyset\). Namely, \(Sp_{2n}(())=\{()\}\), and, for \(1\le l\le n\), \(G=(1,3, \dots,2l-1)\in Sp_{2n}(\varpi_l)\) or \(H=(2,4,\dots, 2l)\in Sp_{2n}(\varpi_l)\). In particular, \(Sp_{2n}(\varpi_1)=SST_{2n}(\varpi_1)\).
\(SpT_{2n}(\lambda)\neq\emptyset\) if and only if \(\ell(\lambda)\le n\).
If \(G'\) is obtained from \(G\in Sp_{2n}(\varpi_l)\) by suppressing \(0\le t_0\le l\) entries then \(G'\in Sp_{2n}(\varpi_{l-t_0})\) and the suppressed part \(G''\in Sp_{2n}(\varpi_{t_0})\).
Let \(\mu\in Par_{\le n}\). If \(G'\) is obtained from \(G\in Sp_{2n}(\mu)\) by reverse column insertion, then \(G'\) is still symplectic and \(G'\in Sp_{2n}(\mu')\) for some partition \(\mu'\subset_{vert}\mu\).
Let \(\lambda\in Par_\le n\), \[SpT_{2n}(\lambda)=\{S\in SST_{2n}(\lambda)| (S(1,1),\cdots,S(\ell(\lambda),1)\in SpT_{2n}(\varpi_{\ell(\lambda)}) \}.\]
Proof. \((a)\) The \(i\)th entry of \(G\) is \(2i-1\), and the \(i\)th entry of \(H\) is \(2i>2i-1\), for \(i=1,\dots,l\).
\((b)\) The "only if part" follows from Proposition 2.
The "if part" is a consequence of \((a)\). Let \(\ell(\lambda)=l\) and let \(T\) with first column \(G\) or \(H\) and the remain columns of \(T\) added according to the semistandard-ness and with entries not exceeding \(2n\). Then \(T\in Sp_{2n}(T)\).
\((c)\) Straightforward from Definition 3.
\((d)\) During the reverse bumping route a box is deleted from the bottom of a column. If the column is the first the process stops and returns a symplectic tableau with one less box. If not, the bumped entry proceeds to the column on its left and looks for the largest entry which smaller or equal and replaces while bumping it. The procedure terminates when an entry is bumped while being replaced by an equal or larger entry. Therefore, the returned tableau \(G'\) is symplectic and \(G'\in Sp_{2n}(\mu')\) for some partition \(\mu'\subset_{vert}\mu\). ◻
Example 3. Let \(G\in SST_{10}(1^6)\) be the column \((1,3,4,6,8,9)\). For the first symplectic fail we consider the pair \(3,4\) where \(3=2\times (3-1)-1\) and symplectic fail occurs in the entry \(4=2(3-1)\). Removing the entries \(3,4\) it remains the symplectic column \((1,6,8,9)\in Sp_{10}(1^6)\).
Let \(G\in SST_{6}(1^6)\) with column \((1,2,4,5,6)\). The first fail occurs in the entry \(2=2(2-1)\). Removing the entries \(1,2\) it remains the symplectic column \((4,5,6)\in Sp_6(1^3)\)
Let \(G\in SST_{14}(1^{10})\) be the column \((1,3,4,5,6,7,11,12,13,14)\). The first symplectic fail occurs in the entry \(4\). Removing the pair \(3,4\) it remains \((1,5,6,7,11,12,13,14)\) still not symplectic, it fails in the entry \(14=2(8-1)\) .
We fix \(\lambda\in Par_{\le 2n}\). The Littlewood–Richardson map of type AII, \({\mathsf{LR}}^{AII}\) is the algorithm [1], [44] which takes \(T \in SST_{2n}(\lambda)\) as input, and returns a pair of tableaux \((P^{II}(T),Q^{II}(T))\in SpT_{2n}(\mu)\times \widetilde{R}ec_{2n}(\lambda/\mu)\).
Set \[\begin{align} \label{rec61rec2}Rec_{2n}(\lambda/\mu) := \{Q^{II}(T) | T \in SST_{2n}(\lambda)such that sh(P^{II}(T)) = \mu \}. \end{align}\tag{10}\]
An explicit description of the set \(Rec_{2n}(\lambda/\mu)\) is given in [Wat25, Theorem 3.1.4 (2)], \[Rec_{2n}(\lambda/\mu)=\widetilde{R}ec_{2n}(\lambda/\mu)\] by providing a combinatorial proof for the inclusion \(Rec_{2n}(\lambda/\mu)\subset \widetilde{R}ec_{2n}(\lambda/\mu)\) while the inclusion \(Rec_{2n}(\lambda/\mu)\supset \widetilde{R}ec_{2n}(\lambda/\mu)\) is concluded via representation theory. In turn it is shown in [1] that \[\widetilde{R}ec_{2n}(\lambda/\mu)\overset{\sim}\rightarrow LRS_{2n}\] whose combinatorial proof for the surjectivity we have given in Theorem 1. Our goal now is to prove combinatorially the inclusion \(Rec_{2n}(\lambda/\mu)\supseteq \widetilde{R}ec_{2n}(\lambda/\mu)\). This is equivalent to provide an algorithm to compute \({{\mathsf{LR}}^{AII}}^{-1}\).
For the reader convenience this section recalls several properties of the removal and reduction maps in [1]. We fix \(l \in [0, 2n]\) and \(\mathbf{a} = (a_1, \dots , a_l)\) a column in \(SST_{2n}(\varpi_l)\). We often regard \(\mathbf{a}\) as a set.
The removal subword of \(\mathbf{a}\) [1] is defined to be the subword \(\mathrm{rem}(\mathbf{a})\) of \(\mathbf{a}\) obtained by the following recursive formula:
\[\begin{align} \label{removal} \mathrm{rem}(\mathbf{a}) := \begin{cases} \emptyset,&ifl\le 1,\\ \mathrm{rem}(a_l,\dots, a_{l-2}) (a_{l-1},a_l),&ifl\ge 2, a_l\in 2\mathbb{Z}, a_{l-1} = a_l-1,\\ & anda_l < 2l -| \mathrm{rem}(a_1,\dots,a_{l-2})|-1,\\ \mathrm{rem}(a_1,\dots,a_{l-1})&otherwise. \end{cases} \end{align}\tag{11}\]
Definition 4. [1] For the column \(\mathbf{a}\), the new column \(\mathrm{red}(\mathbf{a})\), reduction of \(\mathbf{a}\), is defined to be the one obtained from \(\mathbf{a}\) by removing the entries in the set \(\mathrm{rem}(\mathbf{a})\), \[\mathrm{red}( \mathbf{a})=\mathbf{a}\setminus \mathrm{rem}(\mathbf{a})\in SST_{2n}(\varpi_{l-| \mathrm{rem}(\mathbf{a})|}).\]
In fact we shall see in Proposition 5 [1] that \(\mathrm{red}( \mathbf{a})\in SpT_{2n}(\varpi_{l-| \mathrm{rem}(\mathbf{a})|})\) where \(l-| \mathrm{rem}(\mathbf{a})|\) satisfy further conditions.
Some useful properties of removable entries in \(\mathbf{a}\).
Proposition 3. [1]
If \(a_l \in \mathrm{rem}(\mathbf{a})\), then \(a_l\in 2\mathbb{Z}\).
If \(a_l\notin \mathrm{rem}(\mathbf{a})\) then \(\mathrm{rem}(\mathbf{a})= rem(a_1,\dots, a_{l-1})\).
If \(a_l\) is odd then \(a_l\notin \mathrm{rem}(a)\), and \(\mathrm{rem}(\mathbf{a})= \mathrm{rem}(a_1,\dots, a_{l-1})\).
If \(a_l \in \mathrm{rem}(\mathbf{a})\), then \(a_l\in 2\mathbb{Z}\) and \(\mathrm{rem}(a_1,\dots, a_{l-1})= \mathrm{rem}(a_1,\dots, a_{l-2})\).
\(\mathrm{rem}(a_1,\dots, a_k) \subset \mathrm{rem}(\mathbf{a})\) for all \(k \in[0,l]\).
Points \((2)\) and \((3)\) together in the next proposition are an alternative to Definition 4 to compute \(\mathrm{rem}(\boldsymbol{a})\) and allowing its computation easily.
For each \(x\in\mathbb{Z}\), set \(s(x)=x+1\), if \(x\notin 2\mathbb{Z}\), and \(s(x)=x-1\), if \(x\in 2\mathbb{Z}\).
Proposition 4. [1]
If \(i\in [1,l]\) and \(a_j\notin \mathrm{rem}(\mathbf{a})\), for \(j\in[i,l]\) then \(\mathrm{rem}(\mathbf{a})= \mathrm{rem}(a_1,\dots,a_{i-1})\).
for each \(i\in [1,l]\), \(a_i\in \mathrm{rem}(\mathbf{a})\) if and only if one of the following holds
\(a_i\) odd, \(i<l\), \(a_{i+1}=a_i+1\) and \(a_i<2i-| \mathrm{rem}(a_1,\dots,a_{i-1})|\)
\(a_i\) even, \(i>1\), \(a_{i}=a_{i-1}+1\) and \(a_i<2i-| \mathrm{rem}(a_1,\dots,a_{i-2})|-1\)
\(a_i \in \mathrm{rem}(\mathbf{a})\) if and only if \(s(a_i) \in \mathrm{rem}(\mathbf{a})\). Consequently, \(| \mathrm{rem}(\mathbf{a})|\in 2\mathbb{Z}\).
For \(i\in[1,l]\),
\(a_i\) odd, then \(| \mathrm{rem}(a_1,\dots,a_{i-1})|\ge 2i-a_i-1\)
\(a_i\) even, then \(| \mathrm{rem}(a_1,\dots,a_{i})|\ge 2i-a_i\)
Example 4. Let \(n=4\),
\(l=6\), \(\mathrm{rem}(123456)=(123456)\), \(\mathrm{red}(123456)=()\).
\(l=4\), \(\mathrm{rem}(1256)=(1,2)\) since \(5>2\times 3-2\), \(\mathrm{red}(1256)=(56)\).
\(l=5\), \(\mathrm{rem}(4,5,6,7,8)=(7,8)\), \(\mathrm{red}(4,5,6,7,8)=(4,5,6)\).
\(l=5\), \(\mathrm{rem}(1,2,3,4,8)=(1,2,3,4)\), \(\mathrm{red}(1,2,3,4,8)=(8)\).
\(\mathrm{rem}(1245678)=(125678)\), \(7<2\times 6-4\), \(4=6-2\)
\(\mathrm{rem}(12478)=(12)\), \(7\nless 2\times 4-2=6\)
The next corollary follows from the definition of removal subword in 11 and item \((2)\) in the previous proposition.
Corollary 3. For \(l \in [0, 2n]\),
\(l\) even \(\Rightarrow \mathrm{rem}(1,2,\dots,l)=(1,2,\dots,l)=\mathrm{red}(1,2,\dots,l)=\emptyset\).
\(l\) odd \(\Rightarrow \mathrm{rem}(1,2,\dots,l)= \mathrm{rem}(1,2,\dots,l-1)= (1,2,\dots,l-1)\Rightarrow \mathrm{red}(1,2,\dots,l)=\{l\}\).
\(\mathrm{rem}(\mathbf{a})=\mathbf{a}\) if and only if \(l\) is even and \(\mathbf{a}=(1,2,\dots,l)\).
\(\mathrm{red}(\mathbf{a})=\emptyset\) if and only if \(l\) is even and \(\mathbf{a}=(1,2,\dots,l)\).
Proof. "Only if part of (3)". For \(l=0\), there is nothing to prove, \(\mathbf{a}=\emptyset\) and \(rem(\emptyset)=\emptyset\). If \(l=1\), it follows from 11 , \(rem(a)=\emptyset\). Let \(l\ge 2\), and \(\mathbf{a}=(a_1,\dots,a_l)\). Let us prove that \(a_1,a_2\in rem(a) \Leftrightarrow a_1=1\) and \(a_2=2\). From Proposition 4, (2),
\[a_1\in \mathrm{rem}(a)\Leftrightarrow 1\le a_1odd, a_{2}=a_1+1anda_1<2-| \mathrm{rem}(\emptyset)|=2\]
This implies \(a_1=1\) and \(a_2=2\)
\[a_2=2\in \mathrm{rem}(a)\Leftrightarrow 2=a_{2}=a_1+1=1+1even and2<2\time 2-| \mathrm{rem}(\emptyset)|-1=4-1=3\]
By induction assume \(a_1,\dots, a_k\in \mathrm{rem}(a_1,\dots,a_k,\dots, a_{l})\Leftrightarrow\) \(k\) even and \(a_i=i\), for \(i=1,\dots, k\). Then from Proposition 4, (2), \(a_{k+1}\in \mathrm{rem}(a)\Leftrightarrow\) \(a_{k+1}\) odd, \(k+1<l\), \(a_{k+2}=a_{k+1}+1\) and \[k=a_k~even<a_{k+1}~odd <2{k+1}-| \mathrm{rem}(a_1,\dots,a_{k})|=2(k+1)-k=k+2\] (Note that \(a_{k+1}\) even \(\Rightarrow a_{k+1}=a_{k}+1\) odd which is absurd.)
Therefore \(a_{k+1}=k+1\) odd and \(a_{k+2}=k+2\) even. Also \(a_{k+2}=k+2\in \mathrm{rem}(a)\) since \(a_{k+2}\) even, \(k+2>1\), \(k+2=a_{k+2}=a_{k+1}+1\) and \(k+2=a_{k+2}<2(k+2)-| \mathrm{rem}(a_1,\dots,a_{k})|-1=2(k+2)-k-1=k+3\). ◻
Some properties of the reduction and successor maps follow. More precisely, the reduction map on columns is the injective assignment as shown in the next proposition.
Proposition 5. [1], [32] Let \(l\in [0,2n]\) and \(\mathbf{a}=(a_1,\dots,a_l) \in SST_{2n}(\varpi_l)\). Then
\(\mathrm{red}(\mathbf{a})\) is symplectic.
\(\mathrm{red}(\mathbf{a})=\mathbf{a}\) if and only if \(\mathbf{a}\) is symplectic.
[1] The reduction map \(\mathrm{red}\) on \(SST_{2n}(\varpi_l)\) is an injective assignment
\[\begin{align} \mathrm{red}:SST_{2n}(\varpi_l)&\rightarrow\bigsqcup_{\begin{smallmatrix} 0\le t\le min\{l,2n-l\}\\ l-t\in 2\mathbb{Z} \end{smallmatrix}}SpT_{2n}(\varpi_t)\nonumber\\ \mathbf{a}&\mapsto \mathrm{red}( \mathbf{a})=\mathbf{a}\setminus \mathrm{rem}(\mathbf{a}). \label{redumap} \end{align}\qquad{(5)}\]
Definition 5. [1] Let \(S \in SST_{2n}\). Let \(S_1\) denote the first column of \(S\), and \(S_{\ge 2}\) the rest of the tableau \(S\). The new tableau \(\mathrm{suc}(S)\) is defined to be
\[\begin{align} \mathrm{suc}(S) := \mathrm{red}(S_1) \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}_{\ge 2}the Schensted column insertion of \mathrm{red}({S_1}) in S_{\ge 2}. \end{align}\] In addition the successor* map on \(SST_{2n}\) is the assignment \[\begin{align} SST_{2n}(\lambda)&\rightarrow \bigsqcup_{\begin{smallmatrix} \mu\in Par_{\le 2n} \end{smallmatrix}} SST_{2n}(\mu)\\ S&\mapsto \mathrm{suc}(S)=\mathrm{red}(S_1) \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}_{\ge 2} \end{align}\]*
In particular, if \(S\) is a column \(\mathrm{suc}(S)=\mathrm{red}(S)\). From [1] the successor map is also injective.
Proposition 6. [1], [32] Let \(T\in SST_{2n}(\lambda)\) and let \(\mu\) be the shape of \(\mathrm{suc}(T)\). Then
\(\mu\subset_{vert} \lambda\), \(\lambda/\mu\) is a vertical strip.
\(\lambda=\mu\) if and only if \(T= \mathrm{suc}(T)\).
\(T\) is symplectic if and only if \(\mathrm{suc}(T)=T\).
The \(\mathrm{suc}\) map on \(SST_{2n}(\lambda)\) is injective.
For \(T\in SST_{2n}(\lambda)\)define \(\mathrm{suc}^k(T):= \mathrm{suc}( \mathrm{suc}^{k-1}(T))\), for \(k\ge 1\). This sequence stabilizes in a finite number of iterations, that is, there is \(N\ge 0\) such that \(\mathrm{suc}^{N+1}(T)= \mathrm{suc}^{N}(T)\). In other words, \(\mathrm{suc}^{N}(T)\) is symplectic, for some \(N\ge 0\).
Definition 6. [1] Let \(T\in SST_{2n}(\lambda)\) and let \(N\) be a nonnegative integer satisfying \(suc^{N+1}(T)= \mathrm{suc}^{N}(T)\). Set \(P^{AII}(T)= \mathrm{suc}^N(T)\) and \(Q^{AII}(T)\) to be the tableau that records the process of transformations from \(T\) to \(P^{AII}(T)\). Let the shapes of \(T\), \(\mathrm{suc}^1(T)\), \(\mathrm{suc}^2(T),\dots, \mathrm{suc}^N(T)\) be \(\lambda^0=\lambda\supset_{vert}\lambda^1\supset_{vert}\lambda^2\supset_{vert}\dots\supset_{vert}\lambda^N=sh(P^{AII}(T))\) respectively, and define \(Q^{AII}(T)\) as the tableau obtained by placing the entry \(j\) in the vertical strip \(\lambda^{(j-1)}/\lambda^j\) for \(j=1,2,\dots,N\)
With the setting in the previous definition, Watanabe has shown combinatorially in [1] that \(Rec_{2n}(\lambda/\mu)\subset \widetilde{R}ec_{2n}(\lambda/\mu)\) and thus that \[\begin{align} \label{lrIIsurj} {\mathsf{LR}^{AII}}: SST_{2n}(\lambda) &\longrightarrow \bigsqcup_{\begin{smallmatrix}\mu\in Par_{\le n}\\ \mu\subset\lambda\end{smallmatrix}}SpT_{2n}(\mu) \times \widetilde{R}ec_{2n}(\lambda/\mu)\nonumber\\ T&\mapsto (P^{AII}(T)=suc^{N}(T),Q^{AII}(T)) \end{align}\tag{12}\] is an injection. In particular, \(SpT_{2n}(\lambda)\subset SST_{2n}(\lambda)\) and \(LR^{AII}(S)=(\mathrm{suc}^1(S),Q(S)=(S,D(\lambda/\lambda))=(S,\emptyset)\) for all \(S\in SpT_{2n}(\lambda)\).
We now show that the reduction map ?? is also surjective. For a fixed \(n\in\mathbb{N}\), let \(l\in [0,2n]\) and \(t\in[0,n]\) such that \(0\le t\le min\{l,2n-l\}\) and \(l-t\in 2 \mathbb{Z}\). Then we define the expanding map
\[\begin{align} \label{exp} \mathrm{exp}_t:SpT_{2n}(\varpi_t)&\rightarrow SST_{2n}(\varpi_l)\quad\\ \mathbf{a}&\mapsto\mathrm{exp}(\mathbf{a}):=T,\nonumber \end{align}\tag{13}\] such that \(\mathrm{red}(\mathrm{exp}(\boldsymbol{a}))=a\). That is, \(\mathrm{red}_{t}^{-1}:=\mathrm{red}^{-1}_{|SpT_{2n}(\varpi_t)}=\mathrm{exp}_t\).
The next lemma and the two theorems show that the reduction map, \(\mathrm{red}\), ?? , [1], [32], is surjective and explicitly exhibits its inverse in the cases where the symplectic column decomposes into non empty factors \(\boldsymbol{A}_i\) of even length consisting of consecutive integers starting with an odd number, and when the symplectic column is such that consecutive integers occur only as an even number followed with an odd number. For example, for the first case, we mean \((5678)\in SpT_{10}(\varpi_4)\) or \((78)( 13, 14, 15, 16)\in SpT_{20}(\varpi_6)\) or \((7,8)(13,14,15,16) (19,20\), and for the second case \((2,3,6,7)\) or \((1,4,5,7,10,12)\).
Lemma 1. Let \(l\in [0,2n]\) and \(t\in[n]\cap 2 \mathbb{Z}\) such that \(2\le t\le min\{l,2n-l\}\) and \(l-t\in 2 \mathbb{Z}\). Let \(a_1\notin 2 \mathbb{Z}\), and \(\mathbf{a} = (a_1, a_1+1,\dots , a_1+t-2,a_1+t-1)\in SpT_{2n}(\varpi_t)\). Then \(a_1\ge t+1\), and \[\begin{align} \label{lem:inverseredx} \mathrm{red}_{t}^{-1}:=\mathrm{red}^{-1}_{|SpT_{2n}(\varpi_t)}:SpT_{2n}(\varpi_t)&\rightarrow SST_{2n}(\varpi_l)\quad\\ \mathbf{a}&\mapsto\mathrm{red}_{t}^{-1}(\mathbf{a})=T,\nonumber \end{align}\qquad{(6)}\] where \[\begin{align} T=(1,2,\dots, l_1)\mathbf{a}(a_1+t-1+1,\dots, a_1+t-1+l_{t+1})\label{lem:T}. \end{align}\qquad{(7)}\] and \(l_1\), \(l_{t+1}\) are nonnegative even numbers defined by \[\begin{align} \label{lem0} l_1=min\{a_1-t-1,l-t\},\quad l_{t+1}=l-t-l_1. \end{align}\qquad{(8)}\]
Proof. If \({\boldsymbol{a}}\in SpT_{2n}(\varpi_t)\) then \(a_1+t-1\ge 2t-1\Leftrightarrow a_1\ge t\Leftrightarrow a_1\ge t+1\) since \(a_1\notin 2 \mathbb{Z}\) and \(t\in 2 \mathbb{Z}\). This condition guarantees that, for \(0\le j\le t-1\), \[a_1+j\ge t+1+j\ge j+2+j=2(j+1)>2(j+1)-1.\] The entries of column \(\boldsymbol{a}\) are bounded by \(2n\). In the worst case when \(t=n\), one has for \(a_1=t+1\), \(a_1+t-1=t+1+t-1=2t= 2n\). Henceforth in the worst case, for \(a_1=t+1\), \({\boldsymbol{a}}\in SpT_{2n}(\varpi_t)\).
We now show that \(T\in SST_{2n}(\varpi_l)\). Since \(0\le l_1\le l-t\), it follows that \(l_{t+1}=l-t-l_1\ge l-t-(l-t)\ge 0\) is a non negative even number such that \(l_{t+1}=0\) when \(l_1=l-t\). Therefore \(\ell (T)=l_1+t+l_{t+1}=l\), and, in addition the entries are bounded by \(2n\). In fact, if \(l_1=a_1-t-1<l-t\), one has \[a_1+t-1+l_{t+1}=a_1+t-1+l-t-(a_1-t-1)=l+t\le 2n.\]
We next show that \(\mathrm{red}(T)=\boldsymbol{a}\). Note \(a_1+t-2\notin 2 \mathbb{Z}, a_1+t-1\in 2 \mathbb{Z}\) and \[\begin{align} &T=(1,2,\dots, a_1-t-1)\mathbf{a}(a_1+t-1+1,\dots, a_1+t-1+l_{t+1})\nonumber\\ =&(1,2,\dots, a_1-t-1)(a_1,a_1+1,\dots, a_1+t-2)(a_1+t-1,a_1+t-1+1,\dots, a_1+t-1+l_{t+1}). \end{align}\]
From Corollary 3, one has \(\mathrm{rem}(1,2,\dots, a_1-t-1)=(1,2,\dots, a_1-t-1)\). From Proposition 3, \((5)\),and Proposition 4, one has \[\mathrm{rem}(1,2,\dots, a_1-t-1)(a_1, a_1+1,\dots , a_1+t-2,a_1+t-1)=(1,2,\dots, a_1-t-1)\] because for \(0\le j\le t-2\) and \(j\in 2 \mathbb{Z}\), \[\begin{align} a_1+j\nless 2(j+1+a_1-t-1)-(a_1-t-1)-j=a_1+j-t. \end{align}\]
Then, when \(l_{t+1}=l-t-l_1>0\Leftrightarrow l_1=a_1-t-1<l-t\), one has \[\begin{align} & \mathrm{rem}((1,2,\dots, a_1-t-1)(a_1,a_1+1,\dots, a_1+t-2)(a_1+t-1,a_1+t-1+1,\dots, a_1+t-1+l_{t+1}))\\ &=(1,2,\dots, a_1-t-1)\cup \mathrm{rem}(a_1+t-1+1,\dots, a_1+t-1+l_{t+1}) \\ &=(1,2,\dots, a_1-t-1)(a_1+t-1+1,\dots, a_1+t-1+l_{t+1}). \end{align}\] In fact, for \(1\le j\le l_{t+1}\) and \(j\notin 2 \mathbb{Z}\), \[\begin{align} a_1+t-1+j< 2(a_1-1+j)-(a_1-t-1)-(j-1)=2a_1-2+2j-a_1+t+1-j+1=a_1+t+j. \end{align}\] Hence \(\mathrm{red}(T)=\boldsymbol{a}\). ◻
Example 5. Let \(n=7\) and \(t=4\in[6]\cap 2 \mathbb{Z}\)
\(0\le l=8\le 14\) such that \(2\le 4\le min\{l=8,14-8\}\) and \(l-t=8-4\in 2 \mathbb{Z}\). Let \(a_1=7\notin 2 \mathbb{Z}\), and \(a_1=7\ge t+1=5\) such that \(\mathbf{a} = (7, 8, 9,10)\) is a column in \(SpT_{12}(\varpi_4)\). Then \(l_1=a_1-4-1=2\), \(l_5=8-4-2=2\), and \(T=(1,2)(7,8,9,10)(11,12)\in SST_{12}(\varpi_8)\) and \(\mathrm{red}(T)=\mathbf{a}\).
\(0\le l=10\le 14\) such that \(2\le 4\le min\{l=10,14-10\}\) and \(l-t=10-4\in 2 \mathbb{Z}\), and \(a_1=7\ge t+1=5\) such that \(\mathbf{a} = (7, 8, 9,10)\) is a column in \(SpT_{14}(\varpi_4)\). Then \(l_1=a_1-4-1=2\), \(l_5=10-4-2=4\), and \(T=(1,2)(7,8,9,10)(11,12,13,14)\in SST_{14}(\varpi_8)\) and \(\mathrm{red}(T)=\mathbf{a}\). Note \(13<2\times 9-4=14\).
Theorem 2. Let \(l\in [0,2n]\) and \(t_1\neq 0, t_2\neq 0\in 2 \mathbb{Z}\) such that \(t=t_1+t_2\in[n]\), \[4\le t=t_1+t_2\le min\{l,2n-l\}\] and \(l-t\in 2 \mathbb{Z}\). Let \({a_1},{a_2}\notin 2 \mathbb{Z}\) such that \(a_2\ge a_1+t_1+2\), and \[\begin{align} \mathbf{a} &={\boldsymbol{A}_1}{\boldsymbol{A}_2}\in SpT_{2n}(\varpi_{t_1+t_2}) \end{align}\] where \({\boldsymbol{A}_1}=({a_1}, a_1+1,\dots , a_1+t_1-2,a_1+t_1-1)\) and \({\boldsymbol{A}_2}=({a_2}, a_2+1,\dots , a_2+t_2-2,a_2+t_2-1)\). Then,
if \(t_1\ge t_2\), define \[\begin{align} \label{prop:inverseredx2} \mathrm{red}_{t}^{-1}:=\mathrm{red}^{-1}_{|SpT_{2n}(\varpi_t)}:SpT_{2n}(\varpi_t)&\rightarrow SST_{2n}(\varpi_l)\quad\\ \mathbf{a}&\mapsto\mathrm{red}_{t}^{-1}(\mathbf{a}):=T,\nonumber \end{align}\qquad{(9)}\] where \[\begin{align} T&=(1,\dots, l_1){\boldsymbol{A_1}}( a_1+t_1-1+1,\dots, a_1+ t_1-1+l_2){\boldsymbol{A_2}}( {a_2}+ t_2-1+1,\dots, a_2+ t_2-1+l_{t+1}) \end{align}\] and \(l_1, l_2\), \(l_{t+1}\) are nonnegative even numbers defined by \[\begin{align} \label{prop02} l_1=min\{a_1-t_1-1,l-t\},~ l_2=min\{a_2-a_{1}-t_1-2,l-t-l_1\},\quad l_{t+1}=l-t-\sum_{i=1}^{2}l_i. \end{align}\qquad{(10)}\]
if \(t_1<t_2\), define \(\tilde{l}_1=a_2-t_1-t_2-5\in 2 \mathbb{Z}\) and
when \(\tilde{l}_1>a_1-3\) put \(l_1=min\{max\{a_1-3,0\}, l-t\}\), \(l_2=min\{\tilde{l}_1-l_1,l-t-l_1\}\in 2 \mathbb{Z}\), \(l_{t+1}=l-t-\sum_{i=1}^{2}l_i\), and \[\begin{align} \label{aphgbcxy} \mathrm{red}_{t}^{-1}:=\mathrm{red}^{-1}_{|SpT_{2n}(\varpi_t)}:SpT_{2n}(\varpi_t)&\rightarrow SST_{2n}(\varpi_l)\quad\\ \mathbf{a}&\mapsto\mathrm{red}_{t}^{-1}(\mathbf{a}):=T,\nonumber \end{align}\qquad{(12)}\] where \[\begin{align} T&=(1,\dots, l_1){\boldsymbol{A_1}}( a_1+t_1-1+1,\dots, a_1+ t_1-1+l_2){\boldsymbol{A_2}}( {a_2}+ t_2-1+1,\dots, a_2+ t_2-1+l_{t+1}). \end{align}\]
Example 6.
Let \(n=12\), \(l=12\), \(t=6\le min\{12, 24-12\},\) \(l-t=6\), and \({\boldsymbol{a}}=(9,10)(17, 18, 19, 20)\in SpT_{24}(\varpi_6)\), \(t_1=2<t_2=4\), \(\tilde{l}_1=a_2-t_2-t_1-5=17-6-5=6\). Then \(l_1=\tilde{l}_1=6\le a_1-3=6\) and \(l_2=0\), \(T=(123456)(9,10)(17,18,19,20)\in SST_{24}(\varpi_{12})\).
For \({\boldsymbol{a}}=(7,8)(17, 18, 19, 20)\in SpT_{24}(\varpi_6)\), one has \(\tilde{l}_1=6> a_1-3=4\), and
\(T=(1234)(76)(9,10)(17,18,19,20).\)
Proof. \((1)\) If \({\boldsymbol{A}_1}{\boldsymbol{A}_2}\in SpT_{2n}(\varpi_t)\), \(t=t_1+t_2\), with \(a_2\ge a_1+t_1+2\), then \[a_2+t_2-1\ge a_1+t_1-1+3+t_2-1\ge 2(t_1+t_2)-1\Leftrightarrow a_1\ge t_1+t_2-2\Leftrightarrow a_1\ge t_1+t_2-1\] since \(a_1\notin 2 \mathbb{Z}\) and \(t\in 2 \mathbb{Z}\). This condition guarantees that, \(a_1\ge t_1+1\) and for \(0\le j\le t_2-1\Leftrightarrow 1\le j+1\le t_2\), \[\begin{align} a_2+j\ge a_1+t_1-1+3+ j&\ge t_1+t_2-1+t_1-1+3+j\\ & \ge 2t_1+t_2+1+j\ge 2t_1+j+1+j+1=2(t_1+j+1)\\ &>2(t_1+j+1)-1. \end{align}\]
We now show that \(T\in SST_{2n}(\varpi_l)\). Since \(0\le l_1\le l-t\), \(0\le l_2\le l-t-l_1\), it follows that \(l_{t+1}=l-t-l_1-l_2\ge l-t-l_1-(l-t-l_1)\ge 0\) is a non negative even number such that \(l_2=0\) and \(l_{t+1}=0\) when \(l_1=l-t\), and \(l_{t+1}=0\) when \(l_2=l-t-l_1\). Therefore \[\ell (T)=l_1+t_1+l_2+t_2+l_{t+1}=t+l_1+l_2+l_{t+1}=l.\] In addition, \(T\) is a column, which means that
\[\begin{align} a_1+ t_1-1+l_2&=a_1+t_1-1+a_2-a_1-t_1-2,{ by definition of l_2}\\ &=a_2-3<a_2 \quad (a_2\ge 5). \end{align}\]
The entries of \(T\) are bounded by \(2n\). We just need to check the following case. If \(l_1=a_1-t_1-1<l-t\) and \(l_2=a_2-a_1-t_1-2<l-t-l_1\), one has
\[\begin{align} a_2+t_2-1+l_{t+1}=&a_2+t_2-1+l-t-l_1-l_2\\ =&a_2+t_2-1+l-t-(a_1-t_1-1)-(a_2-a_1-t_1-2)\\ =&l+t_1+2\\ \le&2n-t+t_1+2\\ \le&2n-t_2+2\le 2n,~by t_2\ge 2 \end{align}\]
We next show that \(\mathrm{red}(T)=\boldsymbol{a}\). Note \(a_1+t_1-2, a_2-t_2-1\notin 2 \mathbb{Z}\), and \(a_1+t_1-1, a_2+t_2-1\in 2 \mathbb{Z}\). From Lemma 1 one has \[\begin{align} \mathrm{rem}((1,2,\dots, a_1-t_1-1){\boldsymbol{A}_1})&= \mathrm{rem}((1,2,\dots, a_1-t_1-1)(a_1, a_1+1,\dots , a_1+t_1-2,a_1+t_1-1))\\ &=(1,2,\dots, a_1-t_1-1). \end{align}\]
Then, from Proposition 4, \[\begin{align} & \mathrm{rem}((1,2,\dots, a_1-t_1-1)\mathbf{A_1}(a_1+t_1-1+1,\dots, a_1+t_1-1+l_{2}))=\\ &=(1,2,\dots, a_1-t_1-1)(a_1+t_1-1+1,\dots, a_1+t_1-1+l_{2}). \end{align}\] In fact \[\begin{align} &(1,2,\dots, a_1-t_1-1)\mathbf{ A_1}(a_1+t_1-1+1,\dots, a_1+t_1-1+l_{2})\nonumber\\ =&(1,2,\dots, a_1-t_1-1)(a_1,a_1+1,\dots, a_1+t_1-2)\mathpalette\mathbin{\vcenter{\scalebox{\\}{\m@th.5\bullet}}} &(a_1+t_1-1,a_1+t_1-1+1,\dots, a_1+t_1-1+u,\dots,a_1+t_1-1+l_{2}-1, a_1+t_1-1+l_{2}), \end{align}\] where \(1\le u\le l_2-1\) and \(u\notin 2 \mathbb{Z}\), and
\[\begin{align} a_1+t_1-1+u&<2(a_1-1-t_1-1+t_1+u)-(a_1-t_1-1)-(u-1)\\ =&2a_1-2+2u-a_1+t_1+1-u+1\\ =&a_1+t_1+u. \end{align}\] Finally we show that \[\begin{align} & \mathrm{rem}(1,2,\dots, a_1-t_1-1)\mathbf{A_1}(a_1+t_1-1+1,\dots, a_1+t_1-1+l_{2})\mathbf{A_2}( {a_2}+ t_2-1+1,\dots, a_2+ t_2-1+l_{t+1})\\ &=(1,2,\dots, a_1-t_1-1)(a_1+t_1-1+1,\dots, a_1+t_1-1+l_{2})( {a_2}+ t_2-1+1,\dots, a_2+ t_2-1+l_{t+1}) \end{align}\]
We have to consider \[\mathbf{A_2}( {a_2}+ t_2-1+1,\dots, a_2+ t_2-1+l_{t+1})=({a_2}, a_2+1,\dots , a_2+t_2-2,a_2+t_2-1)( {a_2}+ t_2-1+1,\dots, a_2+ t_2-1+l_{t+1}),\] and, for \(0\le u\le t_2-2\), and \(u\in 2 \mathbb{Z}\) \[\begin{align} a_2+u&\nless 2(a_1-t_1-1+t_1+l_2+u+1)-l_2-(a_1-t_1-1)-u\\ &=2a_1+2l_2-a_1+t_1+1-l_2+u=a_1+l_2+t_1+1+u\\ &=a_1+(a_2-a_1-t_1-2)+t_1+1+u,by l_2=a_2-a_1-t_1-2\\ &=a_2+u-1, \end{align}\]
and for \(1\le u\le l_{t+1}-1\) and \(u\notin 2 \mathbb{Z}\), \[\begin{align} &a_2+t_2-1+u< 2(a_1-t_1-1+t_1+l_2+t_2+u)-(a_1-t_1-1)-l_{2}-(u-1)\\ &=a_1+2t_2+t_1+l_2+u\\ &=a_1+2t_2+t_1+(a_2-a_1-t_1-2)+u\\ &=a_2+2t_2+u-2\\ &=a_2+t_2+u+(t_2-2),~~t_2\ge 2 \end{align}\] Hence, from Proposition 4, \(\mathrm{red}(T)=\boldsymbol{a}\). ◻
Example 7. Let \(n=12\).
Let \(t=6\le n\), \(l=12\), \(6\le min\{12, 2\times 2\times 12-12\}\), \(l-t=6\) and \(t_1=4>t_2=2\) and \(t_1=4>t_2=2\). Let \({\boldsymbol{a}}=(9,10,11,12)(17,18)\in SpT_{24}(\varpi_6)\), \(a_2-a_1-t_1-2=17-9-4-2=2\), \(a_1-t_1-1=4\). Then \(T=(1,2,3,4)(9,10,11,12)(13,14)(17,18)\in SST_{24}(\varpi_{12})\) and \(\mathrm{red}(T)=\boldsymbol{a}\). Note \(13<2\times 9-4=14\), and \(17\nless 2\times 11-6=16\).
Let \(t=8\), \(t_1=4=t_2=4\), \(l=12\), \(8\le min\{12, 2\times 2\times 12-12\}\), \(l-t=4\). Let \({\boldsymbol{a}}=(9,10,11,12)(17,18, 19, 20)\in SpT_{24}(\varpi_8)\), \(a_2-a_1-t_1-2=17-9-4-2=2\), \(l_1=a_1-t_1-1=4\), and \(l_2=min\{2, l-t-l_1=4-4=0\}=0\). Then \(T=(1,2,3,4)(9,10,11,12)(17,18, 19,20)\in SST_{24}(\varpi_{12})\) and \(\mathrm{red}(T)=\boldsymbol{a}\).
Theorem 3. Let \(l\in [0,2n]\) and \(t\in[0,n]\) such that \(0\le t\le min\{l,2n-l\}\) and \(l-t\in 2 \mathbb{Z}\). Let \(\mathbf{a} = (a_1, \dots , a_t)\in SpT_{2n}(\varpi_t)\) such that, for every \(1\le i< t\), \[a_i\notin 2 \mathbb{Z}\Rightarrow a_{i+1}>a_i+1~~~(that is,a_i\notin 2\mathbb{Z}, ~~~ a_{i+1}\in 2\mathbb{Z}\Rightarrow a_{i+1}-a_i\ge 3).\] Then
\[\begin{align} \label{redu:no32factors} \mathrm{red}_{t}^{-1}:=\mathrm{red}^{-1}_{|SpT_{2n}(\varpi_t)}:SpT_{2n}(\varpi_t)&\rightarrow SST_{2n}(\varpi_l)\quad\\ \mathbf{a}&\mapsto\mathrm{red}_{t}^{-1}(\mathbf{a}):=T_0T_1\cdots T_t,\nonumber \end{align}\qquad{(13)}\] where \[\begin{align} \label{inverseredu59nofactors} &T_0=(1,2,\dots l_1),\\ &T_i= \begin{cases} ({a}_i)( a_i+1,\dots, a_i+l_{i+1})=:a_i\mathpalette\mathbin{\vcenter{\scalebox{\varepsilon}{\m@th.5\bullet}}}_{i+1}, & ifa_i\in 2\mathbb{Z}\\ ({a}_i)( a_i+1+1,a_i+1+2,\dots, a_i+1+l_{i+1})=:a_i\mathpalette\mathbin{\vcenter{\scalebox{\omega}{\m@th.5\bullet}}}_{i+1}, &ifa_i\notin 2\mathbb{Z}\\ \end{cases},& 1\le i\le t, \label{} \end{align}\] {#eq: sublabel=eq:inverseredu59nofactors,eq:} and \(l_1,\dots, l_t\), \(l_{t+1}\) are nonnegative even numbers defined by \[\begin{align} \label{l:nofactors} &l_1= \begin{cases} min\{a_1-2, l-t\},& ifa_1\in 2\mathbb{Z}\\ min\{ a_1-1, l-t\},&ifa_1\notin 2\mathbb{Z},\\ \end{cases} \end{align}\qquad{(14)}\] \[\begin{align} &l_{i+1}= \begin{cases}min\{a_{i+1}-a_i-2, l-t-\sum_{k=1}^i l_k\}, &ifa_i, a_{i+1}\in 2\mathbb{Z}ora_i, a_{i+1}\notin 2\mathbb{Z} \\ min\{{ a_{i+1}-a_i-1},~~ l-t-\sum_{k=1}^i l_k\},&ifa_i\in 2\mathbb{Z}, a_{i+1}\notin 2\mathbb{Z} \\ min\{a_{i+1}-a_i-3, l-t-\sum_{k=1}^i l_k\},&ifa_i\notin 2\mathbb{Z}, ~~~ a_{i+1}\in 2\mathbb{Z}, \end{cases},\quadfor1\le i<t,\label{ells} \end{align}\qquad{(15)}\] and \(l_{t+1}=l-t-\sum_{i=1}^{t}l_i.\)
Proof. We start by checking that \(l_i\) is a non negative integer for \(i=1,\dots,t,t+1\). By definition \(l-t\ge 0\) and by ?? \(l-t\ge l_1\ge 0\). Hence \(l-t-l_1\ge 0\) and by ?? \(l-t-l_1\ge l_2\ge 0\). Hence \(l-t-l_1-l_2\ge 0\) and \(l-t-l_1-l_2\ge l_3\ge 0\) by ?? . Continuing in this manner, by \[\begin{align} l-(\sum_{k=1}^i l_k+i)-(t-i)=l-\sum_{k=1}^i l_k-i-t+i=l-t-\sum_{k=1}^i l_k\ge l_{i+1}\ge 0,for 0\le i<t. \end{align}\] Therefore, \[l_{t+1}=l-t-\sum_{i=1}^{t}l_i\ge 0\] is well defined and indeed \(T_0T_1\cdots T_t\) has length \[\begin{align} \ell(T_0T_1\cdots T_{t})=\sum_{i=0}^{t}\ell(T_i)=\sum_{i=1}^{t}l_i+l_{t+1}+t =l. \end{align}\]
The entries of \(T_0T_1\cdots T_t\) are also bounded by \(2n\), that is, \(T_0T_1\cdots T_t\in SST_{2n}(\varpi_l)\) as we next show.
Note that \[l-t-\sum_{k=1}^i l_k\ge l-t-\sum_{k=1}^{i+1} l_k\ge l_{t+1},for 0\le i<t,\] and, if for some \(0\le i<t\), \(l-t-\sum_{k=1}^i l_k=0\), it follows that \[l_{i+1}=\cdots=l_t=l_{t+1}=0.\] Therefore, if for some \(1\le i\le t\), one has \(l_i=l-t-\sum_{k=1}^{i-1} l_k\), then \[l_{i+1}\le l-t-\sum_{k=1}^il_k=l-t-(\sum_{k=1}^{i-1} l_k+l_i)=l-t-\sum_{k=1}^{i-1} l_k-l_i=0.\] Hence \(l_{t+1}=0\) and the bottom entry of the column of \(T_0T_1\cdots T_t\) is \(a_l\le 2n\).
By induction on \(t\ge 0\). If \(t=0\), \(l_1=l\in 2 \mathbb{Z}\) and \(\mathrm{red}(1,\dots,l)=()\).
Let \(t=k\ge 1\) and by induction let \(\mathrm{red}(T_0T_1\cdots T_{k-2}\mathpalette\mathbin{\vcenter{\scalebox{a}{\m@th.5\bullet}}}_{k-1})=a_1\cdots a_{k-1}\). (Note if \(1\le k\le min\{l, 2n-l\}\), \(l-k\in 2 \mathbb{Z}\) then \(0\le k-1\le min\{l-1, 2n-(l+1)\}\) and \(l-k=(l-1)-(k-1)\in 2 \mathbb{Z}\).)
Then \[\begin{align} \mathrm{red}(T_0\cdots T_{k-1}T_{k})= \begin{cases} \mathrm{red}(T_0\cdots T_{k-2}T_{k-1}({a}_{k}, a_k+1+1,a_k+1+2,\dots, a_{k}+1+l_{k+1})),&ifa_k\notin 2\mathbb{Z},\\ \mathrm{red}(T_0\cdots T_{k-2}T_{k-1}({a}_{k}, a_k+1,a_k+2,\dots, a_{k}+l_{k+1})),&ifa_k\in 2\mathbb{Z}\\ \end{cases} \end{align}\] where \(l_{k+1}=(l-k)-\sum_{i=1}^kl_i.\)
Let \(a_k\notin 2 \mathbb{Z}\).
If for every \(i\), \(a_i\notin 2 \mathbb{Z}\), then, by induction, \[| \mathrm{rem}(T_0\cdots T_{k-2}(a_{k-1})|=\sum_{i=1}^{k-1}l_i=a_{k-1}-2(k-2)-1,by \eqref{ells}\] and for \(1\le u<l_k\) odd, \[a_{k-1}+1+u<2(\sum_{i=1}^{k-1}l_i+(k-1)+u)-\sum_{i=1}^{k-1}l_i-(u-1)=\sum_{i=1}^{k-1}l_i+2k-2+u+1\] \[=a_{k-1}-2(k-2)-1+2k-2+u+1=a_{k-1}+u+2\]
Hence, \(\mathrm{red}(T_0\cdots T_{k-1}=a_1\cdots a_k\) and \(| \mathrm{rem}(T_0\cdots T_{k-2}T_{k-1}a_k|=\sum_{i=1}^{k}l_i=a_{k}-2(k-1)-1\). Similarly, \(\mathrm{red}(T_0\cdots T_k)=a_1\cdots a_k\).
If for every \(1\le i<k\), \(a_i\in 2 \mathbb{Z}\), then, by induction, \[| \mathrm{rem}(T_0\cdots T_{k-2}(a_{k-1})|=\sum_{i=1}^{k-1}l_i=a_{k-1}-2(k-1),by \eqref{ells}\] and for \(1\le u<l_k\) odd, \[a_{k-1}+1+u<2(\sum_{i=1}^{k-1}l_i+(k-1)+u)-\sum_{i=1}^{k-1}l_i-(u-1)=\sum_{i=1}^{k-1}l_i+2k-2+u+1\] \[=a_{k-1}-2(k-1)+2k-2+u+1=a_{k-1}+u+2\]
If for every \(i\), \(a_i\in 2 \mathbb{Z}\), then, by induction, \[| \mathrm{rem}(T_0\cdots T_{k-2}(a_{k-1})|=\sum_{i=1}^{k-1}l_i=a_{k-1}-2(k-1),by \eqref{ells}\] and for \(1\le u<l_k\) odd, \[a_{k-1}+u<2(\sum_{i=1}^{k-1}l_i+(k-1)+u)-\sum_{i=1}^{k-1}l_i-(u-1)=\sum_{i=1}^{k-1}l_i+2k-2+u+1\] \[=a_{k-1}-2(k-1)+2k-2+u+1=a_{k-1}+u+1\]
Let \(a_k\in 2 \mathbb{Z}\). If \(a_k\in 2 \mathbb{Z}\) and for every \(1\le i<k\), \(a_i\notin 2 \mathbb{Z}\), then, by induction, \[| \mathrm{rem}(T_0\cdots T_{k-2}(a_{k-1})|=\sum_{i=1}^{k-1}l_i=a_{k-1}-2(k-2)-1,by \eqref{ells}\] and for \(1\le u<l_k\) odd, \[a_{k-1}+u<2(\sum_{i=1}^{k-1}l_i+(k-1)+u)-\sum_{i=1}^{k-1}l_i-(u-1)=\sum_{i=1}^{k-1}l_i+2k-2+u+1\] \[=a_{k-1}-2(k-2)-1 +2k-2+u+1=a_{k-1}+u+2\] ◻
Example 8. Let \(n=3\).
\(t=2\), \(l=4\le 6\), \(2\le min\{4,6-4\}\), \(l-t=2\) even. Let \(a=(56)\in SpT_{6 }(\varpi_2)\). Then one has \[\begin{align} \nonumber \mathrm{red}_{t}^{-1}:=\mathrm{red}^{-1}_{|SpT_{6}(\varpi_2)}:SpT_{6}(\varpi_2)&\rightarrow SST_{6}(\varpi_4)\quad\\ (56)&\mapsto\mathrm{red}_{2}^{-1}(56):=T_0T_1 T_{2}=(12)(5)(6)\nonumber \end{align}\]
Then the inverse reduction with respect \(l=4\), is \(\mathrm{red}^{-1}_2(56)=(1256)=T_0T_1T_2\in SST_6(\varpi_4)\) with \(T_0=12\), \(T_1=5\), \(T_2=6\). One has \(\mathrm{red}(1256)=(56)\), \(\mathrm{rem}(1256)=(12)\).
\(t=2\), \(l=2\le 6\), \(2\le min\{2,6-2\}\), \(l-t=0\) even. Let \(a=(56)\in SpT_{6 }(\varpi_2)\). Then the inverse reduction with respect \(l=2\), \(\mathrm{red}^{-1}_2(56)=(56)=T_0T_1T_2\in SST_6(\varpi_2)\) with \(T_0=()\), \(T_1=5=T_2=6\), Then one has \[\begin{align} \nonumber \mathrm{red}_{t}^{-1}:=\mathrm{red}^{-1}_{|SpT_{6}(\varpi_2)}:SpT_{6}(\varpi_2)&\rightarrow SST_{6}(\varpi_2)\quad\\ (56)&\mapsto\mathrm{red}_{2}^{-1}(\mathbf{56}):=T_0T_1 T_{2}=()(5)(6)\nonumber \end{align}\] One has \(\mathrm{red}(56)=(56)\), \(\mathrm{rem}(56)=()\),
Example 9.
Let \(n=8\). \(t=5\), \(l=11\le 16\), \(5\le min\{11,16-11\}\), \(l-t=6\) even. Let \(a=(5,6, 10, 11, 15)\in SpT_{16 }(\varpi_5)\). One has, \(l_1=2,\) \(l_2=0\), \(l_3=2\), \(l_4=0\), \(l_5=2\), \(l_6=0\), \(l_1+l_2+l_3+l_4+l_5+l_6+t=l=11\) \[\begin{align} \nonumber \mathrm{red}_{5}^{-1}:=\mathrm{red}^{-1}_{|SpT_{6}(\varpi_2)}:SpT_{16}(\varpi_{11})&\rightarrow SST_{16}(\varpi_5)\nonumber\\ (56)&\mapsto\mathrm{red}_{2}^{-1}(5,6, 10, 11, 15):=T_0T_1 T_{2}T_3T_4T_5 \nonumber\\ &T_0T_1 T_{2}T_3T_4T_5=(12).(\mathbf{5}).{(\mathbf{6}.{78})}.({\boldsymbol{1}0}).{({\boldsymbol{1}1}.13.14)}.(\boldsymbol{1}5)\nonumber \end{align}\]
The inverse reduction with respect \(l=11\), is \(\mathrm{red}^{-1}_2(56)=(1256)=T_0T_1T_2\in SST_6(\varpi_4)\) with \(T_0=12\), \(T_1=5\), \(T_2=6\). One has \(\mathrm{red}(1256)=(56)\), \(\mathrm{rem}(1256)=(12)\).
Let \(n=9\), \(a=(3,4,9,10,11,15,16)\in SpT_{18}(\varpi_7)\), \(l=11\), \(t=7\le min\{l=11,18-l=7\}\), \(l-t=4\in 2 \mathbb{Z}\)
The inverse reduction with respect \(l=11\), \[\mathrm{red}^{-1}_7(a)=({\boldsymbol{3}},{\boldsymbol{4}},5,6,{\boldsymbol{9}}, {\boldsymbol{1}0},{\boldsymbol{1}1}, {\boldsymbol{1}5}, {\boldsymbol{1}6},17,18)\in SST_{18}(\varpi_{11})\]
With \(l=9\), \(t=7\) in \((2)\), \(l-t=2\), one has the inverse reduction with respect \(l=9\), \[\mathrm{red}^{-1}_7(a)=({\boldsymbol{3}},{\boldsymbol{4}},5,6,{\boldsymbol{9}}, {\boldsymbol{1}0},{\boldsymbol{1}1}, {\boldsymbol{1}5}, {\boldsymbol{1}6})\in SST_{18}(\varpi_{9})\]
With \(l=7=t\) in \((2)\), \(l-t=0\), one has the inverse reduction with respect \(l=7\), \[\mathrm{red}^{-1}_7(a)=({\boldsymbol{3}},{\boldsymbol{4}},{\boldsymbol{9}}, {\boldsymbol{1}0},{\boldsymbol{1}1}, {\boldsymbol{1}5}, {\boldsymbol{1}6})=a\in SST_{18}(\varpi_{7})\]
To prove that \(\mathsf{LR}^{AII}\) is also a surjection we exhibit the right inverse \(\widetilde{R}\) of \(\mathsf{LR}^{AII}\). Recall the reverse column insertion in Subsection 2.1, and the properties of bumping and reverse bumping routes in Remark 1 and Remark 2 in there.
Let \(\mu\subseteq\lambda\in Par_{\le 2n}\) with \(\mu\in Par_{\le n}\). If \(\lambda=\mu\), \(Rec_{2n}(\lambda/\lambda)=Rec_{2n}(())=\widetilde{R}ec_{2n}(())\) and \(\widetilde{R}(S, ())=S\), \[\begin{align} \widetilde{R}:SpT_{2n}(\mu)\times \widetilde{R}ec_{2n}(())&\longrightarrow SpT_{2n}(\mu)\subset SST_{2n}(\mu)\\ (S,())&\mapsto S,~\mathrm{red}(S)=S \end{align}\]
Let us consider \(\mu\subset\lambda\) and fix arbitrarily \(S\in SpT_{2n}(\mu)\). In Theorem 6, based on the slack statistics introduced in [36], one defines
\[\begin{align} \label{Rtilde} &\widetilde{R}_{|S}:\{S\}\times \widetilde{R}ec_{2n}(\lambda/\mu)\longrightarrow SST_{2n}(\lambda)\nonumber\\ &\qquad\qquad\qquad \quad(S,Q)\mapsto S^{\mathbf{r}^{(N)}\cdots\mathbf{r}^{(2)} \mathbf{r}^{(1)}}= c \circ (\mathrm{red}_{t_0^{(1)}}^{-1},\mathrm{id})\circ (\underset{{\mu^{(1)}/{\mu^{(1)}}'}} \leftarrow \bigg(\cdots \nonumber\\ &(\underset{{\mu^{(N-2)}/{\mu^{(N-2)}}'}}\leftarrow\bigg( c \circ (\mathrm{red}_{t_0^{(N-1)}}^{-1},\mathrm{id})\circ (\underset{{\mu^{(N-1)}/{\mu^{(N-1)}}'}} \leftarrow \bigg(c \circ (\mathrm{red}_{t_0^{(N)}}^{-1},\mathrm{id})\circ (\underset{{\mu^{(N)}/{\mu^{(N)}}'}}\leftarrow S)\bigg))\bigg))\cdots)\bigg)) \end{align}\tag{14}\] where \(N>0\) is the number of vertical strips \(\{\mu^{(i-1)}/\mu^{(i)}\}_{i=1}^N\) of \(Q\) with \(|\mu^{(i-1)}/\mu^{(i)}|=Q[i]\), for \(i=1,\dots,N\), \[\lambda=\mu^{(0)}\supset_{vert}\mu^{(1)}\supset_{vert}\cdots \supset_{vert} \mu^{(N)}=\mu,\] \(\underline\mathbf{t}=(t_0^{(N)},\dots,t_0^{(1)})\) is the slack of \(Q\), \(\underline\mathbf{r}=[\mathbf{r}^{(N)},\cdots,\mathbf{r}^{(2)}, \mathbf{r}^{(1)}]\) is the slack vector sequence of \(Q\) that encodes the slackness of its sequence of vertical strips, \({\mu^{(i)}}'=\mu^{(i)}-\delta_{\mathbf{r}_i}\), \(i=1,\dots,N\), as defined in [36] and to be recalled and expanded in the next two subsections, \(\mathrm{red}_{t_0^{(i)}}^{-1}\) is the inverse of the reduction map on \(SpT_{2n}(\varpi_{t^{(i)}_0})\), or the expanding map via the reverse removal, defined in Lemma 1, Theorem 2 and Theorem 3, and \(\mathrm{c}\) is the concatenation in the plactic monoid. The basic operation is defined for \(N=1\), \(\underline\mathbf{t}=(t_0)\), in Theorem 4 in the next subsection, \[\mathrm{c}\circ (\mathrm{red}_{t_0}^{-1},\mathrm{id})\circ (\underset{\mu^{(N)}/{{\mu^{(N)}}'}}\leftarrow S).\]
We now recall from [36] some statistics for a vertical strip , called slack data or slack statistics of a vertical strip \(\lambda/\mu\), with \(\lambda\in Par_{\le 2n}\) and \(\mu\in Par_{\le n}\).
Definition 7. Let \(\mu\subset_{vert}\lambda\) with \(\lambda\in Par_{\le 2n}\) and \(\mu\in Par_{\le n}\). Let \(0\le l_0\le\ell(\mu)\) be the number of cells in the vertical strip \(\lambda/\mu\) with row coordinates in \([1,\ell(\mu)]\). We call \(t_0=\ell(\mu)-l_0\) the slack* of the vertical strip \(\lambda/\mu\).*
In other words, the slack number \(t_0\) is the number of gaps of the vertical strip \(\lambda/\mu\) in the interval \([1,\ell(\mu)]\).
Definition 8. With the same setting as above, let \(t_0\) be the slack of the vertical strip \(\lambda/\mu\). We define \(\mathbf{r}=\{r_1<\dots< r_{t_0}\}\subseteq [1,\ell(\mu)]\) to be the complement of the set of the row coordinates of the cells of the vertical strip \(\lambda/\mu\) in \([1,\ell(\mu)]\). We call to \(\mathbf{r}=(r_1<\dots< r_{t_0})\) the slack* row index vector of the vertical strip \(\lambda/\mu\), and define \(\delta_\mathbf{r}\in\{0,1\}^{\ell(\mu)}\) to be the incidence vector of \(\mathbf{r}\) by considering \(\mathbf{r}\) as a subset of \([1,\ell(\mu)]\), that is, \(\delta_\mathbf{r}=(x_s)_{s\in [1,\ell(\mu)]}\), \(x_s=1\) if \(s\in\mathbf{r}\) and \(0\) otherwise. We call to \(\delta_\mathbf{r}\) the slack incidence vector of the vertical strip \(\lambda/\mu\), and write \(|\delta_\mathbf{r}|:=\sum_{i=1}^{t_0}\delta_{r_i}=t_0\).*
Indeed the vertical strip \(\lambda/\mu\) as a dual incidence vector with respect to the interval \([1,\ell(\lambda)]\). The dual of the slack incidence vector is equal to \(\varpi_{\ell(\lambda)}-\delta_\mathbf{r}\) where here we are adding to \(\delta_\mathbf{r}\) a tail of zeroes of length \(\ell (\lambda)-\ell(\mu)\), and \(\mu+(\varpi_{\ell(\lambda)}-\delta_\mathbf{r})=\lambda\).
Define \(\mu'\subset_{vert} \mu\) such that \(\mu'_i=\mu_i-1\) if \(i\in\mathbf{r}=\{r_1,\dots, r_{t_0}\}\) and \(\mu_i\), otherwise. That is \(\mu'=\mu-\delta_\mathbf{r}\in Par_{\le n}\). Then the vertical strip \(\mu/\mu'\) has \(t_0=\ell(\mu)-l_0\) cells with row coordinates \(\mathbf{r}=(r_1<\dots< r_{t_0})\) the slack row index vector of vertical strip \(\lambda/\mu\); and \[\begin{align} \label{dualslack}\lambda=\mu+(\varpi_{\ell(\lambda)}-\delta_\mathbf{r})=\mu'+\varpi_{\ell(\lambda)}\Leftrightarrow \varpi_{\ell(\lambda)}=\lambda-\mu'. \end{align}\tag{15}\]
The slack data is illustrated in Example 10.
Let us write \(\lambda=(\lambda_0,\varpi_{l-\ell(\mu)})\) where \(\mu\subseteq_{vert}\lambda^0\in Par_{\le n}\) such that \(\ell(\lambda^0)=\ell(\mu)\) and \(l=\ell(\lambda)\). Thus, \(\lambda^0/\mu\) is a vertical strip with \(l_0=|\lambda_0|-|\mu|\le \ell(\mu)\) cells, and slack \(t_0=\ell(\mu)-l_0\).
Let \(Q\in \widetilde{R}ec_{2n}((\lambda_0,\varpi_{l-\ell(\mu)})/\mu)\) defined by the vertical strip sequence \(\mu^{(0)}=\lambda\supset_{vert}\mu^{(1)}=\mu\). Then \(Q[1]+t_0=l\). Furthermore, the conditions \((R3)\), \((R4)\), in Proposition impose relations between \(l=\ell(\lambda)\) and the slack \(t_0\) of \(\lambda/\mu\), \[\begin{align} \label{conditions}&l\le 2n-t_0\Leftrightarrow l+t_0\le 2n\Leftrightarrow t_0\le 2n-l,\\ &~ and \nonumber\\ &~0\le l-t_0 \in 2\mathbb{Z}. \end{align}\tag{16}\]
We define the reverse column Schensted insertion data of \(Q\) to be the slack row index vector \(\mathbf{r}\). For simplicity, we often abuse notation and identify the slack row index vector \(\mathbf{r}\) with \(\mu^{(1)}/{\mu^{(1)}}'\) and use the slack row index vector \(\mathbf{r}\) to apply reverse column insertion to an \(S\in SpT_{2n}(\mu)\). We also often say the slack \(t_0\), the slack row index vector \(\mathbf{r}\) and the slack incidence vector \(\delta_\mathbf{r}\) of \(Q\) in the sense that the vertical strip \(\lambda/\mu^{(1)}\) defines \(Q\).
Consequently, the \(1\)-vertical strip recording tableau of shape \(\lambda/\mu\) with slack \(t_0\) and \(l=\ell(\lambda)\) is characterized as \[\begin{align} \label{rec:1vertical} \widetilde{R}ec_{2n}((\lambda_0,\varpi_{l-\ell(\mu)})/\mu)=\left\{Q= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{},{},{},}, {{},{},{}1}, {\vdots,{}}, {{},1}, {{}}, {{1}}, {{1}}, {{\vdots}}, {{1}}, {{1}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,~\ell(\mu)\le l,~~ 0\le l- t_0\in 2 \mathbb{Z},~~ l\le 2n-t_0\right\} \end{align}\tag{17}\]
Example 10. For \(n=6\), let \[\begin{align} & Q= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{},{},{},{}}, {{},{},,1}, {{},{},}, {{},{},}, {{},{1}}, {{}}, {{1}}, {{1}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in \widetilde{R}ec_{2n}((\lambda_0,\varpi_{l-\ell(\mu)})/\mu),\nonumber \end{align}\] where \(\mu=\mu^{(1)}=(4,3,3,3,1,1)\subset_{vert}\lambda^0=(4,4,3,3,2,1)\subseteq \lambda=(\lambda_0,\varpi_{l-\ell(\mu)})=(\lambda_0,\varpi_{2})\), and \(\ell(\lambda^0)=\ell(\mu)=6\), \(l_0= 2\), slack \(t_0=\ell (\mu)-l_0=4\), \(Q[1]+t_0=\ell(\lambda)=8\le 12-4\).
One has \({\mu^{(1)}}'=\mu^{(1)}-\delta_\mathbf{r}=\mu-(1,0,1,1,0,1)=(3,3,2,2,1,0)\) where \(\mathbf{r}=\{1,3,4,6\}\subseteq [1,\ell(\mu)]\subseteq [1,\ell(\lambda)]=[1,8]\) is the slack row index vector of the vertical strip \(\lambda/\mu^{(1)}\), and \[\begin{array}{llllll}\lambda/\mu^{(1)}={\tiny\ydiagram{0,3+1,0,0,1+1,0,0+1,0+1}}&\quad \mu^{(1)}/{\mu^{(1)}}'={\tiny\ydiagram{3+1,0,2+1,2+1,0,0+1}}&\qquad \mathbf{r}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1}, {3}, {4}, {6}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } &\qquad \delta_\mathbf{r}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1}, {0}, {1}, {1}, {0}, {1} } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \end{array}\]
Indeed \(\varpi_{\ell(\lambda)}=\lambda-{\mu^{(1)}}'\).
Let \(S\in SpT_{2n}(\mu)\) and \(Q\in \widetilde{R}ec_{2n}((\lambda_0,\varpi_{l-\ell(\mu)})/\mu)\). Let \((a_1,\dots,a_{t_0})\in Sp_{2n}(\varpi_{t_0})\) be the set of the \(t_0\) bumped elements from the first column of \(S\) by applying successively the reverse column insertion to the rows of \(S\), as prescribed by slack row index vector \(\mathbf{r}=(r_1<\dots<r_{t_0})\) of \(Q\), going from the largest to the smallest row; and let \(S^1\) be the remained tableau after the application of that reverse column insertion to \(S\). For the returned pair \(((a_1,\dots,a_{t_0}),S^1)\) obtained under the action of the reverse Schensted insertion to the entries in the \(t_0\) cells \(\mu^{(1)}/{\mu^{(1)}}'\) of \(S\), put \[\begin{align} \label{revschensted}((a_1,\dots,a_{t_0}),S^1)=:(\underset{{\mu^{(1)}/{\mu^{(1)}}'}}\leftarrow S),or ((a_1,\dots,a_{t_0}),S^1 )=:(\underset{{\mathbf{r}}}\leftarrow S). \end{align}\tag{18}\]
Since \(S\) is symplectic and \((a_1,\dots,a_{t_0})\) is contained in the first column of \(S\), \((a_1,\dots,a_{t_0})\in SpT_{2n}(\varpi_{t_0})\). (Any subset of a symplectic column is still symplectic.)
For \(Q\in \widetilde{R}ec_{2n}((\lambda_0,\varpi_{l-\ell(\mu)})/\mu)\) 17 \[\begin{align} \label{tildeR2}{\mathsf{LR}^{AII}}^{-1}(S, Q)=\tilde{R}(S,Q)&= \mathrm{c}\circ (\mathrm{red}_{t_0}^{-1},\mathrm{id})\circ (\underset{{\mathbf{r}}}\leftarrow S)\nonumber\\ &= \mathrm{c}\circ (\mathrm{red}_{t_0}^{-1},\mathrm{id})((a_1,\dots,a_{t_0}),S^1 )\nonumber\\ &= \mathrm{c}\circ (\mathrm{red}_{t_0}^{-1}((a_1,\dots,a_{t_0})),S^1)\nonumber\\ &:=S^\mathbf{r} \end{align}\tag{19}\] where \(\mathrm{c}\) means concatenation in the plactic monoid and \(\mathrm{red}_{t_0}^{-1}\) means reverse or the inverse reduction map applied to a column in \(SpT_{2n}(\varpi_{t_0})\) a symplectic column of length the slack number \(t_0\) of \(Q\) according to the rules below. The reduction map \(\mathrm{red}\) on \(SST_{2n}(\varpi_l)\) is injective [1] and returns symplectic columns of length \(0\le k\le \min(l,2n-l)\) and \(l-k\in 2 \mathbb{Z}\). Its surjectivity and inverse has been shown for some patterns in Lemma 1, Theorem 2 and Theorem 3. The inverse \(\mathrm{red}_{t_0}^{-1}\) on \(SpT_{2n}(\varpi_{t_0})\)
\[\begin{align} \mathrm{red}_{t_0}^{-1}: SpT_{2n}(\varpi_{t_0})\rightarrow SST_{2n}(\varpi_{l}) \end{align}\] is exhibited below 21 following the rules in Lemma 1, Theorem 2 and Theorem 3.
Let \((a_1,\dots,a_{t_0})\in SpT_{2n}(\varpi_{t_0})\) be the column of bumped entries as in 18 . In the cases where the symplectic column decomposes into non empty factors \(\boldsymbol{A}_i\) of even length consisting of consecutive integers starting with an odd number, or when the symplectic column is such that consecutive integers occur only as an even number followed with an odd number, define \(l_1\), \(l_2\), \(\dots,l_{t_0},l_{t_0+1}\) nonnegative even numbers as in Lemma 1, Theorem 2 and Theorem 3such that
\[\begin{align} \label{suml} \sum_{i=1}^{t_0}l_i+l_{t_0+1}=l-t_0\Leftrightarrow l_{t_0+1}=l-t_0-\sum_{i=1}^{t_0}l_i \end{align}\tag{20}\]
Then \[\begin{align} \label{inversered} \mathrm{red}_{t_0}^{-1}((a_1,\dots,a_{t_0})):=T_0T_1\cdots T_{t_0}\in SST_{2n}(\varpi_l) \end{align}\tag{21}\] where \[\begin{align} \tag{22} &T_0=(1,2,\dots l_1),\\ &T_i= \begin{cases} ({a}_i, a_i+1,\dots, a_i+l_{i+1}), & ifa_i\in 2\mathbb{Z}\\ ({a}_i, a_i+1+1,a_i+1+2,\dots, a_i+1+l_{i+1}), &ifa_i\notin 2\mathbb{Z}\\ \end{cases}& 1\le i\le t_0. \tag{23} \end{align}\]
The following theorem is partially a consequence of Lemma 1, Theorem 2 and Theorem 3.
Theorem 4. With the set up above where \(S\in SpT_{2n}(\mu)\) and \(Q\in \widetilde{R}ec_{2n}((\lambda^0,\varpi_{l-\ell(\mu)})/\mu)\), \(l=\ell(\lambda)\), has slack row index vector \(\mathbf{r}=\{r_1<\dots< r_{t_0}\}\), one has the following assertion:
\[\begin{align} \label{inverse}{\mathsf{LR}^{AII}}^{-1}(S,Q)={\widetilde{R}}(S, Q)&= c \circ (\mathrm{red}_{t_0}^{-1},\mathrm{id})\circ (\underset{{\mathbf{r}}}\leftarrow S)\nonumber\\ &=c\circ (\mathrm{red}_{t_0}^{-1},\mathrm{id})((a_1,\dots,a_{t_0}),S^1 )\nonumber\\ &=c\circ (\mathrm{red}_{t_0}^{-1}(a_1,\dots,a_{t_0}), S^1) \nonumber\\ &=\mathrm{red}_{t_0}^{-1}((a_1,\dots,a_{t_0}))\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^1=:S^\mathbf{r}\in SST_{2n}(\lambda). \end{align}\qquad{(16)}\] where \(\mathrm{red}_{t_0}^{-1}((a_1,\dots,a_{t_0}))=T_0T_1\cdots T_{t_0}\in SST_{2n}(\varpi_{\ell(\lambda)})\) 22 , 23 , and \(S^1\in SpT_{2n}({\mu^{(1)}}')\) with \({\mu^{(1)}}'=\mu^{(1)}-\delta_\mathbf{r}\) and \(\lambda=\mu^{(1)}-\delta_\mathbf{r}+\varpi_{\ell(\lambda)}\).
Proof. It remains to prove that \(S^\mathbf{r}\) is a semistandard tableau. The entries of \((a_1,\dots,a_{t_0})\), in the first column of \(S\), are included in \(T_0T_1\cdots T_{t_0}\). More precisely, for \(i=1,\dots,t_0\), \(a_i\) belongs to \(T_i\), and since the first column of \(S\) is a symplectic column its row coordinate as an entry of \(T_i\) is larger or equal than \(r_i\). Conditions om the formulas in Lemma 1, Theorem 2 and Theorem 3 force the push down of the \(a_i\) row-coordinates as an entry of \(T_i\). Hence the column insertion of \(T_0T_1\cdots T_{t_0}\) in \(S^1\) is just the concatenation \(T_0T_1\cdots T_{t_0}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^1\). ◻
Remark 6. \((1)\) When \(t_0= 0 \Leftrightarrow l_0=\ell(\mu)\Leftrightarrow \mathbf{r}=()\Leftrightarrow \delta_\mathbf{r}=0\), then \(\mu'=\mu\) and the bumped column \(S'_\mathbf{r}=()\) is empty and \(S'_{\ge 2}=S\). That is, \(((),S):=(\underset{{\mathbf{r}}}\leftarrow S)\). If the slack of \(Q\) is \(t_0=0\), then \(l\in 2 \mathbb{Z}\) and \(\mathrm{red}_{0}^{-1}\) means reverse reduction in \([1,l]\) applied to a column of length \(0\). That is, \(\mathrm{red}_{0}^{-1}(())=(12\dots l)\) with \(l\in 2 \mathbb{Z}\) \(\Leftrightarrow \mathrm{red}(12\dots l)=()\Leftrightarrow \mathrm{rem}(12\dots l)=(12\dots l)\) and \(l\le 2n\). Hence,
\[\begin{align} &\widetilde{R}=(S, Q)= \mathrm{c}\circ (\mathrm{red}_{0}^{-1},\mathrm{id})\circ (\underset{{\mathbf{r}}}\leftarrow S) = \mathrm{c}\circ (\mathrm{red}_{0}^{-1},\mathrm{id})((),S)= \mathrm{c}\circ (\mathrm{red}_{0}^{-1}(),S)\\ &= \mathrm{c}\circ ( \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{1}}, {{2}}, {{\vdots}}, {{l}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,S)= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {{1}}, {{2}}, {{\vdots}}, {{l}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}=T_0\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}\in SST_{2n}(\lambda) \end{align}\]
\(\mu'=\mu-\delta_{()}=\mu\) and \(\lambda=\mu+\varpi_{\ell(\lambda)}\).
\((2)\) When \(l_0=0\Leftrightarrow t_0=\ell(\mu)\Leftrightarrow\mathbf{r}=[1,\ell(\mu)]\Leftrightarrow\delta_\mathbf{r}=(1^{\ell(\mu)})\), then \(\mu'=\mu-(1^{\ell(\mu)})\) and the bumped column \(S_1=(a_1,\dots,a_{\ell(\mu)})\subseteq SpT_{2n}(\varpi_{\ell(\mu)})\) is the first column of \(S\). That is, \(((a_1,\dots,a_{\ell(\mu)}),S^1):=(\underset{{\mathbf{r}}}\leftarrow S)\) where \(S^1\) is the symplectic tableau \(S\in SpT_{2n}(\mu)\) minus its first column \(S_1\). Hence \[\begin{align} &&\widetilde{R} (S, Q)= \mathrm{c}\circ (\mathrm{red}_{\ell(\mu)}^{-1},\mathrm{id})\circ (\underset{\mathbf{r}}\leftarrow S) =c\circ (\mathrm{red}_{\ell(\mu)}^{-1},\mathrm{id})(S_1,S^1)=\mathrm{red}_{\ell(\mu)}^{-1}(S_1)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^1 \end{align}\] In this case, \(l-\ell(\mu)\in 2 \mathbb{Z}\) and \(l\le 2n-\ell(\mu)\), and \(\mathrm{red}_{\ell(\mu)}^{-1}(S_1)\in SST_{2n}(\varpi_l)\).
Let \(n=4\), \(S= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {{1},{2},5}, {4,4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_8(3,2,1)\) and \(Q= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {{},{},}, {,}, {{}}, {1}, {1}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in \widetilde{R}ec_8((\mu,1,1)/\mu)\), \(t_0=3\), \(\mathbf{r}=[1,3]\), \(l=5\), \(l-\ell(\mu)=2\in 2 \mathbb{Z}\), \(l=5\le 2n-\ell(\mu)=8-3\). Then \[\mathrm{red}_{\ell(\mu)}^{-1}(S_1)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}_{\ge 2}=\mathrm{red}_{\ell(\mu)}^{-1}(145)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}_{\ge 2}=14578\mathpalette\mathbin{\vcenter{\scalebox{\YT}{\m@th.5\bullet}}}{0.13in}{}{ {{2},5}, {4}, },\quad T_0T_1T_2T_3=()\mathpalette\mathbin{\vcenter{\scalebox{1}{\m@th.5\bullet}}}\mathpalette\mathbin{\vcenter{\scalebox{4}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{5}{\m@th.5\bullet}}} 78\in SST_8(\varpi_5)\]
For \(S= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {{1},{2},5}, {6,6}, {7}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_8(3,2,1)\)
\[\mathrm{red}_{\ell(\mu)}^{-1}(S_1)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}_{\ge 2}=\mathrm{red}_{\ell(\mu)}^{-1}(167)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^1=1 3467\mathpalette\mathbin{\vcenter{\scalebox{\YT}{\m@th.5\bullet}}}{0.13in}{}{ {{2},5}, {6}, },\quad T_0T_1T_2T_3=()\mathpalette\mathbin{\vcenter{\scalebox{1}{\m@th.5\bullet}}} 34 \mathpalette\mathbin{\vcenter{\scalebox{5}{\m@th.5\bullet}}}\mathpalette\mathbin{\vcenter{\scalebox{7}{\m@th.5\bullet}}}\in SST_8(\varpi_5)\]
For an illustration of Theorem 4 see [36].
We now recall and expand from [36] the slack incidence matrix of a tableau \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) which encodes the information as described in Proposition 7 below to define \({\mathsf{LR}^{AII}}^{-1}\) on \(SpT_{2n}\times \widetilde{R}ec_{2n}(\lambda/\mu)\). We follow closely [36].
Definition 9. Let \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) defined by the sequence of vertical strips \[\begin{align} \label{slack}&\mu^{(0)}=\lambda\supset_{vert} \mu^{(1)}\supset_{vert} \cdots\supset_{vert}\mu^{(N-1)}\supset_{vert}\mu^{(N)} =\mu \end{align}\qquad{(17)}\] such that \(2\le |\mu^{(i-1)}/\mu^{(i)}|=Q[i]\in 2 \mathbb{Z}\), \(1\le i\le N\) and \(\nu=( Q[1],\dots,Q[N])^t\) is an even partition. For \(1\le i\le N\), let \(t_0^{(i)}=\ell(\mu^{(i)})-l_0^{(i)}\) where \(l_0^{(i)}\) is the number of cells in the vertical strip \(\mu^{(i-1)}/\mu^{(i)}\) with row coordinates in \([1,\ell(\mu^{(i)})]\). Then, for \(i=1,\dots,N\),
\(t^{(i)}_0\in\{0,1,\dots,\ell(\mu^{(i)}\}\) is the slack* of the vertical strip \(\mu^{(i-1)}/\mu^{(i)}\), and*
\(\mathbf{r}^{(i)}\subseteq [1,\ell(\mu^{(i)})]\) with cardinal \(|\mathbf{r}^{(i)}|=t_0^{(i)}\), the corresponding slack* (row index) vector. We write \(\mathbf{r}^{(i)}=()\) or \(\emptyset\) when \(t_0^{(i)}=0\), and in this case \(\ell(\mu^{(i)})=Q[i]\in 2 \mathbb{Z}\).*
The sequence of slack* numbers \(\underline\mathbf{t}=(t_0^{(N)},\dots,t_0^{(1)})\), is called the slack sequence of \(Q\) uniquely defined by ?? . Analogously, the corresponding sequence of slack (row index) vectors \(\underline\mathbf{r}=[\mathbf{r}^{(N)},\dots,\mathbf{r}^{(1)}]\) is called the slack (row index) vector sequence of \(Q\).*
We have defined \(\delta_{\mathbf{r}^{(i)}}\in \{0,1\}^{\ell(\mu^{(i)})}\), that is, considering \(\mathbf{r}^{(i)}\subseteq [1,\ell(\mu^{(i)})]\), \(i=1,\dots,N\). Since \([1,\ell(\mu)]\subseteq [1,\ell(\mu^{(N-1)})] \subseteq \cdots \subseteq [1,\ell(\mu^{(1)})]\subseteq [1,\ell(\lambda)]\), we may consider \(\delta_{\mathbf{r}^{(i)}}\in \{0,1\}^{\ell(\lambda)}\) by adding to \(\mathbf{r}^{(i)}\) a tail of \(\ell(\lambda)-\ell(\mathbf{r}^{(i)})\) zeroes. In this sense the \(\ell(\lambda)\times N\) matrix \(\delta_{\underline\mathbf{r}}=[\delta_{\mathbf{r}^{(N)}},\dots,\delta_{\mathbf{r}^{(1)}}]\) is called the slack incidence matrix of \(Q\). We also write \(|\delta_{\underline \mathbf{r}}|:=\sum_{i=1}^N|\delta_{\mathbf{r}^{(i)}}|=\sum_{i=1}^N t_0^{(i)}\).
Given the index matrix \(\delta_{\underline\mathbf{r}}\) of a slack vector sequence \(\underline\mathbf{r}=[\mathbf{r}^{(N)}\le_\mathbf{r}\cdots\le\mathbf{r}^{(1)}]\) of \(Q\in Rec_{2n}(\lambda/\mu)\), we define the submatrix \(\delta_{\underline\mathbf{r}^+}\) to be the largest submatrix \(\ell(\mu)\times k\) with \(N\ge k\ge 1\), of \(\delta_{\underline \mathbf{r}}\) such that \(\mu-\delta_{\underline\mathbf{r}^+}\in \mathbb{Z}_{\ge 0}\), where \(\underline\mathbf{r}^+=[\mathbf{r}^{(N)},{\mathbf{r}^{(N-1)}}',\dots,{\mathbf{r}^{(k)}}']\) and \({\mathbf{r}^{(i)}}'\subseteq \mathbf{r}^{(i)}\), \(k\le i< N\). For \(N=1\), \(\underline\mathbf{r}^+=\underline\mathbf{r}\). We call to \(\underline \mathbf{r}^+\) the nonnegative part of \(\underline\mathbf{r}\)
From the slack incidence matrix, one has \(\mu^{(i-1)}=\mu^{(i)}-\delta_{\mathbf{r}^{(i)}}+\varpi_{\ell(\mu^{(i-1)})}\) with \(\mu^{(i)}-\delta_{\mathbf{r}^{(i)}}\subset_{vert}\mu^{(i)}\in Par_{\le \ell(\mu^{(i)})}\), for \(i=1,\dots,N\). (A generalization of 15 .) Henceforth, for \(i=1,\dots,N\), \[\begin{align} &\mu^{(i-1)}=\mu-\sum_{k=i}^N\delta_{\mathbf{r}^{(k)}}+\sum_{k=i}^N\varpi_{\ell(\mu^{(k-1)})}\\ & and \nonumber\\ &\lambda=\mu-\sum_{k=1}^N\delta_{\mathbf{r}^{(k)}}+\sum_{k=1}^N\varpi_{\ell(\mu^{(k-1)})}\nonumber\\ &=\mu-\delta_{\underline\mathbf{r}}+\sum_{k=1}^N\varpi_{\ell(\mu^{(k-1)})},\nonumber\\ &=(\mu-\delta_{\underline\mathbf{r}^+})+\sum_{\mathbf{r}^{(i)}\notin \mathbf{r}^{+}}\delta_{\mathbf{r}^{(i)}\setminus {\mathbf{r}^{(i)}}'}+\sum_{k=1}^N\varpi_{\ell(\mu^{(k-1)})} \end{align}\] where we convention \(\mu-\delta_{\underline\mathbf{r}}:=\mu-\sum_{k=1}^N\delta_{\mathbf{r}^{(k)}}\).
This slack data is illustrated in the next example and in Example 12. For additional illustrations see [36].
Example 11. In Example 10, \(Q\in \widetilde{R}ec_{12}(\lambda/\mu)\), \(N=1\), \(\mu^{(0)}=\lambda\), \(\mu^{(1)}=\mu\) and \(t_0^{(1)}=\ell(\mu)-l^{(1)}_0=6-2=4\) with \(l_0^{(1)}=2\), and \(Q[1]=4\). Thus \(\ell(\mu^{(0)})=\ell(\lambda)=Q[1]+t_0^{(1)}=4+4\) and \(Q[1]+2t_0^{(1)}\le 12\). The slack row index vector is \(\mathbf{r}=(1,3,4,6)\), and the \(\ell(\mu)\times 1\) slack incidence matrix is \([\delta_{\mathbf{r}^{(1)}}]=[1\;0\;1\;1\;0\;1]^T\).
Given \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\), Definition 2, \((R3)\), \((R4)\), impose conditions to the slack sequence of \(Q\), and \((R5)\) impose conditions to its slack vector sequence \(\underline\mathbf{r}=[\mathbf{r}^{(N)},\dots,\mathbf{r}^{(1)}]\) equivalently to the slack incidence matrix \(\delta_{\underline \mathbf{r}}=[\delta_{\mathbf{r}^{(N)}},\dots,\delta_{\mathbf{r}^{(1)}}]\) of \(Q\).
Given \(x\in [1,m]^p\), \(y\in[1,m]^q\), we write \(x\le_{\mathbf{r}} y\) to mean \(p\ge q\ge 0\), and \(x_i\le y_i\) whenever \(i\in[1,q]\). For \(q=0\), \(x\le_\mathbf{r}()\). This means that \(x\le_{\mathbf{r}} y\) if and only if the \(p\times 2\) incidence matrix \([\delta_x,\delta_y]\) is such that for any \(i\ge 1\), the sum of the first \(i\) entries in the first column is always larger or equal than the sum of the first \(i\) entries in the second column. See Example 12.
Proposition 7. [36]Let \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) with vertical strip decomposition ?? , and put \(t^{(0)}_0:=0\) and \(\mathbf{r}^{(0)}:=()\). Then, for each \(1\le i\le N\),
\(\ell(\mu^{(i-1)})=Q[i]+t_0^{(i)}\ge \ell(\mu^{(i)}\),
\(0\le t^{(i-1)}_0\le t^{(i)}_{0} \le \ell(\mu^{(i)})\le \ell(\mu^{(i-1)})\). The slack sequence \(\underline\mathbf{t}=(t^{(N)}_0\ge \dots \ge t^{(1)}_0)\) is weakly decreasing while the sequence \((\ell(\mu^{(N)})\le \dots \le \ell(\mu^{(1)})\le \ell(\mu^{(0)}) )\) is weakly increasing.
\(2n\ge Q[i]+2t_0^{(i)}=\ell(\mu^{(i-1)})+t_0^{(i)}\Leftrightarrow 2n-t_0^{(i)}\ge Q[i]+t_0^{(i)}=\ell(\mu^{(i-1)})\).
\(\mathbf{r}^{(i)}\le_\mathbf{r}\mathbf{r}^{(i-1)}\). The slack vector sequence \(\underline\mathbf{r}=[\mathbf{r}^{(N)}\le_\mathbf{r}\cdots\le_r\mathbf{r}^{(1)}]\) is weakly increasing.
\(\mathbf{r}^{(i)}\subseteq[1,\ell(\mu^{(i)}]\subseteq [1,\ell(\mu^{(i-1)}]\subseteq[1,2n-t_0^{(i)}]\).
Condition \((\mathrm d)\) above guarantees that reverse Schensted column insertion can correctly be \(\mathbf{r}\)-iterated to define \({\mathsf{LR}^{AII}}^{-1}\) as required in Remark 1 and Remark 2.
Example 12.
Let \(n=3\). Let \(V= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {1}, {4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_6(\varpi_3)\), \(\underline \mathbf{t}=(2,2,1,1,0)\), \(\underline\mathbf{r}=[(2,3);(3,4);4;5;()]\) and \[Q= \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {,5,4,3,2,1}, {,4,3,2,1}, {,3,2,1}, {5,2,1}, {3,1}, {1}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in \widetilde{R}ec_{6}((6,5,4,3,2,1)/(1,1,1)),\] \[\mathrm\delta_{\underline\mathbf{r}}=\begin{bmatrix} 0&0&0&0&0\\ 1&0&0&0&0\\ 1&1&0&0&0\\ 0&1&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0\\ \end{bmatrix}_{6\times 5},\quad \mathbf{r}^{(5)}=(2,3)\le_{\mathbf{r}}\mathbf{r}^{(4)}=(3,4)\le_{\mathbf{r}}\mathbf{r}^{(3)}=4\le_{\mathbf{r}}\mathbf{r}^{(2)}=5\le_{\mathbf{r}}\mathbf{r}^{(1)}=().\]
Let \(\mu=\varpi_3\). In this case \(\underline\mathbf{r}^+=(\mathbf{r}^{(5)}=(2,3))\) and \(\mu-\delta_{\underline \mathbf{r}^+}=(1)\). \[\begin{align} & \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})\circ (\underset{{(2,3)}}\leftarrow \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1}, {4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } = \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})((45), 1)=\mathrm{red}_2^{-1}(45).(1)= \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1}, {2}, {4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{(2,3)}, \\ & \mathrm{shape}(S^{(2,3)})=\mu-\mathrm\delta_{\mathbf{r}^{(5)}}+\varpi_4 =(1)+\varpi_4\\ \end{align}\] \[\begin{align} & \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})\circ (\underset{{(3,4)}}\leftarrow S^{(2,3)}) = \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})((45), \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1}, {2}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } )=\mathrm{red}_{2}^{-1}(45). \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1}, {2}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } = \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1,1}, {2,2}, {4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{(2,3),(3,4)}\\ & \mathrm{shape}(S^{(2,3),(3,4)})=(\mu-\mathrm\delta_{\mathbf{r}^{(5)}}+\varpi_4 )-\mathrm\delta_{\mathbf{r}^{(4)}}+\varpi_4\\ \end{align}\] \[\begin{align} & \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})\circ (\underset{{4}}\leftarrow S^{(2,3),(3,4)}) = \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})(5, \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1,1}, {2,2}, {4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } )=\mathrm{red}_1^{-1}(5). \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1,1}, {2,2}, {4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \\ &= \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1,1,1}, {2,2,2}, {3,4}, {4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{(2,3),(3,4),4}\\ & \mathrm{shape}(S^{(2,3),(3,4),4})=(\mu-\mathrm\delta_{\mathbf{r}^{(5)}}+\varpi_4 )-\mathrm\delta_{\mathbf{r}^{(4)}}+\varpi_4-\mathrm\delta_{\mathbf{r}^{(3)}}+\varpi_5 \end{align}\] \[\begin{align} & \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})\circ (\underset{{5}}\leftarrow S^{(2,3),(3,4),4})= \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})(5, \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1,1,1}, {2,2,2}, {3,4}, {4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } )=\mathrm{red}^{-1}(5).S^{(2,3),(3,4),4}\\ &= \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1,1,1,1}, {2,2,2,2}, {3,3,4}, {4,4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{(2,3),(3,4),4,5}\\ & \mathrm{shape}(S^{(2,3),(3,4),4,5})=(\mu-\mathrm\delta_{\mathbf{r}^{(5)}}+\varpi_4 )-\mathrm\delta_{\mathbf{r}^{(4)}}+\varpi_4-\mathrm\delta_{\mathbf{r}^{(3)}}+\varpi_5-\mathrm\delta_{\mathbf{r}^{(2)}}+\varpi_5 \end{align}\] \[\begin{align} & \mathrm{c}\circ (\mathrm{red}_{0}^{-1},\mathrm{id})\circ (\underset{{()}}\leftarrow S^{(2,3),(3,4),4,5})=\mathrm{red}_0^{-1}(()).S^{(2,3),(3,4),4,5}\\ & = \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1,1,1,1,1}, {2,2,2,2,2}, {3,3,3,4}, {4,4,4}, {5,5}, {6}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{(2,3),(3,4),4,5,()}\in SST_{6}(6,5,4,3,2,1). \end{align}\] That is, \[\begin{align} S^{(2,3),(3,4),4,5,()}=\nonumber\\ =&\mathrm{red}_0^{-1}(())\mathpalette\mathbin{\vcenter{\scalebox{\mathrm{red}}{\m@th.5\bullet}}}_1^{-1}(5)\mathpalette\mathbin{\vcenter{\scalebox{(}{\m@th.5\bullet}}}\mathrm{red}_1^{-1}(5)\setminus \{5\})\mathpalette\mathbin{\vcenter{\scalebox{(}{\m@th.5\bullet}}}\mathrm{red}_2^{-1}(45)\setminus \{5\})\mathpalette\mathbin{\vcenter{\scalebox{(}{\m@th.5\bullet}}}\mathrm{red}_2^{-1}(45)\setminus \{4,5\})\mathpalette\mathbin{\vcenter{\scalebox{(}{\m@th.5\bullet}}}1), \end{align}\] where \((1)=\mu-\delta_{\mathbf{r}^{(5)}}\).
The output tableau is a \(\mathfrak{k}\)-lowest weight tableau in \(SST_{6}(6,5,4,3,2,1)\) of \(\mathfrak{k}\)-weight \(-\varpi_3\) as explained in Section 5. The shape of \(S^{(2,3),(3,4),4,5,()}\) is
\[\begin{align} \lambda&=(((\mu-\mathrm\delta_{\mathbf{r}^{(5)}})+\varpi_4)-\mathrm\delta_{\mathbf{r}^{(4)}}+\varpi_4)-\mathrm\delta_{\mathbf{r}^{(3)}}+\varpi_5)-\mathrm\delta_{\mathbf{r}^{(2)}} +\varpi_5-\mathrm\delta_{\mathbf{r}^{(1)}}+\varpi_6\\ &=\mu-\sum_{i=1}^5\delta_{\mathbf{r}^{(i)}}+2\varpi_4+2\varpi_5+\varpi_6=(\mu- \delta_{\mathbf{r}^{+}})+\sum_{\mathbf{r}^{(i)}\notin \mathbf{r}^{+}}\delta_{\mathbf{r}^{(i)}}+2\varpi_4+2\varpi_5+\varpi_6 \end{align}\]
Let \(n=4\). Let \(V= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {1,1}, {4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_8(2,1,1,0)\), \(\ell(\mu)=3\le 4\), \(\underline \mathbf{t}=(2,2)\), \(\underline\mathbf{r}=[(1,3);(2,6)]\) and \[\begin{align} &Q= \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {,,1}, {,2}, {,1}, {2,1}, {2,1}, {2}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in \widetilde{R}ec_{8}((3,2,2,2,2,1)/(2,1,1)),\quad\mathrm\delta_{\underline\mathbf{r}}=\begin{bmatrix} 1&0\\ 0&1\\ 1&0\\ 0&0\\ 0&0\\ 0&1\\ 0&0\\ 0&0 \end{bmatrix}_{8\times 2},\quad \underline\mathbf{r}=[\mathbf{r}^{(2)}=(1,3)\le_{\mathbf{r}}\mathbf{r}^{(2)}=(2,6)], \end{align}\] \(\underline\mathbf{r}^+=[\mathbf{r}^{(2)}=(1,3),{\mathbf{r}^{(2)}}'=(2)]\) and \(\mathrm\delta_{\underline\mathbf{r}^+}=\begin{bmatrix} 1&0\\ 0&1\\ 1&0\\ \end{bmatrix}_{\ell(\mu)\times 2}\), \(\mu-\delta_{\underline\mathbf{r}^+}=(1)\).
\[\begin{align} & \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})\circ (\underset{{(1,3)}}\leftarrow \vcenter{ \begin{tikzpicture}[x={(0in,-0.12in)},y={(0.12in,0in)}] \foreach \rowi [count=\i] in { {1,1}, {4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } = \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})((15), (1,4))=\mathrm{red}_2^{-1}(15)\mathpalette\mathbin{\vcenter{\scalebox{(}{\m@th.5\bullet}}}1,5)= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {1};},1}, {3,4}, {4}, {\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {5};}}, {7}, {8}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{(1,3)}, \\ \end{align}\]
\[\begin{align} & \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})\circ (\underset{{(2,6)}}\leftarrow S^{(1,3)}) = \mathrm{c}\circ (\mathrm{red}_{2}^{-1},\mathrm{id})((48), \vcenter{ \begin{tikzpicture}[x={(0in,-0.14in)},y={(0.14in,0in)}] \foreach \rowi [count=\i] in { {1,1}, {3}, {4}, {5} {7}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } )=\mathrm{red}_{2}^{-1}(48)\mathpalette\mathbin{\vcenter{\scalebox{\YT}{\m@th.5\bullet}}}{0.15in}{}{ {1,1}, {3}, {4}, {5}, {7}, }= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {1};},1}, {2,3}, {\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {4};},4}, {5,\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {5};}}, {6,7}, {\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {8};}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{(1,3),(2,6)}\\ & \mathrm{shape}(S^{(1,3),(2,6)})=(\mu-\delta_{\underline\mathbf{r}^+})-\mathrm\delta_{\mathbf{r}^{(2)}\setminus {\mathbf{r}^{(2)}}'}+\varpi_6 +\varpi_6 \\ \end{align}\] That is, \[\begin{align} S^{(1,3),(2,6)}=(\mathrm{red}_2^{-1}(48))\mathpalette\mathbin{\vcenter{\scalebox{(}{\m@th.5\bullet}}}\mathrm{red}_2^{-1}(15)\setminus \{8\})\mathpalette\mathbin{\vcenter{\scalebox{(}{\m@th.5\bullet}}}1), \end{align}\]
The following statement considers the case where no bumping is needed, that is, the \(0\)-slack sequence is the null vector, \(\underline\mathbf{t}=(t^{(N)}_0\ge \dots \ge t^{(1)}_0)=\underbrace{(0,\dots,0)}_N\). In this case, then the slack vector sequence is written \(\underline\mathbf{r}=(\underbrace{(),\dots,()}_N)\). When the slack number is \(0\) the corresponding slack vector is just written \(\mathbf{r}=()\). It follows from Proposition 7 and the iteration of Remark 6.
Lemma 2. Let \(S\in SpT_{2n}(\mu)\) and let \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) such that \(\lambda=\mu+\sum_{i=1}^{\nu_1}\varpi_{Q[i]}\) with \(Q[1]\ge\cdots\ge Q[N]\ge \ell(\mu)\) and \(\nu=( Q[1],\dots,Q[N])^t\) an even partition.
Then \[\begin{align} \widetilde{R}(S, Q)=Y(\nu)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}} \in SST_{2n}(\lambda) \end{align}\] where \(Y(\nu)\) is the Yamanouchi tableau of shape \(\nu\).
Next one considers constant \(1\)-\(0\)-slack sequences \(\underline\mathbf{t}=(t^{(N)}_0\ge \dots \ge t^{(1)}_0)=(1,1,\dots,1,0,\dots,0)\) possibly with a tail of zeroes. Henceforth, the slack vector sequence \(\underline\mathbf{r}\) can be thought as an increasing sequence of numbers possibly with a tail of empty sets corresponding to the sequence of zeroes in \(\underline\mathbf{t}\). From Theorem 4, ?? , Proposition 7, Lemma 2 and Remark 1 one has.
Theorem 5. Let \(S\in SpT_{2n}(\mu)\) and let \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) with slack sequence \(\underline\mathbf{t}=(\underbrace{1,1,\dots,1}_N)\) and slack vector sequence \(\underline \mathbf{r}=[\mathbf{r}_{N}\le\cdots\le \mathbf{r}_1]\). Then
for \(1\le i\le N\),
\(\mu^{(i-1)}=(\mu^{(i)}-\delta_{\mathbf{r}_i})+\varpi_{Q[i]+1}\) is such that \[\begin{align} &\lambda=\mu^{(0)}\supset_{vert} \mu^{(1)}\supset_{vert} \cdots\supset_{vert}\mu^{(N-1)}\supset_{vert}\mu^{(N)} =\mu \end{align}\]
\(2n-1\ge Q[i]+1=\ell(\mu^{(i-1)})\) and \(\ell(\mu^{(i-1)})\notin 2 \mathbb{Z}\).
\(1\le \mathbf{r}_i\le \ell(\mu^{(i)})\le\ell(\mu^{(i-1)})\le 2n-1\).
\[\begin{align} \widetilde{R}(S, Q)=S^{\underline \mathbf{r}}=S^{\mathbf{r}_N\cdots\mathbf{r}_2 \mathbf{r}_1}\in SST_{2n}(\lambda), \end{align}\] where \(S^{\mathbf{r}_N\cdots\mathbf{r}_2 \mathbf{r}_1}:= {({(\cdots{((S^{\mathbf{r}_N})}^{\mathbf{r}_{N-1}}){\cdots})}^{\mathbf{r}_2})}^{\mathbf{r}_1}\) and \(S^{\mathbf{r}_N\cdots\mathbf{r}_i}\in SST_{2n}(\mu^{(i-1)})\) for \(i=1,\dots,N\).
Proof. The proof is by induction on \(N\). It is enough to analyse the case \(N=2\). Recall \(t_0^{(i)}=1\), for \(i=1,\dots,N\). Let \(P^{(N)}:=S\), \(Q^{(N)}:=Q\) and define \({\mu^{(N)}}'\subset_{vert} \mu^{(N)}\) such that \({\mu^{(N)}}'=\mu^{(N)}-\delta_{\mathbf{r}_N}\), that is, \({\mu^{(N)}}'_{r_{N}}=\mu_{r_{N}}-1\) and \({\mu_i^{(N)}}'=\mu_i\), \(1\le i\neq r_{N}\le \ell(\mu)\). Then the vertical strip \(\mu^{(N)}/{\mu^{(N)}}'\) has the sole cell with row coordinate \(\{\mathbf{r}_{N}\}\). Define \[\begin{align} &P^{(N-1)}:= \mathrm{c}\circ (\mathrm{red}_{1}^{-1},\mathrm{id})\circ (\underset{\mathbf{r}_N}\leftarrow S)\nonumber\\\ &=c \circ (\mathrm{red}_{1}^{-1},\mathrm{id})\circ (b_{r'_{N}}, S^{N})\nonumber\\ &=c\circ (\mathrm{red}_{1}^{-1}(b_{r'_{N}}), S^{N})\nonumber\\\ &= \mathrm{c}\circ(S_{\mathbf{r}_N}, S^{N})=S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N}=:S^{\mathbf{r}_N} \in SST_{2n}(\mu^{(N-1)}), \tag{24}\\ &~ S^N\in SpT_{2n}({\mu^{(N)}}'),~ S_{\mathbf{r}_N}\in SST_{2n}(\varpi_{\ell(\mu^{(N-1)})}),~\mu^{(N-1)}={\mu^{(N)}}-\delta_{\mathbf{r}_N}+\varpi_{\ell(\mu^{(N-1)})} \tag{25} \end{align}\] where \(S(r'_{N},1)=b_{r'_{N}}\) is the entry bumped from the cell \((r'_{N},1)\) in the first column of \(S\) with
\[\begin{align} r'_{N}\ge \mathbf{r}_{N} \end{align}\] and \(S^{N}\) is the tableau obtained from \(S\) after applying the reverse column insertion to the cell of \(\mu/{\mu^{(N)}}'\) in \(S\). Hence \(S^{N}\) has shape \({\mu^{(N)}}'\), and, in particular, \(S^{N}(r'_{N},1)\ge b_{r'_{N}}\) and \(S^{N}(i,1)=S(i,1)\), for \(i\neq r'_{N}\). Since \(S\) and \(S^{N}\) are symplectic, from Proposition 2, \((2)\),
\[\mathbf{r}_{N}\le r'_{N}\le \lfloor \frac{b_{r'_{N}}+1}{2}\rfloor.\]
The expansion \(\mathrm{red}_{1}^{-1}(b_{r'_{N}})=S_{\mathbf{r}_N}\) is a column of length \(Q[N]+1\) obtained according to the rule in Theorem 3
\[\begin{align} S_{\mathbf{r}_N}=: \begin{cases} (1,2,\dots, \mathbf{b}_{r'_{N}}-2, \mathbf{b}_{r'_{N}}, \mathbf{b}_{r'_{N}}+1,\mathbf{b}_{r'_{N}}+2,\dots, \mathbf{b}_{r'_{N}}+(Q[N]-(\mathbf{b}_{r'_{N}}-2))), & \mathbf{b}_{r'_N}\in 2\mathbb{Z}\\ (1,2,\dots \mathbf{b}_{r'_{N}}-1, \mathbf{b}_{r'_{N}}, \mathbf{b}_{r'_{N}}+1+1,\mathbf{b}_{r'_{N}}+1+2,\dots, \mathbf{b}_{r'_{N}}+1+(Q[N]-(\mathbf{b}_{r'_{N}}-1))),& \mathbf{b}_{r'_{N}}\notin 2\mathbb{Z}. \end{cases} \end{align}\]
Let \(\mathbf{b}_{r'_{N }}\in 2\mathbb{Z}\). Indeed \(S'_{\mathbf{r}_N }(i,1)=i\le S^{N}(i,1)\) for \(1\le i\le \mathbf{b}_{r'_{N}}-2\). Note \(S'_{\mathbf{r}_N}(b_{r'_{N}}-1,1)=\mathbf{b}_{r'_{N}}\le S^{N}(r'_{N},1)\).
To ensure that \(S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N}\) is a semi-standard tableau we need to prove that the index row \(\mathbf{b}_{r'_{N}}-1\) is at least \(r'_{N}\), that is, \(\mathbf{b}_{r'_{N}}-1\ge r'_{N}\) when \(r'_{N}= \lfloor \frac{b_{r'_{N}}+1}{2}\rfloor=\frac{b_{r'_{N}}}{2}\). In fact \(\mathbf{b}_{r'_{N}}-1\ge \frac{b_{r'_{N}}}{2}\Leftrightarrow \mathbf{b}_{r'_{N}}\ge 2.\)
Furthermore, \(S_{\mathbf{r}_N}(i,1)=i+1\) for \(i\ge b_{r'_{N}}\ge r'_{N}+1\), and
\[\begin{align} \mathbf{b}_{r'_{N}}&=S_{\mathbf{r}_N}(b_{r'_{N}}-1,1)\\ &\le S^{N}(r'_{N},1)<S^{N}(r'_{N}+1,1)\\ &\Rightarrow \mathbf{b}_{r'_{N}}+1=S_{\mathbf{r}_N}(b_{r'_{N}},1)\\ &\le S^{N}(r'_{N}+1,1)\\ &<S^{N}(r'_{N}+2,1). \end{align}\] Hence, \(\mathbf{b}_{r'_{N}}+k\ge r'_{N}+k+1\), and \[\mathbf{b}_{r'_{N}}+k+1=S'_{\mathbf{r}_N}(b_{r'_{N}}+k,1)\le S^{N}(r'_{N}+k+1,1),~k\ge 0\]
Let \(\mathbf{b}_{r'_{N}}\notin 2\mathbb{Z}\). Indeed \(S_{\mathbf{r}_N}(i,1)=i\le S^{N}(i,1)\) for \(1\le i\le \mathbf{b}_{r'_{N}}-1\). Note \(S'_{N}(b_{r'_{N}},1)=\mathbf{b}_{r'_{N}}\le S^{N}(r'_{N},1)\).
To ensure that \(S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N}\) is a semi-standard tableau we need to prove that the index row \(\mathbf{b}_{r'_{N}}\) is at least \(r'_{N}\), that is, \(\mathbf{b}_{r'_{N}}\ge r'_{N}\) when \(r'_{N}= \lfloor \frac{b_{r'_{N}}+1}{2}\rfloor=\frac{b_{r'_{N}}+1}{2}\). In fact \(\mathbf{b}_{r'_{N}}\ge \frac{b_{r'_{N}}+1}{2}\Leftrightarrow \mathbf{b}_{r'_{N}}\ge 1.\)
On the other hand, \(S^{N}(r'_{N},1)=\mathbf{b}_{r'_{N}}=r'_{N}\notin 2\mathbb{Z}\Rightarrow S^{N}(i,1)=i\), for \(1\le i\le r'_{N}\). Since \(S^{N}\) is symplectic, one has \(r'_{N}=1=\mathbf{b}_{r'_{N}}\) otherwise \(\mathrm{rem}(1,\dots,r'_{N}-1)=(1,\dots,r'_{N}-1)\neq ()\) which is a contradiction. In this case, indeed, \[\begin{align} S_{\mathbf{r}_N}&=( \mathbf{b}_{r'_{N}}=1, \mathbf{b}_{r'_{N}}+1+1=3,\mathbf{b}_{r'_{N}}+1+2=4,\dots, \mathbf{b}_{r'_{N}}+1+Q[N]=1+1+Q[N])\\ &=(1,3,4,\dots,1+Q[N], 2+Q[N]),with\mathrm{rem}(S_{\mathbf{r}_N})=(3,4,\dots,1+Q[N], 2+Q[N]). \end{align}\] Since \(S^{N}(1,1)=1=S'_{\mathbf{r}_N}(1,1)\) and \(S^{N}\) is symplectic, \(S^{N}(i,1)\ge 2i-1\ge 2i-(i-1)=S'_{N}(i,1)\), \(i\ge 2\)
It remains to consider the case, \(S^{N}(r'_{N},1)>\mathbf{b}_{r'_{N}}=S_{\mathbf{r}_N}(\mathbf{b}_{r'_{N}},1)\ge r'_{N}\)
Furthermore, \(S_{\mathbf{r}_N}(i,1)=i+1\) for \(i\ge b_{r'_{N}}+1\ge r'_{N}+1\), and since
\[\mathbf{b}_{r'_{N}}=S'_{\mathbf{r}_N}(b_{r'_{N}},1)< S^{N}(r'_{N},1)<S^{N}(r'_{N}+1,1)\Rightarrow \mathbf{b}_{r'_{N}}+1=S_{\mathbf{r}_N}(b_{r'_{N}},1)\le S^{N}(r'_{N}+1,1)<S^{N}(r'_{N}+2,1)\] Hence, \(\mathbf{b}_{r'_{N}}+k\ge r'_{N}+k\), and \[\mathbf{b}_{r'_{N}}+k+1=S_{\mathbf{r}_N}(b_{r'_{N}}+k,1)\le S^{N}(r'_{N}+k+1,1),~k\ge 0\]
Define also \(Q^{(N-1)}\in \widetilde{R}ec_{2n}(\lambda/\mu^{(N-1)})\) obtained from \(Q^{(N)}\) by deleting the \(Q[N]\) entries \(N\).
We now start with \(P^{(N-1)}= S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N}\in SST_{2n}(\mu^{(N-1)})\) and \(Q^{(N-1)}\). Note the \(\mathbf{r}_{N-1}\ge \mathbf{r}_{N}\) is the row index of the cell of \(\mu^{(N-1)}/{\mu^{(N-1)}}'\).
Define \[\begin{align} P^{(N-2)}&:= \mathrm{c}\circ (\mathrm{red}_{1}^{-1},\mathrm{id})\circ (\underset{\mathbf{r}_{N-1}}\longleftarrow S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N})\\ &= \mathrm{c}\circ (\mathrm{red}_{1}^{-1},\mathrm{id})\circ (b_{r'_{N-1}}, (S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N})^{N-1})\\ &= \mathrm{c}\circ (\mathrm{red}_{1}^{-1}(b_{r'_{N-1}}), (S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N})^{N-1})\\ &= \mathrm{c}\circ (\mathrm{red}_{1}^{-1}(b_{r'_{N-1}}), (S^{\mathbf{r}_N})^{N-1}),by \eqref{srn} \\ &= \mathrm{c}\circ({(S^{\mathbf{r}_N}})_{\mathbf{r}_{N-1}}, (S^{\mathbf{r}_N})^{N-1})\\ &=(S^{\mathbf{r}_N})_{\mathbf{r}_{N-1}}\mathpalette\mathbin{\vcenter{\scalebox{(}{\m@th.5\bullet}}}S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N})^{N-1}=:S^{\mathbf{r}_N\mathbf{r}_{N-1}} \in SST_{2n}(\mu^{(N-2)})\\ &\mu^{(N-2)}={\mu^{(N)}}-\delta_{\mathbf{r}_N}-\delta_{\mathbf{r}_{N-1}}+\varpi_{\ell(\mu^{(N-1)})}+\varpi_{\ell(\mu^{(N-2)})} \end{align}\] where \(S_{\mathbf{r}_N}(r'_{N-1},1)=b_{r'_{N-1}}\) is the entry bumped from the cell \((r'_{N-1},1)\) in the column \(S_{\mathbf{r}_N}\) with \(r'_{N-1}\ge \mathbf{b}_{r'_{N}}-1\ge r'_{N}\), if \(\mathbf{b}_{r'_{N}}\in 2 \mathbb{Z}\) and \(r'_{N-1}\ge \mathbf{b}_{r'_{N}}\ge r'_{N}\), if \(\mathbf{b}_{r'_{N}}\notin 2 \mathbb{Z}\). Therefore \(b_{r'_{N-1}}\ge b_{r'_{N}}\) Since, from \((i)\) \(\mathbf{r}_{N}\le\cdots\le \mathbf{r}_1\), the previous assertion is guaranteed by Remark 2 on reverse bumping routes.
\((S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N})^{N-1}\) is the tableau obtained from \(S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N}\) after applying the reverse column insertion to the cell of \(\mu^{(N-1)}/\mu^{(N-1)'}\) in \(S_{r_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N}\). Hence \((S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N})^{N-1}\) has shape \(\mu^{(N-1)'}\), and, in particular, \((S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N})^{N-1}(r'_{N-1},1)\ge b_{r'_{N-1}}\) and \((S_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N})^{N-1}(i,1)=S'_{\mathbf{r}_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N}(i,1)\), for \(i\neq r'_{N-1}\).
The expansion \(\mathrm{red}_{1}^{-1}(b_{r'_{N-1}})=S_{\mathbf{r}_{N-1}}\) is a column of length \(Q[N-1]+1\ge Q[N]\) obtained by the expansion of order \([Q[N-1]]\) of the column \((b_{r'_{N-1}})\) of length \(1\) according to the rule
\[\begin{align} &S_{\mathbf{r}_{N-1}}=:\\ &=:\begin{cases}& (1,2,\dots, \mathbf{b}_{r'_{N-1}}-2, \mathbf{b}_{r'_{N-1}}, \mathbf{b}_{r'_{N-1}}+1,\mathbf{b}_{r'_{N-1}}+2,\dots, \mathbf{b}_{r'_{N-1}}+(Q[N-1]-(\mathbf{b}_{r'_{N-1}}-2))), \\ & \mathbf{b}_{r'_{N-1}}\in 2\mathbb{Z}\\ &(1,2,\dots \mathbf{b}_{r'_{N-1}}-1, \mathbf{b}_{r'_{N-1}}, \mathbf{b}_{r'_{N-1}}+1+1,\mathbf{b}_{r'_{N-1}}+1+2,\dots, \mathbf{b}_{r'_{N-1}}+1+( Q[N-1]-(\mathbf{b}_{r'_{N-1}}-1))),\\ & \mathbf{b}_{r'_{N-1}}\notin 2\mathbb{Z} \end{cases} \end{align}\]
Since \(b_{r'_{N-1}}\ge b_{r'_{N}}\), it follows that \(S_{\mathbf{r}_{N-1}}\mathpalette\mathbin{\vcenter{\scalebox{(}{\m@th.5\bullet}}}S_{r_N}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{N})^{N-1} \in SST_{2n}(\mu^{(N-2)})\). ◻
More generally from Theorem 4, ?? , Proposition 7, Lemma 2, Theorem 5 and Remark 1, the next assertion follows.
Theorem 6. [36]Let \(S\in SpT_{2n}(\mu)\) and let \(Q\in \widetilde{R}ec_{2n}(\lambda/\mu)\) with slack sequence \(\underline \mathbf{t}=(t^{(N)}_0\ge \dots \ge t^{(1)}_0)\) and slack (row index) vector sequence \(\underline \mathbf{r}=[\mathbf{r}^{(N)}\le_\mathbf{r}\dots\le_\mathbf{r}\mathbf{r}^{(1)}]\). Then
for \(1\le i\le N\),
\(\mu^{(i-1)}=(\mu^{(i)}-\delta_{\mathbf{r}_i})+\varpi_{Q[i]+t_0^{(i)}}\) is such that
\[\begin{align} &\lambda=\mu^{(0)}\supset_{vert} \mu^{(1)}\supset_{vert} \cdots\supset_{vert}\mu^{(N-1)}\supset_{vert}\mu^{(N)} =\mu. \end{align}\]
\(\mathbf{r}^{(i)}\subseteq[1,\ell(\mu^{(i)}]\subseteq [1,\ell(\mu^{(i-1)}]\subseteq[1,2n-t_0^{(i)}]\).
\(2n-t_0^{(i)}\ge Q[i]+t_0^{(i)}=\ell(\mu^{(i-1)}\).
\[\begin{align} \widetilde{R}(S, Q)=S^{\underline \mathbf{r}}=S^{\mathbf{r}^{(N)}\cdots\mathbf{r}^{(2)} \mathbf{r}^{(1)}}:= {\bigg({\bigg(\cdots{\big(\big(S^{\mathbf{r}^{(N)}}\big)}^{\mathbf{r}^{(N-1)}}\big)\cdots\bigg)}^{\mathbf{r}^{(2)}} \bigg)}^{\mathbf{r}^{(1)}}\in SST_{2n}(\lambda). \end{align}\]
From now on we write \(\widetilde{R}= {\mathsf{LR}^{AII}}^{-1}\) and \(\widetilde{R}ec_{2n}=Rec_{2n}\).
The explicit surjectivity and consequently the explicit inverse of the quantum Littlewood-Richardson map allows to explicitly compute \(\mathfrak{k}\)-highest or lowest weight tableaux in the proof of the Naito–Sagaki conjecture via the branching rule for \(\imath\)quantum groups [32]. We closely follow the notation in [32] and refer to it for further definitions and details.
Let \(n\in\mathbb{N}\) and let us consider the two sequences of positive integers defined in [32]. For \(k=1,\dots,n\),
\[\begin{align} &u_k=2k-\frac{1+(-1)^k}{2}=\begin{cases}2k,&\text{ if } k \notin 2 \mathbb{Z},\\ 2k-1,&\text{ if } k \in 2 \mathbb{Z},\tag{26} \end{cases}\\ &v_k=2k-\frac{1+(-1)^{k+1}}{2}=\begin{cases}2k,&\text{ if } k \in 2 \mathbb{Z},\\ 2k-1,&\text{ if } k \notin 2 \mathbb{Z}. \tag{27} \end{cases} \end{align}\] These two sequences \[\begin{align} \{u_i\}_{i=1}^n=\begin{cases}\{2,3,6,7,10,\dots, 2(n-1)-1,2n\},& n\notin 2 \mathbb{Z},\\ \{2,3,6,7,\dots, 2(n-1),2n-1\},& n\in 2 \mathbb{Z}\end{cases}\label{numbers:uu} \end{align}\tag{28}\] and \[\begin{align} \{v_i\}_{i=1}^n=\begin{cases}\{1,4,5,8,\dots, 2(n-1)-1,2n\},& n\in 2 \mathbb{Z}\\ \{1,4,5,\dots, 2(n-1),2n-1\},& n\notin 2 \mathbb{Z},\end{cases}\label{numbers:vv} \end{align}\tag{29}\] have no common values and its union gives \(\{u_i\}_{i=1}^n\sqcup\{v_i\}_{i=1}^n=[1,2n]\).
Example 13. For \(n=6\), \(u_i=2,3,6,7,10,11\) and \(v_i=1,4,5,8,9,12\), \(1\le i\le 6\), and \(\{u_i\}_{i=1}^6\cup \{v_i\}_{i=1}^6=\{1,2,\dots,12\}\); and for \(n=7\), \(u_i=2,3,6,7,10,11, 14\) and \(v_i=1,4,5,8,9,12,13\), \(1\le i\le 7\), and \(\{u_i\}_{i=1}^6\cup \{v_i\}_{i=1}^6=\{1,2,\dots,14\}\).
For \(S\in SST_{2n}(\lambda)\), its \(\mathfrak{k}\)-weight is [32] \[\begin{align} \mathrm{wt}_\mathfrak{k}(S)=(S[u_1]-S[v_1])\tilde{\varepsilon}_1+(S[u_2]-S[v_2])\tilde{\varepsilon}_2+\dots+(S[u_n]-S[v_n])\tilde{\varepsilon}_n.\label{kweight} \end{align}\tag{30}\]
For example for \(n=3\), \(\mathrm{wt}_\mathfrak{k}(S)=(S[2]-S[1])\tilde{\varepsilon}_1+(S[3]-S[4])\tilde{\varepsilon}_2+(S[6]-S[5])\tilde{\varepsilon}_3\).
The set \(\widetilde{P}^+=\{\mu_1 \tilde{\varepsilon}_1+\cdots+\mu_n \tilde{\varepsilon}_n\in\widetilde{P}:\mu_1\ge\cdots\ge \mu_n\ge 0\}\) can be identified with the set \(Par_{\le n}\) [32].
If \(S\) is the symplectic tableau below on the LHS 31 then the shape is equal to the corresponding \(\mathfrak{k}\)-weight, \(\mathrm{wt}_\mathfrak{k}(S)=(S[u_1]\ge S[u_2]\ge \dots\ge S[u_n])\), and if \(S\) is on the RHS 31 then the \(\mathfrak{k}\)-weight is
\[\begin{align} & \eqref{symphw-lw}RHS, \mathrm{wt}_\mathfrak{k}(S) =-S[v_1]\tilde{\varepsilon}_1-S[v_2]\tilde{\varepsilon}_2-\dots-S[v_n]\tilde{\varepsilon}_n \\ &=-(S[v_1]\tilde{\varepsilon}_1+S[v_2]\tilde{\varepsilon}_2+\dots+S[v_n]\tilde{\varepsilon}_n)\\ & identified with-(S[v_1]\ge S[v_2]\ge \dots\ge S[v_n]) \end{align}\] and the shape is \[w_0 \mathrm{wt}_\mathfrak{k}(S)=w_0(-S[v_1],-S[v_2],\dots,-S[v_n])=(S[v_1]\ge S[v_2]\ge\dots\ge S[v_n])\] where \(w_0\) is the longest element of the Weyl group of the Lie algebra \(\mathfrak{k}\) isomorphic to the symplectic Lie algebra \(\mathfrak{sp}_{2n}(\mathbb{C})\),
\[\begin{align} \label{symphw-lw} \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{u_1},{u_1},\cdots,\cdots,\cdots,\cdots,u_1}, {{u_2},\cdots,\cdots,\cdots,\cdots,{u_2}}, {{\vdots},\vdots,\vdots,{\vdots}}, {u_n,\cdots,u_n}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } &\qquad\qquad \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{v_1},{v_1},\cdots,\cdots,\cdots,\cdots,v_1}, {{v_2},\cdots,\cdots,\cdots,\cdots, {v_2}}, {{\vdots},\vdots,\vdots,{\vdots}}, {v_n,\cdots,v_n}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \\symplectic \mathfrak{k}-highest weight tableau& \quadsymplectic \mathfrak{k}-lowest weight tableau \nonumber \end{align}\tag{31}\]
Let \(SST_{2n}\) denote the set of all semi-standard tableaux in the alphabet \([1,2n]\) and \(SpT_{2n}\) its subset of symplectic tableaux. Next, \(\mathfrak{k}\)-highest weight tableaux and \(\mathfrak{k}\)-lowest weight tableaux in \(SST_{2n}\) are defined.
Definition 10. [32] Let \(S\in SST_{2n}\).
\(S\) is called a \(\mathfrak{k}\)-highest weight tableau if \(P^{AII}(S)\) is a symplectic tableau of the form shown in the LHS of 31 ; let \(SST^{\mathfrak{k}-hw} _{2n}(\lambda)\subseteq SST_{2n}(\lambda)\) denote the set of all \(\mathfrak{k}\)-highest weight tableaux of shape \(\lambda\) in \(SST_{2n}\).
\(S\) is called a \(\mathfrak{k}\)-lowest weight tableau if \(P^{AII}(S)\) is a symplectic tableau of the form shown in RHS of 31 ; let \(SST^{\mathfrak{k}-lw} _{2n}(\lambda)\subseteq SST_{2n}(\lambda)\) denote the set of all \(\mathfrak{k}\)-lowest weight tableaux of shape \(\lambda\) in \(SST_{2n}\).
Since \(P^{AII}(S)=S\) for \(S\in SpT_{2n}\), this definition for \(S\in SpT_{2n}\) obviously restricts to 31 . For \(n=1,2,3\), the symplectic \(\mathfrak{k}\)-highest weight respectively \(\mathfrak{k}\)-lowest weight tableaux are then respectively of the form \[\begin{align} \tag{32}n=1:~~ &S^H= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{2},{2},\cdots,\cdots,\cdots,\cdots,2}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,\qquad S_L= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{1},{1},\cdots,\cdots,\cdots,1}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_2;\\\nonumber \\ n=2:~~&S^H= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{2},{2},\cdots,\cdots,\cdots,\cdots,2}, {{3},\cdots,\cdots,\cdots,\cdots,{3}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,\qquad S_L= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{1},{1},\cdots,\cdots,\cdots,1}, {{4},\cdots,\cdots,\cdots,{4}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_4\tag{33}\\ \nonumber \\ n=3:~~&S^H= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{2},{2},2,\cdots,\cdots,\cdots,\cdots,2}, {{3},3,\cdots,\cdots,\cdots,\cdots,{3}}, {{6},\cdots,\cdots,\cdots,\cdots,{6}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,\qquad S_L= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{1},{1},1,\cdots,\cdots,\cdots,1}, {{4},4,\cdots,\cdots,\cdots,{4}}, {{5},\cdots,\cdots,\cdots,{5}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_6\tag{34} \end{align}\] \[\begin{align} n=4:~~&S^H= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{2},{2},2,2,\cdots,\cdots,\cdots,\cdots,2}, {{3},3,3,\cdots,\cdots,\cdots,\cdots,{3}}, {{6},6,\cdots,\cdots,\cdots,\cdots,{6}}, {{7},\cdots,\cdots,\cdots,{7}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,\qquad S_L= \vcenter{ \begin{tikzpicture}[x={(0in,-0.2in)},y={(0.2in,0in)}] \foreach \rowi [count=\i] in { {{1},{1},1,1,\cdots,\cdots,\cdots,1}, {{4},4,4,\cdots,\cdots,\cdots,{4}}, {{5},5,\cdots,\cdots,\cdots,{5}}, {{8},\cdots,\cdots,\cdots,{8}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_6\label{4symphw-lw4} \end{align}\tag{35}\] In general, for \(n\in\mathbb{N}\), the symplectic \(\mathfrak{k}\)-highest weight tableaux in \(SpT_{2n}\) comprise the symplectic columns \[\begin{align} u_1\cdots u_{n-1}u_n\in SpT_{2n}(\varpi_n),~u_1\cdots u_{n-1}\in SpT_{2n}(\varpi_{n-1}),\dots, u_1u_2\in SpT_{2n}(\varpi_{2}), ~u_1\in SpT_{2n}(\varpi_1) \end{align}\] and the symplectic \(\mathfrak{k}\)-lowest weight tableaux in \(SST_{2n}\), comprise the symplectic columns \[\begin{align} v_1\cdots v_{n-1}v_n\in SpT_{2n}(\varpi_n),~v_1\cdots v_{n-1}\in SpT_{2n}(\varpi_{n-1}),\dots, v_1v_2\in SpT_{2n}(\varpi_{2}), ~v_1\in SpT_{2n}(\varpi_1). \end{align}\] \[\begin{align} n\notin 2 \mathbb{Z}:\nonumber\\ &S^H= \vcenter{ \begin{tikzpicture}[x={(0in,-0.28in)},y={(0.28in,0in)}] \foreach \rowi [count=\i] in { {{2},{2},2,2,\cdots,\cdots,\cdots,\cdots,2}, {{3},3,3,\cdots,\cdots,\cdots,\cdots,{3}}, {{6},6,\cdots,\cdots,\cdots,\cdots,{6}}, {{7},\cdots,\cdots,\cdots,\cdots,{7}}, {{10},\cdots,\cdots,\cdots,\cdots,{10}}, {\vdots,\cdots,\cdots,\cdots,\vdots}, {\scriptstyle {2n-3},\cdots, \cdots,\scriptstyle{2n-3}}, {\scriptstyle{2n},\cdots,\scriptstyle{2n}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_{2n}\label{noddsymphw-hw} \end{align}\tag{36}\]
\[\begin{align} n\in 2 \mathbb{Z}:\nonumber\\ &S^H= \vcenter{ \begin{tikzpicture}[x={(0in,-0.28in)},y={(0.28in,0in)}] \foreach \rowi [count=\i] in { {{2},{2},2,2,\cdots,\cdots,\cdots,\cdots,2}, {{3},3,3,\cdots,\cdots,\cdots,\cdots,{3}}, {{6},6,\cdots,\cdots,\cdots,\cdots,{6}}, {{7},\cdots,\cdots,\cdots,{7}}, {\vdots,\cdots,\cdots,\cdots,\vdots}, {\scriptstyle{2n-5},\cdots, \cdots,\scriptstyle{2n-5}}, {\scriptstyle{2n-2},\cdots, \cdots,\scriptstyle{2n-2}}, {\scriptstyle{2n-1},\cdots,\scriptstyle{2n-1}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_{2n}\label{nevensymphw-hw} \end{align}\tag{37}\]
It would be interesting to have an explicit characterization of \(SST^{\mathfrak{k}-hw} _{2n}(\lambda)\) or \(SST^{\mathfrak{k}-lw} _{2n}(\lambda)\). This is known for \(n=1, 2\) [32]. Indeed one has an algorithm for \({\mathsf{LR}^{AII}}^{-1}\) to compute numerically the elements of those sets. However, taking into account the \(n=2\) case, it is expected that these tableaux satisfy restrictions in the form of inequalities between the multiplicities of some of its columns. We provide an interpretation of this phenomena via the reverse Schensted insertion ruled by the slack data on the quantum recording tableaux with one vertical strip and with \(1\)-\(0\)-slack vector sequences.
From the quantum Littlewood-Richardson bijection 2 , we see that for each \(\mu\in Par_{\le n}\), the recording tableaux \(Rec_{2n}(\lambda/\mu)\overset{\sim}\rightarrow LRS_{2n}(\lambda/\mu)\) determine the pairs consisting of \(\mathfrak{k}\)-highest respectively \(\mathfrak{k}\)-lowest weight tableaux \(S^{H,\mu}\) and \(S_{L,-\mu} \in SST_{2n}(\lambda)\) where \(\mathrm{wt}_{\mathfrak{k}}(S^{H,\mu})=\mu\) respectively \(\mathrm{wt}_{\mathfrak{k}}(S_{L,-\mu})=-\mu\). That is, the set \(Rec_{2n}(\lambda/\mu)\) determine the tableaux in \(SST^{\mathfrak{k}-hw} _{2n}(\lambda)\) respectively \(SST^{\mathfrak{k}-lw} _{2n}(\lambda)\)) such that \(\mathrm{wt}_{\mathfrak{k}}(T)=\mu\) and respectively \(wt_{\mathfrak{k}}(T)=-\mu\). Consequently, our algorithm for \({\mathsf{LR}^{AII}}^{-1}\) compute those sets.
Proposition 8. Let \(S^{H,\mu}, S_{L,-\mu}\in SpT_{2n}(\mu)\) be the symplectic \(\mathfrak{k}-hw\) respectively \(\mathfrak{k}-lw\) weight tableaux of \(SpT_{2n}(\mu)\). Then
\({\mathsf{LR}^{AII}}^{-1}(\{S^{H,\mu} \}\times Rec_{2n}(\lambda/\mu))= \{S\in SST^{\mathfrak{k}-hw} _{2n}(\lambda)| \mathrm{wt}_{\mathfrak{k}}(S)=\mu\} \subseteq SST_{2n}(\lambda)\).
\({\mathsf{LR}^{AII}}^{-1}(\{S_{L,-\mu} \}\times Rec_{2n}(\lambda/\mu))= \{S\in SST^{\mathfrak{k}-lw} _{2n}(\lambda)| \mathrm{wt}_{\mathfrak{k}}(S)=-\mu\} \subseteq SST_{2n}(\lambda)\).
Therefore \[\begin{align} SST^{\mathfrak{k}-hw} _{2n}(\lambda)=\bigsqcup_{\begin{smallmatrix}\mu\in Par_{\le n}\\ \mu\subseteq\lambda\end{smallmatrix}} {\mathsf{LR}^{AII}}^{-1}(\{S^{H,\mu}\}\times Rec_{2n}(\lambda/\mu)) \end{align}\] and \[\begin{align} SST^{\mathfrak{k}-lw} _{2n}(\lambda)=\bigsqcup_{\begin{smallmatrix}\mu\in Par_{\le n}\\ \mu\subseteq\lambda\end{smallmatrix}} {\mathsf{LR}^{AII}}^{-1}(\{S_{L,-\mu}\}\times Rec_{2n}(\lambda/\mu)). \end{align}\]
Let \(\mu\subseteq_{vert}\lambda\), \(t_0\) the slack of the vertical strip \(\lambda/\mu\) and \(\mathbf{r}\) the corresponding slack row index vector. Let \(\mu'=\mu-\delta_\mathbf{r}\) and note \(\lambda=\mu'+\varpi_{\ell(\lambda)}\). For \(S^{H,\mu}\) the \(\mathfrak{k}\)-highest respectively \(S_{L,-\mu}\) \(\mathfrak{k}\)-lowest weight tableau in \(SpT_{2n}(\mu)\), Theorem 4 implies the following assertion.
Corollary 4. [36]Let \(S^{H,\mu}\) be the \(\mathfrak{k}\)-highest and let \(S_{L,-\mu}\) be the \(\mathfrak{k}\)-lowest weight tableaux in \(SpT_{2n}(\mu)\) as in 33 . Let \(Q\in Rec_{2n}(\lambda/\mu)\) with slack row index vector \(\mathbf{r}=(r_1,\dots, r_{t_0})\) and \(\mu'=\mu-\delta_\mathbf{r}\). Let \(u_\mathbf{r}=(u_{r_1},\dots, u_{r_{t_0}})\) and \(v_\mathbf{r}=(v_{r_1},\dots, v_{r_{t_0}})\) be sequence of positive numbers in [1,2n] as in 26 respectively in 27 . Then,
\({LR^{AII}}^{-1}(S^{H,\mu},Q)\) returns the following \(\mathfrak{k}\)-highest weight tableau in \(SST_{2n}(\lambda)\) with \({\mathfrak{k}}\)-weight \(\mu\):
\[\begin{align} \label{TUH}(T^u_0T^u_1\cdots T^u_{t_0})S^{H,\mu'}\in SST_{2n}(\lambda) \end{align}\qquad{(18)}\] where \(S^{H,\mu'}=(S^{H,\mu})^1\), 18 , is the \(\mathfrak{k}\)-highest weight tableau in \(SpT_{2n}(\mu')\) and \((T^u_0T^u_1\cdots T^u_{t_0})\in SST_{2n}(\varpi_{\ell(\lambda)})\), given by \[\begin{align} &T^u_0=(1,2,\dots l_1),\nonumber\\ &T^u_i= \begin{cases} (u_{{r}_i}, u_{{r}_i}+1,\dots, u_{{r}_i}+l_{i+1}), & ifu_{{r}_i}\in 2\mathbb{Z}\\ (u_{{r}_i}, u_{{r}_i}+1+1,u_{{r}_i}+1+2,\dots, u_{{r}_i}+1+l_{i+1}), &ifu_{{r}_i}\notin 2\mathbb{Z}\\ \end{cases}& 1\le i\le t_0, \end{align}\] with \(l_1,\dots,l_{t_0+1}\) as in Theorem 3
\({LR^{AII}}^{-1}(S_{L,-\mu},Q)\) returns the \(\mathfrak{k}\)-lowest weight tableau in \(SST_{2n}(\lambda)\) with \({\mathfrak{k}}\)-weight \(-\mu\):
\[\begin{align} \label{TVL}(T^v_0T^v_1\cdots T^v_{t_0})S_{L,-\mu'}\in SST_{2n}(\lambda) \end{align}\qquad{(19)}\] where \(S_{L,-\mu'}=(S_{L,-\mu})^1\), 18 , is the \(\mathfrak{k}\)-lowest weight tableau in \(SpT_{2n}(\mu')\) and \((T^v_0T_1\cdots T^v_{t_0})\in SST_{2n}(\varpi_{\ell(\lambda)})\), given by \[\begin{align} &T^v_0=(1,2,\dots l_1),\nonumber\\ &T^v_i= \begin{cases} (v_{{r}_i}, v_{{r}_i}+1,\dots, v_{{r}_i}+l_{i+1}), & ifv_{{r}_i}\in 2\mathbb{Z}\\ (v_{{r}_i}, v_{{r}_i}+1+1,v_{{r}_i}+1+2,\dots, v_{{r}_i}+1+l_{i+1}), &ifv_{{r}_i}\notin 2\mathbb{Z}\\ \end{cases}& 1\le i\le t_0, \end{align}\] with \(l_1,\dots,l_{t_0+1}\) as in Theorem 3
The following example illustrates the \(\mathfrak{k}\)-lowest weight case. For the \(\mathfrak{k}\)-highest weight case see [36].
Example 14. Let \(n=3\), \(\mu=(4,2,1)\), \(t_0=1\) and \(\mathbf{r}=\{3\}\).
Let \(v=\{v_3=5\}\subseteq \{v_1=1,v_2=4,v_3=5\}\). Let \(S_{L,-\mu'}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,1,1,1}, {4,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SpT_6(\mu')\) with \(\mu'=\mu-\delta_\mathbf{r}=(4,2,1)-(0,0,1)=(4,2,0)\). Then \(\mathrm{red}_1^{-1} (5)=(12345)\), \(\mathrm{wt}_\mathfrak{k}(12345)=(1-1)\tilde{\varepsilon}_1+(1-1)\tilde{\varepsilon}_2+(0-1)\tilde{\varepsilon}_3=-\tilde{\varepsilon}_3\), and \[S=(12346)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}_{L,-\mu'}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,1,1,1,1}, {2,4,4}, {3}, {4}, {5}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \in SST^{\mathfrak{k}-lw}_{2\times 3}(\lambda), ~~l=\ell(\lambda)=5, ~~ \mathrm{wt}_{\mathfrak{k}}(S)=-\mu= \mathrm{wt}_{\mathfrak{k}}(S_{L,-\mu})\] where \(\lambda=\mu-\delta_\mathbf{r}+\varpi_5\) and \(1=t_0\le l\le 6, ~6-5\) and \(l-t_0=5-1=4\in 2 \mathbb{Z}\).
We now consider quantum recording tableaux with \(1\)-\(0\)-slack sequences. In order to proceed, we need some notation. Given \(T\in SST_{2n}\) and \(\mathbf{b}\) a column in \(SST_{2n}\) we write \(m_\mathbf{b}\) to mean the multiplicity of \(\mathbf{b}\) in \(T\). We also write \(\circledcirc_{q}^p\), to mean the concatenation in the plactic monoid of \(q-p+1\) words, say \(w_q\mathpalette\mathbin{\vcenter{\scalebox{\dots}{\m@th.5\bullet}}}\mathpalette\mathbin{\vcenter{\scalebox{w}{\m@th.5\bullet}}}_p\), with the convention that when \(q-p+1\le 0\) the concatenation is the empty word.
Theorem 7. Let \(1\le n\notin 2 \mathbb{Z}\). Consider \(S^{H,\mu}\) the \(\mathfrak{k}\)-highest weight tableau in \(SpT_{2n}(\mu)\) as in 36 . Let \(Q\in Rec_{2n}(\lambda/\mu)\) with \(1\)-\(0\)-slack sequence of the form \(\underline \mathbf{t}=(1,\dots,1,0^M)\) and corresponding slack row index vector sequence of the form \(\underline \mathbf{r}=(1,\dots,1,2,\dots,2,3,\dots,3,\dots,n,\dots,n,2n-1,\dots,2n-1,()^M)\). Let \(u_1=2,\dots,u_n=2n\) be the numbers in 28 or 36 , and \(\mu'=\mu-\delta_{\underline\mathbf{r}^+}\) . Then,
\(2n-1\notin \underline\mathbf{r}^+\).
for some \(0\le k\le n\), \({\mathsf{LR}^{AII}}^{-1}(S^{H,\mu},Q)\) returns the \(\mathfrak{k}\)-highest weight tableau in \(SST_{2n}(\lambda)\), with \({\mathfrak{k}}\)-weight \(\mu\), of the form
\[\begin{align} &Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\nonumber}{\m@th.5\bullet}}}\\ &\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle}{\m@th.5\bullet}}}\circledcirc_{j=n}^{k+1}\big(\mathrm{red}_1^{-1}({u_j})\big)^{m_{\mathrm{red}^{-1}(u_j)}}\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}({u_{k}})\big)^{m_{\mathrm{red}^{-1}(u_k)}}\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=k}^{1}\big(\mathrm{red}_1^{-1}({u_j})\setminus\{u_n\}\big)^{m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}}}{\m@th.5\bullet}}}\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}, \end{align}\] where \(S^{H,\mu'}\) is the \(\mathfrak{k}\)-highest weight tableau in \(SpT(\mu')\), \[\displaystyle \sum_{j=n}^{k+1} m_{\mathrm{red}^{-1}(u_j)}+m_{\mathrm{red}^{-1}(u_k)}+\sum_{j=1}^k m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}=|\delta_{\underline\mathbf{r}}|,\] and satisfying restrictions on linear inequalities on the multiplicities of the columns
\[\begin{align} &\mathrm{red}^{-1}(u_i),~~ \mathrm{red}^{-1}(u_i)\setminus \{u_n\},i=2,\dots,n, on the LHS of S^{H,\mu'},\\ & and the symplectic columns u_1u_2\cdots u_i\in SpT_{2n}(\varpi_i) \eqref{noddsymphw-hw}, i=1,\dots,n,\nonumber\\ &u_1u_2\cdots u_n,~~ u_1u_2\cdots u_{n-1},\dots,u_1u_2,~~u_1 \end{align}\] as follows
\[\begin{align} &m_{\mathrm{red}^{-1}(u_i)}+m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}\le m_{u_1\cdots u_{i-1}}, ~i=2,\dots,n,\\ &0\le m_{\mathrm{red}^{-1}(u_n)}-\sum_{i=1}^{n-1}m_{\mathrm{red}^{-1}(u_i)\setminus\{u_n\}}\le m_{u_1\cdots u_{n-1}}. \end{align}\]
Proof. It follows from Theorem 5 and a detailed analysis of reverse Schensted insertion for the given slack vector sequence \(\underline\mathbf{r}\). ◻
Corollary 5. Let \(S^{H,u}\) be the \(\mathfrak{k}\)-highest weight tableau and let \(S_{L,-u}\) be the \(\mathfrak{k}\)-lowest weight tableau in \(SpT_2(u)\), with \((u)\) a one-row partition, as in 32 . Let \(Q\in Rec_{2}(\lambda/(u))\) of weight \(\nu=(Q[1]^M)^t=(2^M)^t=(M,M)\).Then for \(n=1\), the \(\mathfrak{k}\)-highest (-lowest) weight tableaux in \(SST_{2}(\lambda)\) are respectively the hook-shape tableaux \[\begin{align} &Y(M,M)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,u},\quad \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,{2},{2},\cdots,\cdots,\cdots,2}, {2,\cdots,2}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} }and \\ & Y(M,M)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}_{L,-u},\quad \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,{1},{1},\cdots,\cdots,\cdots,1}, {2,\cdots,2}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \end{align}\] with \({\mathfrak{k}}\)-weights respectively \((u)\) and \((-u)\).
Proof. The identities follow from Lemma 2 and also from previous theorem with \(n=1\). From the previous theorem with \(n=1\), it means \[\begin{align} &\big(\mathrm{red}_0^{-1}(())\big)^M\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,u}=(1,2)^M\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,u}=Y(M,M)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,u}\\ &\big(\mathrm{red}_0^{-1}(())\big)^M\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}_{L,-u}=Y(M,M)\mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}_{L,-u}. \end{align}\] ◻
Corollary 6. Let \(n=3\). Consider \(S^{H,\mu}\) the \(\mathfrak{k}\)-highest weight tableau in \(SpT_6(\mu)\) as in LHS of 34 . Let \(Q\in Rec_{6}(\lambda/\mu)\) with slack sequence and corresponding slack row index vector sequence of the form \(\underline \mathbf{t}=(1,\dots,1,0^M)\) respectively \(\underline \mathbf{r}=(1,\cdots,1,2,\cdots,2,3,\cdots,3,5,\dots,5,()^M)\). Then, \({\mathsf{LR}^{AII}}^{-1}(S^H,Q)\) returns the \(\mathfrak{k}\)-highest weight tableau in \(SST_{6}(\lambda)\) with \({\mathfrak{k}}\)-weight \(\mu\) in either form, with the following legend, circled elements indicate bumped entries from \(S^{H,\mu}\), blue circle indicates the column \(\mathrm{red}_1^{-1}(2)\), orange circle indicates the column \(\mathrm{red}_1^{-1}(3)\) and brown circle indicates the column \(\mathrm{red}_1^{-1}(6)\):
\[\begin{align} \label{type31} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,(5)^0,()^M}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,\color{brown}{1}, \color{orange}{1},\cdots,\color{orange}{1},\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}}, 2,\cdots, 2,2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,2, \color{orange}{ 2},\cdots,2,\color{blue}{3},\cdots,\color{blue}{3},{3},\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,\color{brown}{3}, \color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{blue}{4},\cdots,\color{blue}{4},{6},\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,\color{brown}{4}, \color{orange}{5},\cdots,\color{orange}{5},\color{blue}{5},\cdots,\color{blue}{5}}, {5,\cdots,5,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}}, \color{orange}{6},\cdots,\color{orange}{6},\color{blue}{6},\cdots,\color{blue}{6}}, {6,\cdots,6}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \nonumber\\ &m_{12356}\le m_2and 0\le m_{12346}\le m_{23} \end{align}\qquad{(20)}\] or \[\begin{align} \label{type32} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,(5)^a, ()^M}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,1,\color{brown}{1},\cdots,1, \color{orange}{1},\cdots,1,\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}}, \cdots,\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},2,\cdots, 2,2,\cdots, 2, 2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,2, \color{brown}{2},\cdots,2, \color{orange}{2},\cdots,2,\color{blue}{3},\cdots,3,{3},\cdots,3,3,\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,3,\color{brown}{3},\cdots,3, \color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{blue}{4},\cdots,4, 4,\cdots,4, 6,\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,4, \color{brown}{4},\cdots,4, \color{orange}{5},\cdots,5,\color{blue}{5},\cdots,5, 5,\cdots,5}, {5,\cdots,5,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},{6},\cdots,6, {6},\cdots,6,{6},\cdots,6}, {6,\cdots,6}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \quad a>0\nonumber\\ &m_{12356}+m_{1235}\le m_2and 0\le m_{12346}-m_{1235}-m_{2345}\le m_{23} \end{align}\qquad{(21)}\] or \[\begin{align} \label{type33} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,5^b,()^M}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,1, \color{brown}{1},\cdots,1,\color{orange}{1},\cdots,1, \color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}}, 2,\cdots, 2, 2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,2, \color{brown}{2},\cdots,2,\color{orange}{ 2},\cdots,2,\color{blue}{3},\cdots,3,3,\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,3,\color{brown}{3},\cdots,3, \color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}}, \color{blue}{4},\cdots,4, 6,\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,4,\color{brown}{4},\cdots,4,\color{orange}{ 5},\cdots,5, \color{blue}{5},\cdots,{5}}, {5,\cdots,5,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}}, 6,\cdots,6,6,\cdots,6}, {6,\cdots,6}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \quad b>0\nonumber\\ &m_{12356}+m_{1235}\le m_2and 0\le m_{12346}-m_{1235}-m_{2345}\le m_{23} \end{align}\qquad{(22)}\] or\[\begin{align} \label{type34} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,5^c,()^m}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,1, \color{brown}{1},\cdots,1,\color{orange}{1},\cdots,1,1,\cdots,1,\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}}, 2,\cdots, 2, 2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,2,\color{brown}{2},\cdots,2, \color{orange}{2},\cdots,2, 2,\cdots,2,\color{blue}{3},\cdots,3,3,\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,3, \color{brown}{3},\cdots,3,\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}}, \color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{blue}{4},\cdots,4, 6,\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,4, \color{brown}{4},\cdots,4,\color{orange}{5},\cdots,5,5,\cdots,5, \color{blue}{5},\cdots,5}, {5,\cdots,5,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}}, {6},\cdots,6,{6},\cdots,6}, {6,\cdots,6}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \quad c>0\nonumber\\ &m_{12356}+m_{1235}\le m_2and 0\le m_{12346}-m_{1235}-m_{2345}\le m_{23} \end{align}\qquad{(23)}\] or\[\begin{align} \label{type35} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,5^d,()^M}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,1,\color{brown}{1},\cdots,1,\color{orange}{1},\cdots,1, \color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}}, 2,\cdots, 2, 2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,2,\color{brown}{2},\cdots,2, \color{orange}{2},\cdots,2,\color{blue}{3},\cdots,3,3,\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,3,\color{brown}{3},\cdots,3, \color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots, \color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{blue}{4},\cdots,4, 6,\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,4,\color{brown}{4},\cdots,4, \color{orange}{5},\cdots,5, \color{blue}{5},\cdots,5}, {5,\cdots,5,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},6,\cdots,6}, {6,\cdots,6}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \quad d>0\nonumber\\ &m_{12{\boldsymbol{3}}56}+m_{1235}\le m_{\boldsymbol{2}}and 0\le m_{1234\boldsymbol{6}}-m_{1235}-m_{2345}\le m_{2\boldsymbol{3}} \end{align}\qquad{(24)}\]
or\[\begin{align} \label{type36} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,5^e,()^M}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,1,\color{brown}{1},\cdots,1,1,\cdots,1,\color{orange}{1},\cdots,1, \color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}}, 2,\cdots, 2, 2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,2,\color{brown}{2},\cdots,2,2,\cdots,2, \color{orange}{2},\cdots,2, \color{blue}{3},\cdots,3,3,\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,3,\color{brown}{3},\cdots,3, 3,\cdots,3,\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{blue}{4},\cdots,4, 6,\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,4,\color{brown}{4},\cdots,4, 4,\cdots,4,\color{orange}{5},\cdots,5, \color{blue}{5},\cdots,5}, {5,\cdots,5,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},6,\cdots,6}, {6,\cdots,6}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \nonumber\\ &\quad e>0\nonumber\\ &m_{12356}+m_{1235}\le m_2and 0\le m_{12346}-m_{1235}-m_{2345}\le m_{23} \end{align}\qquad{(25)}\]
These tableaux satisfy inequalities on the multiplicities of the following columns: if \(m_{12356}\) is the multiplicity of \(\mathrm{red}^{-1}(3)=(12356)\), \(m_{12346}\) the multiplicity of \(\mathrm{red}^{-1}(6)=(12346)\), \(m_{23}\) the multiplicity of \((2,3)\in SpT_6(\varpi_2)\), and \(m_{2}\) the multiplicity of \((2)\in SpT_6(\varpi_1)\), then
\[\begin{align} m_{12{\boldsymbol{3}}56}+m_{1235}\le m_{\boldsymbol{2}}and 0\le m_{1234\boldsymbol{6}}-m_{1235}-m_{2345}\le m_{2\boldsymbol{3}}.\label{ineqn613} \end{align}\qquad{(26)}\]
Proof. From 30 , for \(n=3\), the \(\mathfrak{k}\)-weight of \(S\in SST_{4}\) is equal to \[\mathrm{wt}_\mathfrak{k}(S)=(S[2]-S[1])\tilde{\varepsilon}_1+(S[3]-S[4])\tilde{\varepsilon}_2+(S[6]-S[5])\tilde{\varepsilon}_3.\] The tableau patterns are made of columns \((123456)\), \((12356)\), \((12346)\), \((23456)\), \((1234)\), \((1235)\), \(2345\) \(236\), \(23\) and \((2)\). In fact, \[(123456)=\mathrm{red}^{-1}(())with\mathrm{wt}_\mathfrak{k}()=0\]
\[(12356)=\mathrm{red}^{-1}((3))with\mathrm{wt}_\mathfrak{k}(3)=1\tilde{\varepsilon}_2\]
\[(12346)=\mathrm{red}^{-1}((6))with\mathrm{wt}_\mathfrak{k}(6)=\tilde{\varepsilon}_3\]
\[(23456)=\mathrm{red}^{-1}((2))with\mathrm{wt}_\mathfrak{k}(23456)=\tilde{\varepsilon}_1\] and
\[\mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{5}\leftarrow (12346))=(12346)(1234), \mathrm{wt}_\mathfrak{k}(1234)=(0,0,0)\] \[\mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{5}\leftarrow (12356))=(12346)(1235), \mathrm{wt}_\mathfrak{k}(1235)=1\tilde{\varepsilon}_2-\tilde{\varepsilon}_3\] \[\mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{5}\leftarrow (23456))=(12346)(2345), \mathrm{wt}_\mathfrak{k}(2345)=\tilde{\varepsilon}_1-1\tilde{\varepsilon}_3\] ◻
For an illustration of this corollary see [36].
Theorem 8. Let \(2\le n\in 2 \mathbb{Z}\). Consider \(S^{H,\mu}\) the \(\mathfrak{k}\)-highest weight tableau in \(SpT_{2n}(\mu)\) as in 37 . Let \(Q\in Rec_{2n}(\lambda/\mu)\) with \(0\)-\(1\) slack sequence of the form \(\underline \mathbf{t}=(1,\dots,1,0^M)\) and corresponding slack row index vector sequence \(\underline \mathbf{r}=(1,\dots,1,2,\dots,2,3,\dots,3,\dots,n,\dots,n,2n-1,\dots,2n-1,()^M)\). Let \(u_1=2,\dots,u_n=2n-1\) be the numbers in 28 or 37 , and \(\mu'=\mu-\delta_{\underline \mathbf{r}^+}\). Then,
\(2n-1\notin \underline \mathbf{r}^+\)
for some \(0\le k\le n\), \({\mathsf{LR}^{AII}}^{-1}(S^{H,\mu},Q)\) returns the \(\mathfrak{k}\)-highest weight tableau in \(SST_{2n}(\lambda)\) with \({\mathfrak{k}}\)-weight \(\mu\), in either form:
\[\begin{align} &Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}(2n-1))^{m_{\mathrm{red}^{-1}(2n-1) }}\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=n-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\big)^{m_{\mathrm{red}^{-1}(u_j) }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}\\ =&Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}((12\cdots (2n-3).(2n-2).(2n-1)\big)^{m_{12\cdots (2n-3).(2n-2).2n-1 }}\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=n-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\big)^{m_{\mathrm{red}^{-1}(u_j) }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}, \end{align}\]
or
\[\begin{align} &Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}(2n-1)\big)^{m_{\mathrm{red}^{-1}(2n-1) }} \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\cdot 2n\big)^{m_{12\cdots 2n-2\cdot 2n}} \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\big)^{m_{(12\cdots (2n-2)) }}\mathpalette\mathbin{\vcenter{\scalebox{\nonumber}{\m@th.5\bullet}}}\\ & \mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=n-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\setminus\{2n\}\big)^{m_{\mathrm{red}^{-1}(u_j)\setminus\{2n\} }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}\\ =&Y(M^{2n}) \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}((12\cdots (2n-3).(2n-2).(2n-1)\big)^{m_{12\cdots (2n-3).(2n-2).2n-1 }}\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\cdot 2n\big)^{m_{12\cdots 2n-2\cdot 2n}}\nonumber\\ &\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\big)^{m_{12\cdots (2n-2) }}\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=n-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\setminus\{2n\}\big)^{m_{\mathrm{red}^{-1}(u_j)\setminus\{2n\} }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'} \end{align}\]
or
\[\begin{align} &Y(M^{2n})\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}(2n-1)\big)^{m_{\mathrm{red}^{-1}(2n-1) }} \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(12\cdots (2n-2)\cdot 2n\big)^{m_{12\cdots 2n-2\cdot 2n}} \mathpalette\mathbin{\vcenter{\scalebox{\nonumber}{\m@th.5\bullet}}}\\ &\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle}{\m@th.5\bullet}}}\circledcirc_{j=n-1}^{k+1}\big(\mathrm{red}_1^{-1}({u_j})\big)^{m_{\mathrm{red}^{-1}(u_j)}} \mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}({u_{k}})\big)^{m_{\mathrm{red}^{-1}(u_k)}}\mathpalette\mathbin{\vcenter{\scalebox{\big}{\m@th.5\bullet}}}(\mathrm{red}_1^{-1}(u_k)\setminus\{2n\}\big)^{m_{\mathrm{red}^{-1}(u_k)\setminus\{2n\} }}\nonumber\\ &\mathpalette\mathbin{\vcenter{\scalebox{\displaystyle\circledcirc_{j=k-1}^{1}\big(\mathrm{red}_1^{-1}(u_j)\setminus\{2n\}\big)^{m_{\mathrm{red}^{-1}(u_j)\setminus\{2n\} }}}{\m@th.5\bullet}}} \mathpalette\mathbin{\vcenter{\scalebox{S}{\m@th.5\bullet}}}^{H,\mu'}, \end{align}\]
where \(\displaystyle \sum_{j=n}^{k+1} m_{\mathrm{red}^{-1}(u_j)}+m_{\mathrm{red}^{-1}(u_k)}+\sum_{j=k}^1 m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}=|\underline\delta_\mathbf{r}|-M\), and satisfying the restrictions on linear inequalities on the multiplicities of the columns
\[\begin{align} &\mathrm{red}^{-1}(u_i),~~ \mathrm{red}^{-1}(u_i)\setminus \{2n\},i=1,2,\dots,n, on LHS of S^{H,\mu'},\nonumber\\ & and the symplectic columns u_1u_2\cdots u_i\in SpT_{2n}(\varpi_i) \eqref{nevensymphw-hw}, i=1,\dots,n,\nonumber\\ &u_1u_2\cdots u_n,~ u_1u_2\cdots u_{n-1},\dots,u_1u_2,~~u_1\nonumber \end{align}\] as follows
\[\begin{align} &m_{\mathrm{red}^{-1}(u_i)}+m_{\mathrm{red}^{-1}(u_i)\setminus \{2n\}}\le m_{u_1\cdots u_{i-1}}, ~i=2,\dots,n\\ & m_{12\cdots (2n-2)2n} =\sum_{i=1}^{n-1} m_{\mathrm{red}^{-1}(u_i)\setminus\{2n\}}. \end{align}\]
Proof. Follows from Theorem 5 and a detailed analysis of reverse Schensted insertion for the given slack vector sequence \(\underline\mathbf{r}\). ◻
Corollary 7. Let \(n=2\). Consider \(S^H\) the \(\mathfrak{k}\)-highest weight tableau in \(SpT_4(\mu)\) as in LHS of 33 . Let \(Q\in Rec_{4}(\lambda/\mu)\). Then,
the slack sequence and the corresponding slack row index vector sequence of \(Q\) are of the form \(\underline \mathbf{t}=(1,\dots,1,0,\dots,0)\), respectively \(\underline \mathbf{r}=(1,\cdots,1,2,\cdots,2,3,\cdots,3,(),\dots,())\), and
\({\mathsf{LR}^{AII}}^{-1}(S^H,Q)\) returns the \(\mathfrak{k}\)-highest weight tableau in \(SST_{4}(\lambda)\) with \({\mathfrak{k}}\)-weight \(\mu\) as described in [32]: \[\begin{align} \label{type2} & \vcenter{ \begin{tikzpicture}[x={(0in,-0.16in)},y={(0.16in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,{\color{brown}1},\cdots,{\color{brown}1},{\color{brown}1},\cdots,{\color{brown}1}, {\color{blue}\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {2};}},\cdots,{ \color{blue}\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {2};}}, {\color{blue}\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {2};}},\cdots,{\color{blue}\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {2};}},{2},\cdots,{2},{2},\cdots,{2}}, {2,\cdots,2,{\color{brown}2},\cdots,{\color{brown}2},{\color{brown}2},\cdots,{\color{brown}2}, {\color{blue}3},\cdots,{\color{blue}3}, {\color{blue}3},\cdots,{\color{blue}3},{3},\cdots,{3}}, {3,\cdots,3,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,{\color{brown}\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}}, {{\color{blue}4}},\cdots,{\color{blue}4},{\color{blue}4},\cdots,{\color{blue}4}}, {4,\cdots,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \end{align}\qquad{(27)}\] or
\[\begin{align} \label{type1} & \vcenter{ \begin{tikzpicture}[x={(0in,-0.16in)},y={(0.16in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1, {\color{brown}1},\cdots,1, {\color{brown}1},\cdots,{\color{brown}1},{1},\cdots,1,{\color{blue}\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots, \color{blue}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},{2},\cdots,{2},{2},\cdots,{2}}, {2,\cdots,2,{\color{brown}2},\cdots,{\color{brown}2}, {\color{brown}2},\cdots,{\color{brown}2}, {2},\cdots,2,{{\color{blue}3}},\cdots,{\color{blue}3},{3},\cdots,{3}}, {3,\cdots,3,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{brown}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}}, {\color{blue}4},\cdots,{\color{blue}4}}, {4,\cdots,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \end{align}\qquad{(28)}\]
These tableaux satisfy inequalities on the multiplicities of the following columns: if \(m_{123}\) is the multiplicity of \(\mathrm{red}^{-1}(3)=(1,2,3)\), \(m_{124}\) the multiplicity of \(\mathrm{red}^{-1}(4)=(1,2,4)\), \(m_{23}\) the multiplicity of \((2,3)\in SpT_4(\varpi_2)\) and \(m_2\) the multiplicity of \((2)\in SpT_4(\varpi_1)\), then
\[\begin{align} m_{12\boldsymbol{3}}\le m_{\boldsymbol{2}}andm_{12\boldsymbol{4}}\le m_{2\boldsymbol{3}}.\label{ineqn612} \end{align}\qquad{(29)}\] Tableaux of type ?? are produced with slack vectors where \(\#3\)’s\(>\#1\)’\(\ge 0\), and those of type ?? with slack vectors where \(0\le\#3\)’s\(\le\#1\)’.
Proof. They are particular cases with \(n=2\) of the previous theorem.
The pattern ?? appears after the pattern ?? by increasing enough the number of \(3\)’s in \(\underline \mathbf{r}=(1,\dots,1,2,\dots,2,3,\dots,3)\) once pattern ?? appears:
columns \((234)\) are produced with the \(1\)’s in \(\underline\mathbf{r}\) and then columns \((123)\) are obtained with the \(2\)’s in \(\underline\mathbf{r}\). The number of columns \((123)\) in the pattern just obtained \(S^{(1,\dots,1,2,\dots,2)}\) is equal to the multiplicity of \(2\)’s in \(\underline\mathbf{r}\). Therefore \(m_{123}\le m_{2}\) in \(S^{(1,\dots,1,2,\dots,2)}\). At this point we have \[S^{(1,\dots,1,2,\dots,2)}\] whose pattern ?? is of either form with \(m_{123}\le m_2\) and \(0= m_{124}\le m_{23}.\) \[\begin{align} (A):\quad & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{1}\leftarrow \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {{2}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ) = \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {{2}}, {3}, {4} } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^1of type \eqref{type2} \end{align}\] or \[\begin{align} \nonumber (B):\quad & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{2}\leftarrow \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {{2}}, {{3}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ) = \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,{2}}, {{2}}, {3} } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^2of type \eqref{type2} \end{align}\] \((B)\) indicates that \(0= m_{124}\), \(m_{123}\le m_2\).
or \[\begin{align} \nonumber (C):\quad & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{1}\leftarrow \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {2,2}, {{3}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ) = \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {2,{2}}, {{3},3}, {4} } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{1}\\ & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{2}\leftarrow S^1)= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,{2},2}, {{2},3}, {3,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{1,2}\label{small0}of type \eqref{type2}\\ & 0= m_{124}, \quad m_{123}\le m_2\nonumber \end{align}\tag{38}\] or \[\begin{align} \nonumber (D):\quad & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{1}\leftarrow \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {2,2,2}, {{3}}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ) = \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {2,{2},2}, {{3},3}, {4} } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{1}\nonumber\\ & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{1}\leftarrow S^1)= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {2,{2},2}, {{3},3,3}, {4,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{1,1}\nonumber\\ & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{2}\leftarrow S^{1,1})= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,2,{2},2}, {2,{3},3}, {3,4,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{1,1,2}\label{small00}of type \eqref{type2}\\ & 0= m_{124},\quad m_{123}\le m_2\nonumber \end{align}\tag{39}\]
passing from \(\underline \mathbf{r}=(1,\dots,1,2,\dots,2)\) to \(\underline \mathbf{r}=(1,\dots,1,2,\dots,2,3,\dots,3)\) where \(\# 3\)’s\(>\#1\)’s, we reach pattern ?? with \(m_{123}\le m_2\) and \(0\le m_{124}\le m_{23}.\) Note with \(\# 3\)’s\(=\#1\)’s we remain in pattern ?? . From above we respectively get \[\begin{align} (A):\quad & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{3}\leftarrow \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {{2}}, {3}, {4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^1)= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,{2}}, {2,3}, {4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{1,3},\nonumber\\ & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{3}\leftarrow S^{1,3})= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,1,{2}}, {2,2,3}, {4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } )=S^{1,3,3}\nonumber\\ &1=m_{124}\le m_{23},\quad m_{123}\le m_2 \end{align}\]
\[\begin{align} \nonumber (B):\quad & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{3}\leftarrow \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,{2}}, {{2}}, {3} } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^2) = \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,1,{2}}, {{2},2}, {3}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{2,3} \end{align}\] \[\begin{align} (C):\quad & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{3}\leftarrow S^{1,2}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,{2},2}, {{2},3}, {3,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } )= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,1,{2},2}, {2,{2},3}, {3,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{1,2,3},\nonumber\\ & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{3}\leftarrow S^{1,2,3,3})= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,1,1,{2},2}, {2,2,{2},3}, {3,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \nonumber\\ &1=m_{124}\le m_{23},\quad m_{123}\le m_2 \end{align}\]
\[\begin{align} (D):\quad & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{3}\leftarrow \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,2,{2},2}, {2,{3},3}, {3,4,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{1,1,2})= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,1,2,{2},2}, {2,2,{3},3}, {3,4,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } =S^{1,1,2,3},\nonumber\\ &1=m_{124}\le m_{23}\nonumber\\ & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{3}\leftarrow S^{1,1,2,3})=S^{1,1,2,3,3})= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,1,1,2,{2},2}, {2,2,2,{3},3}, {3,4,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \nonumber\\ &2=m_{124}\le m_{23}, \quad m_{123}\le m_2\\ & \mathrm{c}\circ(\mathrm{red}^{-1},id)\circ(\underset{3}\leftarrow S^{1,1,2,3,3})=S^{1,1,2,3,3,3}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.13in)},y={(0.13in,0in)}] \foreach \rowi [count=\i] in { {1,1,1,1,2,{2},2}, {2,2,2,2,{3},3}, {3,4,4}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \nonumber\\ &2=m_{124}\le m_{23},\quad m_{123}\le m_2 \end{align}\]
From 30 , for \(n=2\), the \(\mathfrak{k}\)-weight of \(S\in SST_{4}\) is equal to \(\mathrm{wt}_\mathfrak{k}(S)=(S[2]-S[1],S[3]-S[4])\). The tableau patterns ?? and ?? are made of columns \((1234)\), \((123)\), \((124)\), \((12)\), \((234)\), \((23)\) and \((2)\). In fact, \[(1234)=\mathrm{red}^{-1}(())with\mathrm{wt}_\mathfrak{k}()=0\] \[(1,2,3)=\mathrm{red}^{-1}((3))with\mathrm{wt}_\mathfrak{k}(3)=1\tilde{e}_2\] \[(1,2,4)=\mathrm{red}^{-1}((4))with\mathrm{wt}_\mathfrak{k}(4)=-1\tilde{e}_2\]
\[(2,3,4)=\mathrm{red}^{-1}((2))withwt_\mathfrak{k}(234)=(1,0)\] and \[\mathrm{wt}_\mathfrak{k}(12)=(0,0)\] ◻
Corollary 8. Let \(n=4\). Consider \(S^H\) the \(\mathfrak{k}\)-highest weight tableau in \(SpT_8(\mu)\) as in LHS of 35 . Let \(Q\in Rec_{8}(\lambda/\mu)\) with slack sequence and corresponding slack row index vector sequence of the form \(\underline \mathbf{t}=(1,\dots,1,0^M)\) respectively \(\underline \mathbf{r}=(1,\dots,1,2,\dots,2,3,\dots,3,4\dots,4,7,\dots,7,()^M)\). Then, \({\mathsf{LR}^{AII}}^{-1}(S^{H,\mu},Q)\) returns the \(\mathfrak{k}\)-highest weight tableau in \(SST_{8}(\lambda)\) with \({\mathfrak{k}}\)-weight \(\mu\) either of the form:
\[\begin{align} \label{type41} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,4,\dots,4,7^0,()^m}= \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,\color{brown}{1},\color{orange}{1},\cdots,\color{orange}{1}, \color{blue}{1},\cdots,1,\color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},2,\cdots, 2, 2,\cdots, 2,2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,\color{brown}{2}, \color{orange}{2}, \cdots,\color{orange}{2}, \color{blue}{2},\cdots,2,\color{red}{3},\cdots,3,3,\cdots,3,{3},\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,\color{brown}{3}, \color{orange}{3},\cdots,\color{orange}{3}, \color{blue}{ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{blue}{ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{red}{4},\cdots,4,6,\cdots,6,{6},\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,\color{brown}{4},\color{orange}{4},\cdots,\color{orange}{4}, \color{blue}{5},\cdots,5,\color{red}{5},\cdots,5,7,\cdots,7}, {5,\cdots,5,\color{brown}{5 },\cdots,\color{brown}{5},\color{orange}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{orange}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}}, \color{blue}{6},\cdots,6,\color{red}{6},\cdots,6}, {6,\cdots,6,\color{brown}{6},\cdots,\color{brown}{6},\color{orange}{7},\cdots,\color{orange}{7}, \color{blue}{7},\cdots,{7},\color{red}{7},\cdots,7}, {7,\cdots,7,\color{brown}{ \boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}}, \cdots,\color{brown}{ \boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}}, \color{orange}{8},\cdots,\color{orange}{8}, \color{blue}{8},\cdots,8,\color{red}{8},\cdots,8}, {8,\cdots,8}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } \nonumber\\ &m_{12{\boldsymbol{3}}5678}\le m_2, \quad 0\le m_{1234{\boldsymbol{6}}78}\le m_{23},\quad 0\le m_{123456\boldsymbol{7}}\le m_{236}\nonumber\\ &m_{1234568}= m_{{\boldsymbol{2}}34567}+m_{12{\boldsymbol{3}}567}+ m_{1234{\boldsymbol{6}}7} \end{align}\qquad{(30)}\] or
\[\begin{align} \label{type42} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,7^a}=\nonumber\\ & \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,\color{brown}{1},\color{brown}{1},\cdots,\color{brown}{1}, \color{orange}{1},\cdots,\color{orange}{1}, \color{blue}{1},\cdots,1, \color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},2,\cdots, 2, 2,\cdots, 2,2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,\color{brown}{2}, \color{brown}{2},\cdots,\color{brown}{2}, \color{orange}{2}, \cdots,\color{orange}{2}, \color{blue}{2},\cdots,2,\color{red}{3},\cdots,3,3,\cdots,3,{3},\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,\color{brown}{3}, \color{brown}{3},\cdots,\color{brown}{3}, \color{orange}{3},\cdots,\color{orange}{3}, \color{blue}{ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{blue}{ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{red}{4},\cdots,4,6,\cdots,6,{6},\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,\color{brown}{4},\color{brown}{4},\cdots,\color{brown}{4},\color{orange}{4},\cdots,\color{orange}{4}, \color{blue}{5},\cdots,5,\color{red}{5},\cdots,5,7,\cdots,7}, {5,\cdots,5,\color{brown}{5 },\cdots,\color{brown}{5},\color{brown}{5},\cdots,\color{brown}{5},\color{orange}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{orange}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}}, \color{blue}{6},\cdots,6,\color{red}{6},\cdots,6}, {6,\cdots,6,\color{brown}{6},\cdots,\color{brown}{6},\color{brown}{6},\cdots,\color{brown}{6},\color{orange}{7},\cdots,\color{orange}{7}, \color{blue}{7},\cdots,{7},\color{red}{7},\cdots,7}, {7,\cdots,7,\color{brown}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}},\cdots,\color{brown}{ \boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}}, {8},\cdots,{8}, {8},\cdots,8,{8},\cdots,8}, {8,\cdots,8}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,\quad \nonumber\\ &m_{12{\boldsymbol{3}}5678}+m_{12{\boldsymbol{3}}567}\le m_2, \quad 0\le m_{1234{\boldsymbol{6}}78}+m_{1234{\boldsymbol{6}}7}\le m_{23},\quad 0\le m_{123456\boldsymbol{7}}\le m_{236}\\ &m_{1234568}= m_{{\boldsymbol{2}}34567}+m_{12{\boldsymbol{3}}567}+ m_{1234{\boldsymbol{6}}7} \end{align}\qquad{(31)}\] or \[\begin{align} \label{type43} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,7^b,()^m}=\nonumber\\ & \vcenter{ \begin{tikzpicture}[x={(0in,-0.14in)},y={(0.14in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,\color{brown}{1},\color{brown}{1},\cdots,\color{brown}{1}, \color{orange}{1},\cdots,\color{orange}{1}, \color{blue}{1},\cdots,1,\color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},2,\cdots, 2, 2,\cdots, 2,2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,\color{brown}{2}, \color{brown}{2},\cdots,\color{brown}{2}, \color{orange}{2}, \cdots,\color{orange}{2}, \color{blue}{2},\cdots,2,\color{red}{3},\cdots,3,3,\cdots,3,{3},\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,\color{brown}{3}, \color{brown}{3},\cdots,\color{brown}{3}, \color{orange}{3},\cdots,\color{orange}{3}, \color{blue}{ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots, \color{blue}{ \boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{red}{4},\cdots,4,6,\cdots,6,{6},\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,\color{brown}{4},\color{brown}{4},\cdots,\color{brown}{4},\color{orange}{4},\cdots,\color{orange}{4}, \color{blue}{5},\cdots,5,\color{red}{5},\cdots,5,7,\cdots,7}, {5,\cdots,5,\color{brown}{5 },\cdots,\color{brown}{5},\color{brown}{5},\cdots,\color{brown}{5},\color{orange}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{orange}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}}, \color{blue}{6},\cdots,6,\color{red}{6},\cdots,6}, {6,\cdots,6,\color{brown}{6},\cdots,\color{brown}{6},\color{brown}{6},\cdots,\color{brown}{6},\color{orange}{7},\cdots,\color{orange}{7}, \color{blue}{7},\cdots,{7},\color{red}{7},\cdots,7}, {7,\cdots,7,\color{brown}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}},\cdots,\color{brown}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}}, {8},\cdots,8,{8},\cdots,8}, {8,\cdots,8}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,\quad b>0 \nonumber\\ &m_{12{\boldsymbol{3}}5678}+m_{12{\boldsymbol{3}}567}\le m_2, \quad 0\le m_{1234{\boldsymbol{6}}78}+m_{1234{\boldsymbol{6}}7}\le m_{23},\quad 0\le m_{123456\boldsymbol{7}}\le m_{236},\quad\\ &m_{1234568}= m_{{\boldsymbol{2}}34567}+m_{12{\boldsymbol{3}}567}+ m_{1234{\boldsymbol{6}}7} \end{align}\qquad{(32)}\]
or\[\begin{align} \label{type44} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,7^c, ()^m}=\nonumber\\ & \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,\color{brown}{1},\color{brown}{1},\cdots,\color{brown}{1}, \color{orange}{1},\cdots,\color{orange}{1}, \color{blue}{1},\cdots,1,\color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots,\color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},2,\cdots, 2, 2,\cdots, 2,2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,\color{brown}{2}, \color{brown}{2},\cdots,\color{brown}{2}, \color{orange}{2}, \cdots,\color{orange}{2}, \color{blue}{2},\cdots,2,\color{red}{3},\cdots,3,3,\cdots,3,{3},\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,\color{brown}{3}, \color{brown}{3},\cdots,\color{brown}{3}, \color{orange}{3},\cdots,\color{orange}{3}, \color{blue}{ \tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots, \color{blue}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{red}{4},\cdots,4,6,\cdots,6,{6},\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,\color{brown}{4},\color{brown}{4},\cdots,\color{brown}{4},\color{orange}{4},\cdots,\color{orange}{4}, \color{blue}{5},\cdots,5,\color{red}{5},\cdots,5,7,\cdots,7}, {5,\cdots,5,\color{brown}{5 },\cdots,\color{brown}{5},\color{brown}{5},\cdots,\color{brown}{5},\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{orange}{\tikz[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}}, \color{blue}{6},\cdots,6,\color{red}{6},\cdots,6}, {6,\cdots,6,\color{brown}{6},\cdots,\color{brown}{6},\color{brown}{6},\cdots,\color{brown}{6},\color{orange}{7},\cdots,\color{orange}{7}, \color{blue}{7},\cdots,{7},\color{red}{7},\cdots,7}, {7,\cdots,7,\color{brown}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}},\cdots,\color{brown}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}}, {8},\cdots,8}, {8,\cdots,8}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,\quad c>0 \nonumber \\ &m_{12{\boldsymbol{3}}5678}+m_{12{\boldsymbol{3}}567}\le m_2, \quad 0\le m_{1234{\boldsymbol{6}}78}+m_{1234{\boldsymbol{6}}7}\le m_{23},\quad 0\le m_{123456\boldsymbol{7}}\le m_{236}\quad\\ &m_{1234568}=m_{{\boldsymbol{2}}34567}+ m_{12{\boldsymbol{3}}567}+ m_{1234{\boldsymbol{6}}7} \end{align}\qquad{(33)}\]
or\[\begin{align} \label{type45} &(S^H)^{1,\dots,1,2,\dots,2,3,\dots,3,7^d,()^m}=\nonumber\\ & \vcenter{ \begin{tikzpicture}[x={(0in,-0.15in)},y={(0.15in,0in)}] \foreach \rowi [count=\i] in { {1,\cdots,1,\color{brown}{1},\cdots,\color{brown}{1}, \color{brown}{1},\cdots,\color{brown}{1}, \color{brown}{1},\cdots,\color{brown}{1}, \color{orange}{1},\cdots,\color{orange}{1}, \color{blue}{1},\cdots,1, \color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},\cdots, \color{red}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{2}};}},2,\cdots, 2, 2,\cdots, 2,2,\cdots,{2},2,\cdots,{2}}, {2,\cdots,2,\color{brown}{2},\cdots,\color{brown}{2}, \color{brown}{2},\cdots,\color{brown}{2}, \color{brown}{2},\cdots,\color{brown}{2}, \color{orange}{2}, \cdots,\color{orange}{2}, \color{blue}{2},\cdots,2,\color{red}{3},\cdots,3,3,\cdots,3,{3},\cdots,{3},3,\cdots,3}, {3,\cdots,3,\color{brown}{3},\cdots,\color{brown}{3}, \color{brown}{3},\cdots,\color{brown}{3}, \color{brown}{3},\cdots,\color{brown}{3}, \color{orange}{3},\cdots,\color{orange}{3}, \color{blue}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\cdots,\color{blue}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{3}};}},\color{red}{4},\cdots,4,6,\cdots,6,{6},\cdots,{6}}, {4,\cdots,4,\color{brown}{4},\cdots,\color{brown}{4},\color{brown}{4},\cdots,\color{brown}{4}, \color{brown}{4},\cdots,\color{brown}{4}, \color{orange}{4},\cdots,\color{orange}{4}, \color{blue}{5},\cdots,5,\color{red}{5},\cdots,5,7,\cdots,7}, {5,\cdots,5,\color{brown}{5 },\cdots,\color{brown}{5},\color{brown}{5},\cdots,\color{brown}{5}, \color{brown}{5},\cdots,\color{brown}{5}, \color{orange}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}},\cdots,\color{orange}{\boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{6}};}}, \color{blue}{6},\cdots,6,\color{red}{6},\cdots,6}, {6,\cdots,6,\color{brown}{6},\cdots,\color{brown}{6},\color{brown}{6},\cdots,\color{brown}{6}, \color{brown}{6},\cdots,\color{brown}{6}, \color{orange}{7},\cdots, \color{orange}{7}, \color{blue}{7},\cdots,{7},\color{red}{7},\cdots,7}, {7,\cdots,7,\color{brown}{ \boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}},\cdots,\color{brown}{ \boldsymbol{\tikz}[baseline=(char.base)]{ \node[shape=circle,draw,inner sep=1pt] (char) {\boldsymbol{7}};}}, {8},\cdots,8}, {8,\cdots,8}, } { \foreach \e [count=\j] in \rowi { \draw (\i,\j) rectangle +(-1,-1); \draw (\i-0.5,\j-0.5) node {\e}; } } \end{tikzpicture} } ,\quad d>0 \nonumber\\ &m_{12{\boldsymbol{3}}5678}+m_{12{\boldsymbol{3}}567}\le m_2, \quad 0\le m_{1234{\boldsymbol{6}}78}+m_{1234{\boldsymbol{6}}7}\le m_{23},\quad 0\le m_{123456\boldsymbol{7}}\le m_{236}\quad\\ &m_{1234568}=m_{{\boldsymbol{2}}34567}+ m_{12{\boldsymbol{3}}567}+ m_{1234{\boldsymbol{6}}7} \end{align}\qquad{(34)}\]
These tableaux satisfy inequalities on the multiplicities of the following columns: \(m_{2345678}\) is the multiplicity of \(\mathrm{red}^{-1}(2)=(2345678)\), \(m_{1235678}\) is the multiplicity of \(\mathrm{red}^{-1}(3)=(1235678)\), \(m_{1234678}\) the multiplicity of \(\mathrm{red}^{-1}(6)=(1234678)\), \(m_{2}\) the multiplicity of \((2)\in SpT_6(\varpi_1)\), \(m_{23}\) the multiplicity of \((23)\in SpT_6(\varpi_2)\), \(m_{236}\) the multiplicity of \((236)\in SpT_6(\varpi_3)\).