Proof. We start with Equation ?? . Let \(D\) be an unoriented involutive diagram, and let \(c\) be a crossing of \(D\) on the axis. Denote by \(D_0\) and \(D_1\) the diagrams obtained by performing a \(0\)- and a \(1\)-smoothing on \(c\), respectively. There are obvious maps \(\mathcal{S}^e(D_0)\to\mathcal{S}^e(D)\) and \(\mathcal{S}^e(D_1)\to\mathcal{S}(D)\), yielding a bijection \[\varphi:\mathcal{S}^e(D_0)\sqcup \mathcal{S}^e(D_1) \longrightarrow \mathcal{S}^e(D)\,.\] It is straightforward to check that for any \(s\in \mathcal{S}^e(D_*)\), and \(*\in \{0,1\}\), we have \(\vert\varphi(s)\vert = \vert s\vert + *\), \(\vert\varphi(s)\vert^\mathrm{on} = \vert s\vert^\mathrm{on} + *\), \(\vert\varphi(s)\vert^\mathrm{off} = \vert s\vert^\mathrm{off}\) and \(h(\varphi(s))=h(s)\). It follows that \[\begin{align} \langle D\rangle^e &= \sum_{s\in \mathcal{S}^e(D)} (-1)^{\vert\, s\, \vert^\mathrm{on} + \tfrac{1}{2}\vert\, s\, \vert^\mathrm{off}} q^{h(s) + \vert\, s\, \vert} \\[0.5em] &= \sum_{s_0\in \mathcal{S}^e(D_0)} (-1)^{\vert\, \varphi(s_0)\, \vert^\mathrm{on} + \tfrac{1}{2}\vert\, \varphi(s_0)\, \vert^\mathrm{off}} q^{h(\varphi(s_0)) + \vert\, \varphi(s_0)\,\vert} \\[0.3em] &\quad + \sum_{s_1\in \mathcal{S}^e(D_1)} (-1)^{\vert\, \varphi(s_1)\, \vert^\mathrm{on} + \tfrac{1}{2}\vert\, \varphi(s_1)\, \vert^\mathrm{off}} q^{h(\varphi(s_1)) + \vert\, \varphi(s_1)\,\vert} \\[0.5em] &= \langle D_0\rangle^e - q\, \langle D_1\rangle^e, \end{align}\] as desired. The proof of Equation ?? is entirely analogous, while Equation ?? is a direct consequence of the definition. Finally, Equation ?? follows easily using the natural identification of \(\mathcal{S}^e(D\sqcup_e D')\) with \(\mathcal{S}^e(D)\times\mathcal{S}^e(D')\). ◻

Proof of Proposition 7. Equalities 7 and 8 are immediate consequences of the definitions together with Equations ?? and ?? , respectively.

Let \(D_{++}\), \(D_{+}\), and \(D_{0}\) be the three strongly involutive diagrams, differing only in a small ball near the axis, involved in Equation 9 (see Figure 8). Given an oriented strongly involutive diagram \(D\), set \[\eta(D) := (-1)^{n_{-}^\mathrm{on}(D) + \tfrac{1}{2}n_{-}^\mathrm{off}(D)} q^{\,n_{+}(D)\, -\, 2n_{-}(D)}\,,\] and observe that \(\eta(D_{++}) = q\,\eta(D_{+}) = q^2 \eta(D_{0})\). The proof of Equation 9 is then a straightforward computation: \[\begin{align} J^e(D_{++}) &= \eta(D_{++}) \langle D_{++} \rangle^e \\[0.3em] &\overset{(\star)}{=} \eta(D_{++}) \bigl(\langle \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/zeroresonaxis.pdf}}}}\rangle^e - q^2(q+q^{-1})\, \langle \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/oneresonaxis.pdf}}}}\rangle^e\bigr) \\[0.3em] &= \eta(D_{++}) \bigl(q(q+q^{-1}) \langle \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/crossingonaxisa.pdf}}}}\rangle^e - q^2 \langle \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/zeroresonaxis.pdf}}}}\rangle^e \bigr) \\[0.3em] &= q^2(q+q^{-1}) J^e(D_{+}) - q^4 J^e(D_{0})\,. \end{align}\] In the equality marked \((\star)\) we used ?? , ?? , and ?? , while the subsequent equality is a consequence of ?? . Equation 11 follows immediately from ?? .

It remains to prove Equation 12 . To each enhanced state \(s\) of \(D\), we associate an enhanced state \(\iota(s)\) of \(D^m\). If we interpret each \(i\)-smoothed crossing in \(s\) as an \((1-i)\)-smoothed crossing in \(\iota(s)\), the circles in the underlying states of \(s\) and \(\iota(s)\) can be naturally identified. The labels on the circles in \(s\) are replaced with complementary labels in \(\iota(s)\). More precisely, every circle labelled by \(x_\pm\) in \(s\) is labelled by \(x_\mp\) in \(\iota(s)\). It is clear that \(\iota\) is a bijection. Moreover, for each \(\tau\)-invariant enhanced state \(s\), we have \[j(s) = - h(\iota(s)) + n - \vert \iota(s)\vert + n_{-}(D^m) - 2n_{+}(D^m) = -\, j(\iota(s))\,,\] and \[k(s) = n^\mathrm{on}(D^m) - \vert \iota(s)\vert^\mathrm{on} - n_{+}^\mathrm{on}(D^m) + \tfrac{1}{2}\bigl(n^\mathrm{off}(D^m) - \vert \iota(s)\vert^\mathrm{off} - n_{+}^\mathrm{off}(D^m)\bigr) \equiv k(\iota(s)) \pmod{2}.\] Now, Equation 12 follows directly from the state-sum definition of \(J^e\). ◻

Proof of Theorem 2. By [24], the knot \(K_n := K(5,3,3,2 n, 5)\) has symmetry group isomorphic to \(\mathbb{Z}/2\mathbb{Z}\). Therefore, this group must be generated by the strong involution \(\tau_n\) of \(K_n\) with axis of fixed points shown in the diagram on the left-hand side of Figure 2.

It follows that \(\tau_n\) is the only strong inversion of \(K_n\) up to equivalence. Now consider the strong involution \(\tau'_n\) of \(K'_n := K(5,3,2 n,3,5)\) with axis of fixed points shown in the diagram on the right-hand side of Figure 2. To prove the first part of the statement it suffices to show that \(J^e(K_n,\tau_n)\neq J^e(K'_n,\tau'_n)\) for every sufficiently large \(n\). Skein Relation 8 applied to one of the \(n\) symmetric pairs of (negative) crossings shows that, setting \(R_n := J^e(K_n,\tau_n)\), we have \[q^4 R_n - q^{-4} R_{n-2} = (q^2 - q^{-2})\, R_{n-1}.\] Thus, \(R_n\) satisfies the homogeneous, second-order difference equation \[\label{e:difference} x_n - q^{-4} (q^2 - q^{-2})\, x_{n-1} - q^{-8} x_{n-2} = 0.\tag{13}\] Each sequence \(\{x_n\}_{n\geq 0}\) is clearly determined by the initial values \(x_0\) and \(x_1\), and moreover an explicit expression for \(x_n\) in terms of \(x_0\) and \(x_1\) can be found with the following well-known procedure – see, for instance, [25]. The characteristic roots of Equation 13 are \[r_{\pm} = \frac{q^2 - q^{-2} \pm (q^2 + q^{-2} )}{q^4} = \pm q^{-4 \pm 2}\,\] and the general solution of Equation 13 is of the form \(R_n = A q^{-2n} + (-1)^n B q^{-6n}\), where \((A,B)\) is the unique solution of the system \[\begin{cases} A + B &= R_0\\ A q^{-2} - B q^{-6} &= R_1\,. \end{cases}\] By Skein Relation 7 , \(R'_n := J^e(K'_n,\tau'_n)\) satisfies \(q^2 R'_n - q^{-2} R'_{n-2} = (q - q^{-1})\, R'_{n-1}\), and arguing as above we conclude \(R'_n = A' q^{-n} + (-1)^n B' q^{-3n}\), where \((A',B')\) is the unique solution of \[\begin{cases} A' + B' &= R'_0\\ A' q^{-1} + B' q^{-3} &= R'_1\,. \end{cases}\] Since the rational functions \(A\), \(B\), \(A'\), \(B'\) are independent of \(n\), if at least one of them is nonzero it follows that \(R_n\neq R'_n\) for every sufficiently large \(n\). Hence, to finish the proof it suffices to check that \(R'_0 = J^e(K'_0,\tau_0)\neq 0\). This is an easy exercise which is left to the reader – cf. Proposition 46. ◻

Proof. We prove the second equality proceeding along the lines of [5]. The first equality can be proved similarly. By definition we have \[\langle\, \raisebox{-14pt}{\begin{tikzpicture}[scale = .75] \draw[red, very thin] (-.5,0) -- (.5,0); \draw[thick] (0,.4) circle (.2); \node[right] at (.1,.6) {}; \draw[thick] (0,-.4) circle (.2); \node[right] at (.1,-.6) {}; \end{tikzpicture}} \rangle_N^e = \sum_{\sigma\in\mathcal{S}_N^e} q^{\mathop{\mathrm{rot}}(\sigma)} = \sum_{x\in X_{N}} q^{\pm 2 x}= [N]_{q^{2}}\, .\] In the penultimate term, the sign depends on the orientation of the two circles—which is the same for both, since the graph is equivariant. However, this sign is irrelevant, as \([N]_{q^2}\) remains unchanged under the substitution \(q\mapsto q^{-1}\). ◻

Proof. The two identities are obtained by “doubling" the proof of [5] as in the proof of Lemma 10. ◻

Proof. Applying first the definition and then Lemmas 10 and 12 to the left-hand side of the first equality we obtain \[\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRtwotopleft.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRtworesoo.pdf}}}}\rangle^e_N - q^{-2}\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRtworesoi.pdf}}}}\rangle^e_N - q^2\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRtworesio.pdf}}}}\rangle^e_N + \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRtworesii.pdf}}}}\rangle^e_N =\] \[= \left([N]_{q^2} - (q^2+q^{-2})[N-1]_{q^2}+[N-2]_{q^2}\right)\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRtworesoob.pdf}}}}\rangle^e_N + \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRtwobottomleft.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRtwobottomleft.pdf}}}}\rangle^e_N\,.\] The second equality is a consequence of the first one and Equation 15 . To prove the third equality we observe that, thanks to the first equality in Lemma 11 and the planar isotopy invariance of the bracket polynomial, we have: \[\begin{align} \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeone.pdf}}}}\rangle^e_N = q^{-6}\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresooo.pdf}}}}\rangle^e_N -q^{-4}\left[ \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresioo.pdf}}}}\rangle^e_N+\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresoio.pdf}}}}\rangle^e_N \right]\\ + q^{-2}\left[ \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresiio.pdf}}}}\rangle^e_N+\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresoii.pdf}}}}\rangle^e_N \right] + \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresoio.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresiiia.pdf}}}}\rangle^e_N\,. \end{align}\] A completely analogous computation yields: \[\begin{align} \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreetwo.pdf}}}}\rangle^e_N = q^{-6}\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresooo.pdf}}}}\rangle^e_N -q^{-4}\left[ \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresioo.pdf}}}}\rangle^e_N+\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresoio.pdf}}}}\rangle^e_N \right]\\ + q^{-2}\left[ \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresiio.pdf}}}}\rangle^e_N+\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresoii.pdf}}}}\rangle^e_N \right] + \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresioo.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeresiiib.pdf}}}}\rangle^e_N\,. \end{align}\] Now the third, and last, equality follows immediately from Lemma 13. ◻

Proof. We make use of results from [26], which identifies specific generating sets for the oriented Reidemeister moves. Adopting the notation from that paper, Propositions 14 and 15 together imply that \(P^e_N(L)\) is invariant under equivariant pairs of Reidemeister moves of types in the set \(S=\{\Omega1b,\Omega1d,\Omega2c,\Omega2d, \Omega3b\}\). The proposition then follows from [26], which shows that the set \(S\) generates all oriented Reidemeister moves. ◻

Proof. The equalities follow from the definition and Lemmas 9 and 10. Since the proofs are similar, we give the argument only for the first equality: \[\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Roneposloop.pdf}}}}\rangle^e_N = q \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Roneorientedres.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Roneedgeresol.pdf}}}}\rangle^e_N = (q [N] - [N-1]) \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/arcuponaxis.pdf}}}}\rangle^e_N = q^N \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/arcuponaxis.pdf}}}}\rangle^e_N\,.\] ◻

Proof. In view of Equation 15 , it suffices to prove the first equality. \[\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwocrossings.pdf}}}}\rangle^e_N = q \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwoorientedresolution.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwoedgeresolution.pdf}}}}\rangle^e_N =\] \[= q \left(q^{-1}\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwodoubleresolution.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwoedgeorientedresolution.pdf}}}}\rangle^e_N\right) - q^{-1} \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwoorientededgeresolution.pdf}}}}\rangle^e_N + \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwodoubleedgeresolution.pdf}}}}\rangle^e_N =\] \[= ([N] - (q+q^{-1})[N-1] + [N-2])\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/arcsdownuponaxis.pdf}}}}\rangle^e_N + \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwonocrossings.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwonocrossings.pdf}}}}\rangle^e_N\] ◻

Proof. Assuming \(Mb\), the following shows that \(Ma\) holds. The converse is proved similarly. \[\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/LemmastatoneA.pdf}}}} \;\raisebox{1pt}{\overset{R2}{\longleftrightarrow}} \;\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/LemmaproofA.pdf}}}} \;\raisebox{1pt}{\overset{Mb}{\longleftrightarrow}} \;\raisebox{2.5pt}{\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/LemmaproofB.pdf}}}}} \;\overset{R2}{\longleftrightarrow} \;\raisebox{1pt}{\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/LemmastatoneB.pdf}}}}}\] ◻

Proof. In view of Equations 14 and 15 and Lemma 19, it suffices to prove the first equality. By the definition of the bracket polynomial we get \[\begin{align} \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Monecrossonfirst.pdf}}}}\rangle^e_N = q^3 \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Moneallresolvedfirst.pdf}}}}\rangle^e_N - q [2] \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/arcedgeon.pdf}}}}\rangle^e_N - q\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/edgeonarc.pdf}}}}\rangle^e_N + \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Moneedgeondoubleedgeoff.pdf}}}}\rangle^e_N\,, \\ \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Monecrossonsecond.pdf}}}}\rangle^e_N = q^3 \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Moneallresolvedsecond.pdf}}}}\rangle^e_N - q \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/arcedgeon.pdf}}}}\rangle^e_N - q [2] \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/edgeonarc.pdf}}}}\rangle^e_N + \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Monedoubleedgeoffedgeon.pdf}}}}\rangle^e_N\,. \end{align}\] The equality follows applying Lemma 13. ◻

Proof. Applying the recursive definition and Lemma 11 we get \[\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mtwocrossingsneg.pdf}}}}\rangle^e_N = q^{-2} \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/negcrossingonaxis.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mtwoedgeresolutions.pdf}}}}\rangle^e_N = q^{-2}\left(q\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/orientedresolutiononaxis.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/edgeresolutiononaxis.pdf}}}}\rangle^e_N\right) - \left(q\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/twoedgesbubbleaxis.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/threeedgesaxisbubbles.pdf}}}}\rangle^e_N\right)\] \[= q^{-1}\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/orientedresolutiononaxis.pdf}}}}\rangle^e_N + \left(-q^{-2} - q [2] + [2]_{q^2}\right) \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/edgeresolutiononaxis.pdf}}}}\rangle^e_N = q^{-1}\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/orientedresolutiononaxis.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/edgeresolutiononaxis.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/poscrossingonaxis.pdf}}}}\rangle^e_N\] This yields the first equality. The second equality can be proved applying Equation 15 . ◻

Proof. By the definition of the equivariant bracket and Proposition 15 we have \[q^2 \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeres.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/edgeIRthreefirst.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeoneLemma.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreetwoLemma.pdf}}}}\rangle^e_N = q^2 \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/IRthreeres.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/edgeIRthreesecond.pdf}}}}\rangle^e_N,\] which immediately implies the first equality. The second equality follows similarly, except that one must apply Proposition 16 in place of Proposition 15, using also the fact that IR3 moves preserve the writhe and thus leave \(\langle\,D\, \rangle^e_N = q^{-N w(D)} P^e_N(D)\) invariant. ◻

Proof. We claim that the following holds: \[[2]\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeedgeonfirst.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreebubblededgefirst.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/MthreebubblededgeRtwo.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeapplylemmafirst.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeedgesymmetric.pdf}}}}\rangle^e_N\,.\] Indeed, the first equality follows from Lemma 11, the second one by Proposition 15, the third equality from Lemma 22 and the last one by Proposition 20. Applying a \(\pi\)-rotation about a vertical axis in the plane to the diagrams above produces a similar sequence of equalities, leading to \[[2] \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeedgeonsecond.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeedgesymmetric.pdf}}}}\rangle^e_N = [2]\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeedgeonfirst.pdf}}}}\rangle^e_N.\] Since the ring \(\mathbb{Z}[q, q^{-1}]\) is a domain, we can cancel the factor  \([2]\), thereby establishing the first equality. The proof of the second equality follows the same reasoning, with the only difference being that Lemma 11 must be used in conjunction with Equation 14 , and Proposition 15 must be replaced by Proposition 16. ◻

Proof. By Proposition 20 and the first identity of Lemma 23 we have \[\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreefirst.pdf}}}}\rangle^e_N = q \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeresonfirst.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeedgeonfirst.pdf}}}}\rangle^e_N = q \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreereson.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeedgeonsecond.pdf}}}}\rangle^e_N =\] \[q \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeresonsecond.pdf}}}}\rangle^e_N - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreeedgeonsecond.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreesecond.pdf}}}}\rangle^e_N\,.\] This completes the proof of the first equality. The second follows in the same way, simply by replacing \(q\) with \(q^{-1}\) in the argument above. To prove the third equality, observe that by Proposition 16, the bracket polynomial \(\langle\,\cdot\, \rangle^e_N = q^{-N w(D)} P^e_N\) is invariant under any oriented IR2 move. Therefore, applying the first equality together with IR2 invariance, we obtain \[\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreefirstb.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreebproof.pdf}}}}\rangle^e_N = \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Mthreesecondb.pdf}}}}\rangle^e_N\,.\] The fourth equality follows from the third by combining Propositions 16 and 21. Specifically, one begins by applying M2 invariance at the rightmost crossing on the axis in the left-hand side diagram. ◻

Proof. The first two identities can be proved in the same way as those in Proposition 24, using Proposition 20 together with Lemma 23. The last two identities then follow directly from the corresponding identities in Proposition 24, by applying Lemma 19. ◻

Proof. Invariance under the oriented moves IR1, IR2, and IR3 is established in Proposition 16. Combining Equations 14 and 15 with Propositions 17, 18, 20, 21, 24, and 25, we conclude that the equivariant \(\mathfrak{sl}_N\) polynomial \(P^e_N(L)\) is invariant under all oriented R1, R2, R3, M1, M2, and M3 moves. This completes the proof. ◻

Proof. Property 20 follows immediately from the definition. Since Properties 17 , 18 , 21 , and 22 are proven exactly as in the non-equivariant case, we concentrate on the remaining Property 19 . Let \(D_{++}\), \(D_{+}\) and \(D_{0}\) be three strongly involutive diagrams differing only in small ball near the axis as illustrated in Figure 8.

Then, we have \[\begin{align} P_{N}^e(D_{++}) &= q^{Nw(D_{++})}\langle \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/ClaspCoer.pdf}}}}\rangle_{N}^e =q^{Nw(D_{++})}\left( q^{-2}\langle \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/orientedresolutiononaxis.pdf}}}}\rangle_{N}^e - \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/twoedgesbubbleaxis.pdf}}}}\rangle^e_N \right) =\\ & =q^{Nw(D_{++})}\left( (q^{-2}- q^{-1}[2])\langle \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/orientedresolutiononaxis.pdf}}}}\rangle_{N}^e + [2] \left(q^{-1}\langle \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/orientedresolutiononaxis.pdf}}}}\rangle_{N}^e- \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/edgeresolutiononaxis.pdf}}}}\rangle^e_N \right)\right) \\ & =q^{Nw(D_{++})}\left( -\langle \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/orientedresolutiononaxis.pdf}}}}\rangle_{N}^e + [2] \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/poscrossingonaxis.pdf}}}}\rangle^e_N \right) = -q^{2N}P_{N}^e(D_{0}) + (q+q^{-1})q^{N}P_{N}^e(D_{+}) \end{align}\] which completes the proof. ◻

Proof. Note that, since \(a,\,q\in R[a^{\pm 1} ,q^{\pm 1}]\) and \(q\in R[q^{\pm 1}]\) are invertible, there is no loss of generality in working with \(a^r q^s P(a,q)\in R[a,q]\) in lieu of \(P(a,q)\). Thus, we may assume that \(P(a,q)\) is a polynomial, and not just a Laurent polynomial, in \(a\) and \(q\). Therefore, assuming by contradiction \(P(a,q) \neq 0\), we can write: \[P(a,q)=\sum_{i=0}^d a^i f_i(q),\qquad f_i(q)\in R[q],\qquad f_d\neq 0.\] Substituting \(a=q^N\), and using the hypothesis, gives \[0=P(q^N,q)=\sum_{i=0}^d q^{Ni} f_i(q)\in R[q] \qquad\text{for all }N\ge 2.\] Let \(M_i=\deg f_i\). The maximal \(q\)-degree of the summand \(q^{Ni}f_i(q)\) equals \(Ni+M_i\). For \(N\) sufficiently large, the maximum of \(Ni+M_i\) over \(i=0,\dots,d\) is achieved uniquely at \(i=d\) (since \(Ni\) grows strictly with \(i\)). Hence, for all such \(N\), the highest-degree term of \(\sum_{i=0}^d q^{Ni} f_i(q)\) comes only from \(q^{Nd}f_d(q)\) and cannot cancel with the other summands. This forces \(f_d(q)=0\), a contradiction. Therefore \(P(a,q)=0\). ◻

Proof. A crossingless strongly involutive diagram \(D\) is an equivariant disjoint union of \(\tau\)-invariant circles on the axis and pairs of symmetric circles disjoint from the axis.

Then, if \(\ell\) and \(2m\) denote, respectively, the number of connected components of \(D\) belonging to \(\mathcal{D}^\mathrm{free}\), respectively \(\mathcal{D}^\mathrm{inv}\), using Property 20 one can check directly that the polynomial \[Q(D) := (a - a^{-1})^{\ell - 1} (a^2 - a^{-2})^{m}\] satisfies the requirements of Definition 30, and moreover \(\varphi(P_N^e(D))=I_N(D)\) – since both \(\varphi\circ P_N^e\) and \(I_N\) satisfy Equations 20 and 21 . Therefore, \(D\) is good. ◻

Proof. We argue by induction on \(n(D_1)\). If \(n(D_1)=0\), then \(D_1\) is crossingless and therefore good by Lemma 34. This proves the base case.

Note that the crossing number does not change during a crossing change sequence. The inductive step follows by induction on the length of the crossing change sequence, using the assumption \(D_k\in\mathcal{D}^\mathrm{good}\), Implications 29 and 30 , and the relations \(n(D_0)=n(D_\pm)-1\) and \(n(D_{00})=n(D_{\pm\pm})-2\). ◻

Proof. We proceed by induction on the number of crossings \(n(D)\). If \(n(D)=0\), then \(D\) is crossingless and therefore good by Lemma 34.

We now assume that the statement holds for each \(D'\in \mathcal{D}^\mathrm{free}\) such that \(n(D')<n(D)\). We claim that, given \(D\in\mathcal{D}^\mathrm{free}\), there exists a crossing change sequence \(D=D_1,\ldots,D_k\in\mathcal{D}^\mathrm{free}\) such that \(D_k\) is good. By Lemma 35, this implies that \(D\) is good. We prove the claim by induction on the number of connected components \(|D|\geq 2\) (note that, as \(D\) is free, \(|D|\) is even).

If \(|D|=2\), then \(D = D'\cup\tau(D')\) for some connected component \(D'\subset D\). Applying a sequence of equivariant crossing changes, we may arrange that at each crossing \(c\in D\) the overstrand belongs to \(D'\). Accordingly, at each crossing \(\tau(c)\in D\), the understrand belongs to \(\tau(D')\). Further equivariant crossing changes turn \(D'\) into the diagram \(D''\) of an unknot, with the property that at every crossing \(c\in D''\cup\tau(D'')\) the overstrand lies in \(D''\). Consequently, \(D''\) also lies above the axis at each intersection with the axis, and hence it represents an unknot \(U\subset\mathbb{R}^3\) disjoint from the \(xz\) plane, bounding a disk \(\Delta\) also disjoint from the \(xz\) plane. Then \(\tau(U)\) bounds the disk \(\tau(\Delta)\), disjoint from \(\Delta\), and there is an equivariant isotopy taking \(U\) to an unknot which projects to a crossingless diagram \(D'''\) disjoint from the \(x\) axis. By Lemma 34 \(D'''\cup\tau(D''')\) is good. Moreover, by [11], \(D''\cup\tau(D'')\) and \(D'''\cup\tau(D''')\) are related by a sequence of Lobb–Watson moves, therefore by Remark 32(b) we have that \(D''\cup\tau(D'')\) is good. Since there is a crossing change sequence from \(D=D'\cup\tau(D')\) to \(D''\cup\tau(D'')\), using the induction hypothesis on the number of crossings, by Lemma 35 the claim holds when \(|D|=2\).

If \(|D|>2\) and \(D'\subset D\) is a connected component, we first apply equivariant crossing changes so that at every crossing \(c\in D'\) the overstrand belongs to \(D'\). It is then straightforward to check that the subdiagram \(D'\cup\tau(D')\neq D\) can be separated from the rest of \(D\) by a sequence of Lobb–Watson moves. Thus, there is a crossing change sequence \(D=D_1,\ldots,D_k\) such that \(D_k\stackrel{\rm LW}{\sim}\tilde{D}\), where \(\tilde{D}\) is a split diagram with \(|\tilde{D}|=|D|\). Hence \(\tilde{D} = \tilde{D}_1\sqcup_e \tilde{D}_2\) with \(|\tilde{D}_s|<|D|\), for \(s=1,2\). By the induction hypotheses, each \(\tilde{D}_s\) is already a good diagram, if \(n(\tilde{D}_s)< n(D)\), or it admits a crossing change sequence to a good diagram – and hence it is a good diagram by Lemma 35. Therefore by Remark 32(c) the diagram \(D\) is good, and the proof is complete. ◻

Proof. Write \(D\) as the (not necessarily disjoint) union \(D_{\rm inv} \cup D_{\rm free}\), where \(D_{\rm inv} \subseteq D\) represents the union strongly invertible components and \(D_{\rm free} \subseteq D\) is the subdiagram of \(D\) where \(\tau\) acts freely. Note that the link represented by \(D_{\rm free}\) has an even number of components; in fact, \(D_{\rm free}\) consists of pairs of sub-diagrams of the form \(D' \cup \tau(D')\), where each \(D'\) represents a knot. Order these pairs of subdiagrams, and fix a distinguished component for each pair. We perform equivariant crossing changes so that, for the first pair, the distinguished component consistently passes over at every crossing, while its symmetric counterpart passes under. We then repeat this procedure for each of the remaining pairs, in the prescribed order, ensuring that no crossing already considered when treating a preceding pair is altered. This process yields a diagram consisting of \(D_{\text{inv}}\) plus a diagram representing the equivariant disjoint union of the components of \(D_{\text{free}}\). Since the latter diagram overpasses at each crossing with \(D_{\text{inv}}\), and it represents a disjoint union, it can then be separated from \(D_{\text{inv}}\) via Lobb-Watson moves. ◻

Proof. Let \(D\in\mathcal{D}^\mathrm{inv}\) be a strongly involutive diagram with \(n(D)\) crossings. Throughout the proof, we assume that every strongly involutive diagram with at most \(n_0\) crossings is good. We proceed by induction on the number \(n(D)\). If \(n(D)=0\), then \(D\) is crossingless and therefore good by Lemma 34.

Now suppose \(1<n(D)\leq n_0\). By the inductive hypothesis, every diagram in \(\mathcal{D}\) with less than \(n(D)\) crossings is good. Let \({\rm sp}_{si} (D)\) denote the minimum length of a crossing change sequence that separates the strongly involutive part of \(D\). To prove the inductive step, we will proceed by induction on \({\rm sp}_{si} (D)\). If \({\rm sp}_{si} (D) = 0\) then \(D=D_1\sqcup_e D_2\) with \(D_1\in\mathcal{D}^\mathrm{inv}\) and \(D_2\in\mathcal{D}^\mathrm{free}\). By Proposition 36, \(D_2\) is a good diagram. Since \(n(D_1) \leq n(D) \leq n_0\) and \(D_1 \in \mathcal{D}^\mathrm{inv}\), the diagram \(D_1\) is good by hypothesis. Therefore \(D\) is good by Remark 32.(c).

Choose a crossing \(c\) or an equivariant pair of crossings \((c,\tau(c))\) occurring in a crossing change sequence realizing \({\rm sp}_{si}(D)\). Smoothing \(c\) (or \(c \cup \tau(c)\)) reduces the number of crossings, and the resulting diagram is good by the inductive hypothesis on \(n(D)\). Similarly, changing \(c\) (or \(c \cup \tau(c)\)) reduces \({\rm sp}_{si}\), and the resulting diagram is good by inductive hypothesis on \({\rm sp}_{si}\). Therefore \(D\) is good by Implications 29 and 30 . This completes the proof. ◻

Proof. Let \(I_N\) be an equivariant skein invariant of type \((R,\varphi,N)\). Clearly, \(\varphi(P_N^e(D'))=I_N(D')\) and \(\varphi(P_N^e(D''))=I_N(D'')\) imply \(\varphi(P_N^e(D))=I_N(D)\). Moreover, we have \(\ell(D)=\ell(D'')=\ell(D')+1\) and \(m(D)=m(D'')\). Multiplying both sides of Equation 19 by \[(q-q^{-1})^{\ell(D)} (q^2 - q^{-2})^{m(D)} = (q - q^{-1})^{\ell(D')} (q^2 - q^{-2})^{m(D')} (q - q^{-1}) (q^2 - q^{-2})^{m(D) - m(D')}\] and using the assumption that \(D'\) and \(D''\) are good, yields \[\begin{gather} (q-q^{-1})^{\ell(D)} (q^2 - q^{-2})^{m(D)} P_N^e(D) = \\ (q^N - q^{-N})\left(q^N (q^2 - q^{-2})^{m(D)-m(D')+1} Q(D')\big|_{a=q^N, z=q^2-q^{-2}} - q^{2N} Q(D'')\big|_{a=q^N, z=q^2-q^{-2}}\right). \end{gather}\] If we define \(Q(D)\in\mathbb{Z}[a^\pm,z]\) as the polynomial \(a z^{m(D)-m(D')+1} Q(D') - a^2 Q(D'')\), then Equation ?? holds for \(D\), so \(D\) is good, and Skein Relation 25 holds as well. ◻

Proof. We argue by induction on \(n^\mathrm{on}(D)\). If \(n^\mathrm{on}(D)=0\) there is nothing to prove. Assume \(n^\mathrm{on}(D)>0\). Then there exists a point \(p\) of \(D\) on the axis and a crossing \(c\) of \(D\) joined to \(p\) by the arcs \(a\) and \(\tau(a)\) emanating from \(p\). By equivariant crossing changes, we may arrange that \(a\) is the understrand and \(\tau(a)\) the overstrand at every crossing other than \(c\). Alternatively, we may arrange the opposite convention, with \(a\) always the overstrand and \(\tau(a)\) the understrand. The appropriate choice (depending on the type of crossing) yields the configuration illustrated in Figure 9, left.

An equivariant isotopy of the associated link then produces a new strongly invertible link with diagram \(\widetilde{D}\) satisfying \[p^\mathrm{on}(\widetilde{D})=p^\mathrm{on}(D),\qquad n(\widetilde{D})=n(D)-1,\qquad n^\mathrm{on}(\widetilde{D})=n^\mathrm{on}(D)-1\] (see Figure 9, right). By [11], \(D\) and \(\widetilde{D}\) are related by a sequence of Lobb–Watson moves. Applying the inductive hypothesis to \(\widetilde{D}\), we obtain a diagram \(D'\in\mathcal{D}^\mathrm{inv}\) with \[p^\mathrm{on}(D')=p^\mathrm{on}(\widetilde{D})=p^\mathrm{on}(D),\quad n(D')=n(\widetilde{D})-n^\mathrm{on}(\widetilde{D})=n(D)-n^\mathrm{on}(D),\quad n^\mathrm{on}(D')=0,\] as required. ◻

Proof. We claim that, by a sequence of equivariant crossing changes and Lobb–Watson equivalences which do not increase the number of crossings, \(D\) can be transformed into a diagram of a strongly invertible unlink.

Write \(D = \Delta_1 \cup...\cup \Delta_\ell\), where \(\Delta_1 ,\, \dots,\, \Delta_\ell\) are subdiagrams of \(D\) representing strongly invertible knots. In view of Lemma 40, we may assume that \(D\) has no crossings on the axis. Each \(\Delta_i\) then decomposes as the union of two arcs \(\gamma_i\) and \(\tau(\gamma_i)\) joining the fixed points \(p_i\) and \(q_i\). Since there are no crossings on the axis, up to relabeling the arcs, we may assume that \(\gamma_i\) lies entirely above the axis and \(\tau(\gamma_i)\) entirely below.

Perform equivariant crossing changes along \(\gamma_1\) so that \(\gamma_1\) is the overpassing strand at every crossing. Concretely, starting at \(p_1\), traverse \(\gamma_1\) and, upon first encountering each crossing, change it so that \(\gamma_1\) passes over. By symmetry, the corresponding crossings along \(\tau(\gamma_1)\) are changed so that \(\tau(\gamma_1)\) is always the underpassing strand. Thus \(\gamma_1\cup\tau(\gamma_1)\) becomes unknotted. We repeat this procedure for the remaining \(\Delta_i\)’s without altering crossings involving previously treated subdiagrams. This yields a crossing change sequence \(D = D_1 ,..., D_k\) where the last diagram represents a strongly involutive link with unknotted components.

We claim that \(D_k\) represents in fact an unlink. Since \(p^{\mathrm{on}}=0\), any two arcs \(\gamma_i\) and \(\gamma_j\) intersect in an even number of points – the same holds, symmetrically, for \(\tau(\gamma_i)\) and \(\tau(\gamma_j)\). Therefore, starting from \(\gamma_1\) and \(\tau(\gamma_1)\), and proceeding in the given order, we can separate and unknot all \(\gamma_i\)’s and, symmetrically, all \(\tau(\gamma_i)\)’s by equivariant pairs of Reidemeister moves. The resulting diagram is a crossingless diagram of a strongly invertible unlink. This proves the claim.

The resulting diagram is crossingless and therefore good by Lemma 34. Furthermore, by Lemma 38, if all strongly involutive diagrams with less than \(n(D)\) crossings are good, then all diagrams in \(\mathcal{D}\) with less than \(n(D)\) crossings are good. Then, we can apply Lemma 35 and obtain that \(D\) is good. ◻

Proof. In view of Remark 32(a) and Lemma 38, it suffices to show that, for each integer \(n_0\geq 0\), every strongly invertible diagram with at most \(n_0\) crossings is good.

We proceed by induction on \(n_0\). If \(n_0=0\), then any strongly invertible diagram with at most \(n_0\) crossings is crossingless and hence good by Lemma 34.

Next, let \(D\in\mathcal{D}^\mathrm{inv}\) with \(n(D)=n_0\), and suppose that all strongly involutive diagrams with at most \(n_0-1\) crossings are good. By Lemma 38, this implies that all diagrams in \(\mathcal{D}\) with at most \(n_0-1\) crossings are good. We prove that \(D\) is good by induction on \(p_0 := p^\mathrm{on}(D)\).

By Lemma 40, \(D\) is related by equivariant crossing changes and Lobb–Watson moves to a diagram with no crossings on the axis, without increasing the total number of crossings and preserving \(p^{\mathrm{on}}\). By Remark 32(b), goodness is preserved under Lobb–Watson equivalence. Moreover, by Lemma 35, the inductive hypothesis on \(n_0\) implies that if the last diagram of a crossing change sequence is good, then also the starting diagram is good. So we may assume that D has no crossings on the axis. If \(p^{\mathrm{on}}(D)=0\), the conclusion follows from Lemma 41.

Now we assume that each \(D'\in \mathcal{D}^\mathrm{inv}\) with \(p^\mathrm{on}(D')< p^\mathrm{on}(D)\) and at most \(n_0\) crossings is good. Let \(p_1 < q_1\) and \(p_2 < q_2\) be two interleaving pairs of connected fixed points in \(D\), with \(p_1 < p_2\). Since the set of connected pairs of fixed points is finite, and the corresponding open intervals are partially ordered by inclusion, we may choose such a pair so that the interval \((p_2,q_1)\) is minimal with respect to inclusion among those arising from interleaving pairs. In particular, the pair \((p_2,q_1)\) does not interleave any other connected pair of fixed points in \(D\).

Let \(D'\) be the smallest subdiagram of \(D\) containing all fixed points in the open interval \((p_2,q_1)\). If \(D' \neq \emptyset\), then \(p^{\mathrm{on}}(D') < p^{\mathrm{on}}(D)\). Indeed, by construction the pair \((p_1,q_1)\) and \((p_2,q_2)\) is no longer present as an interleaving pair inside \(D'\), while no new interleaving pairs are created. Therefore, by the inductive hypothesis on \(p^\mathrm{on}\), \(D'\) is good.

Let \(D'' = D \setminus D'\). Up to applying equivariant crossing changes, we may assume that \(D'\) overpasses at each crossing with \(D''\) above the axis, and underpasses at each crossing with \(D''\) below the axis. Since there are no crossing on the axis, \(D \stackrel{\rm LW}{\sim} D' \sqcup_e D''\).

Since \(D'\) is either good or empty, by Remarks 32(b) and (c) it suffices to show that \(D''\) is good. There exists a sequence \(D = D_1, \dots, D_k \in D_{\mathrm{inv}}\) of equivariant crossing changes and Lobb–Watson moves. These do not increase the total number of crossings and can be chosen so that the arcs of \(D_k\) emanating upwards from \(q_1\) and \(p_2\) meet at their first intersection above the axis. This can be achieved by performing equivariant crossing changes so that these arcs overpass all intermediate strands until they meet. A symmetric argument applies below the axis. Therefore, \(D_k\) contains a “positive” clasp as on the left-hand side of Equation 19 , or a mirror (negative) version of it. In view of Remark 32.(d), we may assume without loss of generality that the clasp is positive. By Lemma 35, it suffices to prove \(D_k\in\mathcal{D}^\mathrm{good}\). Let \(D'\) and \(D''\) be the diagrams appearing on the right-hand side of Skein Relation 19 . Clearly \(n(D'),\, n(D'') \leq n(D_k) -1 \leq n_0-1\). Therefore, by the inductive hypothesis on \(n_0\), \(D',D''\in\mathcal{D}^\mathrm{good}\) and by Lemma 39 it follows that \(D_k\in\mathcal{D}^\mathrm{good}\), completing the inductive step. This shows that \(\mathcal{D}^\mathrm{inv}\subseteq\mathcal{D}^\mathrm{good}\). By Lemma 38 it follows that \(\mathcal{D}= \mathcal{D}^\mathrm{good}\), and hence Equation ?? holds for every strongly involutive diagram. ◻

Proof. By Remark 33, Skein Relations 23 and 24 hold, while Skein Relation 25 holds by Lemma 39. Properties 26 , 27 and 28 follow from Equation ?? and, respectively, Skein Relations 20 , 21 and 22 . This proves the first part of the statement.

Let \(D, D'\) be strongly involutive link diagrams such that \(D\stackrel{\rm LW}{\sim} D'\). Since, by Theorem 42, Equation ?? holds for \(D\) and \(D'\), it follows that the Laurent polynomials \(Q(D)\) and \(Q(D')\) of Theorem 42 satisfy \[Q(D)|_{a=q^N,\;z=q^2-q^{-2}} = Q(D')|_{a=q^N,\;z=q^2-q^{-2}}\] for every \(N\geq 2\). Then, by Remark 32(a), \(Q(D) = Q(D')\). ◻

Proof of Theorem 1. Let \(D, D'\) be strongly involutive link diagrams such that \(D\stackrel{\rm LW}{\sim} D'\). By Corollary 43 we have \(Q(D)=Q(D')\), and since \(m(D)=m(D')\) it follows that \(P^e(D) = P^e(D')\).

By Corollary 43, Properties 23 –28 hold and therefore, by the definition of \(P^e\), so do Properties 1 –6 . Now let \(\widetilde{P}^e\) be any polynomial invariant satisfying Properties 1 –6 . Then, it is straightforward to check that, for each \(N\geq 2\), the map \(I_N:\mathcal{D}\to\mathbb{Q}(q)\) given by \[I_N(D) = (q-q^{-1})^{-\ell(D)} (q^N - q^{-N}) \widetilde{P}^e(D)\big|_{a=q^N,\;z=q^2-q^{-2}}\] is an equivariant skein invariant of type \((\mathbb{Q}(q),\iota,N)\), where \(\iota:\Lambda\to\mathbb{Q}(q)\) is the canonical inclusion, and therefore it coincides with \(\iota\circ P^e_N\) by Theorem 42. Note that, for a sufficiently large integer \(M\), we have \[P(a,z) = z^M P^e(D),\;\widetilde{P}(a,z) = z^M \widetilde{P}^e(D)\in\mathbb{Z}[a,a^{-1},z].\] The equality \(I_N = \iota\circ P^e_N\) implies \[P(a,z)\big|_{a=q^N,\;z=q^2-q^{-2}} = \widetilde{P}(a,z)\big|_{a=q^N,\;z=q^2-q^{-2}}\] for each \(D\in\mathcal{D}\). As in Remark 32(a), applying Lemma 31 to the Laurent polynomials \(P(a,z)\big|_{z=q^2-q^{-2}}\) and \(\widetilde{P}(a,z)\big|_{z=q^2-q^{-2}}\) shows that \(P(a,z)=\widetilde{P}(a,z)\) and therefore \(P^e(D)=\widetilde{P}^e(D)\). This proves the uniqueness of \(P^e\) and concludes the proof of the first part of the theorem.

By Proposition 7, \(J^e\) is an equivariant skein invariant of type \((\Lambda,{\rm id}_\Lambda,2)\). Therefore, by Theorem 42 we have \(J^e=P_2^e\). The second part of the theorem follows at once from Equation ?? in the case \(N=2\) and the defintion of \(P^e\). ◻

Proof. Let \(h(a,z):=g(a,z^2)\). Then, \[f(q^N,q+q^{-1}) = h(q^N,q+q^{-1})\] for each \(N\geq 2\). Then, the polynomial \(L(a,q):=f(a,q+q^{-1}) - h(a,q+q^{-1})\in\mathbb{Z}/2\mathbb{Z}[a^{\pm 1},q^{\pm 1}]\) satisfies \(L(q^N,q)=0\) for each \(N\geq 2\), and by Lemma 31 we have \(L(a,q)=0\). Since \(q+q^{-1}\) is not a zero of a non-vanishing polinomial in \(\mathbb{Z}/2\mathbb{Z}[a^{\pm 1}]\), we conclude \[f(a,z) = h(a,z) = g(a,z^2).\] ◻

Proof of Theorem 3. Let \(D\) be an oriented link diagram and, for \(N\geq 2\), let \(P_N(D)\in\mathbb{Z}[q^{\pm 1/2}]\) the polynomial invariant defined in [5]. Consider the following mod \(2\) evaluation of \(P_N\): \[\widetilde{P}_N(D) := P_N(D)\big|_{q\, \mapsto\, q^2} \bmod 2 \in \mathbb{Z}/2\mathbb{Z}[q^{\pm 1}]\] We claim that \(\widetilde{P}_N(D)\) is an equivariant skein invariant of type \((\mathbb{Z}/2\mathbb{Z}[q^{\pm 1}],\pi,N)\), where the map \(\pi:\mathbb{Z}[q^{\pm 1}]\to\mathbb{Z}/2\mathbb{Z}[q^{\pm 1}]\) is the mod \(2\) reduction. In view of the skein relation [5] and the normalization of \(P_N\) chosen in [5] (see the definition of \(\langle G\rangle_n\) and Lemma 2.1 therein), we have \[\label{e:skein-mod2} q^N \widetilde{P}_N(D_-) + q^{-N} \widetilde{P}_N(D_+) = (q + q^{-1}) \widetilde{P}_N(D_0),\qquad \widetilde{P}_N(U) = [N] \bmod 2,\tag{31}\] where \(U\) is a diagram of the unknot. Since \(P_N\) is invariant under Reidemeister moves so is \(\widetilde{P}_N(D)\), and therefore a fortiori invariant under the Lobb–Watson moves. Moreover, Equations 31 imply that \(\widetilde{P}_N(D)\) satisfies Relations 17 –22 . Indeed, Relation 17 is immediate, while 18 can be deduced as an easy exercise using the symmetry of the link.

Switching and resolving the top crossing of the clasp on the left-hand side of 19 and using 31 we get \[\label{e:claspequiv} q^N \widetilde{P}_N(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/claspR2.pdf}}}}) + q^{-N} \widetilde{P}_N(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/ClaspCoer.pdf}}}}) = (q + q^{-1}) \widetilde{P}_N(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/claspresol.pdf}}}}).\tag{32}\] Since \[\widetilde{P}_N(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/claspR2.pdf}}}}) = \widetilde{P}_N(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/orientedresolutiononaxis.pdf}}}})\quad\text{and}\quad \widetilde{P}_N(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/claspresol.pdf}}}}) = \widetilde{P}_N(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/poscrossingonaxis.pdf}}}}),\] Equation 32 is equivalent to 19 . Moreover, when \(\varphi=\pi\) Relations 20 become \[I_N(\raisebox{-4pt}{\begin{tikzpicture}[scale = .75] \draw[red, very thin] (-.5,0) -- (.5,0); \draw[thick] (0,0) circle (.3); \node[right] at (0,.45) {}; \end{tikzpicture}}) = [N]\bmod 2,\qquad I_N(\raisebox{-14pt}{\begin{tikzpicture}[scale = .75] \draw[red, very thin] (-.5,0) -- (.5,0); \draw[thick] (0,.4) circle (.2); \node[right] at (.1,.6) {}; \draw[thick] (0,-.4) circle (.2); \node[right] at (.1,-.6) {}; \end{tikzpicture}}) = [N]_{q^2} \bmod 2 =[N]^2 \bmod 2,\] which are clearly satisfied by \(\widetilde{P}_N\). Finally, Relations 21 and 22 follow easily. This proves the claim that \(\widetilde{P}_N(D)\) is an equivariant skein invariant of type \((\mathbb{Z}/2\mathbb{Z}[q^{\pm 1}],\pi,N)\). Therefore, by Theorem 42, we have \[\widetilde{P}_N(D) = P^e_N(D) \bmod 2\] for every \(N\geq 2\). Now observe that the following identity holds: \[\label{e:tildePN61homflymod2} \widetilde{P}_N(D) = [N] P(D)\big|_{a=q^N,\, z=q+q^{-1}}\bmod 2,\tag{33}\] where \(P(D)\) is the standard HOMFLY–PT polynomial of the link represented by \(D\), characterized by: \[a^{-1}P(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/poscrossing.pdf}}}}) - a P(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/negcrossing.pdf}}}}) = z P(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/orientedresol.pdf}}}}),\qquad P(\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/circle.pdf}}}}) = 1.\] Identity 33 follows from Theorem 42 together with the fact that \[[N] P(D)\big|_{a=q^N,\, z=q+q^{-1}}\bmod 2\] is also an equivariant skein invariant of type \((\mathbb{Z}/2\mathbb{Z}[q^{\pm 1}],\pi,N)\), which can be checked as we just did for \(\widetilde{P}_N(D)\). By Theorem 42, Equation ?? holds as well, therefore \[(q+q^{-1})^{|D|-1} [N] P(D)\big|_{a=q^N,\, z=q+q^{-1}} = [N] Q(D)\big|_{a=q^N,\, z=(q+q^{-1})^2} \bmod 2.\] Hence, by Lemma 45 and since \(P^e(D) = z^{-m(D)} Q(D)\), we have \[z^{-\ell(D)+1} P(D) = P^e(D)\big|_{z\mapsto z^2}\quad \bmod 2.\] This concludes the proof. ◻

Proof. Let \(D\) be a diagram of \(L\). Note that any equivariant crossing change or smoothing performed on a diagram of the form \(D\#^vD^r\) again yields a diagram of the same form. We proceed by induction on \[\gamma(D) := c(D\#^vD^r) + u(D)\,,\] where \(c(D)\) is the number of crossings of \(D\) and \(u(D)\) the minimal number of crossing changes turning \(D\) into a diagram of the unlink.

If \(\gamma(D)=0\), then \(D\#^vD^r\) is a crossingless diagram of the unlink. The statement follows directly from Proposition 28, using the normalization and multiplicativity properties of \(P_N\) – cf. [27].

Assume the statement holds for each link \(L'\) with a diagram \(D'\) such that \(\gamma(D')<n\). Choose a minimal collection \(S\) of crossings of \(D\) such that performing crossing changes on all of them results in a diagram of the unlink. Let \(D_1\) and \(D_2\) be the diagrams obtained by performing on \(D\) a crossing change and an oriented smoothing, respectively, at one of the crossings in \(S\). Note that \(D_K\) remains a subdiagram of each \(D_i\), and \(\gamma(D) > \gamma(D_i)\) for \(i=1,2\). Then, by Proposition 28 and the inductive hypothesis, \[\begin{align} P_N^e(D\#^vD^r) &= \pm q^{\pm 2N}\, P_N^e(D_1\#^vD_1^r) \mp (q^2 + q^{-2})\, P_N^e(D_2\#^vD_2^r) \\[0.3em] &= [N]\bigl(\pm (q^2)^{\pm N} P_N(D_1)(q^2) \mp [2]_{q^2} P_N(D_2)(q^2)\bigr) \\[0.3em] &= [N]\, P_N(L)(q^2). \end{align}\] In the last equality we used the usual skein relations for the \(\mathfrak{sl}_N\)-polynomials (see, for example, [27]). ◻

Proof. Let \(D\) be a diagram of \(L\). We fix the base-points \(p_1\in D_K \subseteq D\) and \(p_2\in D\#^v D^r\) as illustrated in Figure 10.

Since there is precisely a \(\tau\)-invariant pair of crossings in \(D\#^vD^r\) for each crossing in \(D\), there is a bijective correspondence between \(\tau\)-invariant resolutions of \(D\#^v D^r\) and resolutions of \(D\).

Note that the vertical arcs in \(D\#^v D^r\) always belong to the same circle in each equivariant resolution of \(D\#^v D^r\) and, moreover, this is the unique \(\tau\)-invariant circle in each resolution of \(D\#^v D^r\). All circles not containing \(p_1\) in a resolution of \(D\) yield a \(\tau\)-invariant pair of of circles in the corrspoding resolution of \(D\#^vD^r\), while the circle containing \(p_1\) yields the circle containing \(p_2\). This allows us to define a one-to-one correspondence \[\widetilde{\mathcal{S}}^e(D\#^v D^r,p_2) \overset{1:1}{\longleftrightarrow} \widetilde{\mathcal{S}}(D,p_1)\,.\] as follows; we label each circle that does not contain \(p_1\) as the corresponding \(\tau\)-invariant pair of circles, and the circle containing \(p_1\) as the circle containing \(p_2\).

It is easy to check that this bijection sends \(\tau\)-equivariant states of bi-degree \((j,k)\) to states of bi-degree \((k,\tfrac{1}{2}j)\) (note \(j\) is always even). This gives an isomorphism \[(\widetilde{E}^{2}_{\mathcal{G}})^{j,k}(D\#^v D,p_2) \cong \widetilde{\rm CKh}^{k,\tfrac{1}{2}j}(D,p_1)\,,\] of \(\mathbb{F}_2\)-vector spaces, for each \(j,\,k\in\mathbb{Z}\) with \(j\) even. Comparing the Khovanov differential with that on \(\widetilde{E}^{2}_{\mathcal{G}}\) (see [11]), one notices that these isomorphisms give an isomorphism of chain complexes. Therefore, we have that \[(\widetilde{E}^{3}_{\mathcal{G}})^{j,k} \cong \widetilde{\rm Kh}^{k,\tfrac{1}{2}j}(L,K)\] for each \(j,\,k\in\mathbb{Z}\) with \(j\) even. Since \(L\) is Khovanov-thin, \[\sum_{i,j} \dim \widetilde{\rm Kh}^{i,j}(L,K)\, h^i q^j = \sum_{j} \vert a_j \vert\, h^{\tfrac{1}{2}(j+s(L))} q^j.\] Differentials on pages \(t>2\) increase the \(k\)-degree while preserving the quantum degree, so they must vanish. Therefore the spectral sequence collapses at page \(3\) and the statement follows. ◻

Proof. The second page of the \(\mathcal{F}\)-spectral sequence converging to \(\widetilde{\rm Kh}_\tau(L\#^v L^r,K\#K)\) is the group \(\widetilde{\rm Kh}(L\#^v_K L^r,K\#K)\). The differential on page \(t\geq 3\) decreases the homological degree by \(t-2\) while preserving the quantum degree (see [11]).

By Proposition 47, the connected sum of thin links is thin. Since \(L\) is thin, so is \(L\#^v_K L^r\), and therefore, in each fixed quantum degree, the reduced Khovanov homology is supported in a single homological degree. It follows that the third page of the \(\mathcal{F}\)-spectral sequence is concentrated in a single homological degree in each quantum grading, and hence all higher differentials vanish. Thus the \(\mathcal{F}\)-spectral sequence collapses at the third page.

In particular, \(\widetilde{\rm Kh}_\tau^{i,j,*}(L\#^v_K L^r,K\#K)\) is nontrivial only if \(2i-j = s(L\# L^r) = 2s(L)\).

To compute the Poincaré polynomial of \(\widetilde{\rm Kh}^{*,*,*}_\tau(L\#^v_K L^r,K\#K)\), note that by Proposition 48 the \(\mathcal{G}\)-spectral sequence also collapses at page \(3\). Therefore, for each quantum degree \(2j\) there is a unique summand in \(k\)-degree \((j+s(L))/2\) of dimension \(|a_j|\).

Combining this with the constraint above on the homological grading, namely \(2i-2j = 2s(L)\), we see that each such summand contributes a term \(h^{j+s(L)} q^{2j} t^{(j+s(L))/2}\), and the stated formula follows. ◻

Proof of Theorem 5. Let \(K_{n}\) be the connected sum of \(2n\) copies of the figure-eight knot \(4_1\), and let \(K'_n\) be the connected sum of \(n\) copies of the knot \(8_9\). These knots are alternating and hence Khovanov-thin (see [19]). Since both the Jones polynomial and the \(\mathfrak{sl}_3\) polynomial \(P_3\) are multiplicative under connected sums, being specialisations of the HOMFLY–PT polynomial, it follows that the Jones and \(P_3\) polynomials of \(K_1\) are the squares of those of the figure-eight knot. By consulting [29] and the table at the end of [5], one sees that \(K_1\) and \(K'_1\) have the same Jones polynomial but are distinguished by \(P_3\).

Moreover, both knots are slice, so their Rasmussen invariant vanishes. Since the connected sum of slice knots is slice, the same holds for \(K_n\) and \(K'_n\) for all \(n\). It follows that, for every \(n\), the knots \(K_n\) and \(K'_n\) share the same Jones polynomial, are slice, and are distinguished by \(P_3\).

Therefore, by Proposition 49, the strongly invertible knots \(K_n\#^v K_n\) and \(K'_n\#^v K'_n\) have the same refined Lobb–Watson invariants and isomorphic \(\mathcal{F}\)- and \(\mathcal{G}\)-spectral sequences. On the other hand, by Proposition 46, they are distinguished by \(P_3^e\), and hence, by Theorem 42 and the definition of \(P^e\), also by \(P^e\). ◻

Proof of Lemma 10. To prove the first equality it suffices to repeat word-for-word the proof of [5], the only difference being that our local graphs consist of two copies of the graphs in [5], which yield the evaluations of \([N-1]\) at \(q^2\).

The last two equalities have essentially the same proof, so we only provide the argument for the first one. Partition the states of the MOY graph \(\Gamma\) on the left hand side according to the values they take on the thin edges in the local picture. Let \(\mathcal{S}^e_{x,y}(\Gamma)\) be the collection of equivariant states behaving locally as shown in Figure 11. Each \(\sigma\in \mathcal{S}^e_{x,y}(\Gamma)\) determines a unique state \(\sigma'\) of the graph \(\Gamma'\) on the right hand side of the equation, which assigns \(x\) to the local arc – denote by \(\mathcal{S}_{x}^{e}(\Gamma')\subseteq \mathcal{S}^e_N(\Gamma')\) such states.

To conclude the proof it suffices to compare the two sides of the equation. Using the easily verified equality \(\sum_{y\in X_{N}\setminus\{x\}} q^{\mathop{\mathrm{sign}}(x-y) + y} = [N-1]\) we get \[\begin{align} \langle\, \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Roneedgeresol.pdf}}}}\rangle_N^e & = \sum_{x\in X_N}\, \sum_{y\in X_{N}\setminus\{x\}} \sum_{\sigma \in \mathcal{S}^e_{x,y}(\Gamma)} \left(\prod_{v\in V(\Gamma)}\mathop{\mathrm{wt}}(v,\sigma)\right)q^{\mathop{\mathrm{rot}}(\sigma)} = \\ & = \sum_{x\in X_N}\, \sum_{y\in X_{N}\setminus\{x\}} q^{\mathop{\mathrm{sign}}(x-y) + y} \sum_{\sigma' \in \mathcal{S}^e_{x}(\Gamma)} \left(\prod_{v\in V(\Gamma')}\mathop{\mathrm{wt}}(v,\sigma')\right)q^{\mathop{\mathrm{rot}}(\sigma')} = \\ & = [N-1] \sum_{x\in X_N}\, \sum_{\sigma' \in \mathcal{S}^e_{x}(\Gamma')} \left(\prod_{v\in V(\Gamma')}\mathop{\mathrm{wt}}(v,\sigma')\right)q^{\mathop{\mathrm{rot}}(\sigma')} = \\ & = [N-1]\langle\, \mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/arcuponaxis.pdf}}}}\rangle_N^e \end{align}\] ◻

Proof of Lemma 11. As in Lemma 10, the proof of the first equality is a word-for-word repetition of the proof of [5], except for the evaluation at \(q^2\) which is due to the doubling of the local graphs. The second equality can be also proven similarly. We concentrate on the proof of the rightmost, and last, equality. Denote by \(\Gamma\) and \(\Gamma'\), respectively, the equivariant MOY graphs on the right, respectively on the left hand side of the equation. There is a two-to-one correspondence between the states of \(\Gamma\) and the states of \(\Gamma'\), which is illustrated in Figure 12.

Note that the states \(\sigma_1, \sigma_2\in \mathcal{S}^e(\Gamma')\) and the corresponding state \(\sigma\in \mathcal{S}^e(\Gamma)\) have the same rotation numbers. Moreover, a simple computation shows that \[\prod_{i=0}^{5} \mathop{\mathrm{wt}}(v_i,\sigma_1) = q^{{\rm sgn}(x - y)} = q^{-2{\rm sgn}(y - x)}\cdot \mathop{\mathrm{wt}}(w_0,\sigma)\cdot \mathop{\mathrm{wt}}(w_1,\sigma)\] and \[\prod_{i=0}^{5} \mathop{\mathrm{wt}}(v_i,\sigma_2) = q^{3{\rm sgn}(y - x)} = q^{2{\rm sgn}(y - x)}\cdot \mathop{\mathrm{wt}}(w_0,\sigma)\cdot \mathop{\mathrm{wt}}(w_1,\sigma)\,.\] Therefore, we have \[\left(\prod_{v\in V(\Gamma')}\mathop{\mathrm{wt}}(v,\sigma_1)\right)q^{\mathop{\mathrm{rot}}(\sigma_1)} + \left(\prod_{v\in V(\Gamma')}\mathop{\mathrm{wt}}(v,\sigma_2)\right)q^{\mathop{\mathrm{rot}}(\sigma_2)} = [2]_{q^2}\left(\prod_{v\in V(\Gamma)}\mathop{\mathrm{wt}}(v,\sigma)\right)q^{\mathop{\mathrm{rot}}(\sigma)}\] and the desired equality follows. ◻

Proof of Lemma 12. We give the detail of the proof only for the first equality, the proofs of the second and the third equations, being similar, will be left to the reader.

Denote by \(\Gamma\) the strongly involutive MOY graph on the left hand side of the equality. We start by partitioning the equivariant states in \(\mathcal{S}^e_N(\Gamma)\) into subsets \(\mathcal{S}_{a,b,c}\), with \(a,b,c\in X_N\) and \(a,c\neq b\), according to the labels on the boundary of the local picture – as shown in Figure 13.().

Similarly, denote by \(\Gamma'\) and \(\Gamma''\) the strongly involutive MOY graph on first and second summand, respectively, on the right hand side of the equality. We can partition the equivariant states in \(\mathcal{S}^e_N(\Gamma')\) and \(\mathcal{S}^e_N(\Gamma'')\) into subsets \(\mathcal{S}'_{a}\) and \(\mathcal{S}''_{a,c}\), respectively, where \(a,c\in X_N\) – as in Figure 14. Note that, if \(a\neq c\), the set \(\mathcal{S}''_{a,c}\) can be empty – e.g. if the two local arcs belong to the same circle.

There is an obvious bijection, say \(\varphi\), between states in \(\mathcal{S}''_{a,a}\) and states in \(\mathcal{S}'_a\). An easy check shows that \(\mathop{\mathrm{rot}}(\sigma) =\mathop{\mathrm{rot}}(\varphi(\sigma)) - a\) for each sigma \(\sigma \in \mathcal{S}''_{a,a}\).

We are now ready to prove the statement: \[\begin{align} \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwodoubleedgeresolution.pdf}}}}\rangle^e_N & = \sum_{\scriptstyle \begin{array}{c} a, b, c\in X_{N}\\ b\neq a,c\end{array}} \sum_{\sigma\in\mathcal{S}_{a,b,c}}\left(\prod_{v\in V(\Gamma)}\mathop{\mathrm{wt}}(v,\sigma)\right)q^{\mathop{\mathrm{rot}}(\sigma)} \\ & = \sum_{a, c\in X_{N}} \sum_{\scriptstyle \begin{array}{c} b\in X_{N}\\ b\neq a,c\end{array}} \sum_{\sigma''\in\mathcal{S}''_{a,c}} q^{b + \mathop{\mathrm{sign}}(a-b)+\mathop{\mathrm{sign}}(c-b)}\left(\prod_{v\in V(\Gamma'')}\mathop{\mathrm{wt}}(v,\sigma'')\right)q^{\mathop{\mathrm{rot}}(\sigma'')} \\ & \overset{(\star)}{=} \sum_{a, c\in X_{N}} ([N-2] + \delta_{a,c} q^{a})\sum_{\sigma''\in\mathcal{S}''_{a,c}}\left(\prod_{v\in V(\Gamma'')}\mathop{\mathrm{wt}}(v,\sigma'')\right)q^{\mathop{\mathrm{rot}}(\sigma'')} \end{align}\] where \(\delta_{i,j}\) is Kronecker’s delta, and the equality marked by (\(\star\)) follows from the formula: \[\sum_{x\in X_{N}\setminus \{ y,z\}} q^{x + \mathop{\mathrm{sign}}(y-x) + \mathop{\mathrm{sign}}(z-x) } = [N-2] + \delta_{y,z}q^y, \qquad \forall y,z\in X_N.\] We can now resume our computation: \[\begin{align} \phantom{\langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwodoubleedgeresolution.pdf}}}}\rangle^e_N} & = [N-2] \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/arcsdownuponaxis.pdf}}}}\rangle^e_N + \sum_{a\in X_{N}} q^{a}\sum_{\sigma'\in\mathcal{S}'_{a}}\left(\prod_{v\in V(\Gamma')}\mathop{\mathrm{wt}}(v,\sigma')\right)q^{\mathop{\mathrm{rot}}(\varphi^{-1}(\sigma'))} \\ & = [N-2] \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/arcsdownuponaxis.pdf}}}}\rangle^e_N + \langle\mathord{\vcenter{\adjustbox{valign=c}{\includegraphics[scale=0.25]{figurepdf/Rtwonocrossings.pdf}}}}\rangle^e_N \\ \end{align}\] as desired. ◻

Proof of Lemma 13. We prove the first equality; the second, being analogous, is left to the reader. The equivariant states on both sides of the equality can be grouped into three classes—labeled I through III—as shown in Table 15, where we use the notation \(\varepsilon_{x,y} = 1 - \delta_{x,y}\), with \(\delta_{x,y}\) equal to Kronecker’s delta. To establish the claim, it suffices to show that, for each choice of \(a,b,c\in X_N\), the total contribution to the state sum from the states in the left column equals that from the corresponding states in the right column. Since the rotational contribution is identical across all local configurations, it is enough to consider only the vertex weights.

The difference between the LHS and the RHS contributions is given by \[\label{e:contribs-diff} \begin{align} \varepsilon_{b,c} q^{\mathop{\mathrm{sign}}(c-b)} - \varepsilon_{a,b} q^{\mathop{\mathrm{sign}}(b-a)} + \varepsilon_{a,c}(\varepsilon_{a,b}-\varepsilon_{b,c}) q^{\mathop{\mathrm{sign}}(c-a)}\\ +\varepsilon_{a,b}\varepsilon_{b,c} q^{\mathop{\mathrm{sign}}(b-a)} q^{\mathop{\mathrm{sign}}(c-b)} (q^{\mathop{\mathrm{sign}}(b-a)} - q^{\mathop{\mathrm{sign}}(c-b)})\,. \end{align}\tag{34}\] When \(a,b\) and \(c\) are pairwise distinct, Expression 34 becomes \[q^{\mathop{\mathrm{sign}}(c-b)} - q^{\mathop{\mathrm{sign}}(b-a)} + q^{\mathop{\mathrm{sign}}(b-a)} q^{\mathop{\mathrm{sign}}(c-b)} (q^{\mathop{\mathrm{sign}}(b-a)} - q^{\mathop{\mathrm{sign}}(c-b)})\,.\] Setting \(\eta_1 = \mathop{\mathrm{sign}}(b-a)\) and \(\eta_2 = \mathop{\mathrm{sign}}(c-b)\), the latter expression can be written as \[(q^{\eta_2} - q^{\eta_1}) (1 - q^{\eta_1} q^{\eta_2})\,,\] which vanishes for every \(\eta_1,\eta_2\in\{\pm 1\}\). When \(a=b\), Expression 34 becomes \[\varepsilon_{b,c} (q^{\mathop{\mathrm{sign}}(c-b)} - q^{\mathop{\mathrm{sign}}(b-c)} + q^{\mathop{\mathrm{sign}}(b-c)} - q^{\mathop{\mathrm{sign}}(c-b)}) = 0\,.\] When \(a=c\), Expression 34 reduces to \[\delta_{b,c} (q^{\mathop{\mathrm{sign}}(b-c)} - q^{\mathop{\mathrm{sign}}(c-b)}) = 0\,,\] and when \(b=c\) to \[- \varepsilon_{a,b} q^{\mathop{\mathrm{sign}}(b-a)} + \varepsilon_{a,b} q^{\mathop{\mathrm{sign}}(b-a)} = 0\,.\] ◻