Proof. If \(\Phi(\alpha)=0\), then \(\chi(\alpha\cdot x_\sigma)=0\) for every \(\sigma\in\Sigma\). Since the classes \(x_\sigma\) span \(K(X_\Sigma)\) by 4, the perfectness of the Euler pairing in 5 implies \(\alpha=0\). Thus \(\Phi\) is injective.

Now let \(g\in \mathop{\mathrm{GW}}(\Sigma)\). Since \(g\) vanishes on all linear relations among the classes \(x_\sigma\), the assignment \[\lambda_g\Bigl(\sum_\sigma c_\sigma x_\sigma\Bigr)\coloneq \sum_\sigma c_\sigma g(\sigma)\] defines a well-defined homomorphism \(\lambda_g:K(X_\Sigma)\to {\mathbb{Z}}\). By 5, there exists \(\alpha\in K(X_\Sigma)\) such that \[\lambda_g(\beta)=\chi(\alpha\cdot\beta)\] for all \(\beta\in K(X_\Sigma)\). In particular, \[g(\sigma)=\lambda_g(x_\sigma)=\chi(\alpha\cdot x_\sigma)=\Phi(\alpha)(\sigma)\] for every \(\sigma\in\Sigma\). Hence \(\Phi\) is surjective. ◻

Proof.

1. This is immediate from the definition: subfans of unimodular fans are unimodular, and any pair of cones in \(\Sigma'\) is also a pair of cones in \(\Sigma\).

2. Fix \(\sigma\in\Sigma\). Since \(\Sigma\) is unimodular, \(N_\sigma\) is saturated in \(N\), and the quotient \(N/N_\sigma\) is a lattice. The star fan is unimodular because any cone \(\alpha\supseteq\sigma\) is generated by part of a basis of \(N\), and its image in \(N/N_\sigma\) is generated by the complementary basis vectors.

   It remains to check the index condition. Let \(\overline{\alpha},\overline{\beta}\in\mathop{\mathrm{Star}}_\Sigma(\sigma)\), represented by cones \(\alpha,\beta\in\Sigma\) containing \(\sigma\). In the quotient lattice \(N/N_\sigma\), we have \[N_{\overline{\alpha}}=\frac{N_\alpha}{N_\sigma},\qquad N_{\overline{\beta}}=\frac{N_\beta}{N_\sigma}.\] Thus \[\frac{N/N_\sigma}{N_{\overline{\alpha}}+N_{\overline{\beta}}} \cong \frac{N}{N_\alpha+N_\beta},\] and therefore \[[N/N_\sigma:N_{\overline{\alpha}}+N_{\overline{\beta}}] = [N:N_\alpha+N_\beta].\] The right-hand side is \(1\) or \(\infty\) by strong unimodularity of \(\Sigma\), so \(\mathop{\mathrm{Star}}_\Sigma(\sigma)\) is strongly unimodular.

3. Let \(\Sigma_i\subseteq (N_i)_{\mathbb{R}}\) (\(i=1,2\)) be strongly unimodular. The product fan is unimodular because a product of cones generated by parts of bases is generated by part of the product basis. It remains to check the index condition. Take cones \[\sigma=\sigma_1\times \sigma_2,\qquad \tau=\tau_1\times \tau_2\] in \(\Sigma_1\times\Sigma_2\). Then \[N_\sigma=N_{\sigma_1}\oplus N_{\sigma_2},\qquad N_\tau=N_{\tau_1}\oplus N_{\tau_2},\] so \[N_\sigma+N_\tau=(N_{\sigma_1}+N_{\tau_1})\oplus(N_{\sigma_2}+N_{\tau_2}).\] Hence \[[N_1\oplus N_2:N_\sigma+N_\tau] = [N_1:N_{\sigma_1}+N_{\tau_1}]\cdot [N_2:N_{\sigma_2}+N_{\tau_2}],\] with the convention that if one factor is \(\infty\), then the product is \(\infty\). Since each factor is \(1\) or \(\infty\), the product is \(1\) or \(\infty\). Therefore \(\Sigma_1\times\Sigma_2\) is strongly unimodular.

 ◻

Proof. We only prove (2), since (1) is the special case \(\tau=\{0\}\). Since \(\Sigma\) is strongly unimodular, it is unimodular by definition. Fix \(\tau\in\Sigma\), and write \(r=\dim\tau\). Choose a maximal cone \(\gamma\in\Sigma(n)\) containing \(\tau\). After reindexing rays, we may write \[\tau=\operatorname{cone}(u_1,\dots,u_r),\qquad \gamma=\operatorname{cone}(u_1,\dots,u_n),\] with \(u_1,\dots,u_n\) a \({\mathbb{Z}}\)-basis of \(N\). Let \(v_1,\dots,v_n\) be the dual basis of \(M\), i.e. \[\langle v_i,u_j\rangle=\delta_{ij}.\]

Fix any ray \(\rho\in\Sigma(1)\), and write \[u_\rho=\sum_{i=1}^n a_i u_i,\qquad a_i=\langle v_i,u_\rho\rangle\in{\mathbb{Z}}.\] For each \(i\), let \[\sigma_i:=\operatorname{cone}(u_1,\dots,\widehat{u_i},\dots,u_n)\in\Sigma.\] Then \(N_{\sigma_i}=\bigoplus_{j\ne i}{\mathbb{Z}}u_j\), \(N_\rho={\mathbb{Z}}u_\rho\), and strong unimodularity gives \[[N:N_{\sigma_i}+N_\rho]\in\{1,\infty\}.\] If \(a_i=0\), then \(u_\rho\in (N_{\sigma_i})_{\mathbb{R}}\), so the index is \(\infty\). If \(a_i\neq 0\), then \(N_{\sigma_i}+N_\rho\) has rank \(n\), and \[[N:N_{\sigma_i}+N_\rho]=|a_i|,\] since modulo \(N_{\sigma_i}\), the class of \(u_\rho\) equals \(a_i[u_i]\). Hence \(|a_i|=1\). Therefore \(a_i\in\{0,\pm1\}\) for all \(i\), i.e. \[\langle v_i,u_\rho\rangle\in\{0,\pm1\}\quad\text{for all }i,\rho.\] In particular, each \(v_i\in Q_\Sigma\). Now observe that \(v_{r+1},\dots,v_n\) annihilate \(N_\tau\), hence lie in \(M(\tau)\), and they form an integral basis of \(M(\tau)\). Therefore \[\{v_{r+1},\dots,v_n\}\subseteq Q_\Sigma\cap M(\tau),\] so \(Q_\Sigma\cap M(\tau)\) contains an integral basis of \(M(\tau)\). ◻

Proof. Let \(\Psi\colon{\mathbb{Q}}^\Sigma \to K(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}\) be the linear map \((c_\sigma)_{\sigma\in\Sigma}\mapsto \sum_{\sigma}c_\sigma x_\sigma\). By definition, \(g\colon\Sigma\to{\mathbb{Z}}\) is a Grothendieck weight if and only if \(g\) annihilates every element of \(\ker\Psi\).

For each pair \((q,\tau)\) with \(q\in Q\cap M\) and \(\tau\) a \(q\)-neutral cone, define \(b_{(q,\tau)}\in{\mathbb{Q}}^\Sigma\) by \[(b_{(q,\tau)})_\sigma=\begin{cases} (-1)^{\dim \sigma}, & \text{if } \tau\subsetneq\sigma,\;\sigma(1)\setminus \tau(1)\subseteq \mathbf{P}_q, \\ (-1)^{\dim \sigma+1}, & \text{if } \tau\subsetneq\sigma,\;\sigma(1)\setminus \tau(1)\subseteq \mathbf{N}_q, \\ 0, & \text{otherwise}. \end{cases}\] Since \(\tau\) is \(q\)-neutral, \(q\in M(\tau)=\tau^\perp\cap M\), so no ray of \(\tau\) lies in \(\mathbf{P}_q\cup \mathbf{N}_q\). Note that the vector \(b_{(q,\tau)}\) is exactly the coefficients in 1 . Therefore \(b_{(q,\tau)}\in\ker\Psi\), which proves the only if direction: every Grothendieck weight \(g\) satisfies \(\langle g,b_{(q,\tau)}\rangle=0\) for all admissible pairs \((q,\tau)\).

For the if direction, let \(M_{\mathrm{GW}}\) be the matrix with row vectors \(b_{(q,\tau)}\), rows indexed by admissible pairs \((q,\tau)\) and columns by \(\sigma\in\Sigma\). Since all rows lie in \(\ker\Psi\), \[\mathop{\mathrm{rank}}(M_{\mathrm{GW}})\leq\dim \ker\Psi=|\Sigma|-\dim K(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}.\] It suffices to establish the reverse inequality \[\mathop{\mathrm{rank}}(M_{\mathrm{GW}})\geq|\Sigma|-\dim K(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}.\]

The strategy is to compare \(M_{\mathrm{GW}}\) with a Minkowski-weight balancing matrix, using \(\mathop{\mathrm{gr}}K(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}\cong A^*(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}\). Let \[\Psi'\colon{\mathbb{Q}}^\Sigma \to A^*(X)\otimes_{\mathbb{Z}}{\mathbb{Q}},\qquad (c_\sigma)_{\sigma\in\Sigma}\longmapsto \sum_{\sigma\in\Sigma}c_\sigma [V(\sigma)].\] Here \(A^k(X)\) denotes the Chow group of codimension-\(k\) cycles on \(X\), and \(A^*(X)=\bigoplus_{k=0}^n A^k(X)\) is the graded Chow ring.

For \(k=0,1,\dots,n\), write \(\Sigma^{(k)}:=\{\sigma\in\Sigma:\operatorname{codim}\sigma=k\}\). Fix \(k\in\{0,\dots,n-1\}\), \(\tau\in\Sigma^{(k+1)}\), and \(q\in Q\cap M(\tau)\). In the terminology of Fulton–Sturmfels [1], a Minkowski \(k\)-weight \(c\colon\Sigma^{(k)}\to{\mathbb{Q}}\) satisfies the codimension-one balancing condition along \(\tau\) in direction \(m\in M(\tau)\): \[\sum_{\substack{\sigma\in\Sigma^{(k)}\\ \sigma\supset\tau}} c(\sigma)\langle m, u_{\sigma/\tau}\rangle = 0,\] where \(u_{\sigma/\tau}\) denotes the primitive generator of the unique ray of \(\sigma\) not in \(\tau(1)\) (unique since \(\sigma\supset\tau\) strictly with \(\sigma\in\Sigma^{(k)}\) and \(\tau\in\Sigma^{(k+1)}\), so \(|\sigma(1)\setminus\tau(1)|=\dim\sigma-\dim\tau=1\)). Since \(q\in M(\tau)\cap Q\), 1 gives \(\langle q,u_{\sigma/\tau}\rangle\in\{-1,0,1\}\). Grouping terms by the sign of this pairing, the condition for \(m=q\) becomes \[\sum_{\substack{\sigma\in\Sigma^{(k)}\\ \sigma\supset\tau\\ \sigma(1)\setminus\tau(1)\subseteq \mathbf{P}_q}} c(\sigma) - \sum_{\substack{\sigma\in\Sigma^{(k)}\\ \sigma\supset\tau\\ \sigma(1)\setminus\tau(1)\subseteq \mathbf{N}_q}} c(\sigma)=0.\] By 1, \(Q\cap M(\tau)\) contains an integral basis of \(M(\tau)\); hence letting \(q\) range over \(Q\cap M(\tau)\) yields all codimension-one balancing conditions along \(\tau\). Since \(\dim\sigma=n-k\) is constant on \(\Sigma^{(k)}\), multiplying through by \((-1)^{n-k}\) gives the equivalent form \[\sum_{\substack{\sigma\in\Sigma^{(k)}\\ \sigma\supset\tau\\ \sigma(1)\setminus\tau(1)\subseteq \mathbf{P}_q}} (-1)^{\dim\sigma} c(\sigma) + \sum_{\substack{\sigma\in\Sigma^{(k)}\\ \sigma\supset\tau\\ \sigma(1)\setminus\tau(1)\subseteq \mathbf{N}_q}} (-1)^{\dim\sigma+1} c(\sigma)=0.\] Define a row vector \(b^{\mathrm{MW}}_{(q,\tau)}\in{\mathbb{Q}}^{\Sigma^{(k)}}\) by \[\bigl(b^{\mathrm{MW}}_{(q,\tau)}\bigr)_\sigma= \begin{cases} (-1)^{\dim\sigma}, & \text{if }\sigma\supset\tau,\;\sigma(1)\setminus\tau(1)\subseteq \mathbf{P}_q, \\ (-1)^{\dim\sigma+1}, & \text{if }\sigma\supset\tau,\;\sigma(1)\setminus\tau(1)\subseteq \mathbf{N}_q, \\ 0, & \text{otherwise}, \end{cases}\] where \(\sigma\) ranges over all \(\Sigma^{(k)}\). Let \(M_{\mathrm{MW}}^{(k)}\) be the matrix with row vectors \(b^{\mathrm{MW}}_{(q,\tau)}\), rows indexed by \((q,\tau)\) with \(\tau\in\Sigma^{(k+1)}\), \(q\in Q\cap M(\tau)\), and columns indexed by \(\Sigma^{(k)}\).

By the preceding paragraph and the Fulton–Sturmfels description of Minkowski weights [1], the row span of \(M_{\mathrm{MW}}^{(k)}\) is the space of linear relations among the classes \([V(\sigma)]\), \(\sigma\in\Sigma^{(k)}\), in \(A^{n-k}(X)\). Therefore \[\mathop{\mathrm{rank}}\bigl(M_{\mathrm{MW}}^{(k)}\bigr) = \dim\ker\bigl(\Psi'|_{{\mathbb{Q}}^{\Sigma^{(k)}}}\bigr).\]

Now form one large matrix \(M_{\mathrm{MW}}\) as follows. Order rows by blocks \(k=n-1,n-2,\dots,0\), where the \(k\)-th row block consists of all \((q,\tau)\) with \(\tau\in\Sigma^{(k+1)}\), \(q\in Q\cap M(\tau)\). Order columns by blocks \(\Sigma^{(n)},\Sigma^{(n-1)},\dots,\Sigma^{(0)}\). Each row from the \(k\)-th block is viewed as a vector in \({\mathbb{Q}}^\Sigma\) by extending it by zero outside \(\Sigma^{(k)}\). Thus \(M_{\mathrm{MW}}\) is block diagonal, with diagonal blocks \(M_{\mathrm{MW}}^{(k)}\).

Since \([V(\sigma)]\in A^{n-k}(X)\) for each \(\sigma\in\Sigma^{(k)}\), the map \(\Psi'\) respects the codimension grading. The zero cone contributes no relation, so the preceding rank computation gives \[\mathop{\mathrm{rank}}(M_{\mathrm{MW}}) = \sum_{k=0}^{n-1}\mathop{\mathrm{rank}}\!\bigl(M_{\mathrm{MW}}^{(k)}\bigr) = \dim\ker\Psi' = |\Sigma|-\dim A^*(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}.\]

We claim that \(M_{\mathrm{GW}}\) has the same diagonal blocks as \(M_{\mathrm{MW}}\), with the same row and column order. Indeed, for a row indexed by \((q,\tau)\) with \(\tau\in\Sigma^{(k+1)}\), nonzero entries of \(b_{(q,\tau)}\) occur only at cones \(\sigma\supsetneq\tau\), hence only in column blocks \(\Sigma^{(j)}\) with \(j\le k\). In the chosen order, these are the diagonal block \(\Sigma^{(k)}\) and blocks to its right, so \(M_{\mathrm{GW}}\) is block upper triangular. The diagonal block at level \(k\) keeps precisely the terms with \(\sigma\in\Sigma^{(k)}\), equivalently the terms where \(\sigma\) covers \(\tau\). Those coefficients are exactly the entries of \(b^{\mathrm{MW}}_{(q,\tau)}\). Thus \[\mathop{\mathrm{rank}}(M_{\mathrm{GW}})\ge \mathop{\mathrm{rank}}(M_{\mathrm{MW}}).\] Consequently, \[\mathop{\mathrm{rank}}(M_{\mathrm{GW}}) \geq |\Sigma|-\dim A^*(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}.\] Since \[\mathop{\mathrm{gr}}K(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}\cong A^*(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}\] (see [12]), we have \(\dim K(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}=\dim A^*(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}\), so \[\mathop{\mathrm{rank}}(M_{\mathrm{GW}})\ge |\Sigma|-\dim K(X)\otimes_{\mathbb{Z}}{\mathbb{Q}}.\] Together with the reverse inequality, \(\mathop{\mathrm{rank}}(M_{\mathrm{GW}})=\dim\ker\Psi\), so the vectors \(\{b_{(q,\tau)}\}\) span \(\ker\Psi\). Hence any function \(g\) annihilating all \(b_{(q,\tau)}\) is a Grothendieck weight, completing the proof. ◻

Proof. Let \[g_1\boxtimes g_2 \mathrel{\vcenter{:}}= p_1^*g_1\cdot p_2^*g_2 \in K(X\times X),\] where \(p_1,p_2\colon X\times X\to X\) are the projections. Multiply both sides by \(g_1\boxtimes g_2\) and then apply \(\chi\).

On the left-hand side, by the projection formula, \[\chi\bigl((g_1\boxtimes g_2)\cdot \delta_*[\mathcal{O}_{V(\gamma)}]\bigr) = \chi\bigl(\delta^*(g_1\boxtimes g_2)\cdot [\mathcal{O}_{V(\gamma)}]\bigr).\] By the definition of \(\delta^*\), we have \(g_1\cdot g_2=\delta^*(g_1\boxtimes g_2)\), so the left-hand side equals \[(g_1\cdot g_2)(\gamma) = \chi\bigl((g_1\boxtimes g_2)\cdot \delta_*[\mathcal{O}_{V(\gamma)}]\bigr).\]

On the right-hand side, since \([\mathcal{O}_{V(\sigma\times\tau)}]=[\mathcal{O}_{V(\sigma)}]\boxtimes[\mathcal{O}_{V(\tau)}]\), \[\chi\bigl((g_1\boxtimes g_2)\,[\mathcal{O}_{V(\sigma\times\tau)}]\bigr) = \chi\bigl(g_1[\mathcal{O}_{V(\sigma)}]\bigr)\, \chi\bigl(g_2[\mathcal{O}_{V(\tau)}]\bigr) = g_1(\sigma)\,g_2(\tau).\] Comparing the two sides finishes the proof. ◻

Proof. For \(\sigma,\tau\in\Sigma\), set \(C_{\sigma,\tau}\mathrel{\vcenter{:}}=\tau-\sigma\). Define \[E_1 \mathrel{\vcenter{:}}= \bigcup_{\sigma,\tau\in\Sigma} \partial C_{\sigma,\tau}, \qquad E_2 \mathrel{\vcenter{:}}= \bigcup_{\substack{\sigma,\tau\in\Sigma\\ \dim((N_\sigma)_{\mathbb{R}}+(N_\tau)_{\mathbb{R}})<n}} C_{\sigma,\tau}.\] Set \(U\mathrel{\vcenter{:}}= N_{\mathbb{R}}\setminus(E_1\cup E_2)\). Since \(\Sigma\) is finite, both unions are finite. For each \((\sigma,\tau)\), \(\partial C_{\sigma,\tau}\) is a polyhedral subset of codimension at least \(1\) in \(N_{\mathbb{R}}\). If \(\dim((N_\sigma)_{\mathbb{R}}+(N_\tau)_{\mathbb{R}})<n\), then \[C_{\sigma,\tau}\subseteq (N_\sigma)_{\mathbb{R}}+(N_\tau)_{\mathbb{R}}\] is contained in a proper linear subspace of \(N_{\mathbb{R}}\), so \(C_{\sigma,\tau}\) also has codimension at least \(1\) in \(N_{\mathbb{R}}\). Thus every component in the finite union \(E_1\cup E_2\) has codimension at least \(1\), so \(E_1\cup E_2\neq N_{\mathbb{R}}\). Since \(E_1\cup E_2\) is closed, \(U\) is a nonempty open subset. Because the exceptional set is a finite union of rational polyhedral subsets of positive codimension, \(U\cap N\neq\emptyset\).

For fixed \((\sigma,\tau)\), we have \[(\sigma+v)\cap\tau\neq\emptyset \;\iff\; v\in C_{\sigma,\tau}.\] Also, \[(\sigma^\circ+v)\cap(\tau^\circ)\neq\emptyset \;\iff\; v\in C_{\sigma,\tau}^\circ.\] Hence the first statement is equivalent to avoiding \[\bigcup_{\sigma,\tau\in\Sigma} \partial C_{\sigma,\tau}.\]

For the second statement, if \(\dim((N_\sigma)_{\mathbb{R}}+(N_\tau)_{\mathbb{R}})<n\), then \[C_{\sigma,\tau} \subseteq (N_\sigma)_{\mathbb{R}}+(N_\tau)_{\mathbb{R}}\] and \(C_{\sigma,\tau}\) is contained in a proper linear subspace of \(N_{\mathbb{R}}\). Therefore avoiding \(E_2\) is equivalent to requiring that every nonempty \((\sigma+v)\cap\tau\) satisfies \(\dim((N_\sigma)_{\mathbb{R}}+(N_\tau)_{\mathbb{R}})= n\).

Combining this with the first part, for every pair with \((\sigma+v)\cap\tau\neq\emptyset\), we also have \((\sigma^\circ+v)\cap(\tau^\circ)\neq\emptyset\). Hence \[\operatorname{aff}\bigl((\sigma+v)\cap\tau\bigr) = \operatorname{aff}(\sigma+v)\cap\operatorname{aff}(\tau),\] so \[\dim (\sigma+v)\cap\tau=\dim (\operatorname{aff}(\sigma+v)\cap\operatorname{aff}(\tau))=\dim((N_\sigma)_{\mathbb{R}}\cap(N_\tau)_{\mathbb{R}})=\dim\sigma+\dim\tau-n.\] Here we use the fact that if two polyhedrons \(C\) and \(D\) satisfy \(C^\circ\cap D^\circ\neq\emptyset\), then \(\operatorname{aff}(C\cap D)=\operatorname{aff}(C)\cap \operatorname{aff}(D)\).

Combining the two parts gives the claim. ◻

Proof. The tropicalization of \(\delta(X)\) in \(X\times X\) is the diagonal lattice \(\delta(N)\subset N\oplus N\), with tropical multiplicity \(m_x=1\) everywhere. With \(w=(-v,0)\), the set \(-w+\delta(N_{\mathbb{R}})\) consists of the points \((u+v,u)\). Its intersection with \(-(\sigma^\circ\times\tau^\circ)\) is identified, by \(q=-u\), with \((\sigma^\circ+v)\cap\tau^\circ\). Thus the displacement condition is \((\sigma+v)\cap\tau\neq\emptyset\). With \(Y=X\times X\), \(\Delta=\Sigma\times\Sigma\), \(w=(-v,0)\), and \(\eta=\sigma\times\tau\), Katz’s theorem gives the multiplicity as \[\bigl[M\oplus M : (\delta(N))^\perp + \eta^\perp\bigr],\] where \((\delta(N))^\perp = \{(m,-m)\mid m\in M\}\) and \(\eta^\perp=\sigma^\perp\oplus\tau^\perp\subset M\oplus M\).

To evaluate this index, consider the surjection \(M\oplus M\to M\), \((m_1,m_2)\mapsto m_1+m_2\), whose kernel is \((\delta(N))^\perp\). This map sends \((\delta(N))^\perp+\sigma^\perp\oplus\tau^\perp\) onto \(\sigma^\perp+\tau^\perp\), so \[\frac{M\oplus M}{(\delta(N))^\perp+\sigma^\perp\oplus\tau^\perp} \;\cong\; \frac{M}{\sigma^\perp+\tau^\perp},\] and the multiplicity equals \([M:\sigma^\perp+\tau^\perp]\).

Since \(V(\sigma\times\tau)\) is a component of \(Z\), we have \(\dim\sigma+\dim\tau=n\) and \((\sigma+v)\cap\tau\neq\emptyset\). The latter and 2(2) give \(\mathop{\mathrm{rank}}(N_\sigma+N_\tau)=n\). Hence 4 yields \[[M:\sigma^\perp+\tau^\perp]=[N:N_\sigma+N_\tau].\qedhere\] ◻

Proof. Put \(a=\dim\sigma\), \(b=\dim\tau\), so \(a+b=n\). Choose \(\mathbb{Z}\)-bases \(u_1,\dots,u_a\) of \(N_\sigma\) and \(v_1,\dots,v_b\) of \(N_\tau\), and let \(A\in\mathrm{Mat}_{n\times n}(\mathbb{Z})\) be the matrix with these vectors as columns (in a fixed basis of \(N\)). Since \(\mathrm{im}(A)=N_\sigma+N_\tau\) has rank \(n\), \[[N:N_\sigma+N_\tau]=|\det A|.\]

Since \(N_\sigma\) is a saturated sublattice of \(N\), it is a direct summand, so \(M\to N_\sigma^\vee\cong\mathbb{Z}^a\), \(m\mapsto m|_{N_\sigma}\), is surjective with kernel \(\sigma^\perp\); similarly for \(\tau\). In the given bases, the map \[\phi\colon M\longrightarrow M/\sigma^\perp\oplus M/\tau^\perp\cong\mathbb{Z}^a\oplus\mathbb{Z}^b, \qquad \phi(m)=\bigl(m+\sigma^\perp,\;m+\tau^\perp\bigr)\] has matrix \(A^T\). Its kernel is \(\sigma^\perp\cap\tau^\perp=(N_\sigma+N_\tau)^\perp=0\), so \(|\mathrm{coker}\,\phi|=|\det A^T|=|\det A|\). The surjection \((\bar m_1,\bar m_2)\mapsto\overline{m_1-m_2}\) from \(M/\sigma^\perp\oplus M/\tau^\perp\) to \(M/(\sigma^\perp+\tau^\perp)\) has kernel \(\mathrm{im}\,\phi\), so \[[M:\sigma^\perp+\tau^\perp]=|\mathrm{coker}\,\phi|=|\det A|=[N:N_\sigma+N_\tau].\qedhere\] ◻

Proof. It is enough to work on maximal affine charts \(U_{\sigma_0}\times U_{\tau_0}\). Since \(\Sigma\) is strongly unimodular, it is smooth. Let \(\{a_\rho:\rho\in\sigma_0(1)\}\) and \(\{b_{\rho'}:\rho'\in\tau_0(1)\}\) be the dual bases of \(M\) determined by the ray bases of \(\sigma_0\) and \(\tau_0\). On this chart the diagonal is defined by the toric ideal \[I_{\sigma_0,\tau_0} = \ker\Bigl( \mathbb{C}[x_\rho,y_{\rho'}]\longrightarrow\mathbb{C}[M],\quad x_\rho\mapsto\chi^{a_\rho},\; y_{\rho'}\mapsto\chi^{b_{\rho'}} \Bigr).\] The translate by \(\lambda_{-v}\) gives \[x_\rho\mapsto t^{-\langle a_\rho,v\rangle}\chi^{a_\rho}, \qquad y_{\rho'}\mapsto\chi^{b_{\rho'}}.\] Set \[\eta_v(x_\rho)=\eta_v(a_\rho)=\langle a_\rho,v\rangle, \qquad \eta_v(y_{\rho'})=\eta_v(b_{\rho'})=0 .\] The graph closure over this chart is the Groebner degeneration of \(I_{\sigma_0,\tau_0}\) with respect to \(\eta_v\); hence its special fiber is defined by the minimum-weight initial ideal \(\operatorname{in}_{\eta_v}(I_{\sigma_0,\tau_0})\) [14].

The special fiber uses the minimum \(\eta_v\)-initial ideal. In Sturmfels’ convention for toric initial ideals, this corresponds to using the height vector \(-\eta_v\) on the configuration \[\mathcal{A}= \{a_\rho:\rho\in\sigma_0(1)\} \cup \{b_{\rho'}:\rho'\in\tau_0(1)\} \subset M .\] The resulting regular subdivision is obtained from the lower faces of the lifted configuration \[\{(a_\rho,-\eta_v(a_\rho)):\rho\in\sigma_0(1)\} \cup \{(b_{\rho'},-\eta_v(b_{\rho'})):\rho'\in\tau_0(1)\} \subset M_\mathbb{R}\oplus\mathbb{R}\] [14]. In this convention, a subset of the configuration \(\mathcal{A}\) is a cell exactly when there is some \(p\in N_\mathbb{R}\) such that \[m\longmapsto \langle m,p\rangle-\eta_v(m)\] is zero on that subset and strictly positive on all other vectors. We now see what such a cell must look like. For a lower cell defined by \(p\), the values on the two parts of \(\mathcal{A}\) are \[\langle a_\rho,p-v\rangle,\qquad \langle b_{\rho'},p\rangle .\] Since \(a_\rho\) and \(b_{\rho'}\) are the dual bases, the nonnegativity of these numbers gives \(p-v\in\sigma_0\) and \(p\in\tau_0\). Let \(\sigma\le\sigma_0\) and \(\tau\le\tau_0\) be the unique faces with \[p-v\in\sigma^\circ,\qquad p\in\tau^\circ .\] Then the zero set of the function above is exactly \[\{a_\rho:\rho\in\sigma_0(1)\setminus\sigma(1)\} \cup \{b_{\rho'}:\rho'\in\tau_0(1)\setminus\tau(1)\}.\] Conversely, any \(p\) satisfying these two relative-interior conditions defines this cell. Thus the cells in this chart are precisely the displayed subsets, with \((\sigma^\circ+v)\cap\tau^\circ\neq\emptyset\). By 2, this is the same as \((\sigma+v)\cap\tau\neq\emptyset\).

Under the correspondence above, inclusion of cells is opposite to componentwise inclusion of the pairs \((\sigma,\tau)\). Hence a cell is maximal exactly when \((\sigma,\tau)\) is minimal among the pairs with \((\sigma+v)\cap\tau\neq\emptyset\). In the common refinement, these minimal pairs are exactly the vertices, i.e. those for which \((\sigma+v)\cap\tau\) is a point. Under the genericity condition in 2, every nonempty intersection satisfies \[\dim((\sigma+v)\cap\tau)=\dim\sigma+\dim\tau-n.\] Thus the maximal cells are exactly those with \[(\sigma+v)\cap\tau\neq\emptyset, \qquad \dim\sigma+\dim\tau=n.\]

Note that the vectors \(a_\rho\) with \(\rho\notin\sigma(1)\) are exactly the dual basis vectors vanishing on \(N_\sigma\), so they form a lattice basis of \(\sigma^\perp\), and similarly the corresponding \(b_{\rho'}\)’s form a lattice basis of \(\tau^\perp\). Therefore, for such a maximal cell, strong unimodularity and 4 give its normalized volume as \[[M:\sigma^\perp+\tau^\perp] = [N:N_\sigma+N_\tau] = 1.\] Each such maximal cell contains \(n\) vectors. Hence every maximal cell is a unimodular simplex, so the regular subdivision is in fact a unimodular triangulation. A toric initial ideal is square-free if and only if the corresponding regular triangulation is unimodular [14]. Therefore \(\operatorname{in}_{\eta_v}(I_{\sigma_0,\tau_0})\) is square-free monomial. It follows that \(Z\cap(U_{\sigma_0}\times U_{\tau_0})\) is reduced on every maximal affine chart. These charts cover \(X\times X\), so \(Z\) is reduced. The same maximal cells give the displayed orbit-closure decomposition. ◻

Proof. The proof relies on two facts. First, the Mayer–Vietoris exact sequence for two closed subschemes \(A,B\) of a scheme \(X\): \[0 \to \mathcal{O}_{A\cup B} \to \mathcal{O}_A \oplus \mathcal{O}_B \to \mathcal{O}_{A\cap B} \to 0.\] Second, for three \(Z_i=V(\sigma_i)\), we have \[Z_3 \cap (Z_1 \cup Z_2) = (Z_3\cap Z_1) \cup (Z_3\cap Z_2)\] scheme-theoretically. This holds because in each affine chart of \(X\), the ideals of the \(Z_i\) are generated by monomials. Monomial ideals satisfy the distributive law (see, e.g., [15]): \[I_3 + (I_1 \cap I_2) = (I_3 + I_1) \cap (I_3 + I_2).\]

Using these two facts, the lemma follows by induction on \(r\). ◻

Proof. We verify that the two operations are indeed the meet and join in \(P_v\). The meet is obvious. For the join: if \(\operatorname{cone}(\sigma,\sigma')\notin \Sigma\) or \(\operatorname{cone}(\tau,\tau')\notin \Sigma\), then there is no upper bound of \((\sigma,\tau)\) and \((\sigma',\tau')\) in \(P_v\) other than \(\hat{1}\). If both cones lie in \(\Sigma\), it suffices to show that \(\bigl(\operatorname{cone}(\sigma,\sigma'),\, \operatorname{cone}(\tau,\tau')\bigr)\in P_v\), i.e.that \(\bigl(\operatorname{cone}(\sigma,\sigma')+v\bigr)\cap \operatorname{cone}(\tau,\tau')\neq \emptyset\). This is clear, since \[(\sigma+v)\cap \tau \subset \bigl(\operatorname{cone}(\sigma,\sigma')+v\bigr) \cap \operatorname{cone}(\tau,\tau'),\] and the left-hand side is nonempty. ◻

Proof. Recall that the face lattice of a polyhedral complex is the set of all faces of all polyhedra, together with a minimum element \(\emptyset\) and a maximum element \(\hat{1}\), ordered by inclusion. We construct mutually inverse poset isomorphisms \(\Phi\colon P_v\to F(\mathcal{C})\) and \(\Psi\colon F(\mathcal{C})\to P_v\).

For any \((\sigma,\tau)\in P_v\), define \[\Phi\bigl((\sigma,\tau)\bigr) = (\sigma+v)\cap \tau.\] Set \(\Phi(\hat{0})=\emptyset\) and \(\Phi(\hat{1})=\hat{1}\). Conversely, for any face \(F\in F(\mathcal{C})\), choose any point \(x\) in its interior. Let \(\sigma(F)\in \Sigma\) and \(\tau(F)\in \Sigma\) be the minimal cones such that \(x\in \sigma(F)+v\) and \(x\in \tau(F)\). Define \[\Psi(F) = \bigl(\sigma(F),\,\tau(F)\bigr).\]

We verify:

1. \(\Phi\) and \(\Psi\) are well-defined. For \(\Phi\), this is clear. For \(\Psi\), the choice of \(x\) does not matter because \(\mathop{\mathrm{Relint}}(F)=(\sigma^\circ+v)\cap \tau^\circ\) for some \(\sigma,\tau\in \Sigma\); any interior point of \(F\) determines the same minimal cones.

2. \(\Phi\circ \Psi =\mathrm{Id}_{F(\mathcal{C})}\). For any face \(F\in F(\mathcal{C})\), we need \[\Phi(\Psi(F)) = \bigl(\sigma(F)+v\bigr)\cap \tau(F) = F.\] This follows because \(\mathop{\mathrm{Relint}}(F)=(\sigma(F)^\circ+v)\cap \tau(F)^\circ=\mathop{\mathrm{Relint}}\bigl((\sigma(F)+v)\cap \tau(F)\bigr)\) by 2.

3. \(\Psi\circ \Phi = \mathrm{Id}_{P_v}\). For any \((\sigma,\tau)\in P_v\), choose \(x\in \mathop{\mathrm{Relint}}\bigl((\sigma+v)\cap \tau\bigr)\). The minimal cones containing \(x\) are exactly \(\sigma\) and \(\tau\), again by \(\mathop{\mathrm{Relint}}\bigl((\sigma+v)\cap \tau\bigr)=(\sigma^\circ+v)\cap \tau^\circ\).

4. \(\Phi\) and \(\Psi\) are order-preserving. If \((\sigma,\tau)\leq (\sigma',\tau')\) in \(P_v\), then \(\sigma\subseteq \sigma'\) and \(\tau\subseteq \tau'\), whence \[(\sigma+v)\cap \tau \subseteq (\sigma'+v)\cap \tau',\] so \(\Phi((\sigma,\tau))\leq \Phi((\sigma',\tau'))\). Conversely, if \(F\leq F'\) in \(F(\mathcal{C})\) and \(x\in \mathop{\mathrm{Relint}}(F)\subseteq F'\), the minimal cones for \(F\) are contained in those for \(F'\), giving \(\Psi(F)\leq \Psi(F')\).

 ◻

Proof. The value \(\mu_{L_v}\!\bigl(\hat{0},(\sigma,\tau)\bigr)\) depends only on the interval \([\hat{0},(\sigma,\tau)]\) in \(L_v\). Using the isomorphism \(\Phi\), we reduce to the interval between the empty set and the polyhedron \(P=(\sigma+v)\cap \tau\). The interval \([\hat{0},(\sigma,\tau)]\) is isomorphic to the vertex-facet lattice \(\hat{\mathcal{V}}(P)\) of \(P\) defined in [17].

By [17], the Möbius function of \(\hat{\mathcal{V}}(P)\) is \[\mu_{\hat{\mathcal{V}}(P)}(\hat{0},P) = \begin{cases} (-1)^{\dim P+1}, & \text{if } P \text{ is bounded}, \\ 0, & \text{otherwise}. \end{cases}\] By 2, \[\dim P = \dim\bigl((\sigma+v)\cap \tau\bigr) = \dim \sigma + \dim \tau - n,\] which completes the proof. ◻

Proof. (1) For any \(1\le i<j\le n\), we have \(e_i-e_j\in M\). To show \(e_i-e_j\in Q\), let \(T\subsetneq [n]\) be a nonempty proper subset. The ray \(\rho_T\in\Sigma_{[n]}(1)\) has primitive generator \(e_T\in N\), and \[\langle e_i-e_j,e_T\rangle = \begin{cases} 1, & i\in T,\;j\notin T, \\ -1, & i\notin T,\;j\in T, \\ 0, & \text{otherwise}. \end{cases}\] Hence \(\langle e_i-e_j,e_T\rangle\in\{-1,0,1\}\) for all rays, so \(e_i-e_j\in Q\). Therefore \(S\subseteq Q\cap M\).

(2) Fix \(\tau=\sigma_{\mathcal{F}}\), where \[\mathcal{F}:\emptyset\subsetneq F_1\subsetneq\cdots\subsetneq F_r\subsetneq [n].\] Let \[B_1=F_1,\quad B_2=F_2\setminus F_1,\quad \dots,\quad B_r=F_r\setminus F_{r-1},\quad B_{r+1}=[n]\setminus F_r.\] Then \[M(\tau)=\left\{a=(a_1,\dots,a_n)\in M:\sum_{u\in B_t}a_u=0\;\text{for all }t=1,\dots,r+1\right\}.\] Hence \[M(\tau)=\bigoplus_{t=1}^{r+1}L_t,\qquad L_t:=\{a\in{\mathbb{Z}}^n:\operatorname{supp}(a)\subseteq B_t,\;\sum_{u\in B_t}a_u=0\}.\] For each block \(B_t\), choose \(b_t=\min(B_t)\) and define \[\mathcal{B}_t:=\{\,e_{b_t}-e_u\mid u\in B_t\setminus\{b_t\}\,\}.\] Every vector in \(\mathcal{B}_t\) lies in \(S\), and also in \(M(\tau)\). Moreover, \(\mathcal{B}_t\) is a \({\mathbb{Z}}\)-basis of \(L_t\). Therefore \[\mathcal{B}:=\bigcup_{t=1}^{r+1}\mathcal{B}_t\] is a \({\mathbb{Z}}\)-basis of \(M(\tau)\) and satisfies \(\mathcal{B}\subseteq S\cap M(\tau)\). We conclude that \(S\cap M(\tau)\) contains an integral basis of \(M(\tau)\). ◻

Proof. Let \(\sigma=\sigma_{\mathcal{F}}\) and \(Y:=V(\sigma)\subset X_{[n]}\). By definition, \[g_P(\mathcal{F})=\chi(\mathcal{L}_P|_Y).\] Let \(D_P\) be the torus-invariant Cartier divisor corresponding to \(P\), so \(\mathcal{L}_P=\mathcal{O}_{X_{[n]}}(D_P)\). Let \(h_P|_\sigma\) be the restriction of \(h_P\) to \(\sigma\), and choose \(h_{P,\sigma}^\ast\in M\) such that \[h_P(u)=\langle h_{P,\sigma}^\ast,u\rangle \quad \text{for all }u\in \sigma.\] (Here we use \(M=\mathop{\mathrm{Hom}}(N,{\mathbb{Z}})\) to identify integral linear functions with lattice elements.) The character lattice of the dense torus of \(Y=V(\sigma)\) is (cf. [8]): \[M(\sigma)=M\cap \sigma^\perp.\] Let \(\Sigma_{[n]}/\sigma\) be the star fan in \(N(\sigma)_{\mathbb{R}}:=N_{\mathbb{R}}/\langle \sigma\rangle\), so \(Y=X_{\Sigma_{[n]}/\sigma}\). By [8], applied to the toric morphism \(i_\sigma:Y\hookrightarrow X_{[n]}\), the support function of \(D_P|_Y=i_\sigma^\ast D_P\) is \[h_P^\sigma(\bar u)=h_P(u)-\langle h_{P,\sigma}^\ast,u\rangle,\qquad \bar u\in N(\sigma)_{\mathbb{R}},\] where \(u\) is any lift of \(\bar u\). Equivalently (in the polytope description of §6.2), the polytope of \(D_P|_Y\) is the translated face \[\mathop{\mathrm{face}}_{\mathcal{F}}(P)-h_{P,\sigma}^\ast \subset M(\sigma)_{\mathbb{R}}.\] Hence, by [8], \[h^0(Y,\mathcal{L}_P|_Y)=|(\mathop{\mathrm{face}}_{\mathcal{F}}(P)-h_{P,\sigma}^\ast)\cap M(\sigma)|.\] Since \(\mathop{\mathrm{face}}_{\mathcal{F}}(P)\subset h_{P,\sigma}^\ast+M(\sigma)_{\mathbb{R}}\), translation by \(h_{P,\sigma}^\ast\) gives a bijection \[(\mathop{\mathrm{face}}_{\mathcal{F}}(P)-h_{P,\sigma}^\ast)\cap M(\sigma)\xrightarrow{\sim}\mathop{\mathrm{face}}_{\mathcal{F}}(P)\cap M,\] hence \[h^0(Y,\mathcal{L}_P|_Y)=|\mathop{\mathrm{face}}_{\mathcal{F}}(P)\cap M|.\] Since \(\mathcal{L}_P\) is nef on \(X_{[n]}\), its restriction to \(Y\) is nef. By Demazure vanishing [8] on the complete toric variety \(Y\), \[H^i(Y,\mathcal{L}_P|_Y)=0\quad (i>0).\] Therefore \[g_P(\mathcal{F})=\chi(\mathcal{L}_P|_Y)=h^0(Y,\mathcal{L}_P|_Y)=|\mathop{\mathrm{face}}_{\mathcal{F}}(P)\cap M|.\] ◻

Proof. Write \(u=\sum_{t=1}^r\lambda_t e_{F_t}\in\sigma^\circ_{\mathcal{F}}\) with \(\lambda_t>0\). For \(i\in F_t\setminus F_{t-1}\), \[u_i:=\sum_{s\ge t}\lambda_s\] gives a representative of \(u\) in \({\mathbb{R}}^n\) modulo the line \({\mathbb{R}}(1,\dots,1)\). The values \(u_i\) are constant within each block and strictly decreasing as \(t\) increases. Therefore \[h_{\Delta_I}(u)=\max_{i\in I}(u_i-u_{i_0})\] is linear on \(\sigma_{\mathcal{F}}\), so the normal fan of \(\Delta_I\) coarsens \(\Sigma_{[n]}\). Thus \(\Delta_I\) is a generalized permutohedron. Moreover, for \(u\in\sigma^\circ_{\mathcal{F}}\), the maximum is achieved exactly by \(i\in I\cap F_{t_I(\mathcal{F})}\), giving \[\mathop{\mathrm{face}}_{\mathcal{F}}(\Delta_I)=\operatorname{conv}\!\bigl(\{e_i-e_{i_0}:i\in I\cap F_{t_I(\mathcal{F})}\}\bigr).\]

Since \(\mathop{\mathrm{face}}_{\mathcal{F}}(\Delta_I)\) is a unimodular simplex, its only lattice points are its vertices, hence \[|\mathop{\mathrm{face}}_{\mathcal{F}}(\Delta_I)\cap M|=|I\cap F_{t_I(\mathcal{F})}|.\] ◻

Proof. Since \(K(\mathsf{M})\) is generated by the classes \(x_F\), and \(x_Fx_G=0\) for incomparable flats, the classes \(x_{\mathcal{F}}\) span \(K(\mathsf{M})\). If \(\Phi(a)=0\), then \(\chi(ax_{\mathcal{F}})=0\) for all \(\mathcal{F}\), hence \(\chi(ab)=0\) for every \(b\in K(\mathsf{M})\). The perfectness of the Euler pairing implies \(a=0\), so \(\Phi\) is injective.

Let \(g\in \mathop{\mathrm{GW}}(\mathsf{M})\). By definition, \(g\) annihilates all linear relations among the classes \(x_{\mathcal{F}}\), so the assignment \[\lambda_g\Bigl(\sum_{\mathcal{F}} c_{\mathcal{F}}x_{\mathcal{F}}\Bigr)\coloneq \sum_{\mathcal{F}} c_{\mathcal{F}}g(\mathcal{F})\] defines a well-defined homomorphism \(\lambda_g:K(\mathsf{M})\to{\mathbb{Z}}\). Since the Euler pairing is perfect, there exists a unique \(a\in K(\mathsf{M})\) such that \[\lambda_g(b)=\chi(ab)\] for all \(b\in K(\mathsf{M})\). In particular, \[g(\mathcal{F})=\lambda_g(x_{\mathcal{F}})=\chi(ax_{\mathcal{F}})=\Phi(a)(\mathcal{F})\] for every flag \(\mathcal{F}\). Thus \(\Phi\) is surjective. ◻

Proof. Let \(J\subset K(X_{[n]})\) be the ideal generated by the classes \(x_S\), where \(S\subsetneq E\) is a nonempty proper subset that is not a flat of \(\mathsf{M}\). Comparing the presentation of \(K(X_{[n]})\) in 4 with the presentation of \(K(\mathsf{M})\) above, we obtain a canonical isomorphism \[K(X_{[n]})/J\xrightarrow{\sim} K(\mathsf{M}),\] and we write \(\pi:K(X_{[n]})\twoheadrightarrow K(\mathsf{M})\) for the quotient map.

We also note that every element of \(J\) is an integral linear combination of the classes \(x_{\mathcal{H}}\), where \(\mathcal{H}\) contains a non-flat subset. Indeed, for such a subset \(S\), let \(i_S:V(\rho_S)\hookrightarrow X_{[n]}\) be the corresponding invariant divisor. Since \(x_S=i_{S*}\mathcal{O}_{V(\rho_S)}\), the projection formula gives \[x_S\cdot a=i_{S*}i_S^*(a)\] for every \(a\in K(X_{[n]})\). As \(K(V(\rho_S))\) is additively generated by its invariant subvarieties, the ideal \((x_S)\) is contained in \(\mathrm{span}_{\mathbb{Z}}\{x_{\mathcal{H}}:S\in\mathcal{H}\}\). The reverse inclusion holds since \(x_{\mathcal{H}}=x_S\cdot x_{\mathcal{H}\setminus\{S\}}\in(x_S)\) for any \(S\in\mathcal{H}\). Summing over all non-flat \(S\) gives the claim.

Let \(\widetilde{g}:\Sigma_{[n]}\to{\mathbb{Z}}\) be the zero-extension of \(g\). We claim that \(g\in \mathop{\mathrm{GW}}(\mathsf{M})\) if and only if \(\widetilde{g}\in \mathop{\mathrm{GW}}(\Sigma_{[n]})\). Suppose first that \(\widetilde{g}\in \mathop{\mathrm{GW}}(\Sigma_{[n]})\), and let \[\sum_{\mathcal{F}} c_{\mathcal{F}}x_{\mathcal{F}}=0\] be a linear relation in \(K(\mathsf{M})\), where the sum is over flat flags. Viewing the same sum in \(K(X_{[n]})\), it lies in \(\ker(\pi)=J\). Hence, by the previous paragraph, there exist integers \(d_{\mathcal{H}}\) such that \[\sum_{\mathcal{F}} c_{\mathcal{F}}x_{\mathcal{F}}=\sum_{\mathcal{H}} d_{\mathcal{H}}x_{\mathcal{H}}\] in \(K(X_{[n]})\), where each \(\mathcal{H}\) contains a non-flat subset. Applying \(\widetilde{g}\) to the resulting relation \[\sum_{\mathcal{F}} c_{\mathcal{F}}x_{\mathcal{F}}-\sum_{\mathcal{H}} d_{\mathcal{H}}x_{\mathcal{H}}=0\] gives \(\sum_{\mathcal{F}} c_{\mathcal{F}}g(\mathcal{F})=0\), since \(\widetilde{g}(\mathcal{H})=0\) for every non-flat flag \(\mathcal{H}\). Thus \(g\in\mathop{\mathrm{GW}}(\mathsf{M})\). Conversely, if \(g\in\mathop{\mathrm{GW}}(\mathsf{M})\) and \(\sum_{\mathcal{K}} c_{\mathcal{K}}x_{\mathcal{K}}=0\) in \(K(X_{[n]})\), then applying \(\pi\) gives a relation \(\sum_{\mathcal{F}} c_{\mathcal{F}}x_{\mathcal{F}}=0\) in \(K(\mathsf{M})\), where the sum is over flat flags. Therefore \[\sum_{\mathcal{K}} c_{\mathcal{K}}\widetilde{g}(\mathcal{K})=\sum_{\mathcal{F}} c_{\mathcal{F}}g(\mathcal{F})=0,\] so \(\widetilde{g}\in\mathop{\mathrm{GW}}(\Sigma_{[n]})\).

By 1, \(\widetilde{g}\in \mathop{\mathrm{GW}}(\Sigma_{[n]})\) if and only if it satisfies the \(\{i,j\}\)-balancing condition for every \(\{i,j\}\)-neutral flag of subsets \(\mathcal{G}\). If \(\mathcal{G}\) contains a non-flat subset, then so does every refinement \(\mathcal{F}\succneqq\mathcal{G}\), hence \(\widetilde{g}(\mathcal{F})=0\) and both sides are zero. If \(\mathcal{G}\) is a flag of flats, then \(\widetilde{g}(\mathcal{F})=g(\mathcal{F})\) for flat refinements and \(0\) otherwise, so this is exactly the balancing condition in the statement. ◻

Proof. This is immediate from 1. ◻

Proof. By the projection formula, \[\chi\bigl(\mathop{\mathrm{MC}}_y(U\to X)\cdot i_{I*}\det N_{D_I/X}\bigr) =\chi\bigl(i_I^*\mathop{\mathrm{MC}}_y(U\to X)\cdot\det N_{D_I/X}\bigr).\] We claim that \[\label{eq:claim-MCy-pullback} i_I^*\mathop{\mathrm{MC}}_y(U\to X)\cdot\det N_{D_I/X}=(1+y)^{|I|}\mathop{\mathrm{MC}}_y(D_I^\circ\to D_I).\tag{4}\] Granting 4 , the proposition follows from the standard identity \[\chi\bigl(\mathop{\mathrm{MC}}_y(V\to Y)\bigr)=\chi_y(V)\] applied to \(D_I^\circ\hookrightarrow D_I\).

It remains to prove 4 . Set \(D|_{D_I}=\bigcup_{j\notin I}D_j|_{D_I}\), an SNC divisor on \(D_I\) with \(D_I^\circ=D_I\setminus D|_{D_I}\). Applying 13 gives \[\begin{align} \mathop{\mathrm{MC}}_y(U\to X) & = \lambda_y[\Omega_X(\log D)]\cdot\mathcal{O}_X(-D), \tag{5} \\ \mathop{\mathrm{MC}}_y(D_I^\circ\to D_I) & = \lambda_y[\Omega_{D_I}(\log D|_{D_I})]\cdot\mathcal{O}_{D_I}(-D|_{D_I}). \tag{6} \end{align}\] Since \(D\) is SNC, \(N_{D_I/X}=\bigoplus_{i\in I}\mathcal{O}_X(D_i)|_{D_I}\), so \(\det N_{D_I/X}=i_I^*\mathcal{O}_X(\sum_{i\in I}D_i)\). The line bundle factor in 5 therefore satisfies \[i_I^*\mathcal{O}_X(-D)\cdot\det N_{D_I/X} = i_I^*\mathcal{O}_X\bigl(-\textstyle\sum_{j\notin I}D_j\bigr) = \mathcal{O}_{D_I}(-D|_{D_I}).\] For the logarithmic cotangent bundle, the residue maps \(\mathrm{Res}_{D_i}:\Omega_X(\log D)\to \mathcal{O}_{D_i}\) for \(i\in I\), restricted to \(D_I\), assemble into a short exact sequence \[0\to\Omega_{D_I}(\log D|_{D_I})\to i_I^*\Omega_X(\log D) \xrightarrow{\,\bigoplus_{i\in I}\mathrm{Res}_{D_i}\,} \mathcal{O}_{D_I}^{\oplus|I|}\to 0.\] In \(K(D_I)\), this gives \([i_I^*\Omega_X(\log D)]=[\Omega_{D_I}(\log D|_{D_I})]+|I|[\mathcal{O}_{D_I}]\). Since \(\lambda_y\) is multiplicative in \(K\)-theory and \(\lambda_y[\mathcal{O}_{D_I}]=1+y\), this gives \[\label{eq:lambda-split} \lambda_y[i_I^*\Omega_X(\log D)]=(1+y)^{|I|}\,\lambda_y[\Omega_{D_I}(\log D|_{D_I})].\tag{7}\] Combining 5 7 and the line bundle computation, \[i_I^*\mathop{\mathrm{MC}}_y(U\to X)\cdot\det N_{D_I/X} =(1+y)^{|I|}\,\lambda_y[\Omega_{D_I}(\log D|_{D_I})]\cdot\mathcal{O}_{D_I}(-D|_{D_I}).\] By 6 , the right-hand side equals \((1+y)^{|I|}\mathop{\mathrm{MC}}_y(D_I^\circ\to D_I)\), proving 4 . ◻

Proof. Using the identity \[i_{I*}i_I^*(a)=a\cdot i_{I*}\mathcal{O}_{D_I}\] in \(K(X)\), we get \[\chi\bigl(\mathbb{D}_X(\mathop{\mathrm{MC}}_y(U\to X))\cdot i_{I*}\mathcal{O}_{D_I}\bigr) =\chi\bigl(i_{I*}i_I^*\mathbb{D}_X(\mathop{\mathrm{MC}}_y(U\to X))\bigr) =\chi\bigl(i_I^*\mathbb{D}_X(\mathop{\mathrm{MC}}_y(U\to X))\bigr).\] Since \(X\) is smooth of dimension \(n\), we have \[\mathbb{D}_X(\mathop{\mathrm{MC}}_y(U\to X))=(-1)^n\,\mathop{\mathrm{MC}}_y(U\to X)^\vee\otimes \omega_X\] in \(K(X)[y]\). Therefore \[\chi\bigl(i_I^*\mathbb{D}_X(\mathop{\mathrm{MC}}_y(U\to X))\bigr) =(-1)^n\chi\bigl(i_I^*\mathop{\mathrm{MC}}_y(U\to X)^\vee\otimes i_I^*\omega_X\bigr).\] Since \(D_I\) is smooth, every K-class on \(D_I\) is represented by a perfect complex and Serre duality holds in \(K(D_I)\). Applying it on \(D_I\), of dimension \(n-|I|\), we obtain \[\chi\bigl(i_I^*\mathbb{D}_X(\mathop{\mathrm{MC}}_y(U\to X))\bigr) =(-1)^n(-1)^{n-|I|}\chi\bigl(i_I^*\mathop{\mathrm{MC}}_y(U\to X)\otimes i_I^*\omega_X^{-1}\otimes \omega_{D_I}\bigr).\] By the adjunction formula, \(i_I^*\omega_X^{-1}\otimes \omega_{D_I}\cong \det N_{D_I/X}\). Hence \[\chi\bigl(\mathbb{D}_X(\mathop{\mathrm{MC}}_y(U\to X))\cdot i_{I*}\mathcal{O}_{D_I}\bigr) =(-1)^{|I|}\chi\bigl(i_I^*\mathop{\mathrm{MC}}_y(U\to X)\otimes \det N_{D_I/X}\bigr).\] Applying the projection formula in the opposite direction, \[\chi\bigl(\mathbb{D}_X(\mathop{\mathrm{MC}}_y(U\to X))\cdot i_{I*}\mathcal{O}_{D_I}\bigr) =(-1)^{|I|}\chi\bigl(\mathop{\mathrm{MC}}_y(U\to X)\cdot i_{I*}\det N_{D_I/X}\bigr).\] By 13, the right-hand side equals \((-1)^{|I|}(1+y)^{|I|}\chi_y(D_I^\circ)=(-1-y)^{|I|}\chi_y(D_I^\circ)\), as desired. ◻

Proof. Let \(\mathcal{F}=\emptyset\subsetneq F_1\subsetneq \cdots \subsetneq F_k\subsetneq E\). Applying 5 to \(X=W_L\), \(U=U_L\), and the boundary components \(D_{F_1},\ldots,D_{F_k}\), \[g_{\mathsf{M}}^{\mathbb{D}}(\mathcal{F})=(-1-y)^k\,\chi_y(D_{\mathcal{F}}^\circ).\] The wonderful compactification gives a canonical isomorphism \[D_{\mathcal{F}}^\circ \;\cong\; U_{\mathsf{M}|F_1}\times U_{\mathsf{M}|F_2/F_1}\times\cdots\times U_{\mathsf{M}/F_k},\] where each factor \(U_{\mathsf{N}}\) is the complement of a realizing hyperplane arrangement for \(\mathsf{N}\). Since \(\chi_y\) is multiplicative under products, and for each minor \(\mathsf{N}\) we have \(\chi_y(U_{\mathsf{N}})=\overline{\chi}_{\mathsf{N}}(-y)\) by [20], we obtain \[\chi_y(D_{\mathcal{F}}^\circ)=\overline{\chi}_{\mathsf{M}}(-y)[\mathcal{F}].\] The second equality in the statement follows from \(\chi_{\mathsf{N}}(t)=(t-1)\overline{\chi}_{\mathsf{N}}(t)\) applied to each of the \(k+1\) factors in \(\chi_{\mathsf{M}}(-y)[\mathcal{F}]\). ◻

Proof. It suffices to expand the definitions. By definition of the characteristic polynomial, see, for example, [21], \[\chi_{\mathsf{N}/F}(t)=\sum_{F\subseteq G}\mu(F,G)t^{r-\mathop{\mathrm{rk}}(G)},\] and by applying Weisner’s theorem to \(\mathsf{N}|F\), see, for example, [21], \[(-1)^{\mathop{\mathrm{rk}}(F)-1}\beta(\mathsf{N}|F)=\overline{\chi}_{\mathsf{N}|F}(1)=\sum_{\substack{H\in L(\mathsf{N})\\H\subseteq F,\;i\not\in H}}\mu(\emptyset,H).\] Therefore the right-hand side is \[\sum_{\substack{H\subseteq F\subseteq G\\i\not\in H,\;i\in F}}\mu(\emptyset,H)\mu(F,G)t^{r-\mathop{\mathrm{rk}}(G)}.\] Rearranging the summation gives \[\sum_{\substack{H\subseteq G\\i\not\in H}}\mu(\emptyset,H)t^{r-\mathop{\mathrm{rk}}(G)} \left(\sum_{\substack{F:\,\overline{H\cup\{i\}}\subseteq F\subseteq G}}\mu(F,G)\right),\] where \(\overline{H\cup\{i\}}\) denotes the closure of \(H\cup\{i\}\). By the defining property of the Möbius function on a poset, see, for example, [22], the inner sum is zero unless \(G=\overline{H\cup\{i\}}\), in which case it is equal to \(1\). Hence the above summation simplifies to \[\sum_{i\not\in H}\mu(\emptyset,H)t^{r-\mathop{\mathrm{rk}}(\overline{H\cup\{i\}})}.\] Since \(H\) is a flat and \(i\not\in H\), we have \(\mathop{\mathrm{rk}}(\overline{H\cup\{i\}})=\mathop{\mathrm{rk}}(H)+1\). Thus the previous expression is equal to \[\sum_{i\not\in H}\mu(\emptyset,H)t^{r-\mathop{\mathrm{rk}}(H)-1},\] which is exactly the pointed expression for \(\overline{\chi}_{\mathsf{N}}(t)\) given by Weisner’s theorem. ◻

Proof. We prove the lemma by induction on the rank \(r\). If \(r=1\), then \(\mathsf{N}\) has no nonempty proper flats, so the only flag contributing to the summation is the empty flag. Hence the left-hand side is \[\chi_{\mathsf{N}}(t)=t-1.\] On the other hand, \(\overline{\chi}_{\mathsf{N}}(t)=1\), so \(\beta(\mathsf{N})=1\). Therefore the desired identity holds in this case.

Assume now \(r\ge 2\). Separating the empty flag from the summation, we obtain \[\sum_{\substack{\mathcal{F}\\i\in H,\forall H\in \mathcal{F}}}(-1)^{\ell(\mathcal{F})} \chi_{\mathsf{N}}(t)[\mathcal{F}] = \chi_{\mathsf{N}}(t) + \sum_{\substack{\mathcal{F}\neq\emptyset\\i\in H,\forall H\in \mathcal{F}}}(-1)^{\ell(\mathcal{F})} \chi_{\mathsf{N}}(t)[\mathcal{F}].\]

To further simplify the second summation, let \(F_k\) be the largest proper flat in \(\mathcal{F}\). Since \(i\in H\) for all \(H\in \mathcal{F}\), in particular \(i\in F_k\). Therefore, we separate \(F_k\) out in the summation: \[\begin{align} & \sum_{\substack{\mathcal{F}\neq\emptyset\\i\in H,\forall H\in \mathcal{F}}}(-1)^{\ell(\mathcal{F})} \chi_{\mathsf{N}}(t)[\mathcal{F}] \\ ={} & \sum_{\substack{\mathcal{F}:F_1\subsetneq\cdots\subsetneq F_k,k>0 \\i\in H,\forall H\in \mathcal{F}}}(-1)^k \chi_{\mathsf{N}|F_1}(t)\cdots \chi_{\mathsf{N}|F_k/F_{k-1}}(t)\chi_{\mathsf{N}/F_k}(t)\\ ={} & -\sum_{\substack{F\in L(\mathsf{N})\\i\in F,\;F\neq E}}\left(\sum_{\substack{\mathcal{F}'\\i\in H,\forall H\in \mathcal{F}'}}\!\!(-1)^{\ell(\mathcal{F}')} \chi_{\mathsf{N}|F}(t)[\mathcal{F}']\right)\chi_{\mathsf{N}/F}(t). \end{align}\] Here \(\mathcal{F}'\) ranges over all flags of nonempty proper flats of \(\mathsf{N}|F\), including the empty flag. Applying the induction hypothesis to \(\mathsf{N}|F\), we obtain \[\sum_{\substack{\mathcal{F}\neq\emptyset\\i\in H,\forall H\in \mathcal{F}}}(-1)^{\ell(\mathcal{F})} \chi_{\mathsf{N}}(t)[\mathcal{F}] = -(t-1)\sum_{\substack{F\in L(\mathsf{N})\\i\in F,\;F\neq E}}(-1)^{\mathop{\mathrm{rk}}(F)-1}\beta(\mathsf{N}|F)\chi_{\mathsf{N}/F}(t).\]

By 12, \[\overline{\chi}_{\mathsf{N}}(t) = \sum_{\substack{F\in L(\mathsf{N})\\ i\in F}} (-1)^{\mathop{\mathrm{rk}}(F)-1}\beta(\mathsf{N}|F)\,\chi_{\mathsf{N}/F}(t).\] Separating the term \(F=E\), we get \[\sum_{\substack{F\in L(\mathsf{N})\\i\in F,\;F\neq E}}(-1)^{\mathop{\mathrm{rk}}(F)-1}\beta(\mathsf{N}|F)\chi_{\mathsf{N}/F}(t) = \overline{\chi}_{\mathsf{N}}(t)-(-1)^{r-1}\beta(\mathsf{N}),\] since \(\chi_{\mathsf{N}/E}(t)=1\). Therefore, \[\begin{align} \sum_{\substack{\mathcal{F}\\i\in H,\forall H\in \mathcal{F}}}(-1)^{\ell(\mathcal{F})} \chi_{\mathsf{N}}(t)[\mathcal{F}] &= \chi_{\mathsf{N}}(t)-(t-1)\bigl(\overline{\chi}_{\mathsf{N}}(t)-(-1)^{r-1}\beta(\mathsf{N})\bigr) \\ &= (-1)^{r-1}\beta(\mathsf{N})(t-1), \end{align}\] where the last equality follows from \(\chi_{\mathsf{N}}(t)=(t-1)\overline{\chi}_{\mathsf{N}}(t)\). ◻

Proof of 3. Let \(t=-y\). We prove the theorem by verifying the \(K\)-balancing condition of 1. Since \(t-1=-1-y\) is a nonzero element of the integral domain \({\mathbb{Z}}[y]\), a \({\mathbb{Z}}[y]\)-valued function satisfies the balancing condition if and only if its product with \(t-1\) does. It therefore suffices to check the balancing condition for \((t-1)g_{\mathsf{M}}^{\mathbb{D}}=\chi_{\mathsf{M}}(t)[\cdot]\).

We first prove the balancing condition for \(\mathcal{G}=\emptyset\). In this case, we need to show that for every loopless matroid \(\mathsf{M}\) on \(E=[n]\) and every pair \(i,j\in E\), \[\label{eq:Aij-symmetry} A_{i,j}(\mathsf{M})=A_{j,i}(\mathsf{M}),\tag{8}\] where we set \[A_{i,j}(\mathsf{M}) \coloneq \sum_{\substack{\mathcal{F}\neq\emptyset\\ i\in H,\;j\notin H\\ \forall H\in\mathcal{F}}} (-1)^{\ell(\mathcal{F})}\chi_{\mathsf{M}}(t)[\mathcal{F}].\] To further simplify the summation, we note that the condition \(j\not\in H\) for all \(H\in \mathcal{F}\) is equivalent to saying \(j\not\in F_k\), where \(F_k\) is the largest proper flat in \(\mathcal{F}\). Therefore, we separate \(F_k\) out in the summation: \[\begin{align} A_{i,j}(\mathsf{M}) & =\sum_{\substack{\mathcal{F}:F_1\subsetneq\cdots\subsetneq F_k,k>0 \\i\in F_1,j\not\in F_k}}(-1)^k \chi_{\mathsf{M}|F_1}(t)\cdots \chi_{\mathsf{M}|F_{k}/F_{k-1}}(t)\chi_{\mathsf{M}/F_k}(t)\\ & =\sum_{\substack{F_k\in L(\mathsf{M})\\i\in F_k,\;j\not\in F_k}}\left(\sum_{\substack{\mathcal{F}':F_1\subsetneq\cdots\subsetneq F_{k-1}\\i\in H,\forall H\in \mathcal{F}'}}(-1)^k \chi_{\mathsf{M}|F_1}(t)\cdots \chi_{\mathsf{M}|F_k/F_{k-1}}(t)\right)\chi_{\mathsf{M}/F_k}(t)\\ & =\sum_{\substack{F_k\in L(\mathsf{M})\\i\in F_k,\;j\not\in F_k}}\sum_{\substack{\mathcal{F}'\\i\in H,\forall H\in \mathcal{F}'}}-(-1)^{\ell(\mathcal{F}')} \chi_{\mathsf{M}|F_k}(t)[\mathcal{F}']\chi_{\mathsf{M}/F_k}(t). \end{align}\]

Applying 13 to \(\mathsf{N}=\mathsf{M}|F_k\), and noting that \(\mathsf{M}|F_j/F_{j-1}=(\mathsf{M}|F_k)|F_j/F_{j-1}\), we obtain \[A_{i,j}(\mathsf{M})=-(t-1)\sum_{\substack{F\in L(\mathsf{M})\\i\in F,\;j\not\in F}}(-1)^{\mathop{\mathrm{rk}}(F)-1}\beta(\mathsf{M}|F)\chi_{\mathsf{M}/F}(t).\] Here \(F=F_k\) ranges over all flats of \(\mathsf{M}\) containing \(i\) and not containing \(j\).

Furthermore, \[\begin{align} & \sum_{\substack{F\in L(\mathsf{M})\\i\in F,\;j\not\in F}}(-1)^{\mathop{\mathrm{rk}}(F)-1}\beta(\mathsf{M}|F)\chi_{\mathsf{M}/F}(t) \\ = & \sum_{\substack{F\in L(\mathsf{M})\\i\in F}}(-1)^{\mathop{\mathrm{rk}}(F)-1}\beta(\mathsf{M}|F)\chi_{\mathsf{M}/F}(t) - \sum_{\substack{F\in L(\mathsf{M})\\i,j\in F}}(-1)^{\mathop{\mathrm{rk}}(F)-1}\beta(\mathsf{M}|F)\chi_{\mathsf{M}/F}(t). \end{align}\] By 12, the first summation is \(\overline{\chi}_{\mathsf{M}}(t),\) hence independent of \(i\). Therefore the right-hand side is symmetric in \(i\) and \(j\), and we obtain \(A_{i,j}=A_{j,i}\).

For a general \(\{i,j\}\)-neutral flag \(\mathcal{G}\), write \[\emptyset=G_0\subsetneq G_1\subsetneq\cdots\subsetneq G_m\subsetneq G_{m+1}=E.\] Since \(\mathcal{G}\) is \(\{i,j\}\)-neutral, there is a unique index \(s\) such that \(i,j\not\in G_s\) and \(i,j\in G_{s+1}\). Set \[\mathsf{N}= (\mathsf{M}|G_{s+1})/G_s.\] Then the interval \([G_s,G_{s+1}]\subseteq L(\mathsf{M})\) is naturally identified with \(L(\mathsf{N})\). Under this identification, every refinement \(\mathcal{F}\succneqq\mathcal{G}\) contributing to the left-hand side of the balancing condition in 1 is obtained by inserting a nonempty flag \(\mathcal{H}\) of nonempty proper flats of \(\mathsf{N}\) such that \(i\in H\) and \(j\not\in H\) for every \(H\in \mathcal{H}\). Moreover, \[(-1)^{\ell(\mathcal{F})}\chi_{\mathsf{M}}(t)[\mathcal{F}] = C_{\mathcal{G}}\,(-1)^{\ell(\mathcal{H})}\chi_{\mathsf{N}}(t)[\mathcal{H}],\] where \(C_{\mathcal{G}}\) is the product of the factors \(\chi_{(\mathsf{M}|G_{a+1})/G_a}(t)\) coming from the intervals of \(\mathcal{G}\) other than \([G_s,G_{s+1}]\), multiplied by \((-1)^{\ell(\mathcal{G})}\). Hence the left-hand side of the balancing condition is equal to \(C_{\mathcal{G}}A_{i,j}(\mathsf{N})\), and similarly the right-hand side is equal to \(C_{\mathcal{G}}A_{j,i}(\mathsf{N})\). By the empty-flag case proved above, these two quantities are equal. Therefore \(\chi_{\mathsf{M}}(-y)[\cdot]\), and hence also \(g_{\mathsf{M}}^{\mathbb{D}}\), satisfies the balancing condition of 1. ◻

Proof. By Hirzebruch–Riemann–Roch, \[\chi(\alpha\cdot\beta)=\int_X ch(\beta)\cap td_*(\alpha)\] for \(\alpha\in K(X)[y]\) and \(\beta\in K(X)\). Expanding by dimension gives \[\chi(\alpha\cdot\beta)=\sum_{p\ge 0}\int_X ch^p(\beta)\cap td_p(\alpha).\] Therefore \[\begin{align} \chi(\alpha\cdot\beta) &=\sum_{p\ge 0}\int_X (1+y)^p ch^p(\beta)\cap (1+y)^{-p}td_p(\alpha) \\ &=\int_X ch_{(1+y)}(\beta)\cap td_{(1+y)}(\alpha). \end{align}\] Applying this to \(\alpha=\mathop{\mathrm{MC}}_y(U\to X)\) and \(\beta=i_{I*}\det N_{D_I/X}\), and then using 13, we obtain \[\int_X ch_{(1+y)}\bigl(i_{I*}\det N_{D_I/X}\bigr)\cap T_{y*}(U\to X) =\chi\bigl(\mathop{\mathrm{MC}}_y(U\to X)\cdot i_{I*}\det N_{D_I/X}\bigr) =(1+y)^{|I|}\chi_y(D_I^\circ).\]

To specialize at \(y=-1\), we apply Grothendieck–Riemann–Roch to the regular embedding \(i_I\): \[ch\bigl(i_{I*}\det N_{D_I/X}\bigr) =i_{I*}\bigl(ch(\det N_{D_I/X})\,td(N_{D_I/X})^{-1}\bigr).\] Hence \(ch^p(i_{I*}\det N_{D_I/X})=0\) for \(p<|I|\), and the codimension-\(|I|\) component is \[i_{I*}\bigl(ch^0(\det N_{D_I/X})\,td^0(N_{D_I/X})^{-1}\bigr)=i_{I*}(1)=[D_I].\] Thus \((1+y)^{-|I|}ch_{(1+y)}\bigl(i_{I*}\det N_{D_I/X}\bigr)=[D_I]+(1+y)\gamma_I(y)\) for some \(\gamma_I(y)\in A^*(X)_{\mathbb{Q}}[y]\). Dividing the displayed integral formula above by \((1+y)^{|I|}\) and substituting, \[\int_X \bigl([D_I]+(1+y)\gamma_I(y)\bigr)\cap T_{y*}(U\to X)=\chi_y(D_I^\circ).\] Setting \(y=-1\) and using \(T_{-1,*}=c_{SM}\otimes{\mathbb{Q}}\) and \(\chi_{-1}=\chi\) (topological Euler characteristic) from [4], we obtain \[\int_X [D_I]\cap c_{SM}(1_U)=\chi(D_I^\circ).\qedhere\] ◻

Proof. Applying 15 to \(X=W_L\), \(U=U_L\), and the boundary components \(D_{F_1},\ldots,D_{F_k}\), we obtain \[\deg\!\bigl(c_{SM}(1_{U_L})\cdot [D_{\mathcal{F}}]\bigr)=\chi(D_{\mathcal{F}}^\circ).\] Since \(D_{\mathcal{F}}=W_L\cap V(\sigma_{\mathcal{F}})\), we have \(j^*[V(\sigma_{\mathcal{F}})]=[D_{\mathcal{F}}]\). The projection formula then gives \[\deg\!\bigl(j_*c_{SM}(1_{U_L})\cdot [V(\sigma_{\mathcal{F}})]\bigr)=\chi(D_{\mathcal{F}}^\circ).\] As before, the wonderful compactification gives \[D_{\mathcal{F}}^\circ \cong U_{\mathsf{M}|F_1}\times U_{\mathsf{M}|F_2/F_1}\times\cdots\times U_{\mathsf{M}/F_k}.\] Therefore \[\chi(D_{\mathcal{F}}^\circ)=\overline{\chi}_{\mathsf{M}}(1)[\mathcal{F}],\] because \(\chi(U_{\mathsf{N}})=\overline{\chi}_{\mathsf{N}}(1)\) for every minor \(\mathsf{N}\), which follows from [20] at \(y=-1\). Finally, for every minor \(\mathsf{N}\), \[\overline{\chi}_{\mathsf{N}}(1)=(-1)^{\mathop{\mathrm{rk}}(\mathsf{N})-1}\beta(\mathsf{N}),\] and applying this to each factor in \(\overline{\chi}_{\mathsf{M}}(1)[\mathcal{F}]\) gives the beta-invariant formula. By the identification of Chow classes on \(X_{\Sigma_{\mathsf{M}}}\) with Minkowski weights on \(\Sigma_{\mathsf{M}}\), this is exactly the classical CSM class of \(\mathsf{M}\). ◻

Proof. We first compute the total Euler characteristic. By [6] there is a ring isomorphism \(\zeta : K(X) \xrightarrow{\sim} A^\bullet(X)\) satisfying \(\chi([\mathcal{E}]) = \deg_\alpha(\zeta([\mathcal{E}]))\), where \(\alpha \in A^1(X)\) is the hyperplane class. Since \(\zeta\) is a ring isomorphism and \(\mathcal{S}_{\mathsf{M}}, \mathcal{Q}_{\mathsf{M}}\) have simple Chern roots in the sense of [6], that proposition gives \[\zeta\!\left(\lambda_u(\mathcal{S}_{\mathsf{M}}^\vee)\right) = (u+1)^r\, c\!\left(\mathcal{S}_{\mathsf{M}}^\vee,\tfrac{u}{u+1}\right)\] and \[\zeta\!\left(\lambda_v(\mathcal{Q}_{\mathsf{M}}^\vee)\right) = (v+1)^{n-r}\, c\!\left(\mathcal{Q}_{\mathsf{M}}^\vee,\tfrac{v}{v+1}\right) = (v+1)^{n-r}\, c\!\left(\mathcal{Q}_{\mathsf{M}},-\tfrac{v}{v+1}\right).\] Therefore \[\chi\!\left(\lambda_u(\mathcal{S}_{\mathsf{M}}^\vee)\lambda_v(\mathcal{Q}_{\mathsf{M}}^\vee)\right) = (u+1)^r(v+1)^{n-r} \sum_{i+k+\ell = n} \left(\int_X \alpha^i c_k(\mathcal{S}_{\mathsf{M}}^\vee)\,c_\ell(\mathcal{Q}_{\mathsf{M}})\right) \left(\tfrac{u}{u+1}\right)^k\!\left(-\tfrac{v}{v+1}\right)^\ell.\] By [6], this integral equals \[\begin{align} (u+1)^r(v+1)^{n-r}\,t_{\mathsf{M}}\!\left(1,0,\tfrac{u}{u+1},-\tfrac{v}{v+1}\right) &= (u+1)^r(v+1)^{n-r}\left(\tfrac{u}{u+1}\right)^r\!\left(1-\tfrac{v}{v+1}\right)^{n-r} T_{\mathsf{M}}\!\left(\tfrac{u+1}{u},\,1+v\right) \\ &= u^r T_{\mathsf{M}}\!\left(1+\tfrac{1}{u},\,1+v\right). \end{align}\] For a general flag \(\mathcal{F}\), the Grothendieck weight is \(\chi(\lambda_u(\mathcal{S}_{\mathsf{M}}^\vee)\lambda_v(\mathcal{Q}_{\mathsf{M}}^\vee) \cdot x_{\mathcal{F}})\). By the minor decomposition property [6], the tautological bundles restricted to \(V(\sigma_{\mathcal{F}})\) decompose as products over the successive minors of \(\mathcal{F}\), so the same computation applied to each minor gives \[u^r T_{\mathsf{M}}\!\left(1+\tfrac{1}{u},\,1+v\right)[\mathcal{F}].\] ◻