Proof. We assume $\nu \in {\mathbb {Z}}_{\ge 0}\Delta ^+$ and prove that $A\le A'$ . We may assume $\nu = \alpha \in \Delta ^+$ . Take $n\in {\mathbb {Z}}$ such that $n - 1 < \langle a,\alpha ^\vee \rangle < n$ for any $a\in A$ . For $a\in A$ , we have $\langle s_{\alpha ,n}(a),\alpha ^\vee \rangle = \langle a - (\langle a,\alpha ^\vee \rangle - n)\alpha ,\alpha ^\vee \rangle = 2n - \langle a,\alpha ^\vee \rangle $ . Hence, $n < \langle s_{\alpha ,n}(a),\alpha ^\vee \rangle < n + 1$ . Therefore, $A \le s_{\alpha ,n}(A)\le s_{\alpha ,n + 1}s_{\alpha ,n}(A) = A + \alpha $ .

However, assume that $A\le A'$ . Take $a\in A$ . Then by Lemma 2.2, we have $(a + \nu ) - a \in {\mathbb {R}}_{\ge 0}\Delta ^+$ . Hence, $\nu \in {\mathbb {R}}_{\ge 0}\Delta ^+$ . Since $\nu \in {\mathbb {Z}}\Delta $ , we get $\nu \in {\mathbb {Z}}_{\ge 0}\Delta ^+$ .

Assumption 2.5. In the rest of this section, we assume the following.

1. (1) We have $2\in {\mathbb {K}}^\times $ , and any $\alpha ^\vee \ne \beta ^\vee \in (\Delta ^\vee )^+$ are linearly independent in $X^\vee _{{\mathbb {K}}/\mathfrak {m}}$ for any maximal ideal $\mathfrak {m}\subset {\mathbb {K}}$ . This is the GKM-property of the moment graph attached to the finite Weyl group [Reference FiebigFie11, 9.1].

2. (2) The torsion primes of the root system $(X^\vee ,\Delta ^\vee ,X,\Delta )$ [Reference Juteau, Mautner and WilliamsonJMW14, Definition 2.43] are invertible in ${\mathbb {K}}$ .

Proof. If $w\in W_{\mathrm {f}}$ fixes any element in $X^\vee _{{\mathbb {K}}}$ , it fixes any image of $\alpha \in \Delta $ . By the assumption, $\Delta ^\vee \to X^\vee _{{\mathbb {K}}}$ is injective. Therefore, w fixes any coroot. Hence, w is identity.

Proof. The first part is obvious, and for the second part, the left-hand side is contained in the right-hand side. Take m from the right-hand side and let $f\in S$ such that $fm\in M$ . Then, we have $fm\in M_{I}$ , and m is in the left-hand side. The last assertion is obvious.

Proof. Set $\Omega = W^{\prime }_{\alpha ,{\mathrm {aff}}}A$ and let $I_1,I_2\subset \mathcal {A}$ be closed subsets. We have $\Omega = \{A,\alpha \uparrow A,\alpha \uparrow (\alpha \uparrow A),\dots \}\cup \{\alpha \downarrow A,\alpha \downarrow (\alpha \downarrow A),\dots \}$ and $\Omega $ is a totally ordered subset of $\mathcal {A}$ . Since $\Omega $ is totally ordered, $I_1\cap \Omega \subset I_2\cap \Omega $ or $I_2\cap \Omega \subset I_1\cap \Omega $ . We may assume $I_1\cap \Omega \subset I_2\cap \Omega $ . We can take closed subsets $I^{\prime }_1$ and $I^{\prime }_2$ such that $I^{\prime }_1\subset I^{\prime }_2$ , $I^{\prime }_1\cap \Omega = I_1\cap \Omega $ and $I^{\prime }_2\cap \Omega = I_2\cap \Omega $ . Then, we have $M_{I^{\prime }_1} = M_{I_1}$ , $M_{I^{\prime }_2} = M_{I_2}$ and $M_{I^{\prime }_1\cup I^{\prime }_2} = M_{I_1\cup I_2}$ . Hence, we may assume $I_1 = I^{\prime }_1$ and $I_2 = I^{\prime }_2$ . In this case, (S) obviously holds.

Proof. The proof is divided into 4 steps.

(1) Assume that both $K_1,K_2$ are closed. Then, the lemma follows from the definitions.

(2) Assume that $K_1$ is open and $K_2$ is closed. Set $I_1 = \mathcal {A}\setminus K_1$ . Then, we have

$$\begin{align*}(M_{K_1})_{K_2} = M/M_{I_1}\cap \bigoplus_{A\in K_2}(M/M_{I_1})_{A}^{\emptyset}. \end{align*}$$

Note that $M_{K_2}/(M_{K_2}\cap M_{I_1}) = M_{K_2}/M_{K_2\cap I_1} = M_{K_1\cap K_2}$ . There is a canonical embedding from $M_{K_2}/(M_{K_2}\cap M_{I_1})$ to $(M_{K_1})_{K_2}$ . Let $m\in M$ such that $m + M_{I_1}\in \bigoplus _{A\in K_2}(M/M_{I_1})_{A}^{\emptyset }$ . Then, $M_A^{\emptyset }$ -component $m_{A}$ of m is $0$ for $A\notin I_1\cup K_2$ . Hence, $m\in M_{I_1\cup K_2} = M_{I_1} + M_{K_2}$ . Therefore, the canonical embedding is surjective. We get the lemma.

(3) Assume that $K_2$ is closed. Take a closed subset $I_1$ and an open subset $J_1$ such that $K_1 = I_1\cap J_1$ . Then, by (2), $(M_{J_1})_{I_1} \simeq M_{K_1}$ . Hence, $(M_{K_1})_{K_2} \simeq ((M_{J_1})_{I_1})_{K_2} = (M_{J_1})_{I_1\cap K_2}$ by (1). This is isomorphic to $M_{J_1\cap I_1\cap K_2} = M_{K_1\cap K_2}$ by (2).

(4) Now we prove the lemma in general. Let $I_i$ be a closed subset and $J_i$ be an open subset such that $K_i = I_i \cap J_i$ and put $J_i^c = \mathcal {A}\setminus J_i$ for $i = 1,2$ . Then,

$$\begin{align*}(M_{K_1})_{K_2} = (M_{K_1})_{I_2}/(M_{K_1})_{I_2\cap J_2^c} \simeq M_{K_1\cap I_2}/M_{K_1\cap I_2\cap J_2^c} \end{align*}$$

by (3). We have $M_{K_1\cap I_2} = M_{I_1\cap I_2}/M_{I_1\cap I_2\cap J_1^c}$ and $M_{K_1\cap I_2\cap J_2^c} = M_{I_1\cap I_2\cap J_2^c}/M_{I_1\cap I_2\cap J_2^c\cap J_1^c}$ . Hence,

$$\begin{align*}(M_{K_1})_{K_2} \simeq M_{I_1\cap I_2}/(M_{I_1\cap I_2\cap J_1^c} + M_{I_1\cap I_2\cap J_2^c}). \end{align*}$$

Since $M_{I_1\cap I_2\cap J_1^c} + M_{I_1\cap I_2\cap J_2^c} = M_{(I_1\cap I_2\cap J_1^c)\cup (I_2\cap I_2\cap J_2^c)} = M_{(I_1\cap I_2)\setminus (J_1\cap J_2)}$ , we get the lemma.

Proof. Take a closed subset I and an open subset J such that $K = I \cap J$ .

We prove $M_{K}$ satisfies (S). Let $I_1,I_2$ be closed subsets. Since $(M_{K})_{I_i} = M_{K\cap I_i}$ is a quotient of $M_{I\cap I_i}$ , it is sufficient to prove that $M_{I\cap I_1}\oplus M_{I\cap I_2}\to (M_{K})_{I_1\cup I_2}$ is surjective. The module $(M_{K})_{I_1\cup I_2} = M_{K\cap (I_1\cup I_2)}$ is a quotient of $M_{I\cap (I_1\cup I_2)}$ , and since $M_{I\cap (I_1\cup I_2)} = M_{I\cap I_1} + M_{I\cap I_2}$ , the map is surjective.

We prove $M_{K}$ satisfies (LE). We may assume $M = \bigoplus _{\Omega \in W^{\prime }_{\alpha ,{\mathrm {aff}}}\backslash \mathcal {A}}(\bigoplus _{A\in \Omega }M_{A}^{\emptyset }\cap M^{\alpha })$ . Let $m \in M_{I}^{\alpha }$ . Then, for each $\Omega \in W^{\prime }_{\alpha ,{\mathrm {aff}}}\backslash \mathcal {A}$ , we have $m_{\Omega }\in M^{\alpha }\cap \bigoplus _{A\in \Omega }M_{A}^{\emptyset }$ such that $m = \sum m_{\Omega }$ . Then, for each $A\in \mathcal {A}$ , we have $m_{A} = (m_{\Omega })_{A}$ , where $\Omega $ is the unique $W^{\prime }_{\alpha ,{\mathrm {aff}}}$ -orbit containing A. Therefore, since $m\in M_{I}^{\alpha }$ , we have $m_{\Omega }\in M_{I}^{\alpha }$ . Hence, $m_{\Omega }\in M^{\alpha }_I\cap \bigoplus _{A\in \Omega }(M_I)_{A}^{\emptyset }$ . Namely, $M_{I}$ satisfies (LE). Since $M_{K}$ is a quotient of $M_{I}$ , it also satisfies (LE).

Proof. There exist $A_0^-,A_0^+$ such that $\operatorname {\mathrm {supp}}_{\mathcal {A}}(M_i)\subset [A_0^-,A_0^+]$ for any $i = 1,\dots ,l$ by [Reference LusztigLus80, Proposition 3.7]. Put $I_0 = \{A'\in \mathcal {A}\mid A'\nless A_0^+\}\cap I$ . We enumerate the elements in $(I\setminus \{A\})\cap [A_0^-,A_0^+]$ (resp. $[A_0^-,A_0^+]\setminus I$ ) as $\{A_1,\dots ,A_{k - 1}\}$ (resp. $\{A_{k + 1},\dots ,A_r\}$ ) such that $A_i\ge A_j$ implies $i\le j$ . Put $A_k = A$ . Then, it is easy to see that $I_i = I_0\cup \{A_1,\dots ,A_i\}$ is closed and satisfies the conditions of the lemma.

Proof. Since $M_K\in \widetilde {\mathcal {K}}_\Delta (S_0)$ , we may assume $K = \mathcal {A}$ . Take closed subsets $I_0\subset I_1\subset \dotsm \subset I_r$ such that $I_{i + 1}\setminus I_i = \{A_i\}$ , $M_{I_0} = 0$ and $M_{I_r} = M$ . Then, $M_{I_{i + 1}}/M_{I_{i}} = M_{\{A_i\}}$ is a graded free $S_0$ -module. Hence, $M_{I_r}/M_{I_0} = M$ is also graded free.

Proof. Replacing $M_i$ with $(M_i)_K$ for $i = 1,2,3$ , we may assume $K = \mathcal {A}$ . We can take closed subsets $I_0\subset I_1\subset \dotsb \subset I_r$ such that $(M_i)_{I_0} = 0$ , $(M_i)_{I_r} = M_i$ and $\#(I_{j + 1}\setminus I_j) = 1$ for $i = 1,2,3$ and $j = 0,\dots ,r$ , as in Lemma 2.14. Then, the exactness of $0\to (M_1)_{I_j}\to (M_2)_{I_j}\to (M_3)_{I_j}\to 0$ follows by induction on j and a standard diagram argument.

Proof. Note that $M_{I_1}/M_{I_2} = M_{I_1\setminus I_2}$ . The lemma follows from Lemma 2.12.

Proof. Let $A'\in \mathcal {A}$ and take $\beta \in \Delta ^+$ and $m\in {\mathbb {Z}}$ such that $A's = s_{\beta ,m}A'$ . Take $x\in W^{\prime }_{\mathrm {aff}}$ such that $A' = xA$ . Then, $A's = xAs = xs_{\alpha ,n}A$ . Since the action of $W^{\prime }_{{\mathrm {aff}}}$ on $X_{{\mathbb {R}}}$ preserves the set $\{\{\lambda \in X_{{\mathbb {R}}}\mid \langle \lambda ,\alpha ^{\vee }\rangle = n\}\mid \alpha \in \Delta ,n\in {\mathbb {Z}}\}$ , there exists $(\gamma ,k)\in \Delta \times {\mathbb {Z}}$ such that $xs_{\alpha ,n} = s_{\gamma ,k}x$ . Moreover, $\gamma \in \{\pm \overline {x}(\alpha )\}$ , where $\overline {x}\in W_{\mathrm {f}}$ is the image of x under $W^{\prime }_{{\mathrm {aff}}}\to W_{\mathrm {f}}$ . We may assume $\gamma = \overline {x}(\alpha )$ . We have $A's = s_{\gamma ,k}xA = s_{\gamma ,k}A'$ . Hence, $s_{\gamma ,k} = s_{\beta ,m}$ and therefore, $\beta = \varepsilon \gamma = \varepsilon \overline {x}(\alpha )$ for $\varepsilon = 1$ or $\varepsilon = -1$ . We have $\beta ^{A'} = \varepsilon \overline {x}(\alpha )^{xA} = \varepsilon \alpha ^A$ and $(\beta ^\vee )^{A'} = \varepsilon (\alpha ^\vee )^{A}$ .

Proof. Assume that $B_1\in {\mathcal {S}\mathrm {Bimod}}$ is a direct summand of $B\in {\mathcal {S}\mathrm {Bimod}}$ , and let $e\in \operatorname {\mathrm {End}}_{{\mathcal {S}\mathrm {Bimod}}}(B)$ be the idempotent such that $B_1 = e(B)$ . If B satisfies the lemma, then by putting $(B_1)_x^{\emptyset } = e(B_x^{\emptyset })$ , we see that $B_1$ also satisfies the lemma. Therefore, we may assume $B = B_{s_1}\otimes \cdots \otimes B_{s_l}$ for $s_i\in S_{{\mathrm {aff}}}$ . Note that for $B = B_s$ , the lemma holds as we have seen in the above. Hence, it is sufficient to prove that if $B_1,B_2$ satisfy the lemma, then $B = B_1\otimes B_2$ also satisfies the lemma.

For $x\in W_{{\mathrm {aff}}}$ and $b\in (B_1)_x^{\emptyset }$ , we have $bf = x(f)b$ for $f\in R$ . Since $\{(\alpha ^\vee )^{A}\mid \alpha \in \Delta \}$ is stable under the action of x, the formula says that $(B_1)_x^{\emptyset }$ is also a right $R^{\emptyset }$ -module. Therefore, $B_1^{\emptyset }$ is also a right $R^{\emptyset }$ -module. Hence, $R^{\emptyset }\otimes _{R}B_1\otimes _{R}B_2\simeq B_1^{\emptyset }\otimes _{R}B_2\simeq B_1^{\emptyset }\otimes _{R^{\emptyset }}R^{\emptyset }\otimes _{R}B_2 \simeq B_1^{\emptyset }\otimes _{R^{\emptyset }}B_2^{\emptyset }$ . We put $B_{x}^{\emptyset } = \bigoplus _{yz = x}(B_1)_y^{\emptyset }\otimes _{R^{\emptyset }}(B_2)_z^{\emptyset }$ . Then, we get $B^{\emptyset } = \bigoplus _{x\in W_{{\mathrm {aff}}}}B_x^{\emptyset }$ and we have $\operatorname {\mathrm {Frac}}(R)\otimes _{R^{\emptyset }}B_x^{\emptyset }\simeq B_x^{\operatorname {\mathrm {Frac}}(R)}$ .

Proof. Take $\delta \in \Lambda ^\vee _{{\mathbb {K}}}$ such that $\langle \alpha _s,\delta \rangle = 1$ . As $(S_0,R)$ -bimodules, we have $N*B_s = N\otimes _{R^s}R(1)$ and $M*B_s = M\otimes _{R^s}R(1)$ . For $\varphi \colon M\otimes _{R^s}R(1)\to N$ , define $\psi \colon M\to N\otimes _{R^s}R(1)$ by $\psi (m) = \varphi (m\delta \otimes 1)\otimes 1 - \varphi (m\otimes 1)\otimes s(\delta )$ . We know that if $\varphi $ is an $(S_0,R)$ -bimodule homomorphism, $\psi $ is also an $(S_0,R)$ -bimodule homomorphism and it induces a bijection between the spaces of $(S_0,R)$ -bimodule homomorphisms (ee, for example, [Reference LibedinskyLib08, Lemma 3.3]). We prove that $\varphi $ is a morphism in $\widetilde {\mathcal {K}}'(S_0)$ if and only if $\psi $ is a morphism in $\widetilde {\mathcal {K}}'(S_0)$ .

Set $a(m) = m\delta \otimes 1 - m\otimes s(\delta )$ and $b(m) = ms(\delta ) \otimes 1 - m\otimes s(\delta )$ for $m\in M^\emptyset $ . We also define $a'(n),b'(n)\in N^{\emptyset }\otimes _{R^s}R$ for $n\in N^\emptyset $ in the same way. Then, we have $(M*B_s)_{A}^{\emptyset } = a(M_{A}^{\emptyset }) + b(M_{As}^{\emptyset })$ and the same for N by (2.1) for $A\in \mathcal {A}$ .

Let $A\in \mathcal {A}$ and $m\in M_{A}^{\emptyset }$ . By the definition, $\psi (m) = \varphi (a(m))\otimes 1 + b'(\varphi (m\otimes 1))$ . Since $a(m)\in (M*B_s)^{\emptyset }_A$ , $\varphi (a(m))\otimes 1 = (\alpha _s)_{A}^{-1}\varphi (a(m))\alpha _s\otimes 1 = (\alpha _s)_{A}^{-1}a'(\varphi (a(m))) - (\alpha _s)_{A}^{-1}b'(\varphi (a(m)))$ . However, we have $m\otimes 1 = (\alpha _s)_{A}^{-1}m\alpha _s\otimes 1 = (\alpha _s)_{A}^{-1}a(m) - (\alpha _s)_{A}^{-1}b(m)$ . Since $\varphi $ and $b'$ are left $S_0$ -equivariant, we get $\psi (m) = (\alpha _s)_{A}^{-1}a'(\varphi (a(m))) - (\alpha _s)_{A}^{-1}b'(\varphi (b(m)))$ .

Assume that $\varphi $ is a morphism in $\widetilde {\mathcal {K}}'(S_0)$ . Then, for any $m\in M_{A}^{\emptyset }$ , $\varphi (a(m))\in \bigoplus _{A'\ge A}N_{A'}^{\emptyset }$ . Hence, $a'(\varphi (a(m)))\in \bigoplus _{A' \ge A}(N*B_s)_{A'}^{\emptyset }$ . Since $b(m)\in (M*B_s)_{As}^{\emptyset }$ , we have $\varphi (b(m))\in \bigoplus _{A' \ge As, A'\in As + {\mathbb {Z}}\Delta }N_{A'}^{\emptyset }$ . Therefore, $b'(\varphi (b(m)))\in \bigoplus _{A' \ge As,A' \in As + {\mathbb {Z}}\Delta }(N*B_s)_{A's}^{\emptyset }$ . If $A'\in As + {\mathbb {Z}}\Delta $ satisfies $A'\ge As$ , since $s\colon As + {\mathbb {Z}}\Delta \to A + {\mathbb {Z}}\Delta $ preserves the order, we get $A's \ge A$ . Hence, $b'(\varphi (b(m)))\in \bigoplus _{A' \ge A}(N*B_s)_{A'}^{\emptyset }$ . Therefore, $\psi $ is a morphism in $\widetilde {\mathcal {K}}'(S_0)$ .

However, assume that $\psi $ is a morphism in $\widetilde {\mathcal {K}}'(S_0)$ . Consider the map $\Phi \colon N\otimes _{R^s}R\to N$ defined by $n\otimes f\mapsto nf$ . Then, $\Phi (a'(n)) = n\alpha _s$ and $\Phi (b'(n)) = 0$ . Therefore, $\Phi ((N*B_s)_{A}^{\emptyset }) = \Phi (a'(N_{A}^{\emptyset }) + b'(N_{As}^{\emptyset }))\subset N_{A}^{\emptyset }$ . Let $m\in M_{A}^{\emptyset }$ . Then applying $\Phi $ to $\psi (m) = (\alpha _s)_{A}^{-1}a'(\varphi (a(m))) - (\alpha _s)_{A}^{-1}b'(\varphi (b(m)))$ , we get $(\alpha _s)_{A}^{-1}\varphi (a(m))\alpha _s\in \bigoplus _{A'\in A + {\mathbb {Z}}\Delta ,A'\ge A}M_{A'}^{\emptyset }$ . Hence, $\varphi (a(M_{A}^{\emptyset }))\subset \bigoplus _{A'\ge A}N_{A'}^{\emptyset }$ . Similarly, using $N\otimes _{R^s}R\to N$ defined by $n\otimes f\mapsto ns(f)$ , we get $\varphi (b(M_{As}^{\emptyset }))\subset \bigoplus _{A'\ge A}N_{A'}^{\emptyset }$ . Since $(M*B_s)_{A}^{\emptyset } = a(M_{A}^{\emptyset }) + b(M_{As}^{\emptyset })$ , $\varphi $ is a morphism in $\widetilde {\mathcal {K}}'(S_0)$ .

Proof. We have

$$ \begin{align*} (M*B_s)^{(\Omega)} & = M^{\alpha}*B_s\cap \bigoplus_{A\in \Omega}(M*B_s)_{A}^{\emptyset}\\ & = M^{\alpha}*B_s\cap \left(\bigoplus_{A\in \Omega}M_A^{\emptyset}\otimes (B_s)^{\emptyset}_e\oplus \bigoplus_{A\in \Omega}M_{As}^{\emptyset}\otimes (B_s)^{\emptyset}_s\right). \end{align*} $$

If $\Omega s = \Omega $ , then in the second direct sum, we can replace $As$ with A. Therefore,

$$ \begin{align*} (M*B_s)^{(\Omega)} & = M^{\alpha}*B_s\cap \left(\bigoplus_{A\in \Omega}M_A^{\emptyset}\otimes (B_s)^{\emptyset}_e\oplus \bigoplus_{A\in \Omega}M_{A}^{\emptyset}\otimes (B_s)^{\emptyset}_s\right)\\ & = M^{\alpha}*B_s\cap \bigoplus_{A\in \Omega}M_A^{\emptyset}\otimes (B_s)^{\emptyset}\\ & = (M^{\alpha}\cap \bigoplus_{A\in \Omega}M_A^{\emptyset})\otimes B_s\\ & = M^{(\Omega)}* B_s. \end{align*} $$

Assume that $\Omega s\ne \Omega $ and take $A\in \Omega $ . Set $\beta ^\vee = (\alpha _s^\vee )_A$ . Then the assumption $\Omega s\ne \Omega $ tells us that $\beta ^\vee \ne \pm \alpha ^\vee $ . Hence, $\beta ^\vee $ is invertible in $S^\alpha $ . The element $s_\alpha (\beta ^\vee )$ is also invertible.

Let $\delta \in X^\vee _{{\mathbb {K}}}$ such that $\langle \alpha ,\delta \rangle = 1$ . For $m \in M^{(\Omega )}$ , there exists $m_1\in \bigoplus _{A'\in A + {\mathbb {Z}} \alpha }M_{A'}^{\emptyset }$ and $m_2\in \bigoplus _{A'\in s_{(\alpha ,0)}A + {\mathbb {Z}} \alpha }M_{A'}^{\emptyset }$ such that $m = m_1 + m_2$ . For each $f\in R$ , $m_1f = f_Am_1$ and $m_2f = s_\alpha (f_A)m_2$ . By calculations using this, we have

$$\begin{align*}\left(\frac{1}{\beta^\vee}m + \frac{\langle\alpha,\beta^\vee\rangle}{\beta s_{\alpha}(\beta^\vee)}(\delta m - m\delta^A)\right)\alpha_s^\vee = m. \end{align*}$$

Hence, the right action of $\alpha _s^\vee $ is invertible.

Therefore, we have $(M*B_s)^{(\Omega )} = (M*B_s[\alpha _s^{-1}])^{(\Omega )}$ where $B_s[(\alpha _s^\vee )^{-1}] = B_s\otimes _{R}R[(\alpha _s^\vee )^{-1}]$ . Since $B_s[(\alpha _s^\vee )^{-1}] = R[(\alpha _s^\vee )^{-1}](\delta _s\otimes 1 - 1\otimes s(\delta _s))\oplus R[(\alpha _s^\vee )^{-1}](\delta _s\otimes 1 - 1\otimes \delta _s)$ with $R[(\alpha _s^\vee )^{-1}](\delta _s\otimes 1 - 1\otimes s(\delta _s))\subset (B_s)_e^{\emptyset }$ and $R[(\alpha _s^\vee )^{-1}](\delta _s\otimes 1 - 1\otimes \delta _s)\subset (B_s)_s^{\emptyset }$ , the definition of $(M*B_s)^{(\Omega )}$ implies (b).

(2) Fix $\alpha \in \Delta $ . By replacing $M^{\alpha }$ with an object which is isomorphic to $M^{\alpha }$ , we may assume $M^{\alpha } = \bigoplus _{\Omega \in W_{\alpha ,{\mathrm {aff}}}'\backslash \mathcal {A}}(\bigoplus _{A\in \Omega }M_A^{\emptyset }\cap M^{\alpha })$ . Let $\{\Omega _i\}$ be a complete set of representatives for $\{\Omega \in W^{\prime }_{\alpha ,{\mathrm {aff}}}\backslash \mathcal {A}\mid \Omega s\ne \Omega \}/\{e,s\}$ . Then, we have

$$ \begin{align*} \bigoplus_{\Omega\in W^{\prime}_{\alpha,{\mathrm{aff}}}\backslash \mathcal{A}}(M^{\alpha}*B_s)^{(\Omega)} & = \bigoplus_{\Omega s = \Omega}(M*B_s)^{(\Omega)} \oplus \bigoplus_{i}((M*B_s)^{(\Omega_i)}\oplus (M*B_s)^{(\Omega_i s)})\\ & = \bigoplus_{\Omega s = \Omega}M^{(\Omega)}*B_s \oplus \bigoplus_{i}((M*B_s)^{(\Omega_i)}\oplus (M*B_s)^{(\Omega_i s)}). \end{align*} $$

From the argument of the proof of (1)(b), we have $M^{(\Omega _i)}\otimes (\delta _s \otimes 1 - 1\otimes s(\delta _s))\oplus M^{(\Omega _i)}\otimes (\delta _s\otimes 1 - 1\otimes \delta _s) = M^{(\Omega _i)}\otimes B_s[\alpha _s^{-1}] = M^{(\Omega _i)}\otimes B_s$ . Therefore, by (1)(b), $((M*B_s)^{(\Omega _i)}\oplus (M*B_s)^{(\Omega _i s)}) = M^{(\Omega _i)}\otimes B_s\oplus M^{(\Omega _i s)}\otimes B_s$ . Hence,

$$ \begin{align*} \bigoplus_{\Omega\in W^{\prime}_{\alpha,{\mathrm{aff}}}\backslash \mathcal{A}}(M*B_s)^{(\Omega)} & = \bigoplus_{\Omega s = \Omega}M^{(\Omega)}*B_s \oplus \bigoplus_{i}(M^{(\Omega_i )}*B_s\oplus M^{(\Omega_i s)}*B_s)\\ & = \bigoplus_{\Omega\in W^{\prime}_{\alpha,{\mathrm{aff}}}\backslash \mathcal{A}}M^{(\Omega)}*B_s\\ & = M^{\alpha}*B_s. \end{align*} $$

Hence, $M*B_s$ satisfies (LE).

Proof. We have $(M*B_s)_{I}^{\emptyset } = \bigoplus _{A\in I}M_{A}^{\emptyset }\otimes (B_s)_{e}^{\emptyset }\oplus \bigoplus _{A\in I}M_{As}^{\emptyset }\otimes (B_s)_{s}^{\emptyset }$ . Since I is s-invariant, $\bigoplus _{A\in I}M_{As}^{\emptyset }\otimes (B_s)_{s}^{\emptyset } = \bigoplus _{A\in I}M_{A}^{\emptyset }\otimes (B_s)_s^{\emptyset }$ . Hence, $(M*B_s)_{I}^{\emptyset } = \bigoplus _{A\in I}M_{A}^{\emptyset }\otimes ((B_s)_{e}^{\emptyset }\oplus (B_s)_{s}^{\emptyset }) = \bigoplus _{A\in I}M_{A}^{\emptyset }\otimes B_{s}^{\emptyset } = M_I^{\emptyset }\otimes B_{s}^{\emptyset }$ .

Proof. First we note that $I\setminus \{A,As\} = I\setminus J$ and $I\setminus \{As\} = (I\setminus J)\cup \{A'\in \mathcal {A}\mid A'\ge A\}$ are closed. We have an exact sequence

(2.2) $$ \begin{align} 0\to N_{I\setminus\{As\}}/N_{I\setminus\{A,As\}}\to N_{I}/N_{I\setminus\{A,As\}}\to N_{I}/N_{I\setminus\{As\}}\to 0. \end{align} $$

We have $(N_I/N_{I\setminus \{A,As\}})^{\emptyset } = N_{A}^{\emptyset }\oplus N_{As}^{\emptyset }$ and we have the following commutative diagram:

Therefore, $N_{I\setminus \{As\}}/N_{I\setminus \{A,As\}} = (N_I/N_{I\setminus \{A,As\}})\cap (N_A^{\emptyset }\oplus 0)$ .

Set $L = N_I/N_{I\setminus \{A,As\}}$ . By Lemma 2.25, $L\simeq M_{\{A,As\}}\otimes _{R^s}R(1)$ . We have $L^{\emptyset } = L_{A}^{\emptyset }\oplus L_{As}^{\emptyset }$ . We determine $L\cap (L_{A}^{\emptyset }\oplus 0)$ .

By (2.1), we have $L_{A}^{\emptyset }\simeq M_{A}^{\emptyset }\oplus M_ {As}^{\emptyset }$ and $L_{As}^{\emptyset }\simeq M_{As}^{\emptyset }\oplus M_ {A}^{\emptyset }$ . In general, we write $m_{A'}$ for the image of $m\in M$ in $M_{A'}^{\emptyset }$ , where $A'\in \mathcal {A}$ . The image of $m_1\otimes 1 + m_2\otimes \delta \in L = M_{\{A,As\}}\otimes _{R^s}R(1)$ in each direct summand is

$$ \begin{gather*} m_{1,A} + m_{2,A}\delta\in M_{A}^{\emptyset}\subset L_{A}^{\emptyset},\\ m_{1,As} + m_{2,As}s(\delta)\in M_{As}^{\emptyset}\subset L_{A}^{\emptyset},\\ m_{1,As} + m_{2,As}\delta\in M_{As}^{\emptyset}\subset L_{As}^{\emptyset},\\ m_{1,A} + m_{2,A}s(\delta)\in M_{A}^{\emptyset}\subset L_{As}^{\emptyset}. \end{gather*} $$

Therefore, $m_1\otimes 1 + m_2\otimes \delta \in L_{A}^{\emptyset }$ if and only if $m_{1,As} + m_{2,As}\delta = 0$ , $m_{1,A} + m_{2,A}s(\delta ) = 0$ . Note that $m_{2,As}\delta = (s(\delta ))_Am_{2,As}$ and $m_{2,A}s(\delta ) = (s(\delta ))_{A}m_{2,A}$ . Therefore, $(m_1 + (s(\delta ))_Am_2)_{A'} = 0$ for $A' = A,As$ . Hence, $m_1 + (s(\delta ))_Am_2 = 0$ . Therefore, we have

$$\begin{align*}L\cap (L_{A}^{\emptyset}\oplus 0) = \{m_2\otimes \delta - (s(\delta))_Am_2\otimes 1\mid m_2\in M_{\{A,As\}}\}(1) \end{align*}$$

which is isomorphic to $M_{\{A,As\}}(-1)$ .

The map $L\simeq M_{\{A,As\}}\otimes _{R^s}R(1)\ni m\otimes f\mapsto (s(f))_Am\in M_{\{A,As\}}(1)$ is surjective and, by the above argument, the kernel is $L\cap (L_{A}^{\emptyset }\oplus 0)\simeq N_{I\setminus \{As\}}/N_{I\setminus \{A,As\}}$ . Therefore, by the exact sequence (2.2), we have $N_{I}/N_{I\setminus \{As\}}\simeq M_{\{A,As\}}(1)$ .

Proof. Set $N = M*B_s\in \widetilde {\mathcal {K}}'$ .

(1) Put $I_1 = \{A'\in \mathcal {A}\mid A'\ge As\}$ . This is s-invariant. Since I is closed and contains $As$ , we have $I_1\subset I$ . Hence, $N_{I_1}/N_{I_1\setminus \{As\}}\hookrightarrow N_{I}/N_{I\setminus \{As\}}$ . By Lemma 2.26, we have $N_{I_1}/N_{I_1\setminus \{As\}}\simeq M_{\{A,As\}}(-1)$ . Hence, we have $M_{\{A,As\}}(-1)\hookrightarrow N_{I}/N_{I\setminus \{As\}}$ .

Let $\nu \in X^\vee _{{\mathbb {K}}}$ and write $S_{(\nu )}$ for the localization at the prime ideal $(\nu )$ . Set $N_{(\nu )} = S_{(\nu )}\otimes _S N$ . The algebra $S_{(\nu )}$ is an $S^{\alpha }$ -algebra for a certain $\alpha \in \Delta $ . Therefore, $N_{(\nu )}$ satisfies (S). Hence, the above embedding $(M_{(\nu )})_{\{A,As\}}(-1)\hookrightarrow (N_{(\nu )})_{I}/(N_{(\nu )})_{I\setminus \{As\}}$ is an isomorphism. Since M admits a standard filtration, $M_{\{A,As\}}$ is graded free as an S-module. Therefore, $M_{\{A,As\}}(-1) = \bigcap _{\nu \in X_{{\mathbb {K}}}}(S_{(\nu )}\otimes _S M_{\{A,As\}}(-1)) = \bigcap _{\nu \in X_{{\mathbb {K}}}}((N_{(\nu )})_{I}/(N_{(\nu )})_{I\setminus \{As\}})\supset N_{I}/N_{I\setminus \{As\}}$ . We get the lemma.

(2) First, we prove that there exists an embedding $(M*B_s)_I/(M*B_s)_{I\setminus J}\hookrightarrow M_{\{A,As\}}(-1)$ . We may assume $J = \{A'\in \mathcal {A}\mid A'\le A\}$ since $I\setminus J$ is not changed. Then, J is s-invariant. Put $I_1 = I\cup Is$ . Then $I_1$ is an s-invariant closed subset and $I_1\cap J = (I\cap J)\cup (Is\cap J) = (I\cap J)\cup (I\cap J)s = \{A,As\}$ . We have $I_1\setminus \{As\}\supset I$ . Hence, we have an embedding $N_I/N_{I\setminus J}\hookrightarrow N_{I_1\setminus \{As\}}/N_{I_1\setminus \{A,As\}}\simeq M_{\{A,As\}(-1)}$ . We prove that this embedding is surjective.

First, we assume that ${\mathbb {K}}$ is a field. Take a sequence of closed subsets $I_0 \subset \dotsm \subset I_r$ such that $\#(I_{i + 1}\setminus I_i) = 1$ , $N_{I_0} = 0$ , $N_{I_r} = N$ , and there exists $k = 1,\dots ,r$ such that $I_{k - 1}\cap \operatorname {\mathrm {supp}}_{\mathcal {A}}(N) = I\cap \operatorname {\mathrm {supp}}_{\mathcal {A}}(N)$ and $I_k = I_{k - 1}\cup \{A\}$ (Lemma 2.14). Let $A_i\in \mathcal {A}$ such that $I_i = I_{i - 1}\cup \{A_i\}$ . Since $N_{I_i}$ is a filtration of N, for each l, the l-th graded piece $N^{l}$ satisfies $\dim _{{\mathbb {K}}} N^{l} = \sum _i (N_{I_{i}}/N_{I_{i - 1}})^{l}$ . By the existence of an embedding we have proved, $\dim _{{\mathbb {K}}}(N_{I_{i}}/N_{I_{i - 1}})^{l}\le \dim _{{\mathbb {K}}}(M_{\{A_i,A_is\}})^{l + \varepsilon (A_i)}$ , where $\varepsilon (A_i) = 1$ if $A_is> A_i$ and $\varepsilon (A_i) = -1$ otherwise. We have

$$ \begin{align*} & \dim_{{\mathbb{K}}}(M_{\{A_i,A_is\}})^{l + \varepsilon(A_i)}\\ & = \sum_i(\dim_{{\mathbb{K}}}(M_{\{A_i\}})^{l+ \varepsilon(A_i)} + \dim_{{\mathbb{K}}}(M_{\{A_is\}})^{l + \varepsilon(A_i)})\\ & = \sum_{A_is> A_i}\dim_{{\mathbb{K}}}(M_{\{A_i\}}^{l + 1}) + \sum_{A_is > A_i}\dim_{{\mathbb{K}}}(M_{\{A_is\}}^{l + 1})\\ & \quad + \sum_{A_is < A_i}\dim_{{\mathbb{K}}}(M_{\{A_i\}}^{l - 1})+ \sum_{A_is < A_i}\dim_{{\mathbb{K}}}(M_{\{A_is\}}^{l - 1}). \end{align*} $$

By replacing $A_i$ with $A_is$ in the second and fourth sum, we have

$$ \begin{align*} & \sum_i(\dim_{{\mathbb{K}}}(M_{\{A_i\}})^{l+ \varepsilon(A_i)} + \dim_{{\mathbb{K}}}(M_{\{A_is\}})^{l + \varepsilon(A_i)})\\[5.5pt] & = \sum_{A_is> A_i}\dim_{{\mathbb{K}}}(M_{\{A_i\}}^{l + 1}) + \sum_{A_is < A_i}\dim_{{\mathbb{K}}}(M_{\{A_i\}}^{l + 1}))\\[5.5pt] & \quad + \sum_{A_is < A_i}\dim_{{\mathbb{K}}}(M_{\{A_i\}}^{l - 1}) + \sum_{A_is> A_i}\dim_{{\mathbb{K}}}(M_{\{A_i\}}^{l - 1})\\[5.5pt] & = \sum_i (\dim_{{\mathbb{K}}} M_{\{A_i\}}^{l + 1} + \dim_{{\mathbb{K}}} M_{\{A_i\}}^{l - 1}). \end{align*} $$

Since $\{M_{\{A_i\}}\}$ are subquotients of a filtration $\{M_{I_i}\}$ on M, we have $\sum _i\dim _{{\mathbb {K}}} (M_{\{A_i\}})^{l'} = \dim _{{\mathbb {K}}} M^{l'}$ . Hence, $\sum _i (\dim _{{\mathbb {K}}} M_{\{A_i\}}^{l + 1} + \dim _{{\mathbb {K}}} M_{\{A_i\}}^{l - 1}) = \dim _{{\mathbb {K}}} M^{l + 1} + \dim _{{\mathbb {K}}} M^{l - 1}$ .

However, since $N = M*B_s = M\otimes _{R^s}R(1) = M(1)\otimes 1\oplus M(1)\otimes \delta _s$ where $\delta _s$ satisfies $\langle \delta _s,\alpha _s^\vee \rangle = 1$ , we have $\dim _{{\mathbb {K}}} N^l = \dim _{{\mathbb {K}}} M^{l + 1} + \dim _{{\mathbb {K}}} M^{l - 1}$ . Therefore, we get

$$\begin{align*}\dim_{{\mathbb{K}}} N^l = \sum_i\dim_{{\mathbb{K}}}(N_{I_{i}}/N_{I_{i - 1}})^{l}\le \sum_i\dim_{{\mathbb{K}}}(M_{\{A_i,A_is\}})^{l + \varepsilon(A_i)} = \dim_{{\mathbb{K}}} N^l. \end{align*}$$

Hence, the embedding has to be a bijection.

Now, let ${\mathbb {K}}$ be a general Noetherian integral domain. Assume that we can prove that $(N_{I_i}/N_{I_{i - 1}})\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\simeq (M_{\{A_i,A_is\}}(\varepsilon (A_i)))\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ for each maximal ideal $\mathfrak {m}$ in ${\mathbb {K}}$ . Since $M_{\{A_i,A_is\}}^l$ is finitely generated as a ${\mathbb {K}}$ -module, by Nakayama’s lemma, $(N_{I_i}/N_{I_{i - 1}})^l_{\mathfrak {m}}\to (M_{\{A_i,A_{i}s\}})^{l + \varepsilon (A_i)}_{\mathfrak {m}}$ is surjective, where $(\bullet )_{\mathfrak {m}}$ means the localization at $\mathfrak {m}$ . Since this is true for any maximal ideal $\mathfrak {m}$ , the map $(N_{I_i}/N_{I_{i - 1}})^l\to M^l_{\{A_i,A_{i}s\}}$ is surjective for any $l\in {\mathbb {Z}}$ . Hence, it is an isomorphism. Therefore, it is sufficient to prove $(N_{I_i}/N_{I_{i - 1}})\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\simeq (M_{\{A_i,A_is\}}(\varepsilon (A_i)))\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ . In the rest of the proof, we omit the grading.

To prove this, we need some properties on the base change to ${\mathbb {K}}/\mathfrak {m}$ . Let $L\in \widetilde {\mathcal {K}}'$ . Then, $L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ is an $(S/\mathfrak {m}S,R/\mathfrak {m}R)$ -bimodule and we have $S^{\emptyset }\otimes _{S}L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}) \simeq \bigoplus _{A\in \mathcal {A}}L^{\emptyset }_{A}\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ . Therefore, it defines an object in $\widetilde {\mathcal {K}}^{\prime }_{{\mathbb {K}}/\mathfrak {m}}$ . Here, the suffix ${\mathbb {K}}/\mathfrak {m}$ means that, in the definition of $\widetilde {\mathcal {K}}'$ , we replace ${\mathbb {K}}$ with ${\mathbb {K}}/\mathfrak {m}$ . We also have $B\otimes _{{\mathbb {K}}}{\mathbb {K}}/\mathfrak {m}\in {\mathcal {S}\mathrm {Bimod}}_{{\mathbb {K}}/\mathfrak {m}}$ (the meaning of the suffix ${\mathbb {K}}/\mathfrak {m}$ is the same as above) and we have $(M*B)\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\simeq (M\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))*(B\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))$ . Let $K\subset \mathcal {A}$ be a closed subset. Then, we have a map $L_K\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ . Since $\operatorname {\mathrm {supp}}_{\mathcal {A}}(L_K\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))\subset K$ , the image of this homomorphism is in $(L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_K$ . Hence, we get a map $L_K\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to (L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_K$ . We claim:

1. (a) The map is surjective.

2. (b) If $L/L_K$ is graded free, then this map is an isomorphism.

We prove (a) first. By the exact sequence $0\to L_K\to L\to L/L_K\to 0$ , we have an exact sequence $L_K\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to (L/L_K)\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to 0$ . Since $\operatorname {\mathrm {supp}}_{\mathcal {A}}((L/L_K)\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))\subset \mathcal {A}\setminus K$ , the map $L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to (L/L_K)\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ factors through $L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to (L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))/(L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_K$ . Hence, $(L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_K\subset \operatorname {\mathrm {Ker}} (L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to (L/L_K)\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})) = \operatorname {\mathrm {Im}}(L_K\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))$ . Therefore, we get (a). If $L/L_K$ is graded free, then $L/L_K$ is free as a ${\mathbb {K}}$ -module. Hence, $L_K\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\to L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ is injective. Therefore, we have (b).

In particular, if L satisfies (S), then $L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ also satisfies (S). Indeed, let $K_1,K_2$ be closed subsets. Then, we have a commutative diagram

Here, the horizontal maps are surjective by (a) in the above, and the left vertical map is surjective since L satisfies (S). Hence, the right vertical map is surjective and it means that $L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ satisfies (S).

We also have that if L satisfies (LE), then $L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ satisfies (LE). Let $\alpha \in \Delta $ and decompose $L^{\alpha }$ as $L^{\alpha } \simeq \bigoplus _{\Omega \in W^{\prime }_{\alpha }\backslash \mathcal {A}}L^{(\Omega )}$ such that $\operatorname {\mathrm {supp}} L^{(\Omega )}\subset \Omega $ . Then, $(L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))^{\alpha }\simeq \bigoplus _{\Omega \in W^{\prime }_{\alpha }\backslash \mathcal {A}}L^{(\Omega )}\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ and it gives a desired decomposition in (LE).

Let $K_1\subset K_2\subset \mathcal {A}$ be closed subsets and suppose that $L\in \widetilde {\mathcal {K}}_{\Delta }$ . Since $L\in \widetilde {\mathcal {K}}_{\Delta }$ , $L/L_{K_1}$ and $L/L_{K_2}$ are both graded free. Hence, $L_{K_1}\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\simeq (L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_{K_1}\subset (L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_{K_2}\simeq L_{K_2}\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})$ . By the right exactness of the tensor product, we have $(L_{K_2}/L_{K_1})\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\simeq (L_{K_2}\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))/(L_{K_1}\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))\simeq (L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_{K_2}/(L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_{K_1}$ . Therefore, for any locally closed subset $K\subset \mathcal {A}$ , we have $L_K\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\simeq (L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_K$ . In particular, $L\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\in \widetilde {\mathcal {K}}_{\Delta ,{\mathbb {K}}/\mathfrak {m}}$ .

We return to the proof of the lemma. We have $M\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\in \widetilde {\mathcal {K}}_{\Delta ,{\mathbb {K}}/\mathfrak {m}}$ as we have proved. We have $M_{\{A_i,A_is\}}\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\simeq (M\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_{\{A_i,A_is\}}$ . Hence, we have the following commutative diagram:

Note that the bottom homomorphism is an isomorphism since the lemma is proved if ${\mathbb {K}}$ is a field.

We prove that the left vertical map is an isomorphism by backward induction on i. By inductive hypothesis, $N_{I_{i'}}/N_{I_{i' - 1}}\simeq M_{\{A_{i'},A_{i'}s\}}$ for any $i'> i$ and, in particular, it is graded free. Hence, $N/N_{I_i}$ is also graded free. Therefore, we have $N_{I_i}\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m})\simeq (N\otimes _{{\mathbb {K}}}({\mathbb {K}}/\mathfrak {m}))_{I_i}$ . Now we get the desired result by applying the five lemmas to the following commutative diagram with exact columns:

Proof. Take $A_0,A_1\in \mathcal {A}$ such that $\operatorname {\mathrm {supp}}_{\mathcal {A}}N\subset [A_0,A_1]$ . Replacing $I_1$ with $I_1\cap \{A\in \mathcal {A}\mid A\ge A_0\}$ and $I_2$ with $I_2\cup \{A\in \mathcal {A}\mid A\not \le A_1\}$ , we may assume $I_1\setminus I_2$ is finite. We can take a sequence of closed subsets $I_2 = I^{\prime }_0 \subset I^{\prime }_1\subset \dots \subset I^{\prime }_r = I_1$ such that $\#(I^{\prime }_{i}\setminus I^{\prime }_{i - 1}) = 1$ . Let $A_i$ such that $I^{\prime }_{i} = I^{\prime }_{i - 1}\cup \{A_i\}$ . Then by Lemma 2.27, $N_{I^{\prime }_i}/N_{I^{\prime }_{i - 1}}\simeq M_{\{A_i,A_is\}}(\varepsilon (A_i))$ , where $\varepsilon (A_i)\in \{\pm 1\}$ is as in the proof of Lemma 2.27. In particular, this is graded free and therefore, $M_{I_1}/M_{I_2} = M_{I^{\prime }_r}/M_{I^{\prime }_0}$ is also graded free.

Proof of Proposition 2.24

Set $N = M*B_s$ . We prove that N satisfies (S). Let $I_1,I_2$ be closed subsets, and we prove the surjectivity of $N_{I_1}/N_{I_1\cap I_2}\hookrightarrow N_{I_1\cup I_2}/N_{I_2}$ . For each $\nu \in X^\vee _{{\mathbb {K}}}$ , let $S_{(\nu )}$ be the localization at the prime ideal $(\nu )$ . Then, $S_{(\nu )}$ is an $S^{\alpha }$ -algebra for some $\alpha \in \Delta ^{+}$ . Since $N_{(\nu )} = S_{(\nu )}\otimes _{S}N$ satisfies (LE), $N_{(\nu )}$ satisfies (S) by Lemma 2.10. Hence, this embedding is surjective after applying $S_{(\nu )}\otimes _S$ . Put $L_{(\nu )} = S_{(\nu )}\otimes _{S}L$ for a left S-module L. Since $N_{I_1}/N_{I_1\cap I_2}$ is graded free by Lemma 2.28, we have $N_{I_1}/N_{I_1\cap I_2} = \bigcap _{\nu } (N_{I_1}/N_{I_1\cap I_2})_{(\nu )}$ . Hence, $N_{I_1}/N_{I_1\cap I_2} = \bigcap _{\nu } (N_{I_1}/N_{I_1\cap I_2})_{(\nu )} \simeq \bigcap _{\nu } (N_{I_1\cup I_2}/N_{I_2})_{(\nu )}\supset N_{I_1\cup I_2}/N_{I_2}$ . We get the surjectivity.

Now $N_{\{A\}}$ is well-defined and isomorphic to $M_{\{A,As\}}(\varepsilon (A))$ , where $\varepsilon (A)\in \{\pm 1\}$ is as in the proof of Lemma 2.27. Hence, $N_{\{A\}}$ is graded free; namely, N admits a standard filtration.

Proof. Let $M\in \widetilde {\mathcal {K}}_P$ and $s\in S_{\mathrm {aff}}$ . We prove $M* B_s\in \widetilde {\mathcal {K}}_{P}$ . We have already proved that $M*B_s\in \widetilde {\mathcal {K}}_{\Delta }$ .

Assume that a sequence $M_1\to M_2\to M_3$ in $\widetilde {\mathcal {K}}_{\Delta }$ satisfies (ES). By Lemma 2.17, $0\to (M_1)_{\{A,As\}}\to (M_2)_{\{A,As\}}\to (M_3)_{\{A,As\}}\to 0$ is also exact for any $A\in \mathcal {A}$ . Hence, $0\to (M_1*B_s)_{\{A\}}\to (M_2*B_s)_{\{A\}}\to (M_3*B_s)_{\{A\}}\to 0$ is exact (i.e., $M_1*B_s\to M_2*B_s\to M_3*B_s$ also satisfies (ES)). Since $M\in \widetilde {\mathcal {K}}_P$ , the sequence $0\to \operatorname {\mathrm {Hom}}^{\bullet }(M,M_1*B_s)\to \operatorname {\mathrm {Hom}}^{\bullet }(M,M_2*B_s)\to \operatorname {\mathrm {Hom}}^{\bullet }(M,M_3*B_s)\to 0$ is exact. By Proposition 2.22, $M*B_s\in \widetilde {\mathcal {K}}_P$ .

Proof. Since $S_0$ is flat, we have

$$\begin{align*}S_0\otimes_{S}Q_{\lambda} = \{(z_A)\in S_0^{W_{\mathrm{f}}A_{0}^-}\mid z_A\equiv z_{s_{\alpha,\langle\lambda,\alpha^\vee\rangle}A}\quad\pmod{\alpha^\vee} \text{ for } \alpha\in\Delta \text{ and } A\in W_{\mathrm{f}}A_{\lambda}^-\}. \end{align*}$$

Put $I = \{A'\in \mathcal {A}\mid A'\ge A_{\lambda }^{-}\}$ and $q = (1)_{A\in W_{\mathrm {f}}A_{\lambda }^-}\in S_0\otimes _{S}Q_{\lambda }$ .

Any $(S_0,R)$ -bimodule is regarded as an $S_0\otimes R$ -module. Let $M\in \widetilde {\mathcal {K}}_{\Delta }(S_0)$ and $m\in M$ . According to the decomposition $M^{\emptyset } = \bigoplus _{A\in \mathcal {A}}M_A^{\emptyset }$ , m can be written as $m = \sum _{A\in \mathcal {A}}m_A$ with $m_{A}\in M_{A}^{\emptyset }$ . Consider $S^{W_{\mathrm {f}}} = \{f\in S\mid w(f) = f \text { for all } w\in W_{\mathrm {f}}\}$ . Then, we have the following.

• • For $A\in \mathcal {A}$ and $f\in S^{W_{\mathrm {f}}}$ , $f^A$ does not depend on A.

• • For $f\in S$ , we have $fm = \sum fm_A = \sum m_Af^A$ .

Therefore, we have an embedding $S^{W_{\mathrm {f}}}\hookrightarrow R$ naturally and any M is an $S_0\otimes _{S^{W_{\mathrm {f}}}}R$ -module. Then, we have a map $S\otimes _{S^{W_{\mathrm {f}}}}R\to Q_{\lambda }$ defined by $f\otimes g\mapsto (fg_{w(A_{\lambda }^-)})$ , and by the property of $\mathcal {Z}$ we have remarked, this is an isomorphism. Therefore, $Q_{\lambda }$ is a free $S\otimes _{S^{W_{\mathrm {f}}}}R$ -module of rank one with a basis q. We also remark that $q\in S_{0}\otimes _{S}Q_{\lambda } = (S_{0}\otimes _{S}Q_{\lambda })_{I}$ . Therefore, $\varphi \mapsto \varphi (q)$ gives an embedding

$$\begin{align*}\operatorname{\mathrm{Hom}}^{\bullet}_{\widetilde{\mathcal{K}}_{\Delta}(S_0)}(S_0\otimes_{S}Q_{\lambda},M)\hookrightarrow M_I. \end{align*}$$

Let $m\in M_I$ and $\varphi \colon S_0\otimes _{S}Q_{\lambda }\to M$ be an $(S_0,R)$ -bimodule homomorphism such that $\varphi (q) = m$ . We prove that this is a morphism in $\widetilde {\mathcal {K}}(S_0)$ . Let $A\in W_{\lambda }'A_{\lambda }^-$ . Then, $\varphi ((Q_{\lambda })_{A}^{\emptyset })\subset \bigoplus _{A'\in A + {\mathbb {Z}}\Delta ,A'\in I}M_{A'}^{\emptyset }$ . Therefore, the lemma follows from the following lemma.