Proof. By conjugate-linearity it suffices to consider h of the form $N_w \theta _x$ with $w \in W$ and $x \in X$ . From (1.7) we see that

$$\begin{align*}h^* = N_{w_\Delta} \theta_{-w_\Delta (x)} N_{w_\Delta}^{-1} N_{w^{-1}} \quad \text{and} \quad (h_e^P)^{*_P} = \left\{ \begin{array}{cc} N_{w_P} \theta_{-w_P (x)} N_{w_P}^{-1} N_{w^{-1}} & w \in W_P , \\ 0 & w \notin W_P. \end{array} \right. \end{align*}$$

We recall from [Reference Humphreys16, §1.8] that

$$\begin{align*}\ell (w^{-1}) + \ell (w w_\Delta) = \ell (w_\Delta) = \ell (w_\Delta w) + \ell (w^{-1}). \end{align*}$$

By definition of the multiplication in ${\mathcal {H}} (W,q^\lambda )$ :

(1.9) $$ \begin{align} & N_{w_\Delta w} N_{w^{-1}} = N_{w_\Delta} = N_{w^{-1}} N_{w w_\Delta} , \end{align} $$

(1.10) $$ \begin{align} & N_{w_\Delta} \theta_{-w_\Delta (x)} N_{w_\Delta}^{-1} N_{w^{-1}} = N_{w_\Delta} \theta_{-w_\Delta (x)} N_{w w_\Delta}^{-1}. \end{align} $$

For a simple reflection $s \in W$ , $N_s^{-1} = N_s + (q_s^{-1/2} - q_s^{1/2}) N_e$ . That and the multiplication relations in the Bernstein presentation of ${\mathcal {H}}$ [Reference Lusztig20, §3] show that

(1.11) $$ \begin{align} \theta_y N_s^{-1} - N_s^{-1} \theta_{s (y)} \in \mathcal A \qquad \text{for all } y \in X. \end{align} $$

We denote the Bruhat order on W by $\leq $ . Applying (1.11) recursively, (1.10) can be expressed as $N_{w_\Delta } \sum _{v \in W , v \leq w w_\Delta } N_v^{-1} a_v$ for suitable $a_v \in \mathcal A$ . By (1.9) that equals $\sum _{v \in W ,v \leq w w_\Delta } N_{w_\Delta v^{-1}} a_v$ . Here $v^{-1} \leq w_\Delta w^{-1}$ , so $w_\Delta v^{-1} \geq w^{-1}$ .

Suppose that $w \notin W_P$ . Any reduced expression of $w^{-1}$ contains simple reflections not in $W_P$ , so the same goes for $w_\Delta v^{-1}$ with v as above. Hence $w_\Delta v^{-1} \notin W_P$ , and it can be written as $u w'$ with $u \in W^P \setminus \{e\}$ and $w' \in W_P$ . Thus $N_{w_\Delta v^{-1}} a_v \in N_u {\mathcal {H}}^P$ . That works for every $v \leq w w_\Delta $ , showing that (1.10) lies in ${\mathcal {H}} (W^P \setminus \! \{e\}) {\mathcal {H}}^P$ . In other words, $(h^*)^P_e = 0$ .

Suppose that $w \in W_P$ . We need to show that ${\mathcal {H}} (W^P \setminus \! \{e\}) {\mathcal {H}}^P$ contains

(1.12) $$ \begin{align} N_{w_\Delta} \theta_{-w_\Delta (x)} N_{w_\Delta}^{-1} N_{w^{-1}} - N_{w_P} \theta_{-w_P (x)} N_{w_P}^{-1} N_{w^{-1}}. \end{align} $$

With (1.9) we rewrite this element as

$$\begin{align*}\big( N_{w_\Delta} \theta_{-w_\Delta (x)} N_{w_{P,\Delta}}^{-1} - N_{w_P} \theta_{-w_P (x)} \big) N_{w_P}^{-1} N_{w^{-1}}. \end{align*}$$

Reasoning as above we find

$$\begin{align*}N_{w_\Delta} \theta_{-w_\Delta (x)} N_{w_{P,\Delta}}^{-1} = \sum\nolimits_{v \in W ,v \leq w_{P,\Delta}} N_{w_\Delta v^{-1}} a_v. \end{align*}$$

Here $w_\Delta v^{-1} \geq w_P$ , so this only belongs to $W_P$ if $v = w_{P,\Delta }$ . From (1.11) one obtains

$$\begin{align*}a_{w_{P,\Delta}} = \theta_{w_{P,\Delta} (-w_\Delta (x))} = \theta_{-w_P(x)}. \end{align*}$$

Then (1.12) reduces to

$$\begin{align*}\sum\nolimits_{v \in W ,v < w_{P,\Delta}} N_{w_\Delta v^{-1}} a_v N_{w_P}^{-1} N_{w^{-1}}. \end{align*}$$

The same argument as in the case $w \notin W_P$ shows that this lies in ${\mathcal {H}} (W^P \setminus \! \{e\}) {\mathcal {H}}^P$ .

Proof. (a) The ${\mathcal {H}} (W,q^\lambda )$ -module ${\mathcal {H}} (W,q^\lambda ) p_V^*$ is the Hermitian dual of ${\mathcal {H}} (W,q^\lambda ) p_V$ , with respect to the pairing

$$\begin{align*}\langle h_1 p_V^*, h_2 p_V \rangle = \tau (h_1 p_V^* h_2^*) \qquad h_1, h_2 \in {\mathcal{H}} (W,q^\lambda). \end{align*}$$

We note that ${\mathcal {H}}(W,q^\lambda ) p_V$ is a direct summand of the left regular representation of ${\mathcal {H}} (W,q^\lambda )$ , with complement ${\mathcal {H}}(W,q^\lambda ) (1 - p_V)$ . Applying $\mathrm {ind}_{{\mathcal {H}} (W, q^\lambda )}^{{\mathcal {H}}}$ yields

$$\begin{align*}\mathrm{ind}_{{\mathcal{H}} (W,q^\lambda)}^{{\mathcal{H}}} {\mathcal{H}}(W,q^\lambda) p_V \oplus \mathrm{ind}_{{\mathcal{H}} (W,q^\lambda)}^{{\mathcal{H}}} {\mathcal{H}}(W,q^\lambda) (1 - p_V) = {\mathcal{H}} p_V \oplus {\mathcal{H}} (1 - p_V) = {\mathcal{H}}. \end{align*}$$

We pass to Hermitian dual modules and obtain

(2.6) $$ \begin{align} {\mathcal{H}}^\dagger = {\mathcal{H}}^\dagger p_V^* \oplus {\mathcal{H}}^\dagger (1 - p_V^*). \end{align} $$

Hence the Hermitian dual of $\mathrm {ind}_{{\mathcal {H}} (W,q^\lambda )}^{{\mathcal {H}}} {\mathcal {H}}(W,q^\lambda ) p_V$ is

$$\begin{align*}{\mathcal{H}}^\dagger p_V^* = {\mathcal{H}}^\dagger \underset{{\mathcal{H}} (W,q^\lambda)}{\otimes} {\mathcal{H}} (W,q^\lambda) p_V^* \cong {\mathcal{H}}^\dagger \underset{{\mathcal{H}} (W,q^\lambda)}{\otimes} V^\dagger. \end{align*}$$

The pairing comes from the pairing between ${\mathcal {H}}$ and (2.6):

$$\begin{align*}\langle h_1 p_V^*, h_2 p_V \rangle = \tau (h_1 p_V^* (h_2 p_V)^* ) = \tau (h_1 p_V^* h_2^* ) \end{align*}$$

for $h_1 \in {\mathcal {H}}^\dagger $ and $h_2 \in {\mathcal {H}}$ . (b) Mod $({\mathcal {H}} (W,q^\lambda ))$ is only involved via functors that respect (infinite) direct sums. Further ${\mathcal {H}} (W,q^\lambda )$ is finite dimensional semisimple, so it suffices to consider its irreducible modules. They are all of the form $V^\dagger $ , for some irreducible ${\mathcal {H}} (W,q^\lambda )$ -module V. Let $Y \in \mathrm {Mod} ({\mathcal {H}})$ . With part (a), (2.2) and Frobenius reciprocity we compute

$$ \begin{align*} \mathrm{Hom}_{{\mathcal{H}}} \big( Y, {\mathcal{H}}^\dagger \otimes_{{\mathcal{H}} (W,q^\lambda)} V^\dagger \big) & \cong \mathrm{Hom}_{{\mathcal{H}}} \big( Y, (\mathrm{ind}^{{\mathcal{H}}}_{{\mathcal{H}} (W,q^\lambda)} V )^\dagger \big) \\ & \cong \mathrm{Hom}_{{\mathcal{H}}} \big( \mathrm{ind}^{{\mathcal{H}}}_{{\mathcal{H}} (W,q^\lambda)} V, Y^\dagger \big) \\ & \cong \mathrm{Hom}_{{\mathcal{H}} (W,q^\lambda)} (V,Y^\dagger ) \\ & \cong \mathrm{Hom}_{{\mathcal{H}} (W,q^\lambda)} (Y, V^\dagger ).\\[-37pt] \end{align*} $$

Proof. Take $v \in V_\pi , \lambda \in V_\pi ^\dagger , w \in W_P$ and $x \in X$ . With (1.7) we compute

$$\begin{align*}\begin{aligned} \lambda \big( (\pi \otimes t)(N_w \theta_x) v \big) & = \lambda \big( t(x) \pi (N_w \theta_x) v \big) = \overline{t(x)} \lambda \big( \pi (N_w \theta_x) v \big) \\ & = \overline{t(x)} \big( \pi^\dagger ((N_w \theta_x)^{*_P}) \lambda \big) (v) = \overline{t(x)} \big( \pi^\dagger ((N_w \theta_x)^{*_P}) \lambda \big) (v) \\ & = \overline{t(x)} \big( \pi^\dagger (N_{w_P} \theta_{-w_P (x)} N_{w_P}^{-1} N_{w^{-1}} ) \lambda \big) (v) \\ & = \big( (\pi^\dagger \otimes \overline{w_P t}^{-1}) (N_{w_P} \theta_{-w_P (x)} N_{w_P}^{-1} N_{w^{-1}} ) \lambda \big) (v) \\ & = \big( (\pi^\dagger \otimes \overline{w_P t}^{-1}) ((N_w \theta_x)^{*_P}) \lambda \big) (v). \end{aligned} \end{align*}$$

This shows that $(\pi \otimes t)^\dagger = \pi ^\dagger \otimes \overline {w_P t}^{-1}$ . But $w_P t = t$ because $w_P \in W(R_P)$ and $t \in \mathrm {Hom}_{\mathbb {Z}} (X / X \cap \mathbb Q P, {\mathbb {C}}^\times )$ .

Proof. By definition, as $w \in W_P$ :

$$\begin{align*}* \, *_P (N_w) = * (N_{w^{-1}}) = N_w. \end{align*}$$

From (1.7) and the anti-homomorphism property of * we obtain

(2.7) $$ \begin{align} * \, *_P (\theta_x) = *(N_{w_P}^{-1}) *(\theta_{-w_P (x)}) *(N_{w_P}) = N_{w_P}^{-1} N_{w_\Delta} \theta_{w_{\Delta,P} (x)} N_{w_\Delta}^{-1} N_{w_P}. \end{align} $$

We note that here the lengths of the involved elements of W add up:

$$\begin{align*}\ell (w_P) + \ell (w_{P,\Delta}) = |R_P^+| + |R^+ \setminus R_P^+ | = |R^+| = \ell (w_\Delta). \end{align*}$$

Therefore $N_{w_P} N_{w_{P,\Delta }} = N_{w_\Delta }$ , and the right-hand side of (2.7) simplifies to $N_{w_{P,\Delta }} \theta _{w_{\Delta ,P} (x)} N_{w_{P,\Delta }}^{-1}$ , proving the statement for $\theta _x$ .

To conclude, we use that $* \, *_P$ is an algebra homomorphism.

Proof. (a) Consider the algebra isomorphism

(2.8) $$ \begin{align} \begin{array}{llll} \psi_{w_{P,\Delta}} : & {\mathcal{H}}^{P^{op}} & \to & {\mathcal{H}}^P \\ & N_{w'} \theta_x & \mapsto & N_{w_{P,\Delta} w' w_\Delta wP} \theta_{w_{P,\Delta} (x)} \qquad w' \in W_{P^{op}}, x \in X. \end{array} \end{align} $$

Then $\psi _{\Delta P} \circ \psi _{w_{P,\Delta }} : {\mathcal {H}}^{P^{op}} \to {\mathcal {H}}$ is an injective algebra homomorphism, and by Lemma 2.3:

(2.9) $$ \begin{align} \begin{aligned} \psi_{\Delta P} \circ \psi_{w_{P,\Delta}} (N_{w'} \theta_x) & = \mathfrak c_{N_{w_{P,\Delta}}^{-1}} \big( N_{w_{P,\Delta} w' w_{\Delta,P}} N_{w_{P,\Delta}} \theta_{x} N_{w_{P,\Delta}}^{-1} \big) \\ & = N_{w_{P,\Delta}}^{-1} N_{w_{P,\Delta} w' w_{\Delta,P}} N_{w_{P,\Delta}} \theta_x. \end{aligned} \end{align} $$

Notice that $\psi _{\Delta P} \circ \psi _{w_{P,\Delta }}$ is the identity on $\mathcal A$ and sends ${\mathcal {H}} ( W_{P^{op}}, q^\lambda )$ bijectively to itself. For $\alpha \in P^{op}$ , $N_{s_\alpha }$ commutes with the same elements of $\mathcal A$ as $\psi _{\Delta P} \circ \psi _{w_{P,\Delta }} (N_{s_\alpha })$ . That forces

$$\begin{align*}\psi_{\Delta P} \circ \psi_{w_{P,\Delta}} (N_{s_\alpha}) \in {\mathbb{C}} N_e + {\mathbb{C}} N_{s_\alpha}. \end{align*}$$

Furthermore $\psi _{\Delta P} \circ \psi _{w_{P,\Delta }} (N_{s_\alpha })$ has the same eigenvalues $q_{s_\alpha }^{1/2}$ and $-q_{s_\alpha }^{-1/2}$ as $N_{s_\alpha }$ , so it can only be $N_{s_\alpha }$ or $-N_{s_\alpha }^{-1}$ . The involved constructions work for any $q \in \mathbb R_{>0}$ , and depend continuously on q. For $q=1$ we see directly from (2.9) that $\psi _{\Delta P} \circ \psi _{w_{P,\Delta }} (N_{s_\alpha }) = N_{s_\alpha }$ . Hence $\psi _{\Delta P} \circ \psi _{w_{P,\Delta }} (N_{s_\alpha })$ cannot be $-N_{s_\alpha }^{-1}$ for any $q \in \mathbb R_{>0}$ .

We deduce that $\psi _{\Delta P} \circ \psi _{w_{P,\Delta }} (N_{w'}) = N_{w'}$ for $w'$ any simple reflection in $W_{P^{op}}$ , and then the same follows for all $w' \in W_{P^{op}}$ . Apply that to $w' = w_{\Delta ,P} w w_{P,\Delta }$ . (b) By part (a) and (2.9), $\psi _{\Delta P} \circ \psi _{w_{P,\Delta }}$ is the identity on ${\mathcal {H}}^{P^{op}}$ . As $\psi _{w_{P,\Delta }} : {\mathcal {H}}^{P^{op}} \to {\mathcal {H}}^P$ is an isomorphism, this shows that $\psi _{\Delta P}$ is its inverse. By construction $w_{P^{op}} = w_\Delta w_P w_\Delta $ . From that, part (a) and (2.8) we see that $\psi _{w_{P,\Delta }} = \psi _{\Delta P^{op}}$ .

Further, from (2.8) and the definition of $\theta _x$ we obtain

$$\begin{align*}\psi_{w_{P,\Delta}} (N_{w'}) = N_{w_{P,\Delta} w' w_{\Delta,P}} \qquad \text{for all } w' \in W_{P^{op}} \ltimes X. \end{align*}$$

This shows that $\psi _{w_{P,\Delta }}$ is in fact a *-isomorphism, and hence so is its inverse.

Proof. By definition

$$\begin{align*}\mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^P}(\pi)^\dagger = \pi^\dagger \circ * \, *_P = \pi^\dagger \circ \mathfrak c_{N_{w_{P,\Delta}}} \circ \psi_{\Delta P}. \end{align*}$$

Since $N_{w_{P,\Delta }} \in {\mathcal {H}}^\times $ , multiplication with $\pi ^\dagger (N^{-1}_{w_{P,\Delta }})$ provides an isomorphism from the right-hand side to $\pi ^\dagger \circ \psi _{\Delta P}$ . By Lemma 2.4.b, that can be regarded as $\mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^{P^{op}}} (\pi ^\dagger ) \circ \psi _{\Delta P}$ . (b) Start with part (a) for $P^{op}$ . Composing the representations on both sides with ${\psi _{\Delta P^{op}}^{-1} = \psi _{\Delta P}}$ gives

$$\begin{align*}\mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^P} (\pi^\dagger) \cong \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^{P^{op}}}(\pi)^\dagger \circ \psi_{ \Delta P^{op}}^{-1} = \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^{P^{op}}}(\pi)^\dagger \circ \psi_{\Delta P}.\\[-33pt] \end{align*}$$

Proof. Let $a,b \in W^P, h_1 \in {\mathcal {H}}^P$ and $\lambda \in {\mathcal {H}}^{P\dagger }$ . For $h_2 \in {\mathcal {H}}$ there are elements

$$\begin{align*}m^\dagger \big( h_2 \cdot (m^\dagger)^{-1}(N_a \otimes \lambda) \big) \quad \text{and} \quad h_2 \cdot (N_a \otimes \lambda) \quad \text{in} \quad \mathrm{ind}^{{\mathcal{H}}}_{{\mathcal{H}}^P} ({\mathcal{H}}^{P\dagger}). \end{align*}$$

We can compare them by pairing $N_b \otimes {\mathcal {H}} (W^P) \otimes {\mathcal {H}}^P$ , as in (2.11). In the notation from (1.8), we compute

(2.13) $$ \begin{align} &\langle h_2 \cdot (N_a \otimes \lambda), N_b \otimes h_1 \rangle = \langle h_2 N_a \cdot (1 \otimes \lambda), N_b \otimes h_1 \rangle \\& \quad =\sum\nolimits_{v \in W^P} \langle N_v (h_2 N_a)^P_v \cdot (1 \otimes \lambda), N_b \otimes h_1 \rangle = \langle (h_2 N_a)^P_b \cdot \lambda, h_1 \rangle .\nonumber \end{align} $$

We note that, for any $w \in W_P, v \in W^P$ :

(2.14) $$ \begin{align} \langle N_b^* N_v, N_w \rangle = \langle N_u , N_b N_w \rangle = \langle N_u, N_{bw} \rangle = \left\{ \begin{array}{@{}ll} 1 & v = b, w = e, \\ 0 & \text{otherwise.} \end{array} \right. \end{align} $$

This implies

(2.15) $$ \begin{align} \big( N_b^* N_v (h_2 N_a )_v \big)^P_e = \left\{ \begin{array}{@{}ll} (h_2 N_a)_v & b = v,\\ 0 & b \neq v. \end{array} \right. \end{align} $$

With that the right-hand side of (2.13) can be rewritten as

(2.16) $$ \begin{align} \langle (N_b^* h_2 N_a)^P_e \cdot \lambda, h_1 \rangle = \big\langle \lambda, \big( (N_b^* h_2 N_a)^P_e \big)^{*_P} h_1 \big\rangle. \end{align} $$

On the other hand

(2.17) $$ \begin{align} \begin{aligned} \big\langle m^\dagger \big( h_2 \cdot & (m^\dagger)^{-1}(N_a \otimes \lambda) \big), N_b \otimes h_1 \big \rangle \;=\; \big\langle h_2 \cdot (m^\dagger)^{-1}(N_a \otimes \lambda) , N_b h_1 \big\rangle \\ & =\; \big\langle (m^\dagger)^{-1}(N_a \otimes \lambda) , h_2^* N_b h_1 \big\rangle \;=\; \langle N_a \otimes \lambda , m^{-1}(h_2^* N_b) h_1 \rangle \\ & =\; \sum\nolimits_{v \in W^P} \langle N_a \otimes \lambda, N_v \otimes (h_2^* N_b)^P_v h_1 \rangle \;=\; \langle \lambda, (h_2^* N_b)^P_a h_1 \rangle \end{aligned} \end{align} $$

Using (2.15) we identify the last expression in (2.17) with

$$\begin{align*}\langle \lambda, (N_a^* h_2^* N_b)^P_e h_1 \rangle = \big\langle \lambda, \big( (N_b^* h_2 N_a)^* \big)^P_e h_1 \big\rangle. \end{align*}$$

Lemma 1.1 guarantees that this equals (2.16), which proves that the bijection $m^\dagger $ is an ${\mathcal {H}}$ -module homomorphism.

By construction $m^{-1} : {\mathcal {H}} \to {\mathcal {H}} (W^P) \otimes {\mathcal {H}}^P$ and $m^\dagger $ transfer the pairing between ${\mathcal {H}}$ and ${\mathcal {H}}^\dagger $ to the pairing (2.11). Thus we have realized ${\mathcal {H}}^\dagger = \mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}}({\mathcal {H}}^P)^\dagger $ as $\mathrm {ind}^{{\mathcal {H}}}_{{\mathcal {H}}^P} ({\mathcal {H}}^{P\dagger })$ .

Proof. We abbreviate $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}}$ to ind for the duration of this proof.

Recall that $\{ N_w : w \in W \}$ is an orthonormal basis of ${\mathcal {H}} (W,q^\lambda )$ for the inner product (1.4). Hence ${\mathcal {H}} (W^P)^\dagger \cong {\mathcal {H}} (W^P)$ and (2.11) identifies

$$\begin{align*}{\mathcal{H}} (W^P) \otimes_{\mathbb{C}} V_\rho^\dagger \cong {\mathcal{H}} \otimes_{{\mathcal{H}}^P} V_\rho^\dagger \quad \text{with} \quad ({\mathcal{H}} (W^P) \otimes_{\mathbb{C}} V_\rho )^\dagger \cong ({\mathcal{H}} \otimes_{{\mathcal{H}}^P} V_\rho )^\dagger. \end{align*}$$

It remains to show that

(2.20) $$ \begin{align} \langle \mathrm{ind} (\pi^\dagger) (h^*) x , y \rangle \quad \text{equals} \quad \langle x, \mathrm{ind} (\pi)(h) y \rangle \end{align} $$

for all $h \in {\mathcal {H}}, x \in \mathrm {ind} (V_\rho ^\dagger ), y \in \mathrm {ind} (V_\rho )$ .

Choose a surjective ${\mathcal {H}}^P$ -homomorphism $p : F \otimes {\mathcal {H}}^P \to V_\rho $ , where $F \otimes {\mathcal {H}}^P$ is a free ${\mathcal {H}}^P$ -module. Dually, that yields an injective ${\mathcal {H}}^P$ -homomorphism $p^\dagger : V_\rho ^\dagger \to (F \otimes {\mathcal {H}}^P )^\dagger $ . For $\lambda \in V_\rho ^\dagger $ and $v \in V_\rho $ with a preimage $\tilde v \in F \otimes {\mathcal {H}}^P$ that means

$$\begin{align*}\langle \lambda , v \rangle = \langle \lambda, p(\tilde v) \rangle = \langle p^\dagger (\lambda), \tilde v \rangle. \end{align*}$$

With the functoriality of induction we obtain a surjective ${\mathcal {H}}$ -homomorphism

$$\begin{align*}\mathrm{ind} (p) : F \otimes {\mathcal{H}} = \mathrm{ind} (F \otimes {\mathcal{H}}^P) \to \mathrm{ind} (V_\rho) , \end{align*}$$

and an injective ${\mathcal {H}}$ -homomorphism

$$\begin{align*}\mathrm{ind} (p^\dagger) : \mathrm{ind} (V_\rho^\dagger) \to \mathrm{ind} \big( (F \otimes {\mathcal{H}}^P)^\dagger \big). \end{align*}$$

Now we encounter the minor complication that it is difficult to work with $(F \otimes {\mathcal {H}}^P)^\dagger $ when F has infinite dimension. We overcome that by playing it via finitely generated submodules. Choose a finite dimensional linear subspace $F_y \subset F$ and $\tilde y \in F_y \otimes {\mathcal {H}}$ such that

$$\begin{align*}y = \mathrm{ind} (p) \tilde y \in \mathrm{ind} (p) (F_y \otimes {\mathcal{H}}) = \mathrm{ind} \big( p (F_y \otimes {\mathcal{H}}^P) \big). \end{align*}$$

It follows from Lemma 2.6 that

(2.21) $$ \begin{align} \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} \big( F_y^\dagger \otimes {\mathcal{H}}^{P\dagger} \big) = F_y^\dagger \otimes \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} ( {\mathcal{H}}^{P\dagger}) \cong \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} ( F_y \otimes {\mathcal{H}}^P )^\dagger , \end{align} $$

where the ${\mathcal {H}}$ -invariant pairing with $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}} ( F_y \otimes {\mathcal {H}}^P )$ is given by (2.11) and the pairing between $F_y$ and $F_y^\dagger $ . The transpose of the inclusion $i_y : F_y \otimes {\mathcal {H}}^P \to F \otimes {\mathcal {H}}^P$ is the projection

$$\begin{align*}i_y^\dagger : (F \otimes {\mathcal{H}}^P)^\dagger \to (F_y \otimes {\mathcal{H}}^P )^\dagger = F_y^\dagger \otimes ({\mathcal{H}}^P )^\dagger. \end{align*}$$

To these maps we can also apply ind. In that way (2.20) can be evaluated via the pairing of $\mathrm {ind} (F_y \otimes {\mathcal {H}}^P)$ with $\mathrm {ind} \big ( F_y^\dagger \otimes ({\mathcal {H}}^P )^\dagger \big )$ determined by (2.11). More explicitly:

$$\begin{align*}\langle x, \mathrm{ind} (\pi)(h) y \rangle = \langle x, \mathrm{ind} (\pi)(h) \mathrm{ind} (p \, i_y) \tilde y \rangle = \langle x, \mathrm{ind} (p \, i_y) h \tilde y \rangle = \langle \mathrm{ind} (i_y^\dagger p^\dagger) x , h \tilde y \rangle. \end{align*}$$

By (2.21) the right-hand side equals

$$ \begin{align*} \langle h^* \, \mathrm{ind} (i_y^\dagger p^\dagger) x , \tilde y \rangle & = \langle \mathrm{ind} (i_y^\dagger p^\dagger) \mathrm{ind} (\pi^\dagger) (h^*) x , \tilde y \rangle \\ & = \langle \mathrm{ind} (\pi^\dagger) (h^*) x , \mathrm{ind} (p \, i_y) \tilde y \rangle = \langle \mathrm{ind} (\pi^\dagger) (h^*) x , y \rangle. \end{align*} $$

This establishes (2.20).

Proof. (a) Let $(\pi , V_\pi ) \in \mathrm {Mod} ({\mathcal {H}})$ and $(\rho , V_\rho ) \in \mathrm {Mod} ({\mathcal {H}}^P)$ . By the transposition isomorphism (2.2) and Proposition 2.5

(3.5) $$ \begin{align} \mathrm{Hom}_{{\mathcal{H}}^P} \big( \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^P} (\pi), \rho^\dagger \big) \cong \mathrm{Hom}_{{\mathcal{H}}^P} \big( \rho, \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^P} (\pi)^\dagger \big) \cong \mathrm{Hom}_{{\mathcal{H}}^P} \big( \rho, \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^{P^{op}}} (\pi^\dagger) \circ \psi_{\Delta P} \big). \end{align} $$

We know from Lemma 2.4 that $\psi _{\Delta P} : {\mathcal {H}}^P \to {\mathcal {H}}^{P^{op}}$ is invertible, so that the right-hand side of (3.5) becomes isomorphic with

(3.6) $$ \begin{align} \mathrm{Hom}_{{\mathcal{H}}^{P^{op}}} \big( \rho \circ \psi_{\Delta P}^{-1}, \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^{P^{op}}} (\pi^\dagger) \big). \end{align} $$

Now we apply Frobenius reciprocity in the form (3.1) and again the transposition isomorphism:

(3.7) $$ \begin{align} \cong \mathrm{Hom}_{{\mathcal{H}}} \big( \mathrm{ind}_{{\mathcal{H}}^{P^{op}}}^{{\mathcal{H}}} (\rho \circ \psi_{\Delta P}^{-1}), \pi^\dagger \big) \cong \mathrm{Hom}_{{\mathcal{H}}} \big( \pi, \mathrm{ind}_{{\mathcal{H}}^{P^{op}}}^{{\mathcal{H}}} (\rho \circ \psi_{\Delta P}^{-1})^\dagger \big). \end{align} $$

Using Proposition 2.7, we identify that with

(3.8) $$ \begin{align} \mathrm{Hom}_{{\mathcal{H}}} \big( \pi, \mathrm{ind}_{{\mathcal{H}}^{P^{op}}}^{{\mathcal{H}}} (\rho^\dagger \circ \psi_{\Delta P}^{-1}) \big) = \mathrm{Hom}_{{\mathcal{H}}} \big( \pi, \mathrm{ind}_{{\mathcal{H}}^{P^{op}}}^{{\mathcal{H}}} \circ \psi_{\Delta P} (\rho^\dagger) \big). \end{align} $$

The first isomorphism in (3.5) and the second in (3.7) are conjugate-linear. The other isomorphisms above are complex linear, so the composition of (3.5)–(3.8) is again a complex linear bijection. That proves the desired adjointness relation for $(\pi , \rho ' = \rho ^\dagger )$ , so whenever $\rho '$ is the Hermitian dual of some ${\mathcal {H}}^P$ -module. The same argument as in the analogous situation for reductive p-adic groups [Reference Renard25, p. 232] shows why that implies part (a) for all $(\pi , \rho ')$ . (b) Reverse the roles of P and $P^{op}$ and apply part (a) with $\rho ' = \psi _{\Delta P}^* (\rho ) = \rho \circ \psi _{\Delta P}$ . That gives isomorphisms

$$\begin{align*}\mathrm{Hom}_{{\mathcal{H}}} \big( \pi, \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} (\rho') \big) \cong \mathrm{Hom}_{{\mathcal{H}}^{P^{op}}} \big( \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^{P^{op}}} (\pi), \rho \big). \end{align*}$$

Left composition with $\psi _{\Delta P}^{-1}$ on both terms of the right-hand side makes this isomorphic with $\mathrm {Hom}_{{\mathcal {H}}^{P}} \big ( \psi _{\Delta P}^* \, \mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^{P^{op}}} (\pi ), \rho ' \big )$ .

Proof. Let $s \in T$ be a weight of $\pi ^\dagger $ , with an eigenvector $\lambda \in V_\pi ^\dagger \setminus \{0\}$ . For any $v \in V, x \in X$ we compute, using (1.7),

(4.2) $$ \begin{align} s(x) \lambda (v) = (\pi^\dagger (\theta_x) \lambda) (v) = \lambda (\pi (\theta_x^*) v) = \lambda (\pi (N_{w_\Delta} \theta_{-w_\Delta (x)} N_{w_\Delta}^{-1}) v). \end{align} $$

Write $\pi ' = \pi \circ \mathfrak {c}_{N_{w_\Delta }}$ , so that

$$\begin{align*}\begin{array}{ccc} (\pi, V_\pi) & \to & (\pi', V_\pi) \\ v & \mapsto & \pi (N_{w_\Delta}) v \end{array} \end{align*}$$

is an isomorphism of ${\mathcal {H}}$ -representations. We can rewrite (4.2) as

$$\begin{align*}\lambda (\overline{s(x)} v) = \lambda (\pi' (\theta_{-w_\Delta (x)}) v). \end{align*}$$

Equivalently, for each $x \in X, v \in V_\pi $ the kernel of $\lambda $ contains

$$\begin{align*}\pi' \big( \theta_{-w_\Delta (x)} - \overline{s(x)} \big) v = \pi' \big( \theta_{x'} - \overline{s}(-w_\Delta (x')) \big) v = \pi' \big( \theta_{x'} - \overline{w_\Delta s}^{-1}(x') \big) v , \end{align*}$$

where we abbreviated $x' = -w_\Delta (x)$ . Thus $\pi ' \big ( \theta _{x'} - \overline {w_\Delta s}^{-1}(x') \big )$ is not surjective, for any $x' \in X$ . Since $V_\pi $ has finite dimension, we can use the decomposition, which shows that $\overline {w_\Delta s}^{-1}$ is a weight of $\pi '$ . Via the isomorphism $\pi ' \cong \pi $ , it is also a weight of $\pi $ .

Hence $s \mapsto \overline {w_\Delta s}^{-1}$ maps the weights of $\pi ^\dagger $ to the weights of $\pi $ . As $\dim V_\pi < \infty $ , $\pi ^{\dagger \dagger } = \pi $ and the same arguments apply with the roles of $\pi $ and $\pi ^\dagger $ exchanged. Therefore we have found a bijection between the set of weights of $\pi ^\dagger $ and of $\dagger $ , with inverse $t \mapsto \overline {w_\Delta t}^{-1}$ .

Proof. Since $-w_\Delta $ stabilizes $\Delta $ , $w_{\Delta } s^{-1} \in T^{\Delta -}$ if and only if $s \in T^{\Delta -}$ . Apply that to $s = |t|$ (resp. $s = |t|^{-1}$ ) for a weight t of $V_\pi $ , and use Lemma 4.1.

Proof. (i) The Langlands classification, as in [Reference Evens12, Theorem 2.1] and [Reference Solleveld31, Theorem 2.2.4], states (a) and (b). Although the characterizing property of $L(P,\pi ,t)$ is not made explicit in these sources, it plays an important role in [Reference Evens12, §2.7] and in [Reference Solleveld31, proof of Theorem 2.2.4.a]. (ii) The same proof as for (i) applies, when we rewrite all the arguments in $Y \otimes _{\mathbb {Z}} \mathbb R$ with respect to $-\Delta $ instead of $\Delta $ .

Proof. We assume (i). The proof when (ii) holds is completely analogous, only using the other assumption in Theorem 4.3. (a) By Proposition 2.7 and Lemma 2.2

(4.3) $$ \begin{align} \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}}(\pi \otimes t )^\dagger \cong \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} \big( (\pi \otimes t)^\dagger \big) \cong \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}}(\pi^\dagger \otimes \overline{t}^{-1}). \end{align} $$

Here $\bar t = t$ because it is real-valued, and we know from Lemma 4.2 that $\pi ^\dagger $ is tempered. Theorem 4.3.a says that (4.3) has a unique irreducible quotient $L(P,\pi ^\dagger ,t^{-1})$ , which can be characterized by the existence of an injection

$$\begin{align*}\pi^\dagger \otimes t^{-1} \to \mathrm{Res}_{{\mathcal{H}}^P}^{{\mathcal{H}}} \big( L(P,\pi^\dagger,t^{-1}) \big). \end{align*}$$

Passing to Hermitian duals and using (2.2), we find that $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}}(\pi \otimes t)$ has a unique irreducible subrepresentation $\sigma \cong L(P,\pi ^\dagger ,t^{-1})^\dagger $ . Via Proposition 2.5.a, the characterizing property becomes a surjection $\mathrm {Res}_{{\mathcal {H}}^{P^{op}}}^{{\mathcal {H}}} (\sigma ) \circ \psi _{\Delta P} \to \pi \otimes t$ . (b) Let $\tau \in \mathrm {Irr} ({\mathcal {H}})$ . With Theorem 4.3.b we write $\tau ^\dagger \cong L(P,\pi ,t)$ for suitable $P \subset \Delta $ , tempered $\pi \in \mathrm {Irr} ({\mathcal {H}}^P)$ and $t \in T^{P+}$ . Then (2.2) gives an injection

$$\begin{align*}\tau \cong L(P,\pi,t)^\dagger \to \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} (\pi \otimes t)^\dagger. \end{align*}$$

From the proof of part (a) we know that the right-hand side is isomorphic with ${\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}}(\pi \otimes t^{-1})}$ , where $(P,\pi ,t^{-1})$ is as in (i). Then part (a) says $\tau \cong \tilde L (P,\pi ,t^{-1})$ . The uniqueness in Theorem 4.3.b implies the uniqueness of $(P,\pi ,t^{-1})$ .

Proof. From Proposition 3.2.c we know that $\mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^P} \mathrm {ind}_{{\mathcal {H}}^Q}^{{\mathcal {H}}} (\psi _{\gamma *} \pi )$ has a filtration with successive subquotients

(4.4) $$ \begin{align} \mathrm{ind}_{{\mathcal{H}}^{P \cap d(Q)}}^{{\mathcal{H}}^P} \big( \psi_d \, \mathrm{Res}^{{\mathcal{H}}^Q}_{{\mathcal{H}}^{d^{-1}(P) \cap Q}} (\psi_\gamma \, \pi) \big) = \mathrm{ind}_{{\mathcal{H}}^{P \cap d(Q)}}^{{\mathcal{H}}^P} (\psi_d \psi_\gamma \, \pi ) \qquad d \in W^{P,Q}. \end{align} $$

By construction $\mathrm {Wt}(\psi _{d *} \psi _{\gamma *} \pi ) = d \gamma \mathrm {Wt}(\pi )$ , and with [Reference Opdam23, Proposition 4.20] we obtain

(4.5) $$ \begin{align} \mathrm{Wt} \big( \mathrm{ind}_{{\mathcal{H}}^{P \cap d(Q)}}^{{\mathcal{H}}^P} (\psi_{d *} \psi_{\gamma *} \, \pi ) \big) \: \subset \: W_P d \gamma \mathrm{Wt}(\pi). \end{align} $$

Equivalently, the set of $Z({\mathcal {H}}^P)$ -weights of $\mathrm {ind}_{{\mathcal {H}}^{P \cap d(Q)}}^{{\mathcal {H}}^P} (\psi _{d *} \psi _{\gamma *} \, \pi )$ is contained in $W_P d \gamma \mathrm {Wt}(\pi ) / W_P$ . Suppose that $d,d' \in W^{P,Q}, t, t' \in \mathrm {Wt}(\pi )$ and

$$\begin{align*}W_P d \gamma t \cap W_P d' \gamma t' \neq \emptyset. \end{align*}$$

Pick $w_1,w_2 \in W_P$ such that $w_1 d \gamma t = w_2 d' \gamma t'$ . By the irreducibility of $\pi $ , t and $t'$ belong to the same $W_P$ -orbit. Furthermore we assumed $W_t \subset W_P$ , so

(4.6) $$ \begin{align} w_1 d \gamma W_P = w_2 d' \gamma W_P. \end{align} $$

From that we obtain $\gamma ^{-1} d^{-1} w_1^{-1} w_2 d' \gamma \in W_P$ and

$$\begin{align*}d^{-1} w_1^{-1} w_2 d' \in \gamma W_P \gamma^{-1} = W_Q. \end{align*}$$

We note that now $w_1^{-1} w_2 d' \in W_P d' \cap d W_Q$ . As $W^{P,Q}$ represents $W_P \backslash W / W_Q$ , this shows that $d' = d$ .

Thus, for different $d,d' \in W^{P,Q}$ the ${\mathcal {H}}^P$ -representations (4.4) have disjoint sets of $\mathcal A$ -weights and disjoint sets of $Z({\mathcal {H}}^P)$ -weights. In particular every extension of one of these modules by the other is a trivial extension. It follows that the aforementioned filtration of $\mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^P} \mathrm {ind}_{{\mathcal {H}}^Q}^{{\mathcal {H}}} (\psi _{\gamma *} \pi )$ actually splits, and that $\mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^P} \mathrm {ind}_{{\mathcal {H}}^Q}^{{\mathcal {H}}} (\psi _{\gamma *} \pi )$ is the direct sum of the modules (4.4).

Proof. (a) The element $\gamma ^{-1} \in W$ belongs to $W^{P,Q}$ because $\gamma ^{-1}(Q) \subset R^+$ and $\gamma (P) \subset R^+$ . We follow the proof of Lemma 4.5, with $d' = \gamma ^{-1}$ . This time we cannot conclude (4.6), but our weaker assumption still provides a reasonable substitute. Namely, from $w_1 d \gamma t = w_2 t'$ we get $w_2^{-1} w_1 d \gamma t \in \mathrm {Wt}(\pi )$ , which by the $W,\!P$ -regularity of $\pi $ implies

(4.8) $$ \begin{align} w_2^{-1} w_1 d \gamma \notin W_P (W^{P,P} \setminus \{e\}). \end{align} $$

Notice that $d \gamma (P) = d(Q) \subset R^+$ , which says that $d \gamma \in W^P$ . By [Reference Carter6, Proposition 2.7.5] we can write $d \gamma = a \tilde d$ with $\tilde d \in W^{P,P}$ and $a \in W_P$ . It follows that $W_P d \gamma = W_P \tilde d$ . Then (4.8) forces

$$\begin{align*}\tilde d = e \quad \text{and} \quad d \gamma \in W_P \cap W^P = \{e\}. \end{align*}$$

Hence the representations (4.4) with $d \neq \gamma ^{-1}$ do not have the central character of ${\pi \in \mathrm {Irr} ({\mathcal {H}}^P)}$ as $Z({\mathcal {H}}^P)$ -weight. Like in the proof of Lemma 4.5, this entails that $\pi $ appears with multiplicity one in $\mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^P} \mathrm {ind}_{{\mathcal {H}}^Q}^{{\mathcal {H}}} (\psi _{\gamma *} \pi )$ , as a direct summand. (b) By Frobenius reciprocity

$$\begin{align*}\mathrm{Hom}_{{\mathcal{H}}} \big( \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} (\pi), \mathrm{ind}_{{\mathcal{H}}^Q}^{{\mathcal{H}}} (\psi_{\gamma *} \, \pi) \big) \cong \mathrm{Hom}_{{\mathcal{H}}^P} \big( \pi, \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^P} \mathrm{ind}_{{\mathcal{H}}^Q}^{{\mathcal{H}}} (\psi_{\gamma *} \pi) \big). \end{align*}$$

Now apply part (a).

Proof. (a) Let $\rho $ be any quotient ${\mathcal {H}}$ -representation of $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}}(\pi )$ . The quotient map gives a nonzero element of

$$\begin{align*}\mathrm{Hom}_{{\mathcal{H}}}(\mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}}(\pi), \rho) \cong \mathrm{Hom}_{{\mathcal{H}}^P}(\pi, \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^P} \rho) , \end{align*}$$

so $\pi $ is a subrepresentation of $\mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^P} \rho $ . By Lemma 4.6.a with $\gamma = e$ , $\pi $ is a direct summand of $\mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^P} \rho $ , and appears with multiplicity one. The projection $\mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^P} \rho \to \pi $ and the adjunction from Theorem 3.1.a yield a nonzero ${\mathcal {H}}$ -homomorphism from $\rho $ to $\mathrm {ind}_{{\mathcal {H}}^{P^{op}}}^{{\mathcal {H}}} (\psi _{\Delta P *} \pi )$ . Thus we have ${\mathcal {H}}$ -homomorphisms

(4.10) $$ \begin{align} \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} (\pi) \to \rho \to \mathrm{ind}_{{\mathcal{H}}^{P^{op}}}^{{\mathcal{H}}} (\psi_{\Delta P *} \pi). \end{align} $$

Suppose now that $\rho $ is irreducible. Then the second map in (4.10) is injective, and the first map is surjective by definition, so the composition of the two maps in (4.10) is nonzero. Lemma 4.6.a guarantees that (4.10) is a multiple of $I(w_{\Delta ,P},P,\pi )$ . In particular

$$\begin{align*}\ker I(w_{\Delta,P},P,\pi) = \ker ({4.10}) = \ker \big( \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} (\pi) \to \rho \big). \end{align*}$$

We conclude that $\rho $ equals $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}} (\pi \otimes t) / \ker I(w_{\Delta ,P},P,\pi )$ . (b) Let $\sigma $ be any subrepresentation of $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}}(\pi )$ . The inclusion map and Theorem 3.1.b give a nonzero element of

$$\begin{align*}\mathrm{Hom}_{{\mathcal{H}}}(\sigma, \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}}(\pi)) \cong \mathrm{Hom}_{{\mathcal{H}}^P} \big( \psi_{\Delta P^{op} *} (\mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^{P^{op}}} \sigma), \pi \big) \cong \mathrm{Hom}_{{\mathcal{H}}^{P^{op}}} (\sigma, \psi_{\Delta P *} \pi ). \end{align*}$$

In the setting of Lemma 4.6 we take $P^{op},P,\psi _{\Delta P^{op}},\psi _{\Delta P}\pi $ in the roles of, respectively, $P,Q,\gamma ,\pi $ . Then Lemma 4.6.a says that $\psi _{\Delta P *} \pi $ appears with multiplicity one in

(4.11) $$ \begin{align} \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^{P^{op}}} \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} (\psi_{\Delta P^{op}} \psi_{\Delta P} \pi) = \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^{P^{op}}} \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} (\pi) , \end{align} $$

as a direct summand. Since $\psi _{\Delta P} (\pi )$ appears in the subrepresentation $\mathrm {Res}^{{\mathcal {H}}}_{{\mathcal {H}}^{P^{op}}} \sigma $ of (4.11), it is also a direct summand thereof. In particular there exists a nonzero element of

$$\begin{align*}\mathrm{Hom}_{{\mathcal{H}}^{P^{op}}} (\psi_{\Delta P *} \pi, \mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}}^{P^{op}}} \sigma) \cong \mathrm{Hom}_{{\mathcal{H}}^P} (\mathrm{ind}_{{\mathcal{H}}^{P^{op}}}^{{\mathcal{H}}} (\psi_{\Delta P *} \pi), \sigma). \end{align*}$$

Thus we have nonzero ${\mathcal {H}}$ -homomorphisms

(4.12) $$ \begin{align} \mathrm{ind}_{{\mathcal{H}}^{P^{op}}}^{{\mathcal{H}}} (\psi_{\Delta P *} \pi) \to \sigma \to \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}}(\pi). \end{align} $$

Now we assume that $\sigma $ is irreducible. Then the first map in (4.12) is surjective and the second map is injective, so their composition is nonzero. The same argument as for (4.10) shows that $\sigma $ is isomorphic to

$$\begin{align*}\mathrm{ind}_{{\mathcal{H}}^{P^{op}}}^{{\mathcal{H}}} (\psi_{\Delta P *} \pi) / \ker I ( w_{P,\Delta} , P^{op}, \psi_{w_{\Delta,P} *} \pi ) \cong \mathrm{im}\, I ( w_{P,\Delta} , P^{op}, \psi_{w_{\Delta,P} *} \pi ).\\[-39pt] \end{align*}$$

Proof. First we establish the regularity in all four cases. (i) Recall from [Reference Langlands19, Lemma 4.4] that every $\lambda \in Y \otimes _{\mathbb {Z}} \mathbb R$ can be expressed uniquely as

$$\begin{align*}\lambda = \lambda_- + \lambda_+ , \quad \text{where} \quad \lambda \in (Y \otimes_{\mathbb{Z}} \mathbb R)_{Q-} ,\: \lambda_+ \in (Y \otimes_{\mathbb{Z}} \mathbb R)^{Q+} ,\: Q = Q(\lambda) \subset \Delta. \end{align*}$$

For any $s \in \mathrm {Wt}(\pi )$ we have

(4.13) $$ \begin{align} \log |st|_+ = \log |t| ,\; \log |st|_- = \log |s| \text{ and } Q(\log |st|) = P. \end{align} $$

Assume that $w_1 \in W_P, w_2 \in W^{P,P} \setminus \{e\}$ and $w_1 w_2 s t = s' t \in \mathrm {Wt}(\pi \otimes t)$ . Then

(4.14) $$ \begin{align} w_2 s t = w_1^{-1} s' t \quad \text{and} \quad \log |w_2 s t|_+ = \log |w_1^{-1} s' t|_+. \end{align} $$

By [Reference Kriloff and Ram18, p. 38] $\log |w_1^{-1} s' t|_+ \geq \log |s' t|_+$ by [Reference Kriloff and Ram18, (2.13)] $\log |w_2 s t|_+ < \log |st|_+$ . Together with (4.13) we obtain

$$\begin{align*}\log |w_2 s t|_+ < \log |st|_+ = \log |s' t| \leq \log |w_1^{-1} s' t|_+. \end{align*}$$

That contradicts (4.13) and hence our assumption is untenable. In other words, $\pi \otimes t$ is $W,\!P$ -regular.

(ii) Notice that the proof of (i) only involves root systems and Weyl groups, no Hecke algebras. It can also be applied to the current $\pi \otimes t$ , when we replace the basis $\Delta $ of R by $-\Delta $ .

(iii) As $\psi _{w_{\Delta ,P}} : {\mathcal {H}}^P \to {\mathcal {H}}^{P^{op}}$ is an isomorphism that respects all the structure of these affine Hecke algebras, $\psi _{w_{\Delta ,P}}(\pi )$ is tempered and $w_{\Delta ,P} (t) \in T^{P^{op}}$ . For $\alpha \in \Delta $ there are equalities

(4.15) $$ \begin{align} \langle \alpha , \log |w_\Delta \! w_P t| \rangle = \langle w_\Delta \! (\alpha), \log |w_P t| \rangle = \langle w_\Delta \! (\alpha), \log |t| \rangle = \langle -w_\Delta \! (\alpha), \log |t^{-1}| \rangle , \end{align} $$

where we used $t \in T^P$ in the second step. If $\alpha \in \Delta \setminus P^{op}$ , then $-w_\Delta \alpha \in \Delta \setminus P$ and (4.15) is strictly positive because $t^{-1} \in T^{P+}$ . Therefore $w_{\Delta ,P} (t) \in T^{P^{op}+}$ . Now part (i) says that

$$\begin{align*}\psi_{w_{\Delta,P} *}(\pi \otimes t) = \psi_{w_{\Delta,P} *}(\pi) \otimes w_{\Delta,P} (t) \end{align*}$$

is $W,\!P^{op}$ -regular. (iv) With the same method as for (iii) this can be reduced to part (ii). In the cases (i) and (ii), Theorem 4.7.a says that $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}} (\pi \otimes t)$ has a unique irreducible quotient, which moreover has the given shape. In the cases (iii) and (iv) Theorem 4.7.b says that $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}} (\pi \otimes t)$ has a unique irreducible subrepresentation, namely the image of the indicated intertwining operator.

Condition 5.1. $\mathrm {Res}^M_{M^1}(\sigma )$ is multiplicity-free and $\mathrm {End}_G (\Pi _{\mathfrak s})^{op}$ is isomorphic to an affine Hecke algebra ${\mathcal {H}}$ with q-parameters in $\mathbb R_{\geq 1}$ , via an isomorphism as in [Reference Heiermann13] or [Reference Solleveld34, §10.2].

Proof. First we consider the supercuspidal case, so with $\Pi _{\mathfrak s_M} = \mathrm {ind}_{M^1}^M (\sigma _1)$ . With Frobenius reciprocity we compute

(5.4) $$ \begin{align} \mathrm{Hom}_M \big( \Pi_{\mathfrak s_M}, \Pi_{\mathfrak s_M}^\dagger \big) = \mathrm{Hom}_M \big( \mathrm{ind}_{M^1}^M (\sigma), \mathrm{ind}_{M^1}^M (\sigma)^\dagger \big) \cong \mathrm{Hom}_{M^1} \big( \sigma, \mathrm{ind}_{M^1}^M (\sigma_1)^\dagger \big). \hspace{-4mm} \end{align} $$

Hermitian duals turn ind into Ind [Reference Renard25, Théorème III.2.7], so the right-hand side of (5.4) is also

(5.5) $$ \begin{align} \mathrm{Hom}_{M^1} \big( \sigma_1, \mathrm{Ind}_{M^1}^M (\sigma_1^\dagger) \big) \cong \mathrm{Hom}_{M^1} \big( \sigma_1, \mathrm{Ind}_{M^1}^M (\sigma_1) \big). \end{align} $$

An analogous computation (with ind instead of Ind) applies to $\mathrm {End}_M (\Pi _{\mathfrak s_M})$ . We note that the set

$$\begin{align*}M^\sigma = \{m \in M : m \cdot \sigma_1 \cong \sigma_1\} \end{align*}$$

is a finite index subgroup of M which does not depend on the choice of $\sigma _1$ , see [Reference Roche26, §1.6]. With the Mackey decomposition we obtain

$$\begin{align*}\mathcal A \cong \mathrm{End}_M (\Pi_{\mathfrak s_M})^{op} \cong \bigoplus\nolimits_{m \in M^\sigma / M^1} \mathrm{Hom}_{M^1}(\sigma_1, m \cdot \sigma_1). \end{align*}$$

Here all summands have dimension one, and they multiply like the elements $m \in M^\sigma / M^1$ . In particular $\mathcal A = {\mathbb {C}}[X] \cong {\mathbb {C}}[M^\sigma / M^1]$ . Similarly (5.4) and (5.5) become

(5.6) $$ \begin{align} \mathrm{Hom}_{M^1} \big( \sigma_1, \mathrm{Ind}_{M^1}^M (\sigma_1) \big) \cong \prod\nolimits_{m \in M^\sigma / M^1} \mathrm{Hom}_{M^1}(\sigma_1, m \cdot \sigma_1). \end{align} $$

In view of (2.4), the right-hand side of (5.6) is isomorphic to $\mathcal A^\dagger \cong \prod _{x \in X} {\mathbb {C}} \{x\}$ as ${\mathbb {C}} [X]$ -bimodules. With (5.4)–(5.5) we obtain an isomorphism of $\mathcal A$ -bimodules

(5.7) $$ \begin{align} \mathrm{Hom}_M (\Pi_{\mathfrak s_M}, \Pi_{\mathfrak s_M}^\dagger ) \cong \mathcal A^\dagger. \end{align} $$

In the non-supercuspidal case (5.2) gives an ${\mathcal {H}}$ -isomorphism

(5.8) $$ \begin{align} \mathrm{Hom}_G (\Pi_{\mathfrak s}, \Pi_{\mathfrak s}^\dagger) \cong \mathrm{Hom}_G \big( \Pi_{\mathfrak s}, I_P^G (\mathrm{ind}_{M^1}^M (\sigma_1)^\dagger ) \big). \end{align} $$

By [Reference Roche26, Proposition 1.8.5.1] the right-hand side is naturally isomorphic to

$$\begin{align*}\mathrm{ind}_{\mathcal A}^{{\mathcal{H}}} \: \mathrm{Hom}_M \big( \Pi_{\mathfrak s_M}, \mathrm{ind}_{M^1}^M (\sigma_1)^\dagger \big). \end{align*}$$

From the supercuspidal case we know that this ${\mathcal {H}}$ -module is isomorphic to $\mathrm {ind}_{\mathcal A}^{{\mathcal {H}}} (\mathcal A^\dagger )$ , which by Lemma 2.6 is isomorphic to ${\mathcal {H}}^\dagger $ . It remains to see that the resulting left ${\mathcal {H}}$ -module isomorphism

(5.9) $$ \begin{align} \phi : {\mathcal{H}}^\dagger \to \mathrm{Hom}_G (\Pi_{\mathfrak s}, \Pi_{\mathfrak s}^\dagger) \end{align} $$

is an isomorphism of ${\mathcal {H}}$ -bimodules.

Recall from (2.4) that we identify ${\mathcal {H}}^\dagger $ with $\prod _{w \in W(\mathcal R)} {\mathbb {C}} N_w$ , which contains ${\mathcal {H}}$ as sub-bimodule. On the ${\mathcal {H}}$ -submodule $\mathrm {Hom}_G (\Pi _{\mathfrak s}, \Pi _{\mathfrak s})$ of $\mathrm {Hom}_G (\Pi _{\mathfrak s}, \Pi _{\mathfrak s}^\dagger )$ , the isomorphisms (5.8), (5.4), (5.5) and (5.6) become just the identity. Hence (5.9) extends the given algebra isomorphism ${\mathcal {H}} \cong \mathrm {End}_G (\Pi _{\mathfrak s})^{op}$ .

By (2.18) any element $h^+$ of ${\mathcal {H}}^\dagger $ admits a unique expression as

(5.10) $$ \begin{align} h^+ = \sum\nolimits_{w \in W} \sum\nolimits_{x \in X} c_{w,x} N_w \, \imath (\theta_x) \qquad c_{w,x} \in {\mathbb{C}}, \end{align} $$

with $\imath $ as in (2.19). The element $h_w^+ := \sum _{x \in X} c_{w,x} \theta _x$ belongs to $\mathcal A^\dagger $ , which via (5.7) and $I_P^G$ embeds naturally in $\mathrm {Hom}_G (\Pi _{\mathfrak s}, \Pi _{\mathfrak s}^\dagger )$ . From (5.6) we see that

$$\begin{align*}\phi \, \imath (h_w^+) = \phi \, \imath \big( \sum\nolimits_{x \in X} c_{w,x} \theta_x \big) = \sum\nolimits_{x \in X} c_{w,x} \phi \, \imath (\theta_x). \end{align*}$$

Using the ${\mathcal {H}}$ -linearity of $\phi $ we find

(5.11) $$ \begin{align} \begin{aligned} \phi (h^+) & = \phi \big( \sum_{w \in W} N_w \imath (h_w^+) \big) = \sum_{w \in W} N_w \cdot \phi \, \imath (h_w^+) = \sum_{w \in W} N_w \cdot \sum_{x \in X} c_{w,x} \phi \, \imath (\theta_x) \\ & = \sum_{w \in W} \phi (N_w) \sum_{x \in X} c_{w,x} \phi \, \imath (\theta_x) = \sum_{w \in W} \sum_{x \in X} c_{w,x} \phi (N_w \imath (\theta_x)). \end{aligned} \end{align} $$

This shows that $\phi $ commutes with infinite sums of elements of ${\mathcal {H}}$ in the form (5.10). Hence $\phi $ commutes with limits of sequences in ${\mathcal {H}}$ that converge in ${\mathcal {H}}^\dagger $ . Let $(h_n^+)_{n=1}^\infty $ be a sequence in ${\mathcal {H}}$ with limit $h^+ \in {\mathcal {H}}^\dagger $ . For any $h \in {\mathcal {H}}$ , (5.11) yields

$$\begin{align*}\phi (h^+ h) = \phi \big( \lim_{n \to \infty} h_n^+ h \big) = \lim_{n \to \infty} \phi (h_n^+ h). \end{align*}$$

As $\phi |_{{\mathcal {H}}}$ is an algebra homomorphism, the right-hand side equals

$$\begin{align*}\lim_{n \to \infty} \phi (h_n^+) \phi (h) = \lim_{n \to \infty} \phi (h_n^+) \cdot h = \phi (h^+) \cdot h , \end{align*}$$

where the dot indicates the right action of ${\mathcal {H}}$ on $\mathrm {Hom}_G (\Pi _{\mathfrak s}, \Pi _{\mathfrak s}^\dagger )$ from (5.3).

Proof. First we consider the special case where $\mathrm {Hom}_G (\Pi _{\mathfrak s},\pi )$ is a finitely generated ${\mathcal {H}}$ -module. Since ${\mathcal {H}}$ is Noetherian, there exists a projective resolution

(5.12) $$ \begin{align} \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi) \xleftarrow{d_0} {\mathcal{H}} \otimes F_0 \xleftarrow{d_1} {\mathcal{H}} \otimes F_1 \leftarrow \cdots \end{align} $$

where each ${\mathcal {H}} \otimes F_i$ is a free ${\mathcal {H}}$ -module with a finite dimensional multiplicity space $F_i$ . We note that here $d_i (i>0)$ is determined entirely by the map

$$\begin{align*}d_i |_{F_i} : F_i \cong {\mathbb{C}} 1 \otimes F_i \to {\mathcal{H}} \otimes F_{i-1}. \end{align*}$$

The conjugate-transpose of (5.12) is an injective resolution

(5.13) $$ \begin{align} \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi)^\dagger \xrightarrow{d_0^\dagger} {\mathcal{H}}^\dagger \otimes F_0^\dagger \xrightarrow{d_1^\dagger} {\mathcal{H}}^\dagger \otimes F_1^\dagger \rightarrow \cdots \end{align} $$

Via the equivalence of categories (5.1), (5.12) becomes a projective resolution of G-representations

(5.14) $$ \begin{align} \pi \xleftarrow{d_0} \Pi_{\mathfrak s} \otimes F_0 \xleftarrow{d_1} \Pi_{\mathfrak s} \otimes F_1 \leftarrow \cdots \end{align} $$

Here $d_i (i>0)$ is determined by the map $d_i |_{F_i} : F_i \to \mathrm {End}_G (\Pi _{\mathfrak s}) \otimes F_{i-1}$ given by $d_i |_{F_i}$ above composed with ${\mathcal {H}} \cong \mathrm {End}_G (\Pi _{\mathfrak s})^{op}$ . The conjugate-transpose of (5.14) is the injective resolution

(5.15) $$ \begin{align} \pi^\dagger \xrightarrow{d_0^\dagger} \Pi_{\mathfrak s}^\dagger \otimes F_0^\dagger \xrightarrow{d_1^\dagger} \Pi_{\mathfrak s}^\dagger \otimes F_1^\dagger \rightarrow \cdots \end{align} $$

in $\mathrm {Rep} (G)^{\mathfrak s}$ . Again applying (5.1), we obtain an injective resolution

(5.16) $$ \begin{align} \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi^\dagger) \xrightarrow{\mathrm{Hom}_G(\Pi_{\mathfrak s},d_0^\dagger)} \mathrm{Hom}_G (\Pi_{\mathfrak s}, \Pi_{\mathfrak s}^\dagger) \otimes F_0^\dagger \\ \xrightarrow{\mathrm{Hom}_G (\Pi_{\mathfrak s},d_1^\dagger)} \mathrm{Hom}_G (\Pi_{\mathfrak s}, \Pi_{\mathfrak s}^\dagger) \otimes F_1^\dagger \rightarrow \cdots\nonumber \end{align} $$

The maps $\mathrm {Hom}_G (\Pi _{\mathfrak s},d_i^\dagger )$ are still determined by $d_i |_{F_i}$ . To (5.16) we can apply Proposition 5.2, that yields

(5.17) $$ \begin{align} \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi^\dagger ) \rightarrow {\mathcal{H}}^\dagger \otimes F_0^\dagger \rightarrow {\mathcal{H}}^\dagger \otimes F_1^\dagger \rightarrow \cdots \end{align} $$

The maps in this sequence (except the leftmost) are induced by $d_i |_{F_i}$ , so they equal the maps in (5.13). We deduce isomorphisms of ${\mathcal {H}}$ -modules

(5.18) $$ \begin{align} \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi)^\dagger \cong \ker \big( {\mathcal{H}}^\dagger \otimes F_0^\dagger \rightarrow {\mathcal{H}}^\dagger \otimes F_1^\dagger \big) \cong \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi^\dagger ). \end{align} $$

Now we consider the general case. Write $\mathrm {Hom}_G (\Pi _{\mathfrak s},\pi )$ as the direct limit of its finitely generated submodules $\mathrm {Hom}_G (\Pi _{\mathfrak s},\pi _i)$ , where i runs through some index set. Then $\pi \cong \varinjlim \pi _i$ and $\pi ^\dagger \cong \varprojlim \pi _i^\dagger $ , which gives

$$\begin{align*}\mathrm{Hom}_G (\Pi_{\mathfrak s},\pi^\dagger) \cong \varprojlim \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi_i^\dagger). \end{align*}$$

By (5.18), the right-hand side is isomorphic to

$$\begin{align*}\varprojlim \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi_i )^\dagger \cong \big( \varinjlim \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi_i) \big)^\dagger \cong \mathrm{Hom}_G (\Pi_{\mathfrak s},\pi)^\dagger.\\[-37pt] \end{align*}$$

Proof. Recall that $\Pi _{\mathfrak s } = I_P^G (\mathrm {ind}_{M^1}^M (\sigma _1))$ for an irreducible subrepresentation $\sigma _1$ of $\mathrm {Res}^M_{M^1} (\sigma _1)$ . By [Reference Bushnell and Henniart4, Theorem 2.2] there is a natural isomorphism $J^G_{\overline P} \mathrm {ind}_U^G (\xi ) \cong \mathrm {ind}_{U \cap M}^M (\xi )$ . With Bernstein’s second adjointness we find

(6.3) $$ \begin{align} \begin{aligned} \mathrm{Hom}_G (\Pi_{\mathfrak s}, \mathrm{ind}_U^G (\xi)) & = \mathrm{Hom}_G \big( I_P^G (\mathrm{ind}_{M^1}^M (\sigma_1)), \mathrm{ind}_U^G (\xi) \big) \\ & \cong \mathrm{Hom}_M \big( \mathrm{ind}_{M^1}^M (\sigma_1), \mathrm{ind}_{U \cap M}^M (\xi) \big). \end{aligned} \end{align} $$

(a) The non-genericity of $\sigma $ implies, by [Reference Bushnell and Henniart4, Corollary 4.2], that the component of $\mathrm {ind}_{U \cap M}^M (\xi )$ in $\mathrm {Rep} (M)^{\mathfrak s_M}$ is zero. Hence (6.3) reduces to 0. That and (6.1)–(6.2) imply the second claim of part (a). (b) Since $U \cap M \subset M^1$ , there is a unique irreducible constituent $\sigma _1$ of $\mathrm {Res}^M_{M^1} (\sigma )$ such that $\mathrm {Hom}_{U \cap M} (\sigma _1,\xi )$ is nonzero. Then

(6.4) $$ \begin{align} \dim_{\mathbb{C}} \mathrm{Hom}_{U \cap M} (\sigma_1, \xi) = 1 \text{ and } \sigma_1 \text{ appears with multiplicity 1 in } \mathrm{Res}^M_{M^1} (\sigma). \end{align} $$

Now [Reference Bushnell and Henniart4, Proposition 9.2] says that

$$\begin{align*}\mathrm{ind}_{U \cap M}^{M} (\xi) \cong \mathrm{ind}_{M^1}^M (\sigma_1). \end{align*}$$

By Condition 5.1 $\mathrm {End}_M (\mathfrak s_M) \cong \mathcal A \cong {\mathbb {C}} [X]$ . Hence (6.3) is isomorphic to

(6.5) $$ \begin{align} \mathrm{End}_M (\mathrm{ind}_{M^1}^M (\sigma_1)) \cong \mathrm{End}_M (\Pi_{\mathfrak s_M}) \cong {\mathbb{C}} [X] \end{align} $$

as modules for $\mathrm {End}_M (\mathfrak s_M) \cong {\mathbb {C}} [X]$ . That makes our setup almost the same as in [Reference Solleveld35, §2], which means that the arguments from there remain valid in our setting. Then [Reference Solleveld35, Lemma 3.1] proves the theorem.

Proof. (a) The first isomorphism is an instance of Theorem 5.3, the second is Theorem 6.1 and the third is Lemma 2.1.a for $V = \mathrm {St} = \mathrm {St}^\dagger $ .

(b) The first isomorphism is a version of Frobenius reciprocity and the third is Lemma 2.1.b. The second isomorphism follows from the equivalence of categories (5.1) and part (a).

(c) This follows from part (b) and the semisimplicity of ${\mathcal {H}} (W,q^\lambda )$ .

Proof. (a) The Bernstein presentation of ${\mathcal {H}}$ shows that

$$\begin{align*}\mathrm{Res}^{{\mathcal{H}}}_{{\mathcal{H}} (W,q^\lambda)} \big( \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} V \big) = \mathrm{ind}_{{\mathcal{H}} (W_P ,q^\lambda)}^{{\mathcal{H}} (W,q^\lambda)} \big( \mathrm{Res}^{{\mathcal{H}}^P}_{{\mathcal{H}} (W_P,q^\lambda)} V \big). \end{align*}$$

Then by Frobenius reciprocity

(7.1) $$ \begin{align} \mathrm{Hom}_{{\mathcal{H}} (W,q^\lambda)} (\mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} V, \mathrm{St}) \cong \mathrm{Hom}_{{\mathcal{H}} (W_P,q^\lambda)} (V, \mathrm{St}). \end{align} $$

(b) By [Reference Solleveld35, Lemma 3.5]

$$\begin{align*}\dim \mathrm{Hom}_{{\mathcal{H}} (W,q^\lambda)} (\mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} V, \mathrm{St}) \leq 1 , \end{align*}$$

and by part (a) it is not 0. In view of the semisimplicity of ${\mathcal {H}} (W,q^\lambda )$ , this shows that $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}} V$ contains a unique copy of St, say ${\mathbb {C}} v$ . It follows that $\mathrm {ind}_{{\mathcal {H}}^P}^{{\mathcal {H}}} V$ has a generic irreducible subquotient, which appears with multiplicity one. It can be described as ${\mathcal {H}} v$ modulo the maximal submodule that does not contain v.

Proof. (a) This follows from Lemmas 2.1.b and 7.2.b. (b) For any weight s of $\pi $ , Frobenius reciprocity yields a nonzero (and hence surjective) ${\mathcal {H}}^P$ -homomorphism $\mathrm {ind}_{\mathcal A}^{{\mathcal {H}}^P} (s) \to \pi $ . Hence the unique (up to scalars) ${\mathcal {H}}^P$ -homomorphism $\pi \to {\mathcal {H}}^\dagger \otimes _{{\mathcal {H}} (W,q^\lambda )} \mathrm {St}$ can be inflated to a nonzero ${\mathcal {H}}$ -homomorphism $\phi _s : \mathrm {ind}_{\mathcal A}^{{\mathcal {H}}^P} (s) \to {\mathcal {H}}^\dagger \otimes _{{\mathcal {H}} (W,q^\lambda )} \mathrm {St}$ . Similarly the unique (up to scalars) homomorphism $\pi \otimes t \to {\mathcal {H}}^\dagger \otimes _{{\mathcal {H}} (W,q^\lambda )} \mathrm {St}$ inflates to

$$\begin{align*}\phi_{st} : \mathrm{ind}_{\mathcal A}^{{\mathcal{H}}^P} (st) \to {\mathcal{H}}^\dagger \otimes_{{\mathcal{H}} (W,q^\lambda)} \mathrm{St}. \end{align*}$$

By (7.3) we may assume that

$$\begin{align*}\phi_{st}(h \otimes 1) = h (f_{st} \otimes 1). \end{align*}$$

Let $v_s$ be the image of $N_e \otimes 1 \in \mathrm {ind}_{\mathcal A}^{{\mathcal {H}}^P}(s)$ in $\pi $ , by irreducibility it generates $\pi $ . Then $\phi _{st}$ factors through a homomorphism $\pi \otimes t \to {\mathcal {H}}^\dagger \otimes _{{\mathcal {H}} (W,q^\lambda )} \mathrm {St}$ that sends $v_s$ to $f_{st} \otimes 1$ . Frobenius reciprocity produces a homomorphism

(7.4) $$ \begin{align} \begin{array}{cccc} \mathfrak{Wh} (P,\pi \otimes t,v_s) : & \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} (\pi \otimes t) & \to & {\mathcal{H}}^\dagger \otimes_{{\mathcal{H}} (W,q^\lambda)} \mathrm{St} \\ & \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}} (\pi \otimes t)(h) v_s & \mapsto & h (f_{st} \otimes 1) \end{array}. \end{align} $$

Clearly the image of $\mathfrak {Wh} (P,\pi \otimes t,v_s)$ is generated by $f_{st} \otimes 1$ .

Proof. Suppose that $C(w,P,\pi ,t) = \infty $ . Then

$$\begin{align*}\mathfrak{Wh}(w(P),\psi_w (\pi \otimes t),v_s) \circ I(w,P,\pi ,t) = 0, \end{align*}$$

so $\text {im}\, I(w,P,\pi ,t)$ is not generic. In the short exact sequence

$$\begin{align*}0 \to \ker I(w,P,\pi,t) \to \mathrm{ind}_{{\mathcal{H}}^P}^{{\mathcal{H}}}(\pi \otimes t) \to \text{im}\, I(w,P,\pi,t) \to 0, \end{align*}$$

the middle term is generic by Lemma 7.3. Hence $\mathfrak {Wh}(w,P,\pi \otimes t,v_s)$ does not vanish on $\ker I(w,P,\pi ,t)$ , and the latter is generic.

When $C(w,P,\pi ,t) \in {\mathbb {C}}^\times $ , the equivalence is clear from (7.8).

Suppose that $C(w,P,\pi ,t) = 0$ . Then $I(w,P,\pi ,t)$ has a pole at $t \in T^P$ , caused by a pole of $\imath ^\circ _{w^{-1}}$ . Let f be a holomorphic function on a neighbourhood U of t in $T^P$ , such that $f(t') I(w,P,\pi ,t')$ is regular and nonzero on U. We can replace (7.8) by

(7.10) $$ \begin{align} \mathfrak{Wh}(P,\pi \otimes t,v_s) = C \, \mathfrak{Wh}(w(P),\psi_w (\pi \otimes t),v_s) \circ f (t) I(w,P,\pi, t) , \end{align} $$

for some $C \in {\mathbb {C}} \cup \{\infty \}$ . As $\mathfrak {Wh}(P,\pi \otimes t,v_s)$ is nonzero, so is C. Then (7.10) shows that $\mathfrak {Wh}(w(P),\psi _w (\pi \otimes t), v_s)$ is nonzero on $\mathrm {im}\, f(t) I(w,P,\pi , t)$ . Therefore $C \neq \infty $ , and we conclude that $\mathfrak {Wh}(P,\pi \otimes t,v_s)$ factors through $f(t) I(w,P,\pi , t)$ . In particular

$$\begin{align*}\ker I(w,P,\pi,t) \:\subset\: \ker f(t) I(w,P,\pi,t) \:\subset\: \ker \mathfrak{Wh}(P,\pi \otimes t, v_s).\\[-37pt] \end{align*}$$

Proof. (a) In view of Proposition 4.8.ii and Lemma 7.4, it suffices to check that $C(w_{\Delta ,P},P,\pi ,?)$ does not have a pole at t. By (7.9), it is equivalent to show that

(7.11) $$ \begin{align} C(w_{P,\Delta},P^{op},\psi_{w_{\Delta,P}}(\pi), w_{\Delta,P} (t)) \text{ is nonzero.} \end{align} $$

Suppose that (7.11) is zero. By (7.8), $I(w_{P,\Delta },P^{op},\psi _{w_{\Delta ,P}}(\pi ), w_{\Delta ,P} (t))$ has a pole at t. It is known from [Reference Opdam23, Theorem 4.33.i] that every such t is a zero of

$$\begin{align*}\big( \alpha (w_{\Delta,P} (rt)) - q^{(\lambda (\alpha) + \lambda^* (\alpha))/2} \big) \big( \alpha (w_{\Delta,P} (rt)) + q^{(\lambda (\alpha) - \lambda^* (\alpha))/2} \big), \end{align*}$$

for some weight r of $\pi $ and $-\alpha , w_{P,\Delta } (\alpha ) \in R^+$ . Equivalently, such a t satisfies

(7.12) $$ \begin{align} \big( \beta (rt) - q^{(\lambda (\alpha) + \lambda^* (\alpha))/2} \big) \big( \beta (rt) + q^{(\lambda (\alpha) - \lambda^* (\alpha))/2} \big) = 0 \end{align} $$

for some weight r of $\pi $ and $\beta , -w_{\Delta ,P} (\beta ) \in R^+$ . The eligible $\beta $ are precisely the roots in $R^+ \setminus R_P^+$ . From $t^{-1} \in T^{P+}$ we get $\beta (t) \in (0,1)$ for all $\beta \in R^+ \setminus R_P^+$ . By the temperedness and anti-temperedness of $\pi $ , $|r| = 1$ . Hence $|\beta (rt)| < 1$ , which in combination with $\lambda (\alpha ) \geq \lambda ^* (\alpha ) \geq 0$ implies that (7.12) never holds under our assumptions. Hence (7.11) is valid.