2.1 Pro-p-Iwahori Hecke algebra

Let $\mathcal {H}$ be a pro-p-Iwahori Hecke algebra over a commutative ring C [Reference Vignéras20]. We study modules over $\mathcal {H}$ in this paper. In this paper, a module means a right module. The algebra $\mathcal {H}$ is defined with combinatorial data $(W_{\mathrm {aff}},S_{\mathrm {aff}},\Omega ,W,W(1),Z_{\kappa })$ and a parameter $(q,c)$ .

We recall the definitions. The data satisfy the following:

• • $(W_{\mathrm {aff}},S_{\mathrm {aff}})$ is a Coxeter system.

• • $\Omega $ acts on $(W_{\mathrm {aff}},S_{\mathrm {aff}})$ .

• • $W = W_{\mathrm {aff}}\rtimes \Omega $ .

• • $Z_{\kappa }$ is a finite commutative group.

• • The group $W(1)$ is an extension of W by $Z_{\kappa }$ , namely we have an exact sequence $1\to Z_{\kappa }\to W(1)\to W\to 1$ .

The subgroup $Z_{\kappa }$ is normal in $W(1)$ . Hence, the conjugate action of $w\in W(1)$ induces an automorphism of $Z_{\kappa }$ , hence of the group ring $C[Z_{\kappa }]$ . We denote it by $c\mapsto w\cdot c$ .

Let $\mathrm {Ref}(W_{\mathrm {aff}})$ be the set of reflections in $W_{\mathrm {aff}}$ and $\mathrm {Ref}(W_{\mathrm {aff}}(1))$ the inverse image of $\mathrm {Ref}(W_{\mathrm {aff}})$ in $W(1)$ . The parameter $(q,c)$ is maps $q\colon S_{\mathrm {aff}}\to C$ and $c\colon \mathrm {Ref}(W_{\mathrm {aff}}(1))\to C[Z_{\kappa }]$ with the following conditions. (Here, the image of s by q (resp. c) is denoted by $q_{s}$ (resp. $c_{s}$ ).)

• • For $w\in W$ and $s\in S_{\mathrm {aff}}$ , if $wsw^{-1}\in S_{\mathrm {aff}}$ , then $q_{wsw^{-1}} = q_{s}$ .

• • For $w\in W(1)$ and $s\in \mathrm {Ref}(W_{\mathrm {aff}}(1))$ , $c_{wsw^{-1}} = w\cdot c_{s}$ .

• • For $s\in \mathrm {Ref}(W_{\mathrm {aff}}(1))$ and $t\in Z_{\kappa }$ , we have $c_{ts} = tc_{s}$ .

Let $S_{\mathrm {aff}}(1)$ be the inverse image of $S_{\mathrm {aff}}$ in $W(1)$ . For $s\in S_{\mathrm {aff}}(1)$ , we write $q_{s}$ for $q_{\bar {s}}$ , where $\bar {s}\in S_{\mathrm {aff}}$ is the image of s. The length function on $W_{\mathrm {aff}}$ is denoted by $\ell $ , and its inflation to W and $W(1)$ is also denoted by $\ell $ .

The C-algebra $\mathcal {H}$ is a free C-module and has a basis $\{T_{w}\}_{w\in W(1)}$ . The multiplication is given by

• • (Quadratic relations) $T_{s}^{2} = q_{s}T_{s^{2}} + c_{s}T_{s}$ for $s\in S_{\mathrm {aff}}(1)$ .

• • (Braid relations) $T_{vw} = T_{v}T_{w}$ if $\ell (vw) = \ell (v) + \ell (w)$ .

We extend $q\colon S_{\mathrm {aff}}\to C$ to $q\colon W\to C$ as follows. For $w\in W$ , take $s_{1},\dots ,s_{l}$ and $u\in \Omega $ such that $w = s_{1}\dotsm s_{l} u$ and $l = \ell (w)$ . Then put $q_{w} = q_{s_{1}}\dotsm q_{s_{l}}$ . From the definition, we have $q_{w^{-1}} = q_{w}$ . We also put $q_{w} = q_{\overline {w}}$ for $w\in W(1)$ with the image $\overline {w}$ in W.

2.2 The data from a group

Let F be a non-Archimedean local field, $\kappa $ its residue field, p its residue characteristic and G a connected reductive group over F. We can get the data in the previous subsection from G as follows. See [Reference Vignéras20], especially 3.9 and 4.2 for the details.

Fix a maximal split torus S, and denote the centralizer of S in G by Z. Let $Z^{0}$ be the unique parahoric subgroup of Z and $Z(1)$ its pro-p radical. Then the group $W(1)$ (resp. W) is defined by $W(1) = N_{G}(Z)/Z(1)$ (resp. $W = N_{G}(Z)/Z^{0}$ ), where $N_{G}(Z)$ is the normalizer of Z in G. We also let $Z_{\kappa } = Z^{0}/Z(1)$ . Let $G^{\prime }$ be the group generated by the unipotent radical of parabolic subgroups [Reference Abe, Henniart, Herzig and Vignéras5, II.1] and $W_{\mathrm {aff}}$ the image of $G^{\prime }\cap N_{G}(Z)$ in W. Then this is a Coxeter group. Fix a set of simple reflections $S_{\mathrm {aff}}$ . The group W has the natural length function, and let $\Omega $ be the set of length zero elements in W. Then we get the data $(W_{\mathrm {aff}},S_{\mathrm {aff}},\Omega ,W,W(1),Z_{\kappa })$ .

Consider the apartment attached to S and an alcove surrounded by the hyperplanes fixed by $S_{\mathrm {aff}}$ . Let $I(1)$ be the pro-p-Iwahori subgroup attached to this alcove. Then with $q_{s} = \#(I(1)\widetilde {s}I(1)/I(1))$ for $s\in S_{\mathrm {aff}}$ with a lift $\widetilde {s}\in N_{G}(Z)$ and suitable $c_{s}$ , the algebra $\mathcal {H}$ is isomorphic to the Hecke algebra attached to $(G,I(1))$ [Reference Vignéras20, Proposition 4.4].

When the data come from the group G, let $W_{\mathrm {aff}}(1)$ be the image of $G^{\prime }\cap N_{G}(Z)$ in $W(1)$ and put $\mathcal {H}_{\mathrm {aff}} = \bigoplus _{w\in W_{\mathrm {aff}}(1)}CT_{w}$ . This is a subalgebra of $\mathcal {H}$ .

In this paper, except Section 4, we assume that the data come from a connected reductive group.

2.3 The root system and the Weyl groups

Let $W_{0} = N_{G}(Z)/Z$ be the finite Weyl group. Then this is a quotient of W. Recall that we have the alcove defining $I(1)$ . Fix a special point $\boldsymbol {x}_{0}$ from the border of this alcove. Then $W_{0}\simeq \operatorname {\mathrm {Stab}}_{W}\boldsymbol {x}_{0}$ , and the inclusion $\operatorname {\mathrm {Stab}}_{W}\boldsymbol {x}_{0}\hookrightarrow W$ is a splitting of the canonical projection $W\to W_{0}$ . Throughout this paper, we fix this special point and regard $W_{0}$ as a subgroup of W. Set $S_{0} = S_{\mathrm {aff}}\cap W_{0}\subset W$ . This is a set of simple reflections in $W_{0}$ . For each $w\in W_{0}$ , we fix a representative $n_{w}\in W(1)$ such that $n_{w_{1}w_{2}} = n_{w_{1}}n_{w_{2}}$ if $\ell (w_{1}w_{2}) = \ell (w_{1}) + \ell (w_{2})$ .

The group $W_{0}$ is the Weyl group of the root system $\Sigma $ attached to $(G,S)$ . Our fixed alcove and special point give a positive system of $\Sigma $ , denoted by $\Sigma ^{+}$ . The set of simple roots is denoted by $\Delta $ . As usual, for $\alpha \in \Delta $ , let $s_{\alpha }\in S_{0}$ be a simple reflection for $\alpha $ .

The kernel of $W(1)\to W_{0}$ (resp. $W\to W_{0}$ ) is denoted by $\Lambda (1)$ (resp. $\Lambda $ ). Then $Z_{\kappa }\subset \Lambda (1)$ , and we have $\Lambda = \Lambda (1)/Z_{\kappa }$ . The group $\Lambda $ (resp. $\Lambda (1)$ ) is isomorphic to $Z/Z^{0}$ (resp. $Z/Z(1)$ ). Any element in $W(1)$ can be uniquely written as $n_{w} \lambda $ , where $w\in W_{0}$ and $\lambda \in \Lambda (1)$ . We have $W = W_{0}\ltimes \Lambda $ .

2.4 The map $\nu $

The group W acts on the apartment attached to S, and the action of $\Lambda $ is by the translation. Since the group of translations of the apartment is $X_{*}(S)\otimes _{\mathbb {Z}}\mathbb {R}$ , we have a group homomorphism $\nu \colon \Lambda \to X_{*}(S)\otimes _{\mathbb {Z}}\mathbb {R}$ . The compositions $\Lambda (1)\to \Lambda \to X_{*}(S)\otimes _{\mathbb {Z}}\mathbb {R}$ and $Z\to \Lambda \to X_{*}(S)\otimes _{\mathbb {Z}}\mathbb {R}$ are also denoted by $\nu $ . The homomorphism $\nu \colon Z\to X_{*}(S)\otimes _{\mathbb {Z}}\mathbb {R}\simeq \operatorname {\mathrm {Hom}}_{\mathbb {Z}}(X^{*}(S),\mathbb {R})$ is characterized by the following: For $t\in S$ and $\chi \in X^{*}(S)$ , we have $\nu (t)(\chi ) = -\operatorname {\mathrm {val}}(\chi (t))$ , where $\operatorname {\mathrm {val}}$ is the normalized valuation of F.

We call $\lambda \in \Lambda (1)$ dominant (resp. anti-dominant) if $\nu (\lambda )$ is dominant (resp. anti-dominant).

Since the group $W_{\mathrm {aff}}$ is a Coxeter system, it has the Bruhat order denoted by $\le $ . For $w_{1},w_{2}\in W$ , we write $w_{1} < w_{2}$ if there exists $u\in \Omega $ such that $w_{1}u,w_{2}u\in W_{\mathrm {aff}}$ and $w_{1}u < w_{2}u$ . Moreover, for $w_{1},w_{2}\in W(1)$ , we write $w_{1} < w_{2}$ if $w_{1}\in W_{\mathrm {aff}}(1)w_{2}$ and $\overline {w}_{1} < \overline {w}_{2}$ , where $\overline {w}_{1},\overline {w}_{2}$ are the image of $w_{1},w_{2}$ in W, respectively. We write $w_{1}\le w_{2}$ if $w_{1} < w_{2}$ or $w_{1} = w_{2}$ .

2.5 Other basis

For $w\in W(1)$ , take $s_{1},\dotsm ,s_{l}\in S_{\mathrm {aff}}(1)$ and $u\in W(1)$ such that $l = \ell (w)$ , $\ell (u) = 0$ and $w = s_{1}\dotsm s_{l} u$ . Set $T_{w}^{*} = (T_{s_{1}} - c_{s_{1}})\dotsm (T_{s_{l}} - c_{s_{l}})T_{u}$ . Then this does not depend on the choice, and $\{T_{w}^{*}\}_{w\in W(1)}$ is a basis of $\mathcal {H}$ . In $\mathcal {H}[q_{s}^{\pm 1}]$ , we have $T_{w}^{*} = q_{w}T_{w^{-1}}^{-1}$ .

For a spherical orientation o, there is a basis $\{E_{o}(w)\}_{w\in W(1)}$ of $\mathcal {H}$ introduced in [Reference Vignéras20, Definition 5.22]. This satisfies the following product formula [Reference Vignéras20, Theorem 5.25].

(2.1) $$ \begin{align} E_{o}(w_{1})E_{o\cdot w_{1}}(w_{2}) = q_{w_{1}w_{2}}^{-1/2}q_{w_{1}}^{1/2}q_{w_{2}}^{1/2}E_{o}(w_{1}w_{2}). \end{align} $$

2.6 Parabolic induction

Since we have a positive system $\Sigma ^{+}$ , we have the minimal parabolic subgroup B with a Levi part Z. In this paper, parabolic subgroups are always standard, namely containing B. Note that such parabolic subgroups correspond to subsets of $\Delta $ .

Let P be a parabolic subgroup. Attached to the Levi part of P containing Z, we have the data $(W_{\mathrm {aff},P},S_{\mathrm {aff},P},\Omega _{P},W_{P},W_{P}(1),Z_{\kappa })$ and the parameters $(q_{P},c_{P})$ . Hence, we have the algebra $\mathcal {H}_{P}$ . The parameter $c_{P}$ is given by the restriction of c; hence, we denote it just by c. The parameter $q_{P}$ is defined as in [Reference Abe3, 4.1].

For the objects attached to this data, we add the suffix P. We have the set of simple roots $\Delta _{P}$ , the root system $\Sigma _{P}$ and its positive system $\Sigma _{P}^{+}$ , the finite Weyl group $W_{0,P}$ , the set of simple reflections $S_{0,P}\subset W_{0,P}$ , the length function $\ell _{P}$ and the base $\{T^{P}_{w}\}_{w\in W_{P}(1)}$ , $\{T^{P*}_{w}\}_{w\in W_{P}(1)}$ and $\{E^{P}_{o}(w)\}_{w\in W_{P}(1)}$ of $\mathcal {H}_{P}$ . Note that we have no $\Lambda _{P}$ , $\Lambda _{P}(1)$ and $Z_{\kappa ,P}$ since they are equal to $\Lambda $ , $\Lambda (1)$ and $Z_{\kappa }$ .

An element $n_{w}\lambda \in W_{P}(1)$ , where $w\in W_{P,0}$ and $\lambda \in \Lambda (1)$ is called P-positive (resp. P-negative) if for any $\alpha \in \Sigma ^{+}\setminus \Sigma ^{+}_{P}$ we have $\langle \alpha ,\nu (\lambda )\rangle \le 0$ (resp. $\langle \alpha ,\nu (\lambda )\rangle \ge 0$ ). Set $\mathcal {H}_{P}^{+} = \bigoplus _{w} CT^{P}_{w}$ , where $w\in W_{P}(1)$ runs P-positive elements, and define $\mathcal {H}_{P}^{-}$ by the similar way. Then these are subalgebras of $\mathcal {H}_{P}$ . The linear maps $j^{\pm }_{P}\colon \mathcal {H}_{P}^{\pm }\to \mathcal {H}$ and $j^{\pm *}_{P}\colon \mathcal {H}_{P}^{\pm }\to \mathcal {H}$ defined by $j^{\pm }_{P}(T_{w}^{P}) = T_{w}$ and $j^{\pm *}_{P}(T_{w}^{P*}) = T_{w}^{*}$ are algebra homomorphisms.

Now, for an $\mathcal {H}_{P}$ -module $\sigma $ , we define the parabolically induced module $I_{P}(\sigma )$ by

$$ \begin{align*} I_{P}(\sigma) = \operatorname{\mathrm{Hom}}_{(\mathcal{H}_{P}^{-},j_{P}^{-*})}(\mathcal{H},\sigma). \end{align*} $$

This satisfies:

• • $I_{P}$ is an exact functor.

• • $I_{P}$ has the left adjoint functor $L_{P}$ . The functor $L_{P}$ is exact.

• • $I_{P}$ has the right adjoint functor $R_{P}$ .

For the existence and explicit descriptions of adjoint functors $L_{P},R_{P}$ , see [Reference Abe4, 5.1].

For parabolic subgroups $P\subset Q$ , we also defines $\mathcal {H}_{P}^{Q\pm }\subset \mathcal {H}_{P}$ and $j_{P}^{Q\pm }\colon \mathcal {H}_{P}^{Q\pm }\to \mathcal {H}_{Q}$ and $j_{P}^{Q\pm *}\colon \mathcal {H}_{P}^{Q\pm }\to \mathcal {H}_{Q}$ . This defines the parabolic induction $I_{P}^{Q}$ from the category of $\mathcal {H}_{P}$ -modules to the category of $\mathcal {H}_{Q}$ -modules.

2.7 Twist by $n_{w_{G}w_{P}}$

For a parabolic subgroup P, let $w_{P}$ be the longest element in $W_{0,P}$ . In particular, $w_{G}$ is the longest element in $W_{0}$ . Let $P^{\prime }$ be a parabolic subgroup corresponding to $-w_{G}(\Delta _{P})$ ; in other words, $P^{\prime } = n_{w_{G}w_{P}}P^{\mathrm {op}}n_{w_{G}w_{P}}^{-1}$ , where $P^{\mathrm {op}}$ is the opposite parabolic subgroup of P with respect to the Levi part of P containing Z. Set $n = n_{w_{G}w_{P}}$ . Then the map $P^{\mathrm {op}}\to P^{\prime }$ defined by $p\mapsto npn^{-1}$ is an isomorphism which preserves the data used to define the pro-p-Iwahori Hecke algebras. Hence, $T^{P}_{w}\mapsto T^{P^{\prime }}_{nwn^{-1}}$ gives an isomorphism $\mathcal {H}_{P}\to \mathcal {H}_{P^{\prime }}$ . This sends $T_{w}^{P*}$ to $T_{nwn^{-1}}^{P^{\prime }*}$ and $E^{P}_{o_{+,P}\cdot v}(w)$ to $E^{P^{\prime }}_{o_{+,P^{\prime }}\cdot nvn^{-1}}(nwn^{-1})$ , where $v\in W_{0,P}$ .

Let $\sigma $ be an $\mathcal {H}_{P}$ -module. Then we define an $\mathcal {H}_{P^{\prime }}$ -module $n_{w_{G}w_{P}}\sigma $ via the pull-back of the above isomorphism: $(n_{w_{G}w_{P}}\sigma )(T^{P^{\prime }}_{w}) = \sigma (T^{P}_{n_{w_{G}w_{P}}^{-1}wn_{w_{G}w_{P}}})$ . For an $\mathcal {H}_{P^{\prime }}$ -module $\sigma ^{\prime }$ , we define $n_{w_{G}w_{P}}^{-1}\sigma ^{\prime }$ by $(n_{w_{G}w_{P}}^{-1}\sigma ^{\prime })(T_{w}^{P}) = \sigma ^{\prime }(T_{n_{w_{G}w_{P}}wn_{w_{G}w_{P}}^{-1}}^{P^{\prime }})$ .

2.8 The extension and the generalized Steinberg modules

Let P be the parabolic subgroup and $\sigma $ an $\mathcal {H}_{P}$ -module. For $\alpha \in \Delta $ , let $P_{\alpha }$ be a parabolic subgroup corresponding to $\Delta _{P}\cup \{\alpha \}$ . Then we define $\Delta (\sigma )\subset \Delta $ by

$$ \begin{align*} \Delta(\sigma) = \{\alpha\in\Delta\mid \langle \Delta_{P},\alpha^{\vee}\rangle = 0,\ \sigma(T^{P}_{\lambda}) = 1\text{ for any }\lambda\in W_{\mathrm{aff},P_{\alpha}}(1)\cap \Lambda(1)\}\cup \Delta_{P}. \end{align*} $$

Let $P(\sigma )$ be the parabolic subgroup corresponding to $\Delta (\sigma )$ .

A typical example of the extension is the trivial representation $\boldsymbol {1} = \boldsymbol {1}_{G}$ . This is a one-dimensional $\mathcal {H}$ -module defined by $\boldsymbol {1}(T_{w}) = q_{w}$ , or equivalently $\boldsymbol {1}(T_{w}^{*}) = 1$ . We have $\Delta (\boldsymbol {1}_{P}) = \{\alpha \in \Delta \mid \langle \Delta _{P},\alpha ^{\vee }\rangle = 0\}\cup \Delta _{P}$ , and if Q is a parabolic subgroup between P and $P(\boldsymbol {1}_{P})$ , we have $e_{Q}(\boldsymbol {1}_{P}) = \boldsymbol {1}_{Q}$ .

Let $P(\sigma )\supset P_{0}\supset Q_{1}\supset Q\supset P$ . Then as in [Reference Abe3, 4.5], we have $I^{P_{0}}_{Q_{1}}(e_{Q_{1}}(\sigma ))\subset I^{P_{0}}_{Q}(e_{Q}(\sigma ))$ . Define

$$ \begin{align*} \mathrm{St}_{Q}^{P_{0}}(\sigma) = \operatorname{\mathrm{Coker}}\left(\bigoplus_{Q_{1}\supsetneq Q}I_{Q_{1}}^{P_{0}}(e_{Q_{1}}(\sigma))\to I^{P_{0}}_{Q}(e_{Q}(\sigma))\right). \end{align*} $$

When $P_{0} = G$ , we write $\mathrm {St}_{Q}(\sigma )$ and call it generalized Steinberg modules.

2.9 Supersingular modules

In this subsection, we assume that C is a field of characteristic p. Let $\mathcal {O}$ be a conjugacy class in $W(1)$ which is contained in $\Lambda (1)$ . For a spherical orientation o, set $z_{\mathcal {O}} = \sum _{\lambda \in \mathcal {O}}E_{o}(\lambda )$ . Then this does not depend on o and $z_{\mathcal {O}}\in \mathcal {Z}$ , where $\mathcal {Z}$ is the center of $\mathcal {H}$ [Reference Vignéras18, Theorem 5.1]. The length of $\lambda \in \mathcal {O}$ does not depend on $\lambda $ . We denote it by $\ell (\mathcal {O})$ . For $\lambda \in \Lambda (1)$ and $w\in W(1)$ , we put $w\cdot \lambda = w\lambda w^{-1}$ .

The simple supersingular $\mathcal {H}$ -modules are classified in [Reference Ollivier14, Reference Vignéras18]. We recall their results. Let $W^{\mathrm {aff}}(1)$ be the inverse image of $W_{\mathrm {aff}}$ in $W(1)$ .

Put $\mathcal {H}^{\mathrm {aff}} = \bigoplus _{w\in W^{\mathrm {aff}}(1)}CT_{w}$ . Let $\chi $ be a character of $Z_{\kappa }$ and put $S_{\mathrm {aff},\chi } = \{s\in S_{\mathrm {aff}}\mid \chi (c_{\widetilde {s}}) \ne 0\}$ , where $\widetilde {s}\in W(1)$ is a lift of $s\in S_{\mathrm {aff}}$ . Note that, if $\widetilde {s}^{\prime }$ is another lift, then $\widetilde {s}^{\prime } = t\widetilde {s}$ for some $t\in Z_{\kappa }$ . Hence, $\chi (c_{\widetilde {s}^{\prime }}) = \chi (t)\chi (c_{\widetilde {s}})$ . Therefore, the condition does not depend on a choice of a lift. Let $J\subset S_{\mathrm {aff},\chi }$ . Then the character $\Xi = \Xi _{J,\chi }$ of $\mathcal {H}^{\mathrm {aff}}$ is defined by

$$ \begin{align*} \Xi_{J,\chi}(T_{t}) & = \chi(t) \quad (t\in Z_{\kappa}),\\ \Xi_{J,\chi}(T_{\widetilde{s}}) & = \begin{cases} \chi(c_{\widetilde{s}}) & (s\in S_{\mathrm{aff},\chi}\setminus J)\\ 0 & (s\notin S_{\mathrm{aff},\chi}\setminus J) \end{cases} = \begin{cases} \chi(c_{\widetilde{s}}) & (s\notin J),\\ 0 & (s\in J), \end{cases} \end{align*} $$

where $\widetilde {s}\in W^{\mathrm {aff}}(1)$ is a lift of s and the last equality easily follows from the definition of $S_{\mathrm {aff},\chi }$ . Let $\Omega (1)_{\Xi }$ be the stabilizer of $\Xi $ and V a simple $C[\Omega (1)_{\Xi }]$ -module such that $V|_{Z_{\kappa }}$ is a direct sum of $\chi $ . Put $\mathcal {H}_{\Xi } = \mathcal {H}^{\mathrm {aff}} C[\Omega (1)_{\Xi }]$ . This is a subalgebra of $\mathcal {H}$ . For $X\in \mathcal {H}^{\mathrm {aff}}$ and $Y\in C[\Omega (1)_{\Xi }]$ , we define the action of $XY$ on $\Xi \otimes V$ by $x\otimes y\mapsto xX\otimes yY$ . Then this defines a well-defined action of $\mathcal {H}_{\Xi }$ on $\Xi \otimes V$ . Set $\pi _{\chi ,J,V} = (\Xi \otimes V)\otimes _{\mathcal {H}_{\Xi }}\mathcal {H}$ .

The construction of $\pi _{\chi ,J,V}$ is still valid even if we do not assume that the data come from a group. In Section 4, we do not assume it, and we calculate the extension between the modules constructed as above.

2.10 Simple modules

Definition 2.9. We consider a triple $(P,\sigma ,Q)$ which satisfies the following:

• • P is a parabolic subgroup of G.

• • $\sigma $ is a supersingular finite-dimensional $\mathcal {H}_{P}$ -module.

• • Q is a parabolic subgroup contained in $P(\sigma )$ .

Then we define an $\mathcal {H}$ -module $I(P,\sigma ,Q)$ by

$$ \begin{align*} I(P,\sigma,Q) = I_{P(\sigma)}(\mathrm{St}_{Q}^{P(\sigma)}(\sigma)). \end{align*} $$

3.1 Central character

We prove the following lemma.

To prove this lemma, we calculate the action of the center $\mathcal {Z}$ on simple modules. To do it, we need to calculate the action of $\mathcal {Z}$ on a parabolic induction.

Proof. Since $\Sigma ^{+}\setminus \Sigma _{P}^{+}$ is stable under the action of $W_{0,P}$ , (1) follows from the definition of P-negative.

Let $\varphi \in I_{P}(\sigma ) = \operatorname {\mathrm {Hom}}_{(\mathcal {H}_{P}^{-},j_{P}^{-*})}(\mathcal {H},\sigma )$ . Then for $X\in \mathcal {H}$ , we have

$$ \begin{align*} (\varphi z_{\mathcal{O}})(X) = \varphi(z_{\mathcal{O}} X) = \varphi(Xz_{\mathcal{O}}) \end{align*} $$

since $z_{\mathcal {O}}$ is in the center of $\mathcal {H}$ . Hence, by the definition of $z_{\mathcal {O}}$ , we have

$$ \begin{align*} (\varphi z_{\mathcal{O}})(X) = \sum_{\lambda\in \mathcal{O}}\varphi(X E(\lambda)) = \sum_{i}\sum_{\lambda\in \mathcal{O}_{i}} \varphi(X E(\lambda)) + \sum_{\lambda\in\mathcal{O},\ \text{not }P\text{-negative}}\varphi(XE(\lambda)) \end{align*} $$

We prove the vanishing of the second term.

Let $\lambda \in \mathcal {O}$ , which is not P-negative. Then there exists $\alpha \in \Sigma ^{+}\setminus \Sigma _{P}^{+}$ such that $\langle \alpha ,\nu (\lambda )\rangle < 0$ . Let $\lambda _{P}^{-}$ be as in Proposition 2.2. Then $\langle \alpha ,\nu (\lambda _{P}^{-})\rangle> 0$ . Hence, $\nu (\lambda )$ and $\nu (\lambda _{P}^{-})$ does not belongs to the same closed Weyl chamber. Therefore, we have $E(\lambda )E(\lambda _{P}^{-}) = 0$ in $\mathcal {H}_{C}$ by [Reference Abe4, (2.1), Lemma 2.11]. Hence, by [Reference Abe4, Lemma 2.6],

$$ \begin{align*} \varphi(XE(\lambda)) & = \varphi(X E(\lambda))E^{P}(\lambda_{P}^{-})E^{P}(\lambda_{P}^{-})^{-1}\\ & = \varphi(X E(\lambda)j_{P}^{-*}(E^{P}(\lambda_{P}^{-})))E^{P}(\lambda_{P}^{-})^{-1}\\ & = \varphi(X E(\lambda)E(\lambda_{P}^{-}))E^{P}(\lambda_{P}^{-})^{-1} = 0. \end{align*} $$

If $\lambda \in \mathcal {O}_{i}$ , then $E(\lambda )\in \mathcal {H}_{P}^{-}$ . Hence, we have $E(\lambda ) = j_{P}^{-*}(E^{P}(\lambda ))$ by [Reference Abe4, Lemma 2.6]. Therefore,

$$ \begin{align*} \sum_{\lambda\in \mathcal{O}_{i}}\varphi(XE(\lambda)) = \varphi(X)\sum_{\lambda\in \mathcal{O}_{i}}\sigma(E^{P}(\lambda)) = \varphi(X)\sigma(z^{P}_{\mathcal{O}_{i}}). \end{align*} $$

We get the lemma.

3.2 Reduction to generalized Steinberg modules

By Lemma 3.1, to calculate the extension between $I(P_{1},\sigma _{1},Q_{1})$ and $I(P_{2},\sigma _{2},Q_{2})$ , we may assume $P_{1} = P_{2}$ . We prove the following proposition.

Proposition 3.4. The extension group $\operatorname {\mathrm {Ext}}^{i}_{\mathcal {H}}(I(P,\sigma _{1},Q_{1}),I(P,\sigma ,Q_{2}))$ is isomorphic to

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P(\sigma_{1})\cap P(\sigma_{2})}}(\mathrm{St}^{P(\sigma_{1})\cap P(\sigma_{2})}_{Q_{1}\cap P(\sigma_{2})}(\sigma_{1}),\mathrm{St}^{P(\sigma_{1})\cap P(\sigma_{2})}_{Q_{2}}(\sigma_{2})). \end{align*} $$

if $Q_{2}\subset P(\sigma _{1})$ and $\Delta (\sigma _{1})\subset \Delta _{Q_{1}}\cup \Delta (\sigma _{2})$ . Otherwise, the extension group is zero.

Hence, for the calculation of the extension, it is sufficient to calculate the extensions between generalized Steinberg modules. For an $\mathcal {H}$ -module $\pi $ , set $\pi ^{*} = \operatorname {\mathrm {Hom}}_{C}(\pi ,C)$ . The right $\mathcal {H}$ -module structure on $\pi ^{*}$ is given by $(fX)(v) = f(v\zeta (X))$ for $f\in \pi ^{*}$ , $v\in \pi $ and $X\in \mathcal {H}$ . Here, the anti-involution $\zeta \colon \mathcal {H}\to \mathcal {H}$ is defined by $\zeta (T_{w}) = T_{w^{-1}}$ .

Proof. The exactness of $I_{P}$ and $L_{P}$ implies (1).

Put $P^{\prime } = n_{w_{G}w_{P}}P^{\mathrm {op}}n_{w_{G}w_{P}}^{-1}$ Define the functor $I^{\prime }_{P^{\prime }}$ by

$$ \begin{align*} I^{\prime}_{P^{\prime}}(\sigma^{\prime}) = \operatorname{\mathrm{Hom}}_{(\mathcal{H}_{P^{\prime}}^{-},j_{P^{\prime}}^{-})}(\mathcal{H},\sigma^{\prime}) \end{align*} $$

for an $\mathcal {H}_{P^{\prime }}$ -module $\sigma ^{\prime }$ . Then this has the left adjoint functor $L^{\prime }_{P^{\prime }}$ defined by $L^{\prime }_{P^{\prime }}(\pi ) = \pi \otimes _{(\mathcal {H}_{P^{\prime }}^{-},j_{P^{\prime }}^{-})}\mathcal {H}_{P^{\prime }}$ . This is exact since $\mathcal {H}_{P^{\prime }}$ is a localization of $\mathcal {H}_{P^{\prime }}^{-}$ by Proposition 2.2. Set $\sigma _{\ell - \ell _{P}}(T^{P}_{w}) = (-1)^{\ell (w) - \ell _{P}(w)}\sigma (T_{w})$ [Reference Abe4, 4.1]. Using [Reference Abe2, Proposition 4.2], for an $\mathcal {H}_{P}$ -module $\sigma $ , we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(I_{P}(\sigma),\pi^{*}) & \simeq\operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\pi,I_{P}(\sigma)^{*})\\ & \simeq\operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\pi,I^{\prime}_{P^{\prime}}(n_{w_{G}w_{P}}\sigma^{*}_{\ell - \ell_{P}}))\\ & \simeq \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P}}(n_{w_{G}w_{P}}^{-1}L^{\prime}_{P^{\prime}}(\pi),\sigma^{*}_{\ell - \ell_{P}})\\ & \simeq\operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P}}(\sigma_{\ell - \ell_{P}},n_{w_{G}w_{P}}^{-1}L^{\prime}_{P^{\prime}}(\pi)^{*})\\ & \simeq \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P}}(\sigma,(n_{w_{G}w_{P}}^{-1}L^{\prime}_{P^{\prime}}(\pi))^{*}_{\ell - \ell_{P}}). \end{align*} $$

Put $i = 0$ . Then we get $(n_{w_{G}w_{P}}^{-1}L^{\prime }_{P^{\prime }}(\pi ))^{*}_{\ell - \ell _{P}} \simeq R_{P}(\pi ^{*})$ by $\operatorname {\mathrm {Hom}}_{\mathcal {H}}(I_{P}(\sigma ),\pi ^{*})\simeq \operatorname {\mathrm {Hom}}_{\mathcal {H}_{P}}(\sigma ,R_{P}(\pi ^{*}))$ . Hence, we get (2).

If $\pi $ is finite-dimensional, then $\pi = (\pi ^{*})^{*}$ . Hence, we get $\operatorname {\mathrm {Ext}}^{i}_{\mathcal {H}}(I_{P}(\sigma ),\pi )\simeq \operatorname {\mathrm {Ext}}^{i}_{\mathcal {H}_{P}}(\sigma ,R_{P}(\pi ))$ applying (3) to $\pi ^{*}$ .

Proof of Proposition 3.4

Since $I(P,\sigma _{2},Q_{2})$ is finite-dimensional, we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(I(P,\sigma_{1},Q_{1}),I(P,\sigma_{2},Q_{2})) = \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P(\sigma_{1})}}(\mathrm{St}^{P(\sigma_{1})}_{Q_{1}}(\sigma_{1}),R_{P(\sigma_{1})}(I(P,\sigma_{2},Q_{2}))). \end{align*} $$

We have $R_{P(\sigma _{1})}(I(P,\sigma _{2},Q_{2})) = 0$ if $Q_{2}\not \subset P(\sigma _{1})$ by [Reference Abe4, Theorem 5.20]. If $Q_{2}\subset P(\sigma _{1})$ , then $R_{P(\sigma _{1})}(I(P,\sigma _{2},Q_{2})) = I_{P(\sigma )}(P,\sigma ,Q_{2})$ . Hence, the extension group is isomorphic to

$$ \begin{align*} & \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P(\sigma_{1})}}(\mathrm{St}^{P(\sigma_{1})}_{Q_{1}}(\sigma_{1}),I_{P(\sigma_{1})}(P,\sigma_{2},Q_{2}))\\ & = \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P(\sigma_{1})}}(\mathrm{St}^{P(\sigma_{1})}_{Q_{1}}(\sigma_{1}),I_{P(\sigma_{2})\cap P(\sigma_{1})}^{P(\sigma_{1})}(\mathrm{St}_{Q_{2}}(\sigma_{2})))\\ & = \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P(\sigma_{1})\cap P(\sigma_{2})}}(L_{P(\sigma_{2})\cap P(\sigma_{1})}^{P(\sigma_{1})}(\mathrm{St}^{P(\sigma_{1})}_{Q_{1}}(\sigma_{1})),\mathrm{St}^{P(\sigma_{1})\cap P(\sigma_{2})}_{Q_{2}}(\sigma_{2})) \end{align*} $$

We have $L_{P(\sigma _{2})\cap P(\sigma _{1})}^{P(\sigma _{1})}(\mathrm {St}^{P(\sigma _{1})}_{Q_{1}}(\sigma _{1})) = 0$ if $\Delta (\sigma _{1}) \ne \Delta (Q_{1})\cup \Delta _{P(\sigma _{1})\cap P(\sigma _{2})}$ or $P\not \subset P(\sigma _{1})\cap P(\sigma _{2})$ by [Reference Abe4, Proposition 5.10, Proposition 5.18]. If it is not zero, then the extension group is isomorphic to

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P(\sigma_{1})\cap P(\sigma_{2})}}(\mathrm{St}^{P(\sigma_{1})\cap P(\sigma_{2})}_{Q_{1}\cap P(\sigma_{2})}(\sigma_{1}),\mathrm{St}^{P(\sigma_{1})\cap P(\sigma_{2})}_{Q_{2}}(\sigma_{2})). \end{align*} $$

This holds if $Q_{2}\subset P(\sigma _{1})$ , $\Delta (\sigma _{1}) = \Delta (Q_{1})\cup \Delta _{P(\sigma _{1})\cap P(\sigma _{2})}$ and $P\subset P(\sigma _{1})\cap P(\sigma _{2})$ , and otherwise the extension group is zero. Note that we always have $P\subset P(\sigma _{1})\cap P(\sigma _{2})$ since both $P(\sigma _{1})$ and $P(\sigma _{2})$ contain P. Since $Q_{1}\subset P(\sigma _{1})$ , $\Delta (Q_{1})\cup \Delta _{P(\sigma _{1})\cap P(\sigma _{2})} = (\Delta (\sigma _{1})\cap \Delta (Q_{1}))\cup (\Delta (\sigma _{1})\cap \Delta (\sigma _{2})) = \Delta (\sigma _{1})\cap (\Delta _{Q_{1}}\cup \Delta (\sigma _{2}))$ . (Recall that $P(\sigma _{1})$ is the parabolic subgroup corresponding to $\Delta (\sigma _{1})$ .) Hence, we have $\Delta (\sigma _{1}) = \Delta (Q_{1})\cup \Delta (P(\sigma _{1})\cap P(\sigma _{2}))$ if and only if $\Delta (\sigma _{1})\subset \Delta _{Q_{1}}\cup \Delta (\sigma _{2})$ . We get the proposition.

Therefore, to calculate the extension groups, we may assume $P(\sigma _{1}) = P(\sigma _{2}) = G$ .

3.3 Extensions between generalized Steinberg modules

We assume that $P(\sigma _{1}) = P(\sigma _{2}) = G$ , and we continue the calculation of the extension groups.

Lemma 3.7. Let $Q_{11},Q_{12},Q_{2}$ be parabolic subgroups and $\alpha \in \Delta _{Q_{12}}$ such that $\Delta _{Q_{11}} = \Delta _{Q_{12}}\setminus \{\alpha \}$ . Then we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\mathrm{St}_{Q_{11}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})) \simeq \begin{cases} \operatorname{\mathrm{Ext}}^{i - 1}_{\mathcal{H}}(\mathrm{St}_{Q_{12}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})) & (\alpha\in\Delta_{Q_{2}}),\\ \operatorname{\mathrm{Ext}}^{i + 1}_{\mathcal{H}}(\mathrm{St}_{Q_{12}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})) & (\alpha\notin\Delta_{Q_{2}}). \end{cases} \end{align*} $$

Proof. Let $P_{1}$ be a parabolic subgroup corresponding to $\Delta \setminus \{\alpha \}$ . First, we prove that there exists an exact sequence

(3.1) $$ \begin{align} 0\to \mathrm{St}_{Q_{12}}(\sigma_{1})\to I_{P_{1}}(\mathrm{St}_{Q_{11}}^{P_{1}}(\sigma_{1}))\to \mathrm{St}_{Q_{11}}(\sigma_{1})\to 0. \end{align} $$

We start with the following exact sequence.

$$ \begin{align*} 0\to \sum_{P_{1}\supset Q\supsetneq Q_{11}}I^{P_{1}}_{Q}(e_{Q}(\sigma_{1}))\to I_{Q_{11}}^{P_{1}}(e_{Q_{11}}(\sigma_{1}))\to \mathrm{St}_{Q_{11}}^{P_{1}}(\sigma_{1})\to 0, \end{align*} $$

Apply $I_{P_{1}}$ to this exact sequence. Then we have

$$ \begin{align*} 0\to \sum_{P_{1}\supset Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma_{1}))\to I_{Q_{11}}(e_{Q_{11}}(\sigma_{1}))\to I_{P_{1}}(\mathrm{St}_{Q_{11}}^{P_{1}}(\sigma_{1}))\to 0. \end{align*} $$

Hence, we get the following commutative diagram with exact columns:

Hence, $I_{P_{1}}(\mathrm {St}_{Q_{11}}^{P_{1}}(\sigma _{1}))\to \mathrm {St}_{Q_{11}}(\sigma _{1})$ is surjective, and the kernel is isomorphic to

$$ \begin{align*} \left.\sum_{Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma))\middle/\sum_{P_{1}\supset Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma_{1})) \right. \end{align*} $$

by the snake lemma.

We prove:

1. (1) $I_{Q_{12}}(e_{Q_{12}}(\sigma _{1})) + \sum _{P_{1}\supset Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma _{1})) = \sum _{Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma ))$ .

2. (2) $I_{Q_{12}}(e_{Q_{12}}(\sigma _{1})) \cap \sum _{P_{1}\supset Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma _{1})) = \sum _{Q\supsetneq Q_{12}}I_{Q}(e_{Q}(\sigma ))$ .

We prove (1). Since $Q_{12}\supsetneq Q_{11}$ , $I_{Q_{12}}(e_{Q_{12}}(\sigma _{1}))$ is contained in the right-hand side. Obviously, $\sum _{P_{1}\supset Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma _{1}))$ is also contained in the right-hand side. Hence, $I_{Q_{12}}(e_{Q_{12}}(\sigma _{1})) + \sum _{P_{1}\supset Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma _{1})) \subset \sum _{Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma ))$ . Take $Q_{\supsetneq } Q_{11}$ , and we prove that $I_{Q}(e_{Q}(\sigma _{1}))\subset I_{Q_{12}}(e_{Q_{12}}(\sigma _{1})) + \sum _{P_{1}\supset Q\supsetneq Q_{11}}I_{Q}(e_{Q}(\sigma _{1}))$ . If $P_{1}\supset Q$ , then it is obvious. We assume that $P_{1}\not \supset Q$ . Since $\Delta _{P_{1}} = \Delta \setminus \{\alpha \}$ , this is equivalent to $\alpha \in \Delta _{Q}$ . Hence, $\Delta _{Q}\supset \Delta _{Q_{11}}\cup \{\alpha \} = \Delta _{Q_{12}}$ . Therefore, we have $Q\supset Q_{12}$ . Hence, $I_{Q}(e_{Q}(\sigma _{1}))\subset I_{Q_{12}}(e_{Q_{12}}(\sigma _{1}))$ .

We prove (2). By [Reference Abe2, Lemma 3.10], the left-hand side is

$$ \begin{align*} \sum_{P_{1}\supset Q\supsetneq Q_{11}}I_{\langle Q,Q_{12}\rangle}(e_{\langle Q,Q_{12}\rangle}(\sigma_{1})), \end{align*} $$

where $\langle Q,Q_{12}\rangle $ is the subgroup generated by Q and $Q_{12}$ . We prove

$$ \begin{align*} \{\langle Q,Q_{12}\rangle\mid P_{1}\supset Q\supsetneq Q_{11}\} = \{Q\mid Q\supsetneq Q_{12}\}. \end{align*} $$

If Q satisfies $P_{1}\supset Q\supsetneq Q_{11}$ , then there exists $\beta \in \Delta _{Q}\setminus \Delta _{Q_{11}}$ . We have $\beta \in \Delta _{Q}\subset \Delta _{P_{1}} = \Delta \setminus \{\alpha \}$ . Therefore, we have $\beta \ne \alpha $ . Hence, $\beta \notin \Delta _{Q_{11}}\cup \{\alpha \} = \Delta _{Q_{12}}$ . On the other hand, $\beta \in \Delta _{Q}\subset \Delta _{\langle Q,Q_{12}\rangle }$ . Namely, we have $\beta \in \Delta _{\langle Q,Q_{12}\rangle }\setminus \Delta _{Q_{12}}$ . Obviously, $\langle Q,Q_{12}\rangle \supset Q_{12}$ . Therefore, we get $\langle Q,Q_{12}\rangle \supsetneq Q_{12}$ .

On the other hand, assume that $Q\supsetneq Q_{12}$ . Then $\alpha \in \Delta _{Q}$ since $\alpha \in \Delta _{Q_{12}}$ . Let $Q^{\prime }$ be the parabolic subgroup corresponding to $\Delta _{Q}\setminus \{\alpha \}$ . Then we have $\Delta _{Q^{\prime }}\subset \Delta \setminus \{\alpha \} = \Delta _{P_{1}}$ and $\Delta _{Q^{\prime }} = \Delta _{Q}\setminus \{\alpha \} \supsetneq \Delta _{Q_{12}}\setminus \{\alpha \} = \Delta _{Q_{11}}$ . Hence, $P_{1}\supset Q^{\prime }\supsetneq Q_{11}$ . We have $\Delta _{\langle Q^{\prime },Q_{12}\rangle } = \Delta _{Q^{\prime }}\cup \Delta _{Q_{12}} = (\Delta _{Q}\setminus \{\alpha \})\cup \Delta _{Q_{11}}\cup \{\alpha \} = \Delta _{Q}\cup \Delta _{Q_{11}}\cup \{\alpha \}$ . This is $\Delta _{Q}$ since $\Delta _{Q}\supset \Delta _{Q_{12}} = \Delta _{Q_{11}}\cup \{\alpha \}$ . Hence, $Q = \langle Q^{\prime },Q_{12}\rangle $ . We get the existence of the exact sequence (3.1).

Assume that $\alpha \in \Delta _{Q_{2}}$ . Then $\alpha \in \Delta _{Q_{2}}$ and $\alpha \notin \Delta _{Q_{2}\cap P_{1}}$ . Hence, $\Delta _{Q_{2}}\ne \Delta _{Q_{2}\cap P_{1}}\cup \Delta _{P}$ . Therefore, $R_{P_{1}}(\mathrm {St}_{Q_{2}}(\sigma _{2})) = 0$ by [Reference Abe4, Proposition 5.11]. We have an exact sequence

$$ \begin{align*} & \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(I_{P_{1}}(\mathrm{St}_{Q_{11}}^{P_{1}}(\sigma_{1})),\mathrm{St}_{Q_{2}}(\sigma_{2}))\to \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\mathrm{St}_{Q_{12}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2}))\\ & \to \operatorname{\mathrm{Ext}}^{i + 1}_{\mathcal{H}}(\mathrm{St}_{Q_{11}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2}))\to \operatorname{\mathrm{Ext}}^{i + 1}_{\mathcal{H}}(I_{P_{1}}(\mathrm{St}_{Q_{11}}^{P_{1}}(\sigma_{1})),\mathrm{St}_{Q_{2}}(\sigma_{2})). \end{align*} $$

Since $R_{P_{1}}(\mathrm {St}_{Q_{2}}(\sigma _{2})) = 0$ , for any j, we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{j}_{\mathcal{H}}(I_{P_{1}}(\mathrm{St}_{Q_{12}}(\sigma_{1})),\mathrm{St}_{Q_{2}}(\sigma_{2})) = \operatorname{\mathrm{Ext}}^{j}_{\mathcal{H}}(\mathrm{St}_{Q_{12}}(\sigma_{1}),R_{P_{1}}(\mathrm{St}_{Q_{2}}(\sigma_{2}))) = 0. \end{align*} $$

Therefore, we get

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\mathrm{St}_{Q_{12}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})) \simeq \operatorname{\mathrm{Ext}}^{i + 1}_{\mathcal{H}}(\mathrm{St}_{Q_{11}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})). \end{align*} $$

Next assume that $\alpha \notin \Delta _{Q_{2}}$ . Let $Q_{11}^{c}$ (resp. $Q_{12}^{c},Q_{2}^{c}$ ) be the parabolic subgroup corresponding to $(\Delta \setminus \Delta _{Q_{11}})\cup \Delta _{P}$ (resp. $(\Delta \setminus \Delta _{Q_{12}})\cup \Delta _{P},(\Delta \setminus \Delta _{Q_{2}})\cup \Delta _{P}$ ). Let $\iota = \iota _{G}\colon \mathcal {H}\to \mathcal {H}$ be the involution defined by $\iota (T_{w}) = (-1)^{\ell (w)}T_{w}^{*}$ , and set $\pi ^{\iota } = \pi \circ \iota $ for an $\mathcal {H}$ -module $\pi $ . Then we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\mathrm{St}_{Q_{11}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2}))& \simeq \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}((\mathrm{St}_{Q_{11}}(\sigma_{1}))^{\iota},(\mathrm{St}_{Q_{2}}(\sigma_{2}))^{\iota})\\ &\simeq \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\mathrm{St}_{Q_{11}^{c}}(\sigma_{1,\ell - \ell_{P}}^{\iota_{P}}),\mathrm{St}_{Q_{2}^{c}}(\sigma_{2,\ell - \ell_{P}}^{\iota_{P}})) \end{align*} $$

by [Reference Abe2, Theorem 3.6]. Now we have $\alpha \in \Delta _{Q_{2}^{c}}$ . Applying the lemma (where $Q_{11} = Q_{12}^{c}$ and $Q_{12} = Q_{11}^{c}$ ), we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\mathrm{St}_{Q_{11}^{c}}(\sigma_{1,\ell - \ell_{P}}^{\iota_{P}}),\mathrm{St}_{Q_{2}^{c}}(\sigma_{2,\ell - \ell_{P}}^{\iota_{P}})) & \simeq \operatorname{\mathrm{Ext}}^{i - 1}_{\mathcal{H}}(\mathrm{St}_{Q_{12}^{c}}(\sigma_{1,\ell - \ell_{P}}^{\iota_{P}}),\mathrm{St}_{Q_{2}^{c}}(\sigma_{2,\ell - \ell_{P}}^{\iota_{P}}))\\ & \simeq \operatorname{\mathrm{Ext}}^{i - 1}_{\mathcal{H}}((\mathrm{St}_{Q_{12}}(\sigma_{1}))^{\iota},(\mathrm{St}_{Q_{2}}(\sigma_{2}))^{\iota})\\ & \simeq \operatorname{\mathrm{Ext}}^{i - 1}_{\mathcal{H}}(\mathrm{St}_{Q_{12}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})). \end{align*} $$

We get the lemma.

For sets $X,Y$ , let $X\bigtriangleup Y = (X\setminus Y)\cup (Y\setminus X)$ be the symmetric difference.

Theorem 3.8. We have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\mathrm{St}_{Q_{1}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})) \simeq \operatorname{\mathrm{Ext}}^{i - \#(\Delta_{Q_{1}}\bigtriangleup\Delta_{Q_{2}})}_{\mathcal{H}}(e_{G}(\sigma_{1}),e_{G}(\sigma_{2})). \end{align*} $$

Proof. By applying Lemma 3.7 several times, we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\mathrm{St}_{Q_{1}}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})) \simeq \operatorname{\mathrm{Ext}}^{i - r_{1}}_{\mathcal{H}}(e_{G}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})), \end{align*} $$

where $r_{1} = \#\{\alpha \in \Delta \setminus \Delta _{Q_{1}}\mid \alpha \in \Delta _{Q_{2}}\} - \#\{\alpha \in \Delta \setminus \Delta _{Q_{1}}\mid \alpha \notin \Delta _{Q_{2}}\}$ . Set $Q^{\prime }_{2} = n_{w_{G}w_{Q_{2}}}Q_{2}^{\mathrm {op}}n_{w_{G}w_{Q_{2}}}^{-1}$ . Then by Lemma 3.5, we get

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i - r_{1}}_{\mathcal{H}}(e_{G}(\sigma_{1}),\mathrm{St}_{Q_{2}}(\sigma_{2})) & \simeq \operatorname{\mathrm{Ext}}^{i - r_{1}}_{\mathcal{H}}((\mathrm{St}_{Q_{2}}(\sigma_{2}))^{*},e_{G}(\sigma_{1})^{*})\\ & \simeq \operatorname{\mathrm{Ext}}^{i - r_{1}}_{\mathcal{H}}(\mathrm{St}_{Q^{\prime}_{2}}(\sigma_{2}^{*}),e_{G}(\sigma_{1}^{*})). \end{align*} $$

Again using Lemma 3.7, we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i - r_{1}}_{\mathcal{H}}(\mathrm{St}_{Q^{\prime}_{2}}(\sigma_{2}^{*}),e_{G}(\sigma_{1}^{*})) \simeq \operatorname{\mathrm{Ext}}^{i - r_{1} - r_{2}}_{\mathcal{H}}(e_{G}(\sigma_{2}^{*}),e_{G}(\sigma_{1}^{*})), \end{align*} $$

where $r_{2} = \#(\Delta \setminus \Delta _{Q^{\prime }_{2}}) = \#(\Delta \setminus \Delta _{Q_{2}})$ . Applying Lemma 3.5 again, we get

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i - r_{1} - r_{2}}_{\mathcal{H}}(e_{G}(\sigma_{2}^{*}),e_{G}(\sigma_{1}^{*})) \simeq \operatorname{\mathrm{Ext}}^{i - r_{1} - r_{2}}_{\mathcal{H}}(e_{G}(\sigma_{1}),e_{G}(\sigma_{2})). \end{align*} $$

Since $r_{1} + r_{2} = \#(\Delta _{Q_{1}}\bigtriangleup \Delta _{Q_{2}})$ , we get the lemma.

Recall that the trivial module $\boldsymbol {1}$ is defined by $\boldsymbol {1}(T_{w}) = q_{w}$ . We denote the restriction of $\boldsymbol {1}$ to $\mathcal {H}_{\mathrm {aff}}$ by $\boldsymbol {1}_{\mathcal {H}_{\mathrm {aff}}}$ .

3.4 Extension between extensions

Let P be a parabolic subgroup and $\sigma $ an $\mathcal {H}_{P}$ -module which has the extension $e_{G}(\sigma )$ to $\mathcal {H}$ . In particular, $\Delta _{P}$ and $\Delta \setminus \Delta _{P}$ are orthogonal to each other. Let $P_{2}$ be a parabolic subgroup corresponding to $\Delta \setminus \Delta _{P}$ . Let $J\subset \mathcal {H}$ be an ideal generated by $\{T_{w}^{*} - 1\mid w\in W_{\mathrm {aff},P_{2}}(1)\}$ . Then $e_{G}(\sigma )(J) = 0$ . Hence, for any module $\pi $ of $\mathcal {H}$ , we have

$$ \begin{align*} \operatorname{\mathrm{Hom}}_{\mathcal{H}}(e(\sigma),\pi) = \operatorname{\mathrm{Hom}}_{\mathcal{H}/J}(e(\sigma),\{v\in \pi\mid vJ = 0\}). \end{align*} $$

Note that $T_{w}^{P_{2}}\mapsto T_{w}$ defines the injection $\mathcal {H}_{\mathrm {aff},P_{2}}\to \mathcal {H}$ since the restriction of $\ell $ on $W_{\mathrm {aff},P_{2}}(1)$ is $\ell _{P_{2}}$ . Since any generator of J is in $\mathcal {H}_{\mathrm {aff},P_{2}}$ , we have $\{v\in \pi \mid vJ = 0\} = \{v\in \pi \mid v(J\cap \mathcal {H}_{\mathrm {aff},P_{2}}) = 0\}$ . Since the trivial representation $\boldsymbol {1}_{\mathcal {H}_{\mathrm {aff},P_{2}}}$ of $\mathcal {H}_{\mathrm {aff},P_{2}}$ is isomorphic to $\mathcal {H}_{\mathrm {aff},P_{2}}/(J\cap \mathcal {H}_{\mathrm {aff},P_{2}})$ , we get

$$ \begin{align*} \{v\in \pi\mid v(J\cap \mathcal{H}_{\mathrm{aff},P_{2}}) = 0\} = \operatorname{\mathrm{Hom}}_{\mathcal{H}_{\mathrm{aff},P_{2}}}(\boldsymbol{1}_{\mathcal{H}_{\mathrm{aff},P_{2}}},\pi). \end{align*} $$

Hence, we get

$$ \begin{align*} \operatorname{\mathrm{Hom}}_{\mathcal{H}}(e_{G}(\sigma),\pi) = \operatorname{\mathrm{Hom}}_{\mathcal{H}/J}(e_{G}(\sigma),\operatorname{\mathrm{Hom}}_{\mathcal{H}_{\mathrm{aff},P_{2}}}(\boldsymbol{1}_{\mathcal{H}_{\mathrm{aff},P_{2}}},\pi)). \end{align*} $$

This isomorphism can be generalized as

$$ \begin{align*} \operatorname{\mathrm{Hom}}_{\mathcal{H}}(\pi_{1},\pi) = \operatorname{\mathrm{Hom}}_{\mathcal{H}/J}(\pi_{1},\operatorname{\mathrm{Hom}}_{\mathcal{H}_{\mathrm{aff},P_{2}}}(\boldsymbol{1}_{\mathcal{H}_{\mathrm{aff},P_{2}}},\pi)) \end{align*} $$

for any $\mathcal {H}/J$ -module $\pi _{1}$ . In particular, $\pi \mapsto \operatorname {\mathrm {Hom}}_{\mathcal {H}_{\mathrm {aff},P_{2}}}(\boldsymbol {1}_{\mathcal {H}_{\mathrm {aff},P_{2}}},\pi )$ from the category of $\mathcal {H}$ -modules to the category of $\mathcal {H}/J$ -modules preserves injective modules. Hence, we have a spectral sequence

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}/J}(e_{G}(\sigma),\operatorname{\mathrm{Ext}}^{j}_{\mathcal{H}_{\mathrm{aff},P_{2}}}(\boldsymbol{1}_{\mathcal{H}_{\mathrm{aff},P_{2}}},\pi))\Rightarrow \operatorname{\mathrm{Ext}}^{i + j}_{\mathcal{H}}(e(\sigma),\pi). \end{align*} $$

Now let $\sigma _{1},\sigma _{2}$ be $\mathcal {H}_{P}$ -modules such that both have the extensions $e_{G}(\sigma _{1}),e_{G}(\sigma _{2})$ to $\mathcal {H}$ . Since $e_{G}(\sigma _{2})|_{\mathcal {H}_{\mathrm {aff},P_{2}}}$ is a direct sum of the trivial representations, we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{j}_{\mathcal{H}_{\mathrm{aff},P_{2}}}(\boldsymbol{1}_{\mathcal{H}_{\mathrm{aff},P_{2}}},e(\sigma_{2})) = 0 \end{align*} $$

for $j> 0$ by Corollary 3.9. Hence,

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(e_{G}(\sigma_{1}),e_{G}(\sigma_{2}))\simeq \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}/J}(e_{G}(\sigma_{1}),e_{G}(\sigma_{2})). \end{align*} $$

Therefore, we get

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(e_{G}(\sigma_{1}),e_{G}(\sigma_{2}))\simeq \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{P}/I}(\sigma_{1},\sigma_{2}). \end{align*} $$

Proof. Since $\Delta = \Delta _{P}\cup \Delta _{P_{2}}$ is the orthogonal decomposition, we have $W_{\mathrm {aff}} = W_{\mathrm {aff},P}\times W_{\mathrm {aff},P_{2}}$ and $S_{\mathrm {aff}} = S_{\mathrm {aff},P}\cup S_{\mathrm {aff},P_{2}}$ . The pair $(W^{\prime }_{\mathrm {aff}},S^{\prime }_{\mathrm {aff}}) = (W_{\mathrm {aff},P},S_{\mathrm {aff},P})$ is a Coxeter system, and $\Omega _{P}$ acts on it. Since $W_{\mathrm {aff},P_{2}}$ commutes with $W_{\mathrm {aff},P}$ , this gives the action of $\Omega ^{\prime }$ on $(W^{\prime }_{\mathrm {aff}},S^{\prime }_{\mathrm {aff}})$ . We have $W^{\prime }_{\mathrm {aff}}\subset W_{P}$ , and since $W_{\mathrm {aff},P}\cap W_{\mathrm {aff},P_{2}}$ is trivial, we have the embedding $W^{\prime }_{\mathrm {aff}}\subset W^{\prime }$ . We also have $\Omega _{P}\subset W^{\prime }$ . Since $W_{P} = W_{\mathrm {aff},P}\Omega _{P}$ , we have $W^{\prime } = W^{\prime }_{\mathrm {aff}}\Omega ^{\prime }$ . Since $W_{\mathrm {aff},P}\cap \Omega _{P} = \{1\}$ , we have $W^{\prime }_{\mathrm {aff}}\cap \Omega ^{\prime } = \{1\}$ in $W^{\prime }$ . Hence, $W^{\prime } = W^{\prime }_{\mathrm {aff}}\rtimes \Omega ^{\prime }$ . Since $Z_{\kappa }$ is finite and commutative, $Z^{\prime }_{\kappa }$ is also a finite commutative group. The existence of the exact sequence

$$ \begin{align*} 1\to Z^{\prime}_{\kappa}\to W^{\prime}(1)\to W^{\prime}\to 1 \\[-15pt]\end{align*} $$

is obvious. Note that the length function $\ell ^{\prime }\colon W^{\prime }(1)\to \mathbb {Z}_{\ge 0}$ is given by $\ell _{P}\colon W_{P}^{\prime }(1)\to \mathbb {Z}_{\ge 0}$ since $S^{\prime }_{\mathrm {aff}} = S_{\mathrm {aff},P}$ and $\Omega ^{\prime }$ is the image of $\Omega _{P}$ .

We put $q^{\prime }_{s} = q_{s}$ for $s\in S_{\mathrm {aff},P}$ . (Note that $q_{s} = q_{s,P}$ since $\Delta = \Delta _{P}\cup \Delta _{P_{2}}$ is an orthogonal decomposition.) For $s\in \mathrm {Ref}(W^{\prime }(1))$ , take its lift $\widetilde {s}\in \mathrm {Ref}(W_{P}(1))$ and let $c^{\prime }_{s}$ be the image of $c_{\widetilde {s}}$ in $C[Z^{\prime }_{\kappa }]$ . We prove that this is well-defined. Let $\widetilde {s}^{\prime }$ be another lift, and take $\lambda \in \Lambda (1)\cap W_{\mathrm {aff},P_{2}}(1)$ such that $\widetilde {s}^{\prime } = \widetilde {s}\lambda $ . The image of $\widetilde {s}$ in W is in $\mathrm {Ref}(W_{P})\subset W_{P,\mathrm {aff}}$ since $S_{P,\mathrm {aff}}\subset W_{P,\mathrm {aff}}$ and $W_{P,\mathrm {aff}}$ is normal. (Recall that a reflection is an element which is conjugate to a simple reflection.) Let $\overline {\lambda }$ be the image of $\lambda $ in $\Lambda $ . Since $\widetilde {s},\widetilde {s}^{\prime }\in W_{P,\mathrm {aff}}(1)$ , we have $\overline {\lambda }\in \Lambda \cap W_{\mathrm {aff},P_{2}}\cap W_{\mathrm {aff},P} = \{1\}$ . Hence, $\lambda \in Z_{\kappa }$ . Since $\lambda \in W_{\mathrm {aff},P_{2}}(1)$ , we have $\lambda \in Z_{\kappa }\cap W_{\mathrm {aff},P_{2}}(1)$ . Hence, the image of $c_{\widetilde {s}^{\prime }} = c_{\widetilde {s}\lambda } = c_{\widetilde {s}}\lambda $ is the same as that of $c_{\widetilde {s}}$ in $C[Z^{\prime }_{\kappa }]$ .

We get the parameter $(q^{\prime },c^{\prime })$ and let $\mathcal {H}^{\prime } = \bigoplus _{w\in W^{\prime }(1)}T^{\prime }_{w}$ be the attached algebra. Consider the linear map $\Phi \colon \mathcal {H}_{P}\to \mathcal {H}^{\prime }$ defined by $T^{P}_{w}\mapsto T^{\prime }_{\overline {w}}$ , where $w\in W_{P}(1)$ and $\overline {w}\in W^{\prime }(1)$ is the image of w.

First, we prove that the map $\Phi $ preserves the relations. Let $s\in W_{P}(1)$ be a lift of an affine simple reflection in $S_{\mathrm {aff},P}$ . Then we have $(T^{P}_{s})^{2} = q_{s}T^{P}_{s^{2}} + c_{s}T^{P}_{s}$ . Let $\overline {s}$ be the image of s in $W^{\prime }(1)$ . Then we have $(T^{\prime }_{\overline {s}})^{2} = q^{\prime }_{\overline {s}}T^{\prime }_{\overline {s}^{2}} + c^{\prime }_{\overline {s}}T^{\prime }_{\overline {s}}$ . The definition of $(q^{\prime },c^{\prime })$ says $q^{\prime }_{\overline {s}} = q_{s}$ and $\Phi (c_{s}) = c^{\prime }_{\overline {s}}$ . Hence, $\Phi $ preserves the quadratic relations. The compatibility between $\ell _{P}$ and $\ell ^{\prime }$ implies that $\Phi $ preserves the braid relations.

Obviously, $\Phi $ is surjective. We prove that $\operatorname {\mathrm {Ker}}\Phi = I$ . Clearly, we have $I\subset \operatorname {\mathrm {Ker}}\Phi $ . Let $\sum _{w\in W_{P}(1)}c_{w}T_{w}\in \operatorname {\mathrm {Ker}}\Phi $ , where $c_{w}\in C$ . Fix a section x of $W_{P}(1)\to W^{\prime }(1)$ . Then we have $\sum _{w\in W_{P}(1)}c_{w}T^{P}_{w} = \sum _{w\in W^{\prime }(1)}\sum _{\lambda \in \Lambda (1)\cap W_{\mathrm {aff},P_{2}}(1)}c_{x(w)\lambda }T^{P}_{x(w)\lambda }$ . Hence,

$$ \begin{align*} 0 = \Phi\left(\sum_{w\in W_{P}(1)}c_{w}T^{P}_{w}\right) = \sum_{w\in W^{\prime}(1)}\left(\sum_{\lambda\in \Lambda(1)\cap W_{\mathrm{aff},P_{2}}(1)}c_{x(w)\lambda}\right)T^{\prime}_{w}.\\[-15pt] \end{align*} $$

Therefore, for each $w\in W^{\prime }(1)$ , we have $\sum _{\lambda \in \Lambda (1)\cap W_{\mathrm {aff},P_{2}}(1)}c_{x(w)\lambda } = 0$ . Hence,

$$ \begin{align*} & \sum_{w\in W_{P}(1)}c_{w}T_{w}^{P}\\[-2pt] & = \sum_{w\in W^{\prime}(1)}\sum_{\lambda\in \Lambda(1)\cap W_{\mathrm{aff},P_{2}}(1)}c_{x(w)\lambda}T_{x(w)\lambda}^{P}\\[-2pt] & = \sum_{w\in W^{\prime}(1)}\left(\sum_{\lambda\in \Lambda(1)\cap W_{\mathrm{aff},P_{2}}(1)}c_{x(w)\lambda}T^{P}_{x(w)\lambda} - \sum_{\lambda\in \Lambda(1)\cap W_{\mathrm{aff},P_{2}}(1)}c_{x(w)\lambda}T^{P}_{x(w)}\right)\\[-2pt] & = \sum_{w\in W^{\prime}(1)}\left(\sum_{\lambda\in \Lambda(1)\cap W_{\mathrm{aff},P_{2}}(1)}c_{x(w)\lambda}T^{P}_{x(w)}(T^{P}_{\lambda} - 1)\right)\in I. \\[-15pt]\end{align*} $$

Finally, $\Omega ^{\prime }$ is commutative since $\Omega _{P}$ is commutative.

3.5 Example

Let $G = \mathrm {PGL}_{2}$ . We have $\Lambda (1) \simeq F^{\times } / (1 + (\varpi ))\simeq \mathbb {Z}\times \kappa ^{\times }$ . Consider $\widetilde {G} = \mathrm {SL}_{2}$ . Then $G^{\prime }$ is the image of $\widetilde {G}\to G$ [Reference Abe, Henniart, Herzig and Vignéras5, II.4 Proposition]. By this description, we have $\Lambda (1)\cap W_{\mathrm {aff}}(1) = \{\lambda ^{2}\mid \lambda \in \Lambda (1)\}$ . Therefore, with the notation in Proposition 3.11, we have $W^{\prime }_{\mathrm {aff}} = \{1\}$ , $S^{\prime }_{\mathrm {aff}} = \emptyset $ , $W^{\prime }(1) = \Lambda (1)/\{\lambda ^{2}\mid \lambda \in \Lambda (1)\}$ , $Z_{\kappa } = Z_{\kappa }/\{t^{2}\mid t\in Z_{\kappa }\}$ . We have $\mathcal {H}_{B}/I = C[W^{\prime }(1)]$ .

Consider the trivial module $\boldsymbol {1}_{G}$ . Then we have $\boldsymbol {1}_{G} = e_{G}(\boldsymbol {1}_{B})$ , and we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}}(\boldsymbol{1}_{G},\boldsymbol{1}_{G}) \simeq \operatorname{\mathrm{Ext}}^{i}_{\mathcal{H}_{B}/I}(\boldsymbol{1}_{B},\boldsymbol{1}_{B}) \simeq \operatorname{\mathrm{Ext}}^{i}_{C[W^{\prime}(1)]}(\boldsymbol{1}_{B},\boldsymbol{1}_{B}) = H^{i}(W^{\prime}(1),C). \end{align*} $$

Here, C is the trivial $W^{\prime }(1)$ -module. Since the group $W^{\prime }(1)$ is a $2$ -group, this cohomology is zero if the characteristic of C is not $2$ . However, if the characteristic of C is $2$ , since $W^{\prime }(1)\simeq \mathbb {Z}/2\mathbb {Z}$ ( $p = 2$ ) or $(\mathbb {Z}/2\mathbb {Z})^{\oplus 2}$ ( $p\ne 2$ ), $H^{i}(W^{\prime }(1),C)\ne 0$ if i is even. Therefore, we have infinitely many i with $\operatorname {\mathrm {Ext}}^{i}_{\mathcal {H}}(\boldsymbol {1}_{G},\boldsymbol {1}_{G})\ne 0$ . This recovers Koziol’s example [Reference Karol12, Example 6.2].

3.6 Summary

Now we get a reduction. The $\operatorname {\mathrm {Ext}}^{1}$ between simple modules is equal to $\operatorname {\mathrm {Ext}}^{1 - r}$ between supersingular simple modules for some $r\ge 0$ or zero. In particular, if $r\ge 2$ , then $\operatorname {\mathrm {Ext}}^{1}$ between simple modules is zero. If $r = 1$ , then $\operatorname {\mathrm {Ext}}^{1 - r} = \operatorname {\mathrm {Hom}}$ , so it is zero or one-dimensional. If $r = 0$ , we have to calculate $\operatorname {\mathrm {Ext}}^{1}$ between supersingular simple modules. Therefore, the only remaining task is to calculate $\operatorname {\mathrm {Ext}}^{1}$ between supersingular simple modules.

4.1 $\operatorname {\mathrm {Ext}}^{1}$ for $\mathcal {H}^{\mathrm {aff}}$

Let $\chi ,\chi ^{\prime }$ be characters of $Z_{\kappa }$ and $J\subset S_{\mathrm {aff},\chi },J^{\prime }\subset S_{\mathrm {aff},\chi ^{\prime }}$ subsets. Then we have characters $\Xi = \Xi _{J,\chi }$ , $\Xi ^{\prime } = \Xi _{J^{\prime },\chi ^{\prime }}$ of $\mathcal {H}^{\mathrm {aff}}$ . We calculate $\operatorname {\mathrm {Ext}}^{1}_{\mathcal {H}^{\mathrm {aff}}}(\Xi ,\Xi ^{\prime })$ .

To express the space of extensions, we need some notation. For each $s\in S_{\mathrm {aff}}$ , let $C_{s}$ be the set of functions a on $\{\widetilde {s}\in W(1)\mid \widetilde {s}\mapsto s\in W\}$ such that $a(t\widetilde {s}) = \chi ^{\prime }(t)a(\widetilde {s})$ , $a(\widetilde {s}t) = a(\widetilde {s})\chi (t)$ for any $t\in Z_{\kappa }$ . Then $C_{s}\ne 0$ if and only if $\chi ^{\prime } = s\chi $ and if $\chi ^{\prime } = s\chi $ then $\dim _{C} C_{s} = 1$ .

Now we define some subsets of $S_{\mathrm {aff}}$ . First, consider the sets

$$ \begin{align*} A_{1}(\Xi,\Xi^{\prime}) & = \{s\in S_{\mathrm{aff}}\mid \Xi(T_{\widetilde{s}}) = \Xi^{\prime}(T_{\widetilde{s}}) = 0\},\\ A_{2}(\Xi,\Xi^{\prime}) & = \{s\in S_{\mathrm{aff}}\mid \Xi(T_{\widetilde{s}}) \ne 0,\ \Xi^{\prime}(T_{\widetilde{s}}) = 0\},\\ A_{3}(\Xi,\Xi^{\prime}) & = \{s\in S_{\mathrm{aff}}\mid \Xi(T_{\widetilde{s}}) = 0,\ \Xi^{\prime}(T_{\widetilde{s}}) \ne 0\},\\ A_{4}(\Xi,\Xi^{\prime}) & = \{s\in S_{\mathrm{aff}}\mid \Xi(T_{\widetilde{s}}) \ne 0,\ \Xi^{\prime}(T_{\widetilde{s}}) \ne 0\}, \end{align*} $$

where $\widetilde {s}$ is a lift of s. We define

$$ \begin{align*} S_{2}(\Xi,\Xi^{\prime}) & = A_{2}(\Xi,\Xi^{\prime})\cup A_{3}(\Xi,\Xi^{\prime}),\\ S_{1}(\Xi,\Xi^{\prime}) & = \{s\in A_{1}(\Xi,\Xi^{\prime})\setminus S_{\mathrm{aff},\chi}\mid s\chi = \chi^{\prime},\ (ss_{1})^{2} \ne 1\text{ for any }s_{1}\in S_{2}(\Xi,\Xi^{\prime})\}. \end{align*} $$

If $s\in S_{\mathrm {aff},\chi }$ and $a\in C_{s}$ , then $a(\widetilde {s})\chi (c_{\widetilde {s}})^{-1}\in C$ does not depend on a lift $\widetilde {s}$ of s. We denote it by $a\chi (c_{s})^{-1}$ . We also have that if $s\in S_{\mathrm {aff},\chi ^{\prime }}$ , then $a(\widetilde {s})\chi ^{\prime }(c_{\widetilde {s}})^{-1}$ does not depend on a lift $\widetilde {s}$ . We denote it by $a\chi ^{\prime }(c_{s})^{-1}$ . If $a\ne 0$ , then $\chi ^{\prime } = s \chi $ . Hence, if $s\in S_{\mathrm {aff},\chi }$ , then $s\in S_{\mathrm {aff},\chi ^{\prime }}$ and $a\chi (c_{s})^{-1} = a\chi ^{\prime }(c_{s})^{-1}$ .

For the Hecke algebra attached to a finite Coxeter system, the following proposition is [Reference Fayers8, Theorem 5.1], and we use a similar proof.

Proposition 4.1. Consider the subspace $E_{2}(\Xi ,\Xi ^{\prime })$ of $\bigoplus _{s\in S_{2}(\Xi ,\Xi ^{\prime })}C_{s}$ consisting $(a_{s})$ such that

• • If $s_{1},s_{2}\in A_{2}(\Xi ,\Xi ^{\prime })$ , then $a_{s_{1}}\chi (c_{s_{1}})^{-1} = a_{s_{2}}\chi (c_{s_{2}})^{-1}$ .

• • If $s_{1},s_{2}\in A_{3}(\Xi ,\Xi ^{\prime })$ , then $a_{s_{1}}\chi ^{\prime }(c_{s_{1}})^{-1} = a_{s_{2}}\chi ^{\prime }(c_{s_{2}})^{-1}$ .

• • If $s_{1}\in A_{2}(\Xi ,\Xi ^{\prime })$ , $s_{2}\in A_{3}(\Xi ,\Xi ^{\prime })$ and $(s_{1}s_{2})^{2} = 1$ , then $a_{s_{1}}\chi (c_{s_{1}})^{-1} + a_{s_{2}}\chi ^{\prime } (c_{s_{2}})^{-1} = 0$ ,

and put $E_{1}(\Xi ,\Xi ^{\prime }) = \bigoplus _{s\in S_{1}(\Xi ,\Xi ^{\prime })}C_{s}$ , $E(\Xi ,\Xi ^{\prime }) = E_{1}(\Xi ,\Xi ^{\prime })\oplus E_{2}(\Xi ,\Xi ^{\prime })$ . For $(a_{s})\in E(\Xi ,\Xi ^{\prime })$ , consider the linear map $\mathcal {H}\to M_{2}(C)$ defined by

$$ \begin{align*} T_{\widetilde{s}} \mapsto \begin{pmatrix}\Xi(T_{\widetilde{s}}) & 0\\ a_{s}(\widetilde{s}) & \Xi^{\prime}(T_{\widetilde{s}})\end{pmatrix}, \end{align*} $$

where $a_{s} = 0$ if $s\notin S_{1}(\Xi ,\Xi ^{\prime })\cup S_{2}(\Xi ,\Xi ^{\prime })$ . Then this gives an extension of $\Xi $ by $\Xi ^{\prime }$ , and it gives a surjective map $E(\Xi ,\Xi ^{\prime })\to \operatorname {\mathrm {Ext}}_{\mathcal {H}^{\mathrm {aff}}}^{1}(\Xi ,\Xi ^{\prime })$ . The kernel is

(4.1) $$ \begin{align} \left\{(a_{s})\in E(\Xi,\Xi^{\prime})\;\middle|\; \begin{array}{l} a_{s_{1}}\chi(c_{s_{1}})^{-1} + a_{s_{2}}\chi^{\prime}(c_{s_{2}})^{-1} = 0\\ \quad\quad\quad\quad (s_{1}\in A_{2}(\Xi,\Xi^{\prime}),s_{2}\in A_{3}(\Xi,\Xi^{\prime}))\\ a_{s} = 0\ (s\in S_{1}(\Xi,\Xi^{\prime})) \end{array}\right\}. \end{align} $$

Proof. Let M be an extension of $\Xi $ by $\Xi ^{\prime }$ . Since $\#Z_{\kappa }$ is prime to p, the representation of $Z_{\kappa }$ over C is completely reducible. Hence, we can take a basis $e_{1},e_{2}$ such that $T_{t}e_{1} = \chi (t)e_{1}$ and $T_{t}e_{2} = \chi ^{\prime }(t)e_{2}$ . With this basis, the action of $T_{\widetilde {s}}$ where $\widetilde {s}\in S_{\mathrm {aff}}(1)$ with the image $s\in S_{\mathrm {aff}}$ is described as

$$ \begin{align*} T_{\widetilde{s}} = \begin{pmatrix}\Xi(T_{\widetilde{s}}) & 0\\ a_{s}(\widetilde{s}) & \Xi^{\prime}(T_{\widetilde{s}})\end{pmatrix}. \end{align*} $$

for some $a_{s}(\widetilde {s})\in C$ . The action of $T_{t}$ where $t\in Z_{\kappa }$ is given by

$$ \begin{align*} \begin{pmatrix} \chi(t) & 0\\0 & \chi^{\prime}(t). \end{pmatrix} \end{align*} $$

Since $T_{t}T_{\widetilde {s}} = T_{t\widetilde {s}}$ , we have

$$ \begin{align*} \begin{pmatrix} \chi(t) & 0\\0 & \chi^{\prime}(t). \end{pmatrix} \begin{pmatrix}\Xi(T_{\widetilde{s}}) & 0\\ a_{s}(\widetilde{s}) & \Xi^{\prime}(T_{\widetilde{s}})\end{pmatrix} = \begin{pmatrix}\Xi(T_{t\widetilde{s}}) & 0\\ a_{s}(t\widetilde{s}) & \Xi^{\prime}(T_{t\widetilde{s}})\end{pmatrix}. \end{align*} $$

Hence, $a_{s}(t\widetilde {s}) = \chi ^{\prime }(t)a_{s}(\widetilde {s})$ . Similarly, we have $a_{s}(\widetilde {s}t) = a_{s}(\widetilde {s})\chi (t)$ . Hence, $a_{s}\in C_{s}$ .

Now we check the conditions that the map defines an action of $\mathcal {H}^{\mathrm {aff}}$ . Since we have

$$ \begin{align*} \begin{pmatrix} \Xi(T_{\widetilde{s}}) & 0\\ a_{s}(\widetilde{s}) & \Xi^{\prime}(T_{\widetilde{s}}) \end{pmatrix}^{2} = \begin{pmatrix} \Xi(T_{\widetilde{s}})^{2} & 0\\ a_{s}(\widetilde{s})(\Xi(T_{\widetilde{s}}) + \Xi^{\prime}(T_{\widetilde{s}})) & \Xi^{\prime}(T_{\widetilde{s}})^{2} \end{pmatrix}, \end{align*} $$

this satisfies the quadratic relation $T_{\widetilde {s}}^{2} = T_{\widetilde {s}}c_{\widetilde {s}}$ if and only if

(4.2) $$ \begin{align} a_{s}(\widetilde{s})(\Xi(T_{\widetilde{s}}) + \Xi^{\prime}(T_{\widetilde{s}})) = a_{s}(\widetilde{s})\chi(c_{\widetilde{s}}). \end{align} $$

If $s\in A_{1}(\Xi ,\Xi ^{\prime })$ , then $a_{s}(\widetilde {s}) = 0$ or $\chi (c_{\widetilde {s}}) = 0$ , namely $a_{s} = 0$ or $s\notin S_{\mathrm {aff},\chi }$ .

If $s\in A_{2}(\Xi ,\Xi ^{\prime })$ , then $a_{s} =0$ or $\Xi (T_{\widetilde {s}}) = \chi (c_{\widetilde {s}})$ . Since $\Xi (T_{\widetilde {s}}) \ne 0$ , we always have $\Xi (T_{\widetilde {s}}) = \chi (c_{\widetilde {s}})$ . Hence, equation (4.2) is always satisfied.

If $s\in A_{3}(\Xi ,\Xi ^{\prime })$ , then $a_{s} =0$ or $\Xi ^{\prime }(T_{\widetilde {s}}) = \chi (c_{\widetilde {s}})$ . Note that if $a_{s}\ne 0$ , then $s \chi = \chi ^{\prime }$ ; hence, $S_{\mathrm {aff},\chi } = S_{\mathrm {aff},\chi ^{\prime }}$ and $\chi (c_{\widetilde {s}}) = \chi ^{\prime }(c_{\widetilde {s}})$ . Therefore, under $a_{s}\ne 0$ , we have $\Xi ^{\prime }(T_{\widetilde {s}}) = \chi (c_{\widetilde {s}})$ if and only if $\Xi ^{\prime }(T_{\widetilde {s}}) = \chi ^{\prime }(c_{\widetilde {s}})$ . This always hold since $\Xi ^{\prime }(T_{\widetilde {s}}) \ne 0$ . Hence, equation (4.2) is always satisfied.

If $s\in A_{4}(\Xi ,\Xi ^{\prime })$ , then we have $\Xi (T_{\widetilde {s}}) = \chi (c_{\widetilde {s}})$ . Hence, we have $a_{s}(\widetilde {s})\Xi ^{\prime }(T_{\widetilde {s}}) =0$ . Therefore, we have $a_{s} = 0$ since $\Xi ^{\prime }(T_{\widetilde {s}}) \ne 0$ .

Consequently, the quadratic relation holds if and only if $a_{s} = 0$ or $s\in (A_{1}(\Xi ,\Xi ^{\prime })\setminus S_{\mathrm {aff},\chi })\cup A_{2}(\Xi ,\Xi ^{\prime })\cup A_{3}(\Xi ,\Xi ^{\prime })$ . The action of $T_{\widetilde {s}}$ is given by one of the following matrix:

(4.3) $$ \begin{align} \begin{pmatrix} 0 & 0\\ a_{s}(\widetilde{s}) & 0 \end{pmatrix}, \begin{pmatrix} \chi(c_{\widetilde{s}}) & 0\\ a_{s}(\widetilde{s}) & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0\\ a_{s}(\widetilde{s}) & \chi^{\prime}(c_{\widetilde{s}}) \end{pmatrix}, \begin{pmatrix} \chi(c_{\widetilde{s}}) & 0\\0 & \chi^{\prime}(c_{\widetilde{s}}) \end{pmatrix}. \end{align} $$

Here, each $\chi (c_{\widetilde {s}})$ and $\chi ^{\prime }(c_{\widetilde {s}})$ is not zero, and in the first matrix, we assume that $s\notin S_{\mathrm {aff},\chi }$ if $a_{s}\ne 0$ .

Now we check the braid relations. Let $s_{1},s_{2}\in S_{\mathrm {aff}}$ and $\widetilde {s}_{1},\widetilde {s}_{2}$ their lifts. We consider a braid relation $s_{1}s_{2}\dotsm = s_{2}s_{1}\dotsm $ . It is easy to see that the action satisfies the braid relation for some lifts $\widetilde {s}_{1},\widetilde {s}_{2}$ if and only if it is satisfied for any lifts $\widetilde {s}_{1},\widetilde {s}_{2}$ . Take $\widetilde {s}_{1},\widetilde {s}_{2}$ such that $\widetilde {s}_{1}\widetilde {s}_{2}\dotsm = \widetilde {s}_{2}\widetilde {s}_{1}\dotsm $ . It is easy to see that if $s_{1}\in A_{4}(\Xi ,\Xi ^{\prime })$ or $s_{2}\in A_{4}(\Xi ,\Xi ^{\prime })$ , then the braid relations hold automatically. So we assume that $s_{1},s_{2}\in A_{1}(\Xi ,\Xi ^{\prime })\cup A_{2}(\Xi ,\Xi ^{\prime })\cup A_{3}(\Xi ,\Xi ^{\prime })$ .

Assume that $s_{1}\in A_{1}(\Xi ,\Xi ^{\prime })$ . We have

$$ \begin{gather*} \begin{pmatrix}0 & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} \begin{pmatrix}\Xi(T_{\widetilde{s}_{2}}) & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \Xi^{\prime}(T_{\widetilde{s}_{2}})\end{pmatrix}= \begin{pmatrix} 0 & 0\\ a_{s_{1}}(\widetilde{s}_{1})\Xi(T_{\widetilde{s}_{2}}) & 0 \end{pmatrix},\\ \begin{pmatrix}0 & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} \begin{pmatrix}\Xi(T_{\widetilde{s}_{2}}) & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \Xi^{\prime}(T_{\widetilde{s}_{2}})\end{pmatrix}\begin{pmatrix}0 & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} = \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix},\\ \begin{pmatrix}\Xi(T_{\widetilde{s}_{2}}) & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \Xi^{\prime}(T_{\widetilde{s}_{2}})\end{pmatrix}\begin{pmatrix}0 & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ \Xi^{\prime}(T_{\widetilde{s}_{2}})a_{s_{1}}(\widetilde{s}_{1}) & 0 \end{pmatrix},\\ \begin{pmatrix}\Xi(T_{\widetilde{s}_{2}}) & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \Xi^{\prime}(T_{\widetilde{s}_{2}})\end{pmatrix}\begin{pmatrix}0 & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} \begin{pmatrix}\Xi(T_{\widetilde{s}_{2}}) & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \Xi^{\prime}(T_{\widetilde{s}_{2}})\end{pmatrix}= \begin{pmatrix} 0 & 0 \\ \Xi^{\prime}(T_{\widetilde{s}_{2}})\Xi(T_{\widetilde{s}_{2}})a_{s_{1}}(\widetilde{s}_{1}) & 0 \end{pmatrix},\\ \begin{pmatrix}\Xi(T_{\widetilde{s}_{2}}) & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \Xi^{\prime}(T_{\widetilde{s}_{2}})\end{pmatrix}\begin{pmatrix}0 & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} \begin{pmatrix}\Xi(T_{\widetilde{s}_{2}}) & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \Xi^{\prime}(T_{\widetilde{s}_{2}})\end{pmatrix}\begin{pmatrix}0 & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix}= \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}. \end{gather*} $$

Hence, the braid relation is satisfied if and only if

• • $a_{s_{1}} = 0$

• • or $\Xi (T_{\widetilde {s}_{2}}) = \Xi ^{\prime }(T_{\widetilde {s}_{2}})$ and the order of $s_{1}s_{2}$ is $2$

• • or $\Xi (T_{\widetilde {s}_{2}})\Xi ^{\prime }(T_{\widetilde {s}_{2}}) = 0$ and the order of $s_{1}s_{2}$ is $3$

• • or the order of $s_{1}s_{2}$ is greater than $3$ .

If $s_{2}\in A_{1}(\Xi ,\Xi ^{\prime })$ , then the condition always holds. If $s_{2}\in A_{2}(\Xi ,\Xi ^{\prime })\cup A_{3}(\Xi ,\Xi ^{\prime })$ , then the condition holds if and only if $a_{s_{1}} = 0$ or the order of $s_{1}s_{2}$ is not $2$ , namely $(s_{1}s_{2})^{2} \ne 1$ .

Replacing $s_{1}$ with $s_{2}$ , if $s_{2}\in A_{1}(\Xi ,\Xi ^{\prime })$ , we have the similar condition.

Assume that $s_{1},s_{2}\in A_{2}(\Xi ,\Xi ^{\prime })$ . We have

$$ \begin{align*} \begin{pmatrix}\chi(c_{\widetilde{s}_{1}}) & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} \begin{pmatrix}\chi(c_{\widetilde{s}_{2}}) & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & 0\end{pmatrix} = \begin{pmatrix}\chi(c_{\widetilde{s}_{1}})\chi(c_{\widetilde{s}_{2}}) & 0\\ a_{s_{1}}(\widetilde{s}_{1})\chi(c_{\widetilde{s}_{2}}) & 0\end{pmatrix} = \begin{pmatrix}\chi(c_{\widetilde{s}_{1}}) & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix}\chi(c_{\widetilde{s}_{2}}). \end{align*} $$

By this calculation, the braid relation is satisfied if and only if $a_{s_{1}}(\widetilde {s}_{1})\chi (c_{\widetilde {s}_{2}})\dotsm = a_{s_{2}}(\widetilde {s}_{2})\chi (c_{\widetilde {s}_{1}})\dotsm $ . By [Reference Vignéras20, Proposition 4.13 (6)], we have $c_{\widetilde {s}_{1}}c_{\widetilde {s}_{2}}\dotsm = c_{\widetilde {s}_{2}}c_{\widetilde {s}_{1}}\dotsm $ . Hence, the braid relation is satisfied if and only if $a_{s_{1}}(\widetilde {s}_{1})\chi (c_{\widetilde {s}_{1}})^{-1} = a_{s_{2}}(\widetilde {s}_{2})\chi (c_{\widetilde {s}_{2}})^{-1}$ , namely $a_{s_{1}}\chi (c_{s_{1}})^{-1} = a_{s_{2}}\chi (c_{s_{2}})^{-1}$ . By a similar calculation, if $s_{1},s_{2}\in A_{3}(\Xi ,\Xi ^{\prime })$ , then the braid relation is satisfied if and only if $a_{s_{1}}\chi ^{\prime }(c_{s_{1}})^{-1} = a_{s_{2}}\chi ^{\prime }(c_{s_{2}})^{-1}$ .

Finally, we assume that $s_{1}\in A_{2}(\Xi ,\Xi ^{\prime })$ and $s_{2}\in A_{3}(\Xi ,\Xi ^{\prime })$ . We have

$$ \begin{gather*} \begin{pmatrix}\chi(c_{\widetilde{s}_{1}}) & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} \begin{pmatrix}0 & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \chi^{\prime}(c_{\widetilde{s}_{2}})\end{pmatrix} = \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix},\\ \begin{pmatrix}0 & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \chi^{\prime}(c_{\widetilde{s}_{2}})\end{pmatrix} \begin{pmatrix}\chi(c_{\widetilde{s}_{1}}) & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} = \begin{pmatrix} 0 & 0\\ a_{s_{2}}(\widetilde{s}_{2})\chi(c_{\widetilde{s}_{1}}) + a_{s_{1}}(\widetilde{s}_{1})\chi^{\prime}(c_{\widetilde{s}_{2}}) & 0 \end{pmatrix},\\ \begin{pmatrix}0 & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \chi^{\prime}(c_{\widetilde{s}_{2}})\end{pmatrix} \begin{pmatrix}\chi(c_{\widetilde{s}_{1}}) & 0\\ a_{s_{1}}(\widetilde{s}_{1}) & 0\end{pmatrix} \begin{pmatrix}0 & 0\\ a_{s_{2}}(\widetilde{s}_{2}) & \chi^{\prime}(c_{\widetilde{s}_{2}})\end{pmatrix} = \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}. \end{gather*} $$

Hence, the braid relation is satisfied if and only if

• • $a_{s_{2}}(\widetilde {s}_{2})\chi (c_{\widetilde {s}_{1}}) + a_{s_{1}}(\widetilde {s}_{1})\chi ^{\prime }(c_{\widetilde {s}_{2}}) = 0$ and the order of $s_{1}s_{2}$ is $2$ .

• • the order of $s_{1}s_{2}$ is greater than $2$ .

We notice that $a_{s_{2}}(\widetilde {s}_{2})\chi (c_{\widetilde {s}_{1}}) + a_{s_{1}}(\widetilde {s}_{1})\chi ^{\prime }(c_{\widetilde {s}_{2}}) = 0$ if and only if $a_{s_{1}}\chi (c_{s_{1}})^{-1} + a_{s_{2}}\chi ^{\prime }(c_{s_{2}})^{-1} = 0$ .

We get the following table which shows the condition for the braid relation:

Now we assume that $(a_{s})\in \bigoplus _{s \in S_{\mathrm {aff}}}C_{s}$ defines an action of $\mathcal {H}$ . First, recall that $C_{s}\ne 0$ if and only if $s\chi = \chi ^{\prime }$ . Hence, $a_{s}\ne 0$ implies $s\chi = \chi ^{\prime }$ . Since the quadratic relations hold, if $a_{s}\ne 0$ , then $s\in (A_{1}(\Xi ,\Xi ^{\prime })\setminus S_{\mathrm {aff},\chi })\cup S_{2}(\Xi ,\Xi ^{\prime })$ . If $s\in A_{1}(\Xi ,\Xi ^{\prime })\setminus S_{\mathrm {aff},\chi }$ and $(s s_{1})^{2} = 1$ for some $s_{1}\in S_{2}(\Xi ,\Xi ^{\prime })$ , then the table says that $a_{s} = 0$ . Therefore, if $a_{s}\ne 0$ , then $s\in S_{1}(\Xi ,\Xi ^{\prime })\cup S_{2}(\Xi ,\Xi ^{\prime })$ . Hence, again by the table, $(a_{s})$ belongs to $E(\Xi ,\Xi ^{\prime })$ .

Conversely, if $(a_{s})\in E(\Xi ,\Xi ^{\prime })$ , then $a_{s}\ne 0$ implies $s\in S_{1}(\Xi ,\Xi ^{\prime })\cup S_{2}(\Xi ,\Xi ^{\prime })\subset (A_{1}(\Xi ,\Xi ^{\prime })\setminus S_{\mathrm {aff},\chi })\cup S_{2}(\Xi ,\Xi ^{\prime })$ . Hence, each $T_{\widetilde {s}}$ satisfies the quadratic relation. Let $s_{1},s_{2}\in S_{\mathrm {aff}}$ . If $s_{1}\in A_{1}(\Xi ,\Xi ^{\prime })$ and $s_{2}\in A_{2}(\Xi ,\Xi ^{\prime })\cup A_{3}(\Xi ,\Xi ^{\prime })$ , then the definition of $S_{1}(\Xi ,\Xi ^{\prime })$ says that $(s_{1}s_{2})^{2} \ne 1$ or $a_{s_{1}} = 0$ . Then by the table, the braid relation for $s_{1},s_{2}$ holds. For other cases, the condition on $E_{2}(\Xi ,\Xi ^{\prime })$ and the table imply that the braid relation holds too.

Therefore, the map $E(\Xi ,\Xi ^{\prime })\to \operatorname {\mathrm {Ext}}^{1}_{\mathcal {H}^{\mathrm {aff}}}(\Xi ,\Xi ^{\prime })$ is well-defined and surjective.

Assume that the extension given by $(a_{s})$ splits, namely each matrix in equation (4.3) is simultaneous diagonalizable. If $s\in A_{1}(\Xi ,\Xi ^{\prime })$ , then the matrix corresponding to s is diagonalizable if and only if $a_{s} = 0$ . Let $s_{1},s_{2}\in A_{2}(\Xi ,\Xi ^{\prime })$ . Then by $a_{s_{1}}\chi (c_{s_{1}})^{-1} = a_{s_{2}}\chi (c_{s_{2}})^{-1}$ , the matrices corresponding to $s_{1},s_{2}$ commute with each other. Hence, these matrices are simultaneous diagonalizable. Similarly, matrices corresponding to $A_{3}(\Xi ,\Xi ^{\prime })$ are simultaneous diagonalizable.

If $s_{1}\in A_{2}(\Xi ,\Xi ^{\prime })$ and $s_{2}\in A_{3}(\Xi ,\Xi ^{\prime })$ , then the corresponding matrices are

$$ \begin{align*} \begin{pmatrix} \chi(c_{\widetilde{s}_{1}}) & 0\\ a_{s}(\widetilde{s}_{1}) & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0\\ a_{s}(\widetilde{s}_{2}) & \chi^{\prime}(c_{\widetilde{s}_{2}}) \end{pmatrix}, \end{align*} $$

and these commute with each other if and only if $a_{s_{1}}\chi (c_{s_{1}})^{-1} + a_{s_{2}}\chi (c_{s_{2}})^{-1} = 0$ . Hence, the kernel is equation (4.1).

4.2 Semidirect product

The argument in this subsection is general. Let A be a C-algebra and $\Gamma $ a group acting on A. We assume that a finite commutative normal subgroup $\Gamma ^{\prime }\subset \Gamma $ and an embedding $C[\Gamma ^{\prime }]\hookrightarrow A$ are given. Here, we assume that for $\gamma ^{\prime }\in \Gamma ^{\prime }$ , the action of $\gamma ^{\prime }$ on A as an element in $\Gamma $ is given by $a\mapsto \gamma ^{\prime }a(\gamma ^{\prime })^{-1}$ . We put $B = C[\Gamma ]\otimes _{C[\Gamma ^{\prime }]}A$ and define a multiplication by $(\gamma _{1}\otimes a_{1})(\gamma _{2}\otimes a_{2}) = \gamma _{1}\gamma _{2}\otimes (\gamma _{2}^{-1}\cdot a_{1})a_{2}$ for $a_{1},a_{2}\in A$ and $\gamma _{1},\gamma _{2}\in \Gamma $ . Of course, the example in our mind is $A = \mathcal {H}^{\mathrm {aff}}$ , $\Gamma = \Omega (1)$ and $\Gamma ^{\prime } = Z_{\kappa }$ . We have $B = \mathcal {H}$ .

Let $M_{1},M_{2}$ be right B-modules. Then $\operatorname {\mathrm {Hom}}_{A}(M_{1},M_{2})$ has the structure of a $\Gamma $ -module defined by $(f\gamma )(m) = f(m\gamma ^{-1})\gamma $ . This action factors through $\Gamma \to \Gamma /\Gamma ^{\prime }$ , and we have $\operatorname {\mathrm {Hom}}_{B}(M_{1},M_{2}) = \operatorname {\mathrm {Hom}}_{A}(M_{1},M_{2})^{\Gamma /\Gamma ^{\prime }}$ . Let N be a $\Gamma /\Gamma ^{\prime }$ -module and $\varphi \in \operatorname {\mathrm {Hom}}_{\Gamma /\Gamma ^{\prime }}(N,\operatorname {\mathrm {Hom}}_{A}(M_{1},M_{2}))$ . Set $f\colon N\otimes M_{1}\to M_{2}$ by $f(n\otimes m) = \varphi (n)(m)$ for $n\in N$ and $m\in M_{1}$ . Then for $\gamma \in \Gamma $ , we have $f(n\gamma \otimes m\gamma ) = \varphi (n\gamma )(m\gamma ) = \varphi (n)(m)\gamma = f(n\otimes m)\gamma $ . Namely, f is $\Gamma $ -equivariant. We define an action of $a\in A$ on $N\otimes M_{1}$ by $(n\otimes m)a = n\otimes ma$ . Then it coincides with the action of $\Gamma $ on $C[\Gamma ^{\prime }]$ , and it gives an action of B. This correspondence gives an isomorphism

$$ \begin{align*} \operatorname{\mathrm{Hom}}_{\Gamma/\Gamma^{\prime}}(N,\operatorname{\mathrm{Hom}}_{A}(M_{1},M_{2}))\simeq \operatorname{\mathrm{Hom}}_{B}(N\otimes M_{1},M_{2}). \end{align*} $$

In particular, if $M_{2}$ is an injective B-module, then $\operatorname {\mathrm {Hom}}_{A}(M_{1},M_{2})$ is an injective $\Gamma /\Gamma ^{\prime }$ -module. Therefore, from $\operatorname {\mathrm {Hom}}_{B}(M_{1},M_{2}) = \operatorname {\mathrm {Hom}}_{A}(M_{1},M_{2})^{\Gamma /\Gamma ^{\prime }}$ , we get a spectral sequence

$$ \begin{align*} E_{2}^{ij} = H^{i}(\Gamma/\Gamma^{\prime},\operatorname{\mathrm{Ext}}^{j}_{A}(M_{1},M_{2}))\Rightarrow \operatorname{\mathrm{Ext}}^{i + j}_{B}(M_{1},M_{2}). \end{align*} $$

In particular, we have an exact sequence

(4.4) $$ \begin{align} 0\to H^{1}(\Gamma/\Gamma^{\prime},\operatorname{\mathrm{Hom}}_{A}(M_{1},M_{2}))\to \operatorname{\mathrm{Ext}}^{1}_{B}(M_{1},M_{2})\to \operatorname{\mathrm{Ext}}^{1}_{A}(M_{1},M_{2})^{\Gamma/\Gamma^{\prime}} \end{align} $$

Moreover, we assume the following situation. Let $\Gamma _{1}$ be a finite index subgroup of $\Gamma $ which contains $\Gamma ^{\prime }$ , and put $B_{1} = A\otimes _{C[\Gamma ^{\prime }]}C[\Gamma _{1}]$ . Then this is a subalgebra of B, and B is a free left $B_{1}$ -module with a basis given by a complete representative of $\Gamma _{1}\backslash \Gamma $ . Assume that $M_{1}$ has a form $L_{1}\otimes _{B_{1}}B$ for some $B_{1}$ -module $L_{1}$ . We have $M_{1} = \bigoplus _{\gamma \in \Gamma _{1}\backslash \Gamma }L_{1}\otimes \gamma $ . Since B is flat over $B_{1}$ , we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}_{B}^{1}(M_{1},M_{2})\simeq \operatorname{\mathrm{Ext}}_{B_{1}}^{1}(L_{1},M_{2}). \end{align*} $$

We have a $B_{1}$ -module embedding $L_{1}\hookrightarrow M_{1}$ . This is in particular an A-homomorphism, and we get

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{A}(M_{1},M_{2})\to \operatorname{\mathrm{Ext}}^{1}_{A}(L_{1},M_{2}). \end{align*} $$

Since $L_{1}\hookrightarrow M_{1}$ is a $B_{1}$ -homomorphism, this is a $\Gamma _{1}$ -homomorphism. Hence, this induces

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{A}(M_{1},M_{2})\to \operatorname{\mathrm{Ind}}_{\Gamma_{1}}^{\Gamma}(\operatorname{\mathrm{Ext}}^{i}_{A}(L_{1},M_{2})). \end{align*} $$

The decomposition $M_{1} = \bigoplus _{\gamma \in \Gamma _{1}\backslash \Gamma }L_{1}\otimes \gamma $ respects the A-action. Hence,

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{A}(M_{1},M_{2}) = \bigoplus_{\gamma\in \Gamma_{1}\backslash\Gamma} \operatorname{\mathrm{Ext}}^{i}_{A}(L_{1}\otimes \gamma,M_{2}) = \bigoplus_{\gamma\in \Gamma_{1}\backslash\Gamma} \operatorname{\mathrm{Ext}}^{i}_{A}(L_{1},M_{2})\gamma. \end{align*} $$

Therefore, the above homomorphism is an isomorphism

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{A}(M_{1},M_{2})\simeq \operatorname{\mathrm{Ind}}_{\Gamma_{1}}^{\Gamma}(\operatorname{\mathrm{Ext}}^{i}_{A}(L_{1},M_{2})). \end{align*} $$

This implies

$$ \begin{gather*} H^{1}(\Gamma/\Gamma^{\prime},\operatorname{\mathrm{Hom}}_{A}(M_{1},M_{2})) \simeq H^{1}(\Gamma_{1}/\Gamma^{\prime},\operatorname{\mathrm{Hom}}_{A}(L_{1},M_{2})),\\ \operatorname{\mathrm{Ext}}^{1}_{A}(M_{1},M_{2})^{\Gamma/\Gamma^{\prime}} \simeq \operatorname{\mathrm{Ext}}^{1}_{A}(L_{1},M_{2})^{\Gamma_{1}/\Gamma^{\prime}} \end{gather*} $$

and a commutative diagram

We also assume that there exists a finite index subgroup $\Gamma _{2}$ of $\Gamma $ which contains $\Gamma ^{\prime }$ and $M_{2} = L_{2}\otimes _{B_{2}}B$ , where $B_{2} = A\otimes _{C[\Gamma ^{\prime }]}C[\Gamma _{2}]$ . Let $\{\gamma _{1},\dots ,\gamma _{r}\}$ be a set of complete representatives of $\Gamma _{2}\backslash \Gamma /\Gamma _{1}$ . Then the decomposition $M_{2} = \bigoplus _{\gamma \in \Gamma _{2}\backslash \Gamma }L_{2}\otimes \gamma = \bigoplus _{i}\bigoplus _{\gamma \in (\Gamma _{1}\cap \gamma _{i}^{-1}\Gamma _{2}\gamma _{i})\backslash \Gamma _{1}}L_{2}\otimes \gamma _{i}\gamma $ gives

$$ \begin{align*} M_{2}|_{B_{1}} = \bigoplus_{i} L_{2}\gamma_{i}\otimes_{B_{1}\cap \gamma_{i}^{-1}B_{2}\gamma_{i}}B_{1}, \end{align*} $$

where $L_{2}\gamma _{i}$ is a $\gamma _{i}^{-1}B_{2}\gamma _{i}$ -module defined by: $L_{2}\gamma _{i} = L_{2}$ as a vector space and the action is given by $l (\gamma _{i}^{-1}b\gamma _{i}) = l\cdot b$ for $l\in L_{2}$ and $b\in B_{2}$ , here $\cdot $ is the original action of $b\in B_{2}$ on $L_{2}$ . From this isomorphism, we get

$$ \begin{align*} H^{i}(\Gamma_{1}/\Gamma^{\prime},\operatorname{\mathrm{Ext}}^{j}_{A}(L_{1},M_{2})) \simeq \bigoplus_{i} H^{i}((\Gamma_{1}\cap \gamma_{i}^{-1}\Gamma_{2}\gamma_{i})/\Gamma^{\prime},\operatorname{\mathrm{Ext}}^{j}_{A}(L_{1},L_{2}\gamma_{i})) \end{align*} $$

and

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{i}_{B_{1}}(L_{1},M_{2}) = \bigoplus_{i}\operatorname{\mathrm{Ext}}^{i}_{B_{1}\cap \gamma_{i}^{-1}B_{2}\gamma_{i}}(L_{1},L_{2}\gamma_{i}) \end{align*} $$

which is compatible with the exact sequence in equation (4.4).

Set $A = \mathcal {H}^{\mathrm {aff}}$ , $\Gamma = \Omega (1)$ , $\Gamma ^{\prime } = Z_{\kappa }$ , $\Gamma _{1} = \Omega (1)_{\Xi }$ and $\Gamma _{2} = \Omega (1)_{\Xi ^{\prime }}$ . Then we get the following lemma. Recall that $\Omega $ is assumed to be commutative. For a character $\Xi $ of $\mathcal {H}$ , we defined $\Omega (1)_{\Xi }$ as the stabilizer of $\Xi $ in $\Omega (1)$ and $\mathcal {H}_{\Xi } = \mathcal {H}^{\mathrm {aff}} C[\Omega (1)_{\Xi }]$ in subsection 2.9.

Lemma 4.4. Let $\chi ,\chi ^{\prime }$ be characters of $Z_{\kappa }$ and $J\subset S_{\mathrm {aff},\chi },J^{\prime }\subset S_{\mathrm {aff},\chi ^{\prime }}$ . Put $\Xi = \Xi _{\chi ,J}$ , $\Xi ^{\prime } = \Xi _{\chi ^{\prime },J^{\prime }}$ , and let $V,V^{\prime }$ be irreducible $C[\Omega (1)_{\Xi }]$ , $C[\Omega (1)_{\Xi ^{\prime }}]$ -modules, respectively. Let $\{\omega _{1},\dots ,\omega _{r}\}$ be a set of complete representatives of $\Omega _{\Xi }\backslash \Omega /\Omega _{\Xi ^{\prime }} = \Omega /\Omega _{\Xi }\Omega _{\Xi ^{\prime }}$ , and define $\Xi ^{\prime }_{i}$ by $\Xi ^{\prime }_{i}(X) = \Xi ^{\prime }(\omega _{i} X\omega _{i}^{-1})$ . Consider the representation of $C[\Omega (1)_{\Xi ^{\prime }_{i}}] = \omega _{i}^{-1}C[\Omega (1)_{\Xi ^{\prime }}]\omega _{i}$ twisting $V^{\prime }$ by $\omega _{i}$ , and we denote it by $V^{\prime }_{i}$ . Put $\Omega _{\Xi ,\Xi ^{\prime }} = \Omega _{\Xi }\cap \Omega _{\Xi ^{\prime }}$ and $\mathcal {H}_{\Xi ,\Xi ^{\prime }} = \mathcal {H}_{\Xi }\cap \mathcal {H}_{\Xi ^{\prime }}$ . Then we have a commutative diagram with exact rows:

The following theorem will be proved in subsection 4.4.

Combining with Lemma 4.4, we get the surjectivity of $\operatorname {\mathrm {Ext}}^{1}_{\mathcal {H}}(\pi _{\chi ,J,V},\pi _{\chi ^{\prime },J^{\prime },V^{\prime }})\to \operatorname {\mathrm {Ext}}^{1}_{\mathcal {H}^{\mathrm {aff}}}(\pi _{\chi ,J,V},\pi _{\chi ^{\prime },J^{\prime },V^{\prime }})^{\Omega }$ stated in the introduction of this paper.

4.3 $\operatorname {\mathrm {Ext}}^{1}_{\mathcal {H}^{\mathrm {aff}}}(\Xi \otimes V,\Xi ^{\prime }\otimes V^{\prime })^{\Omega _{\Xi ,\Xi ^{\prime }}}$

To prove Theorem 4.5, we analyze $\operatorname {\mathrm {Ext}}^{1}_{\mathcal {H}^{\mathrm {aff}}}(\Xi \otimes V,\Xi ^{\prime }\otimes V^{\prime })^{\Omega _{\Xi ,\Xi ^{\prime }}}$ . First, we have

$$ \begin{align*} \operatorname{\mathrm{Ext}}^{1}_{\mathcal{H}^{\mathrm{aff}}}(\Xi\otimes V,\Xi^{\prime}\otimes V^{\prime})\simeq \operatorname{\mathrm{Ext}}^{1}_{\mathcal{H}^{\mathrm{aff}}}(\Xi,\Xi^{\prime})\otimes\operatorname{\mathrm{Hom}}_{C}(V,V^{\prime}) \end{align*} $$

and the surjective homomorphism $E(\Xi ,\Xi ^{\prime })\to \operatorname {\mathrm {Ext}}^{1}_{\mathcal {H}^{\mathrm {aff}}}(\Xi ,\Xi ^{\prime })$ . We have the decomposition $E(\Xi ,\Xi ^{\prime }) = E_{1}(\Xi ,\Xi ^{\prime })\oplus E_{2}(\Xi ,\Xi ^{\prime })$ . Let $E^{\prime }_{1}(\Xi ,\Xi ^{\prime })$ (resp. $E^{\prime }_{2}(\Xi ,\Xi ^{\prime })$ ) be the image of $E_{1}(\Xi ,\Xi ^{\prime })$ (resp. $E_{2}(\Xi ,\Xi ^{\prime })$ ). By the description of the kernel (4.1), we have:

• • $\operatorname {\mathrm {Ext}}^{1}_{\mathcal {H}^{\mathrm {aff}}}(\Xi ,\Xi ^{\prime }) = E^{\prime }_{1}(\Xi ,\Xi ^{\prime })\oplus E^{\prime }_{2}(\Xi ,\Xi ^{\prime })$ .

• • $E_{1}(\Xi ,\Xi ^{\prime })\xrightarrow {\sim } E^{\prime }_{1}(\Xi ,\Xi ^{\prime })$ .

• • the dimension of the kernel of $E_{2}(\Xi ,\Xi ^{\prime })\to E^{\prime }_{2}(\Xi ,\Xi ^{\prime })$ is at most 1.

Define $E_{i}\ (i = 2,3)$ by $E_{i} = E_{2}(\Xi ,\Xi ^{\prime })\cap \bigoplus _{s\in A_{i}(\Xi ,\Xi ^{\prime })}C_{s}$ . Then $\dim E_{2},\dim E_{3}\le 1$ and $E_{i}\ne 0$ if and only if for any $s\in A_{i}(\Xi ,\Xi ^{\prime })$ we have $s\chi = \chi ^{\prime }$ .

Assume that $s\chi = \chi ^{\prime }$ for any $s\in A_{2}(\Xi ,\Xi ^{\prime })$ . Fix $s_{0}\in A_{2}(\Xi ,\Xi ^{\prime })$ . Then $a = (a_{s})\mapsto a_{s_{0}}\chi (c_{s_{0}})^{-1}$ gives an isomorphism $E_{2}\simeq C$ . Let $\omega \in \Omega _{\Xi ,\Xi ^{\prime }}(1)$ . Then

$$ \begin{align*} (a\omega)_{s_{0}}\chi(c_{s_{0}})^{-1} = a_{\omega s\omega^{-1}}(\omega\widetilde{s_{0}}\omega^{-1})\chi(c_{\widetilde{s_{0}}})^{-1}. \end{align*} $$

Since $\omega $ stabilizes $\chi $ , we have $\chi (c_{\widetilde {s_{0}}}) = (\omega ^{-1}\chi )(c_{\widetilde {s_{0}}}) = \chi (\omega \cdot c_{\widetilde {s_{0}}}) = \chi (c_{\omega \widetilde {s_{0}}\omega ^{-1}})$ . Therefore, we have

$$ \begin{align*} (a\omega)_{s_{0}}\chi(c_{s_{0}})^{-1} & = a_{\omega s\omega^{-1}}(\omega \widetilde{s_{0}}\omega^{-1})\chi(c_{\omega\widetilde{s_{0}}\omega^{-1}})^{-1}\\ & = a_{\omega s\omega^{-1}}\chi(c_{\omega s\omega^{-1}})^{-1}\\ & = a_{s}\chi(c_{s})^{-1}. \end{align*} $$

Here, the last equality follows from the definition of $E^{\prime }_{2}(\Xi ,\Xi ^{\prime })$ . Namely, $\Omega _{\Xi ,\Xi ^{\prime }}$ acts trivially on $E_{2}$ . By the same argument, $\Omega _{\Xi ,\Xi ^{\prime }}$ also acts trivially on $E_{3}$ . Therefore, it also acts trivially on $E_{2}(\Xi ,\Xi ^{\prime })$ , hence on $E^{\prime }_{2}(\Xi ,\Xi ^{\prime })$ . Hence,