Proof We are going to show that there exists a finite extension $R'$ of R in $\bar {R}$ such that the result is true over $R'$ . Our goal is to prove that the representation $\pi _1 \otimes _R R'$ is a sum of absolutely irreducible representations with the same multiplicity and obtained as Galois conjugate of one another. First of all, D is a division algebra by Schur’s lemma and R is contained in its center E. This division algebra has finite dimension over R because $\pi $ is admissible and irreducible. In particular, there exists a (separable) extension $E'$ of E of degree m, where $m^2 = \text {dim}_E(D)$ , such that $D \otimes _E E' \simeq M_{m}(E')$ is split.

We embed $E'$ , and therefore E, in $\bar {R}$ and take the Galois closure $R'$ of $E'$ in $\bar {R}$ . We see $R'$ as an irreducible representation of the group $\mathbb {Z}$ endowed with the discrete topology in the following way. Thanks to the primitive element theorem, there exists a nonzero $\beta \in R'$ with minimal polynomial P such that $R' = R[\beta ] \simeq R[X]/(P(X))$ . The representation $(\pi ',R') \in \text {Rep}_R(\mathbb {Z})$ defined by $\pi '( 1 ) = \beta \in \text {GL}_R(R')$ is irreducible.

According to Lemma 4.1, the subrepresentations of $\pi _1 \otimes _R \pi _2 = \pi _1 \otimes _R R'$ correspond to right sub- $(D \otimes _R R')$ -modules of $D \otimes _R R' = \text {End}_G(\pi ) \otimes _R R'$ . Since

$$ \begin{align*}E \otimes_R R' \simeq \prod_{\text{Hom}_R(E,R')} R'\end{align*} $$

and $|\text {Hom}_R(E,R')| = n$ , we deduce that

$$ \begin{align*}D \otimes_R R' \simeq \prod_{1 \leq k \leq n} M_m(R').\end{align*} $$

Moreover, $\pi \otimes _R R'$ is semisimple by [Reference Vignéras12, Section II.4.2]. Since $D \otimes _R R' = \text {End}_G(\pi \otimes _R R')$ by Lemma 4.1 again, there exist non-isomorphic irreducible representations $(\tau _k)$ in $\text {Rep}_{R'}(G)$ such that

$$ \begin{align*}\pi \otimes_R R' \simeq \bigoplus_{1 \leq k \leq n} (m \tau_k).\end{align*} $$

Moreover, $\text {End}_G(\tau _k) = R'$ , so each $\tau _k$ is absolutely irreducible. As a result, $\rho _k = \tau _k \otimes _{R'} \bar {R}$ is irreducible. Hence, $\pi \otimes _R \bar {R} \simeq m (\bigoplus _{1 \leq k \leq n} \rho _k)$ .

It remains to show that $\text {Gal}_R(R')$ acts simply transitively on the n isomorphism classes defined by the family $(\tau _k)$ . Because $\pi $ is defined over R, the representations ${}^\sigma (\pi \otimes _R R')$ and $\pi \otimes _R R'$ are isomorphic for all $\sigma \in \text {Gal}_R(R')$ . Let $e_k$ be the idempotent in $D \otimes _R R'$ which cuts out the $\tau _k$ -isotypic part of $\pi \otimes _R R'$ . Note that $e_k \in E \otimes _R R'$ belongs to the center of $D \otimes _R R'$ . The Galois group $\text {Gal}_R(R')$ acts transitively on $\text {Hom}_R(E,R')$ . It also induces an action on $E \otimes _R R'$ which acts transitively on the $e_k$ ’s. Let $\sigma \in \text {Gal}_R(R')$ and let $k'$ be such that ${}^\sigma e_k = e_{k'}$ . We obtain

$$ \begin{align*}{}^\sigma(m \tau_k) \simeq {}^\sigma(e_k (\pi \otimes_R R')) \simeq e_{k'} {}^\sigma(\pi \otimes_R R') \simeq e_{k'} (\pi \otimes_R R') \simeq m \tau_{k'}.\end{align*} $$

So the $\tau _k$ form a single orbit under $\text {Gal}_R(R')$ . Furthermore, there exists a unique $\tau $ among the $\tau _k$ ’s such that $E \to D \otimes _R R' \to \text {End}_{R'[G]}(\tau ) = R'$ corresponds to the natural containment $E \subseteq R'$ . We have $\mathcal {O}_\tau \simeq \text {Hom}_R(E,R')$ by associating to $\tau _k$ the embedding $E \to D \otimes _R R' \to \text {End}_{R'[G]}(\tau _k) =R'$ . This bijection is compatible with the natural action of $\text {Gal}_R(R')$ on each side. Let $\sigma \in \text {Gal}_R(R')$ . Then ${}^\sigma \tau \simeq \tau $ if and only if $E \subseteq R'$ is preserved by $\sigma ,$ that is, if $i_1 : E \subseteq R'$ is the natural containment, we have $\sigma \circ i_1 = i_1$ . We deduce that $E(\tau ) = E$ in $R'$ . This completes the proof.

Proof We start by proving the uniqueness statement. For all irreducible admissible $\pi \in \text {Rep}_R(G)$ , the representation $(\pi \otimes _R \bar {R})|_R$ is $\pi $ -isotypic. Therefore, for any subquotient W of $\pi \otimes _R \bar {R}$ , the representation $W|_R$ is $\pi $ -isotypic. In particular, this holds when $W = \rho $ . As $\rho |_R$ is isotypic, this ensures the uniqueness of the representation $\pi $ assuming it exists.

We realize $\rho $ over a finite extension $R'$ of R with $\tau \otimes _{R'} \bar {R} = \rho $ . It is clear that $\tau $ is admissible because $\rho $ itself is. This implies that $\tau |_R$ is admissible. Since $\tau |_R$ is finitely generated, it admits an irreducible quotient $\pi ,$ that is, there exists a nonzero morphism $f : \tau |_R \to \pi $ . Let $\phi : R[G] \to R'[G]$ be the obvious inclusion. Then the forgetful functor $\text {Rep}_{R'}(G) \to \text {Rep}_R(G)$ has the functor $\text {Hom}_{R[G]}(R'[G],-)$ as a right adjoint. By adjunction, there corresponds to f a nonzero morphism $f' : \tau \to \text {Hom}_{R[G]}(R'[G],\pi )$ so $f'$ is injective. Hence, $\tau |_R$ is a subrepresentation of

$$ \begin{align*}(\text{Hom}_{R[G]}(R'[G],\pi))|_R \simeq \bigoplus_{[R':R]} \pi.\end{align*} $$

So $\tau |_R$ is $\pi $ -isotypic with finite multiplicity. We deduce that $\pi $ is admissible because $\tau |_R$ is and the functor of invariants for compact open subgroups is left exact. Therefore, $\pi = \pi (\rho ,R)$ exists and is admissible.

It remains to show that $\pi (\rho ,R) \simeq \tau |_R$ if the extension $R'/R$ has minimal degree. We let $\pi $ denote $\pi (\rho ,R)$ below to lighten notations. We have already shown that $\tau |_R$ is $\pi $ -isotypic with finite multiplicity. So we simply need to show that it is irreducible. By Theorem 4.3, there exists a subfield $E'$ of $R'$ isomorphic to the center E of $D = \text {End}_G(\pi )$ such that $R'/E'$ has degree m. In particular,

$$ \begin{align*}\pi \otimes_{E'} R' \simeq m \tau \text{ and } (\pi \otimes_{E'} R')|_R = \oplus_{[R':E']} \pi = m \pi.\end{align*} $$

Then $m \tau |_R \simeq m \pi ,$ that is, $\tau |_R \simeq \pi $ .

Proof We first assume that G is connected. On the one hand, it is well-known [Reference Vignéras12, Section II.2.8] that an irreducible representation $\rho $ in $\text {Rep}_{\bar {R}}(G)$ is admissible. On the other hand, any irreducible representation $\rho $ in $\text {Rep}_{\bar {R}}(G)$ can be realized over a finite extension of R by [Reference Vignéras12, Section II.4.7]. We can apply Lemma 4.4 to conclude. In general, the identity component of G is a subgroup of finite index, so we can reduce the problem to the connected case by restriction to the identity component.

Proof a) Note that $D = \text {End}_{G_1}(\pi _1) \otimes _R \text {End}_{G_2}(\pi _2)$ has finite dimension over R as $\pi _1$ and $\pi _2$ are admissible. We first assume that D is a simple R-algebra, that is, D is isomorphic to a matrix algebra over a division algebra. Then D is an isotypic right-D-module, so Lemma 4.1 ensures that $\pi _1 \otimes _R \pi _2$ is isotypic.

Therefore, by combining Lemma 4.1 and Wedderburn–Artin theorem, the result holds if D is a semisimple R-algebra, or equivalently, if the center E of D is a reduced ring [Reference Bourbaki1, A VIII.217, Proposition 7], that is, if E does not contain any nonzero nilpotent elements. Let $E_1$ be the center of $D_1$ and let $E_2$ be the center of $D_2$ . Then $E = E_1 \otimes _R E_2$ , where $E_1$ and $E_2$ are finite field extensions of the perfect field R, so E itself is a finite product of fields. In particular, E is reduced. Therefore, D is semisimple.

b) Same proof as [Reference Vignéras13, Theorem A.4], which follows [Reference Flath3] and uses [Reference Bourbaki1, A VIII.208, Proposition 2].

Proof Let $v = \sum v_1^i \otimes _{D_2} v_2^i \in \pi _1 \otimes _{D_2} \pi _2$ . Since $D_2= \text {End}_{G_2}(\pi _2)$ is a division algebra, we can choose a finite family $v_2^i$ that is free over $D_2$ and such that $v = \sum v_1^i \otimes _{D_2} v_2^i$ with $v_1^i \neq 0$ . We are going to show that $V_1 \otimes _{D_2} v_2^1$ is contained in the subrepresentation generated by v, which is enough to show that v generates $\pi _1 \otimes _{D_2} \pi _2$ . To do so, we choose an element $f_2 \in R[G_2]$ such that

$$ \begin{align*}\pi_2(f_2) v_2^i = \left\{ \begin{array}{@{}cc} 0 & \text{ if } i\neq 1 \\ v_2^1 & \text{ if } i=1. \end{array} \right.\end{align*} $$

Such an element always exists. Indeed, let A be the image of the map $R[G_2] \to \text {End}_R(\pi _2)$ . Then A is contained in $\text {End}_{\text {End}_{G_2}(\pi _2)}(\pi _2) = \text {End}_{\text {comm}(A)}(\pi _2) = \text {comm}(\text {comm}(A)),$ where $\text {comm}$ means the centralizer of an algebra in $\text {End}_R(\pi _2)$ . We consider the subspace $W_2$ of $\pi _2$ generated by the $v_2^i$ ’s. The map we want to interpolate is the projection on $v_2^1$ with kernel generated by the other $v_2^i$ ’s. By the basis extension theorem, there exists $b \in \text {End}_{D_2}(\pi _2) = \text {comm}(\text {comm}(A))$ whose restriction $b|_{W_2}$ to $W_2$ realizes this projection. As the family $v_2^i$ is finite, there exists $a \in A$ such that $a|_{W_2} = b|_{W_2}$ by the Jacobson density theorem [Reference Vignéras12, Section I.B.6]. Hence, the existence of $f_2$ . Therefore, the subrepresentation generated by v contains $V_1 \otimes _{D_2} v_2^1$ and the irreducibility follows.

Proof By Corollary 5.5, the map

$$ \begin{align*}V \otimes_R \bar{R} \to \bigoplus_{w \in \text{Hom}_R( R(\rho_1),\bar{R})} (V\otimes_R \bar{R})_{{}^w \rho_1}\end{align*} $$

is surjective and has kernel $\bigcap _w (V \otimes _R \bar {R}) [{}^w \rho _1]$ . We are going to show that

$$ \begin{align*}V[\pi_1] \otimes_R \bar{R} = \bigcap_{w \in \text{Hom}_R(R(\rho_1),\bar{R})} (V \otimes_R \bar{R})[{}^w \rho_1].\end{align*} $$

Since $(V \otimes _R \bar {R})/ (V[\pi _1] \otimes _R \bar {R}) \simeq V_{\pi _1} \otimes _R \bar {R}$ , the latter is also the largest isotypic quotient of V associated with the finite family $({}^w \rho _1)_w$ in the sense of Corollary 5.5.

The direct inclusion is the easiest part. Let $v \in V[\pi _1]$ . We want to prove that for all $w \in \text {Hom}_R(R(\rho _1),\bar {R})$ and all $f \in \text {Hom}_{\bar {R}[G_1]}(V \otimes _R \bar {R},{}^w \rho _1)$ , we have $v \otimes _R 1 \in \text {Ker}(f)$ . In particular, such an f defines a morphism of $R[G_1]$ -modules $(V \otimes _R \bar {R})|_R \to ({}^w \rho _1)|_R$ by restriction of scalars. Moreover, the morphism f is nonzero if and only if its restriction to $V\otimes _R 1 = \{ v \otimes _R 1 \ | \ v \in V\}$ is nonzero. However, according to the first lines of the proof of Lemma 4.4, we know that the representation $({}^w \rho _1)|_R$ is $\pi _1$ -isotypic. Therefore, $f|_{V \otimes _R 1} : V \simeq V \otimes _R 1 \to ({}^w \rho _1)|_R \simeq \oplus \pi _1$ . So $f(v \otimes _R 1) =0$ by definition of $V[\pi _1]$ .

Regarding the reverse inclusion, we know thanks to Lemma 5.4 that there exists a smooth $R[G_2]-D_1$ -bimodule $V_2$ such that $V_{\pi _1} \simeq V_2 \otimes _{D_1} V_1,$ where $D_1 = \text {End}_{G_1}(\pi _1)$ is a division algebra of degree $m_1$ over its center. Moreover, we have an isomorphism of representations

$$ \begin{align*}V_{\pi_1} \otimes_R \bar{R} \simeq (V_2 \otimes_R \bar{R}) \otimes_{D_1 \otimes_R \bar{R}} (V_1 \otimes_R \bar{R}).\end{align*} $$

Thanks to Theorem 4.3, the ring $D_1 \otimes _R \bar {R}$ is isomorphic to $\prod _w e_w (D_1 \otimes _R \bar {R}) \simeq \prod _w M_{m_1}(\bar {R}),$ where $(e_w)_w$ is a system of primitive central idempotents in $D_1 \otimes _R \bar {R}$ . We deduce that

$$ \begin{align*}V_{\pi_1} \otimes_R \bar{R} \simeq \bigoplus_w V_w \otimes_{M_{m_1}(\bar{R})} (m_1 {}^w \rho_1),\end{align*} $$

where $m_1 {}^w \rho _1 = e_w (\pi _1 \otimes _R \bar {R})$ and $V_w = V_2 e_w$ .

We now prove that there exists a representation $V_{2,w} \in \text {Rep}_{\bar {R}}(G_2)$ such that

$$ \begin{align*}V_w \otimes_{M_{m_1}(\bar{R})} (m_1 ({}^w \rho_1)) \simeq V_{2,w} \otimes_{\bar{R}} ({}^w \rho_1).\end{align*} $$

Indeed, by denoting $e_{i,j}$ the elementary matrix in $M_{m_1}(\bar {R})$ , we have $e_{1,1} + \dots + e_{m_1,m_1} = \text {Id}_{m_1}$ which is a decomposition of the unit as a sum of idempotents. These idempotents are not necessarily central in $M_{m_1}(\bar {R})$ . Nevertheless, each $e_{i,i}$ defines a map $v \in V_w \mapsto v e_{i,i} \in V_w$ which is a morphism of $\bar {R}[G_2]$ -modules. In addition, $e_{i,i} V_w \simeq e_{1,1} V_w$ for all i. Denoting $V_{2,w} = V_w e_{1,1} \in \text {Rep}_{\bar {R}}(G_2)$ , we have $V_w \simeq m_1 V_{2,w}$ and $(m_1 V_{2,w}) \otimes _{M_{m_1}(\bar {R})} (m_1 ({}^w \rho _1)) \simeq V_{2,w} \otimes _{\bar {R}} ({}^w \rho _1)$ . Therefore, the quotient $V \otimes _R \bar {R} \to V_{2,w} \otimes _{\bar {R}} ({}^w \rho _1)$ factors through $(V \otimes _R \bar {R})_{{}^w \rho _1}$ by definition of the largest ${}^w \rho _1$ -isotypic quotient. So the quotient $V \otimes _R \bar {R} \to V_{\pi _1} \otimes _R \bar {R} \simeq \oplus _w V_{2,w} \otimes _{\bar {R}} ({}^w \rho _1)$ factors through $\oplus _w (V \otimes _R \bar {R})_{{}^w \rho _1}$ . In other words, $\cap _w (V\otimes _R \bar {R})[{}^w \rho _1] \subseteq V[\pi _1] \otimes _R \bar {R}$ .

Proof Theorem 5.6 ensures that

$$ \begin{align*}V_{\pi_1} \otimes_R \bar{R} \simeq \bigoplus_{w \in \text{Hom}_R(R(\rho_1),\bar{R})} (V \otimes_R \bar{R})_{{}^w \rho_1}.\end{align*} $$

Now we can use Theorem 6.1 to obtain the first isomorphism of the lemma.

Any representation $V_{\pi _1} \otimes _{E_1,w'} \bar {R}$ with $w' \in \text {Hom}_R(E_1, \bar {R})$ is isomorphic to one and only one of $(V \otimes _R \bar {R})_{{}^w \rho _1}$ with $w \in \text {Hom}_R(R(\rho _1),\bar {R})$ . This defines the bijection $\varphi $ .

To end the proof, the isomorphism $(\Theta (\pi _1) \otimes _{D_1} \pi _1) \otimes _{E_1,\varphi (w)} \bar {R} \simeq \Theta ({}^w \rho _1) \otimes _{\bar {R}} {}^w \rho _1$ induces the desired isomorphism thanks to the fact that $D_1' = D_1 \otimes _{E_1,\varphi (w)} \otimes \bar {R} \simeq \mathcal {M}_{m_1}(\bar {R})$ and

$$ \begin{align*} (\Theta(\pi_1) \otimes_{D_1} \pi_1) \otimes_{E_1,\varphi(w)} \bar{R} & \simeq (\Theta(\pi_1) \otimes_{E_1,\varphi(w)} \bar{R}) \otimes_{D_1'} (\pi_1 \otimes_{E_1,\varphi(w)} \bar{R}) \\ & \simeq (\Theta(\pi_1) \otimes_{E_1,\varphi(w)} \bar{R}) \otimes_{D_1'} (m_1 {}^w\rho_1) \\ & \simeq \Theta({}^w \rho_1) \otimes_{\bar{R}} {}^w \rho_1. \end{align*} $$

By Morita equivalence of $\mathcal {M}_{m_1}(\bar {R})$ and $\bar {R}$ , we obtain $\Theta (\pi _1) \otimes _{E_1,\varphi (w)} \bar {R} \simeq m_1 \Theta ({}^w\rho _1)$ .

Proof We only need to show that a) $\Rightarrow $ b) and c) $\Rightarrow $ a), since b) $\Rightarrow $ c) is obvious.

For a) $\Rightarrow $ b), we first recall that irreducible representations are admissible thanks to Corollary 4.5 and the remark thereafter. Therefore, if the $R[H_2]-D_1$ -bimodule $\Theta (\pi _1)$ has finite length, Lemma 5.2 implies that $\Theta (\pi _1) \otimes _{D_1} \pi _1$ is an admissible representation of finite length in $\text {Rep}_R(H_1 \times H_2)$ . By extending scalars to $\bar {R}$ as in Theorem 4.3, the representation $(\Theta (\pi _1)\otimes _{D_1} \pi _1 ) \otimes _R \bar {R}$ has finite length in $\text {Rep}_{\bar {R}}(H_1 \times H_2)$ . Then Lemma 6.4 allows us to conclude that $\Theta ({}^w \rho _1)$ has finite length for all ${}^w \rho _1$ .

For the last implication c) $\Rightarrow $ a), we use Lemma 6.4 again. The isomorphism

$$ \begin{align*}\Theta(\pi_1) \otimes_{E_1,\varphi(w)} \bar{R} \simeq m_1 \Theta({}^w \rho_1)\end{align*} $$

is compatible with the right action of $D_1 \otimes _{E_1,\varphi (w)} \bar {R} \simeq \mathcal {M}_{m_1}(\bar {R})$ , therefore the $\bar {R}[H_2]-\mathcal {M}_{m_1}(\bar {R})$ -bimodule $\Theta (\pi _1) \otimes _{E_1,\varphi (w)} \bar {R}$ has finite length provided that $\Theta (\rho _1^w)$ has finite length in $\text {Rep}_{\bar {R}}(H_2)$ . This implies that $\Theta (\pi _1)$ is an $R[H_2]-D_1$ -bimodule of finite length.

Proof As in the previous proof, the implication b) $\Rightarrow $ c) is obvious.

For a) $\Rightarrow $ b), we prove the contraposition. We then suppose there exists ${}^w \rho _1$ such that $\theta ({}^w \rho _1)$ is not irreducible. We want to show that there exist two irreducible representations $\tau _2$ and $\tau _2'$ in $\text {Rep}_{\bar {R}}(H_2)$ such that $\Theta ({}^w \rho _1) \otimes _{\bar {R}} ({}^w \rho _1)$ admits $(\tau _2 \otimes _{\bar {R}} ({}^w \rho _1)) \oplus (\tau _2' \otimes _{\bar {R}} ({}^w \rho _1))$ as a quotient.

On the one hand, the representation $(\tau _2 \otimes _{\bar {R}} ({}^w \rho _1))|_R$ of $H_1$ is $\pi _1$ -isotypic. According to Lemma 5.4, there exists an $R[H_2]-D_1$ -bimodule $\sigma _2$ such that $(\tau _2 \otimes _{\bar {R}} ({}^w \rho _1))|_R \simeq \sigma _2 \otimes _{D_1} \pi _1$ . Moreover, it is semisimple as a representation in $\text {Rep}_R(H_1 \times H_2)$ . So $\sigma _2$ is isotypic and we let $\pi _2$ be any irreducible factor of $\sigma _2$ . We use similar notations for $\tau _2'$ and $\sigma _2'$ and $\pi _2'$ .

On the other hand, Lemma 6.4 implies $\Theta ({}^w \rho _1) \otimes _{\bar {R}} ({}^w \rho _1) \simeq (\Theta (\pi _1) \otimes _{D_1} \pi _1) \otimes _{E_1,\varphi (w)} \bar {R}$ . We deduce that the obvious morphism of $\Theta (\pi _1) \otimes _{D_1} \pi _1$ in the right-hand side guarantees that $\Theta (\pi _1) \otimes _{D_1} \pi _1$ admits $(\pi _2 \otimes _{D_1} \pi _1) \oplus (\pi _2' \otimes _{D_1} \pi _1)$ as a quotient. The kernel of this quotient map is of the form $\sigma _2" \otimes _{D_1} \pi _1,$ where $\sigma _2"$ is a sub- $R[H_2]-D_1$ -bimodule of $\Theta (\pi _1)$ by Lemma 5.4. As a result, this quotient map induces a quotient $\Theta (\pi _1) \to \pi _2 \oplus \pi _2'$ of $R[H_2]-D_1$ -bimodules whose kernel is precisely $\sigma _2"$ . Hence, $\theta (\pi _1)$ is not irreducible.

Finally, the implication c) $\Rightarrow $ a), we will prove the contraposition again. We then suppose that $\theta (\pi _1)$ is not irreducible. We need to show that for all ${}^w \rho _1$ , the representation $\theta ({}^w \rho _1)$ is not irreducible. However, $\Theta (\pi _1) \otimes _{D_1} \pi _1$ admits $\theta (\pi _1) \otimes _{D_1} \pi _1$ as a quotient. As a result, $(\Theta (\pi _1) \otimes _{D_1} \pi _1)\otimes _R \bar {R}$ admits $(\theta (\pi _1) \otimes _{D_1} \pi _1) \otimes _R \bar {R}$ as a quotient. But $(\Theta (\pi _1) \otimes _{D_1} \pi _1)\otimes _{E_1,\varphi (w)} \bar {R}$ admits $(\theta (\pi _1) \otimes _{D_1} \pi _1) \otimes _{E_1,\varphi (w)} \pi _1$ and $\theta (\pi _1) \otimes _{E_1,\varphi (w)} \bar {R} = m_1 \theta ({}^w \rho _1)$ as a quotient. Hence, $\theta ({}^w \rho _1)$ is not irreducible because $\theta (\pi _1)$ is not and the scalar extension functor is exact, therefore $\theta (\pi _1) \otimes _{E_1,\varphi (w)} \bar {R}$ has length at least $2 m_1$ .

Proof The implication b) $\Rightarrow $ c) is still obvious.

For a) $\Rightarrow $ b), the isomorphism $\theta (\pi _1) \otimes _{E_1,\varphi (w)} \bar {R} \simeq m_1 \theta ({}^w \rho _1)$ is an isomorphism of $\bar {R}[H_2]-(D_1 \otimes _{E_1,\varphi (w)} \bar {R})$ -bimodules. And likewise for $\theta (\pi _1')$ and some $\theta ({}^{w'}\rho _1')$ . If $\theta (\pi _1) \simeq \theta (\pi _1')$ , then $\theta ({}^w \rho _1) \simeq \theta ({}^{w'} \rho _1')$ in $\text {Rep}_{\bar {R}}(H_2)$ .

Regarding the last implication c) $\Rightarrow $ a), there exists thanks to the previous paragraph an isomorphism $\theta (\pi _1) \otimes _{E_1,\varphi (w)} \bar {R} \simeq \theta (\pi _1') \otimes _{E_1,\varphi '(w')} \bar {R}$ because $\theta ({}^w \rho _1) \simeq \theta ({}^{w'} \rho _1')$ . Let $\sigma \in \text {Gal}_R(\bar {R})$ . Then ${}^{\sigma }(\theta (\pi _1) \otimes _{E_1,\varphi (w)} \bar {R}) \simeq \theta (\pi _1) \otimes _{E_1 , \varphi (\sigma w)} \bar {R}$ . In particular, this implies that $\theta ({}^{\sigma w} \rho _1) \simeq \theta ({}^{\sigma w'} \rho _1)$ . Therefore, there exists a bijection $\psi $ of $\text {Hom}_R(R(\rho _1),\bar {R})$ such that for all $w_0 \in \text {Hom}_R(R(\rho _1),\bar {R})$ , we have $\theta ({}^{w_0} \rho _1) \simeq \theta ({}^{\psi (w_0)} \rho _1')$ . We thus have isomorphisms of $\bar {R}[H_2]-(D_1\otimes _R \bar {R})$ -bimodules

$$ \begin{align*}\theta(\pi_1) \otimes_R \bar{R} \simeq \oplus_{w_0} \theta(\pi_1) \otimes_{E_1 , \varphi(w_0)} \bar{R} \simeq \theta(\pi_1') \otimes_R \bar{R}.\end{align*} $$

By restriction of scalars, $(\theta (\pi _1) \otimes _R \bar {R})|_R$ is an $R[H_2]-D_1$ -bimodule which is at the same time $\theta (\pi _1)$ -isotypic and $\theta (\pi _1')$ -isotypic. Therefore, $\theta (\pi _1) \simeq \theta (\pi _1')$ .

Proof The morphism $V \twoheadrightarrow V_{\pi _1}$ induces ${}^\sigma V \twoheadrightarrow ({}^\sigma V)_{{}^\sigma \pi _1}$ and ${}^\sigma (V_{\pi _1}) \simeq ({}^\sigma V)_{{}^\sigma \pi _1}$ . Therefore, ${}^\sigma (\Theta _V(\pi _1) \otimes _R \pi _1) \simeq {}^\sigma \Theta _V(\pi _1) \otimes _R {}^\sigma \pi _1 \simeq \Theta _{{}^\sigma V}({}^\sigma \pi _1) \otimes _R {}^\sigma \pi _1$ , so ${}^\sigma \Theta _V(\pi _1) \simeq \Theta _{{}^\sigma V}({}^\sigma \pi _1)$ by Theorem 6.1 and the uniqueness of the lift.

Proof This theorem can be seen as an instance of Morita equivalences. We claim that B is a progenerator of the category of A-modules. By definition of G-Galois rings, it is finitely generated and projective. Moreover, $\text {Spec}(B) \to \text {Spec}(A)$ is surjective because $A \hookrightarrow B$ is faithfully flat by [Reference Greither5, Chapter 0, Lemma 1.9], so B is a generator. Therefore, $V \mapsto V \otimes _A B$ provides a Morita equivalence between the categories of A-modules and $\text {End}_A(B)$ -modules.

We need to explain why $V^G \otimes _A B \to V$ is an isomorphism. As a consequence of the previous equivalence of categories, there exists an A-submodule W of V such that $V = W \otimes _A B$ . But $(W \otimes _A B)^G = W$ , so we deduce that $W = V^G$ .

Proof Let $\sigma \in G$ . Since the representation ${}^\sigma \rho _\psi $ is irreducible and has central character $\psi ^\sigma $ , it is isomorphic to $\rho _{\psi ^\sigma }$ by Stone–von Neumann Theorem. Then ${}^\sigma \rho _\psi \simeq \rho _\psi $ only if $\psi ^\sigma = \psi ,$ that is, only if $\sigma = \text {id}$ since the image of $\psi $ consists of all p-roots of unity if $\text {char}(F)>0$ and all p-power roots of unity if $\text {char}(F)=0$ . So $H(\rho _\psi ) = \{ \text {id} \}$ and $R(\rho _\psi )= \mathcal {K}$ . Since $\rho _\psi $ is already realized over $R(\rho _\psi )$ , we obtain $m(\rho _\psi ) = 1$ .

Proof We have $\Omega _{1,\text {det}(a)}^{\psi ^\gamma } = (\gamma ,\text {det}(a))_F \Omega _{1,\text {det}(a)}^\psi $ by the properties of the non-normalized Weil factor according to [Reference Trias11, Section 4], so this proves the first formula. The second one is very straightforward. The last one comes from a simple change of variables.

Proof In the metaplectic group, we have $(m_\gamma ,\mu ) (g,\lambda ) (m_\gamma ,\mu )^{-1} = (m_\gamma g m_\gamma ^{-1},\lambda )$ . Recall that the quadratic Hilbert symbol $( - , - )_F$ is trivial on squares and that the non-normalized Weil factor satisfies $\Omega _\mu (\psi ^{\gamma ^2} \circ Q_w) = \Omega _\mu (\psi \circ Q_w \circ \gamma ) = |\gamma |_F^m \Omega _\mu (\psi \circ Q_w)$ when $\gamma \in F^\times $ . Therefore, when $\gamma \in \mathcal {O}_F^\times $ , we obtain from Proposition 8.2 three equalities

(8.4) $$ \begin{align} \omega_{\psi^{\gamma^2},X}(m,1) =&\ \omega_{\psi,X}(m,1) = \omega_{\psi,X}(m_\gamma m m_\gamma^{-1},1) \end{align} $$

(8.5) $$ \begin{align} \omega_{\psi^{\gamma^2},X}(n,1) =&\ \omega_{\psi,X}(m_\gamma n m_\gamma^{-1},1) \end{align} $$

(8.6) $$ \begin{align} \omega_{\psi^{\gamma^2},X}(w,1) =&\ \omega_{\psi,X}(m_\gamma w m_\gamma^{-1},1). \end{align} $$

Since the symplectic group $\text {Sp}(W)$ is generated by elements of the form $m, n, w$ , the genuine representations $\omega _{\psi ^{\gamma ^2},X}$ and $\omega _{\psi ,X}^{m_\gamma }$ of $\text {Mp}(W)$ must be equal.

Proof We define an $\text {Mp}(W)$ -equivariant semilinear action of $G_{2'}$ on $(\omega _{\psi ,X},C_c^\infty (Y,\mathcal {K}))$ . For $\sigma \in G_{2'}$ , there exists a unique element $\gamma \in {(\mathcal {O}_F^\times )}_{2'}$ in the prime-to- $2$ part of $\mathcal {O}_{F^\times }$ such that $(\psi ^{\gamma ^2})^\sigma = \psi $ . We set

$$ \begin{align*}(r_\sigma \cdot f )(y)= \sigma\bigg( (\omega_{\psi,X}(m_\gamma,1) \cdot f) (y)\bigg).\end{align*} $$

Corollary 8.3 ensures that $\omega _{\psi ,X}(g,\lambda )$ and $r_\sigma $ commute for all $(g,\lambda ) \in \text {Mp}(W)$ and all $\sigma \in G_{2'}$ . It is easy to see that the semilinear action thus defined is smooth. As $\text {Spec}(\mathcal {K}) \to \text {Spec}(\mathcal {L})$ is proétale, we are in the situation of an effective descent data of Theorem 7.3 so the subspace $\mathcal {V}$ of $G_{2'}$ -fixed points in $C_c^\infty (Y,\mathcal {K})$ satisfies $\mathcal {V} \otimes _{\mathcal {L}} \mathcal {K} \overset {\sim }{\to } C_c^\infty (Y,\mathcal {K})$ . This $\mathcal {V}$ clearly inherits a smooth action of $\text {Mp}(W)$ .

Proof Via the identification of G with a subgroup of $\mathcal {O}_F^\times $ , let $\sigma \in G$ and let $\gamma \in \mathcal {O}_F^\times $ be the unique element such that $\psi ^\sigma = \psi ^\gamma $ . The map $\sigma \mapsto \lambda $ induces a group isomorphism with $\mathbb {Z}/(p-1)\mathbb {Z}$ or $\mathbb {Z}_p^\times $ , as embedded subgroups in $\mathcal {O}_F^\times $ , according to whether $\text {char}(F)$ is positive or not. We know by Section 8.4 that $\omega _{\psi ^\gamma ,X}^\pm \simeq \omega _{\psi ,X}^\pm $ if and only if $\gamma \in \mathcal {O}_F^{\times 2}$ .

Therefore, $H(\omega _{\psi ,X}^\pm ) = \{ \sigma \in G \ | \ \omega _{\psi ^\sigma ,X}^\pm \simeq \omega _{\psi ,X}^\pm \} = G \cap \mathcal {O}_F^{\times 2}$ . The subgroup $G \cap \mathcal {O}_F^{\times 2}$ of G is equal to G if and only if $q \in p^{2 \mathbb {N}}$ , and is a subgroup of index $2$ otherwise. As a result, the fixed field of $G \cap \mathcal {O}_F^{\times 2}$ gives the character field, which is either $\mathbb {Q}$ or $\mathbb {Q}[\sqrt {p^*}]$ according to whether q is a square or not.

Proof We first consider the case $q \equiv 3 [4]$ or equivalently $p \equiv 3 [4]$ and $q \in p^{2 \mathbb {N}+1}$ . Note that we already have that $\mathcal {L} = \mathbb {Q}[\sqrt {-p}] = R(\omega _{\psi ,X}^+)$ and therefore the theorem holds.

We can assume $-1 \in \mathcal {O}_F^{\times 2}$ as this is equivalent to $q \equiv 1 [4]$ . Therefore, $\Omega _{1,-1}^\psi = 1$ and $(-1,\gamma )_F=1$ for all $\gamma \in F^\times $ . In particular, $\omega _{\psi ,X}^+((-\text {id}_W,1)) = \text {id}_{S_{\psi ,X}^+}$ and

$$ \begin{align*}\omega_{\psi,X}^+(m_{- \gamma},1) = \omega_{\psi,X}^+(m_\gamma,1) \text{ for all } \gamma \in F^\times.\end{align*} $$

we recall that $G \cap \mathcal {O}_F^{\times 2} \subseteq G$ is a subgroup of index $1$ or $2$ according to whether q is a square or not. Let $\sigma \in G \cap \mathcal {O}_F^{\times 2} \mapsto \gamma \in \mathcal {O}_F^\times $ be any map subject to the relation $\sigma = \gamma ^2$ .

Because p is odd, the restriction of the quadratic Hilbert symbol to $\mathcal {O}_F^\times \times \mathcal {O}_F^\times $ is trivial. Since $\omega _{\psi ,X}^+(m_{-\gamma },1) = \omega _{\psi ,X}^+(m_\gamma ,1)$ , the earlier descent formula

$$ \begin{align*}(r_\sigma \cdot f )(y)= \sigma\bigg( (\omega_{\psi,X}^+(m_\gamma,1) \cdot f) (y)\bigg)\end{align*} $$

still defines a continuous semilinear action of $G \cap \mathcal {O}_F^{\times 2}$ on $\omega _{\psi ,X}^+$ . So the representation can be realized over the fixed field of $G \cap \mathcal {O}_F^{\times 2}$ which is $R(\omega _{\psi ,X}^+)$ .

Proof The easiest case to deal with is $q \equiv 3 [4]$ or equivalently $p \equiv 3 [4]$ and $q \in p^{2\mathbb {N}+1}$ . In this case, $\mathcal {L} = \mathbb {Q}[\sqrt {-p}]$ already, and Theorem 9.2 ensures this is the character field.

To deal with the other two cases, we use information about the endomorphism ring of $\omega _{\psi ,X}^-$ . First of all, for all $\mathbb {Q} \subseteq \mathcal {M} \subseteq \mathcal {L}$ , we have

$$ \begin{align*}\omega_{\psi,X}^-|_{\mathcal{M}} \otimes_{\mathcal{M}} \mathcal{L} \simeq \bigoplus_{\sigma \in \text{Gal}(\mathcal{L}/\mathcal{M})} \omega_{\psi^\sigma,X}^-.\end{align*} $$

Recall that $G_2 = \text {Gal}(\mathcal {L}/\mathbb {Q})$ is a finite cyclic group of order $2^k,$ where $k=\text {val}_2(p-1) \geq 2$ . According to Theorem 9.2, two cases now arise:

• • $\omega _{\psi ^\sigma ,X}^- \simeq \omega _{\psi ,X}^-$ for all $\sigma \in G_2,$ that is, $G \cap \mathcal {O}_F^{\times 2} = G,$ that is, q is a square;

• • there exists $\sigma \in G_2$ such that $\omega _{\psi ^\sigma ,X}^-$ is not isomorphic to $\omega _{\psi ,X}^-$ .

In particular, this distinction implies that the character field is $\mathbb {Q}$ or $\mathbb {Q}[\sqrt {p^*}]$ .

We first prove a negative result: it is impossible to descend further than $\mathcal {L}$ along $\mathcal {L}$ .

Proof Since $\mathcal {L}/\mathbb {Q}$ is cyclic, the extension $\mathcal {L}/\mathcal {L}'$ is cyclic for any subextension $\mathcal {L}'$ of $\mathcal {L}$ . By cyclicity, if any descent can be achieved along $\mathcal {L}$ , then it can be achieved over the totally real subextension $\mathcal {L}_0$ of index $2$ of the CM-field $\mathcal {L}$ . Let $\tau \in \text {Gal}(\mathcal {L}/\mathcal {L}_0) \subseteq \text {Gal}(\mathcal {L}/\mathbb {Q})$ be the complex conjugation, which is the unique element of order $2$ .

If $\omega _{\psi ,X}^-$ descends to $\mathcal {L}_0$ , there is a $\tau $ -linear $\text {Mp}(W)$ -automorphism of $\omega _{\psi ,X}^-$ of order $2$ . Note that $\tau $ -linear $\text {Mp}(W)$ -automorphisms are unique up to a scalar thanks to Schur’s lemma. Choose $i \in \mathcal {O}_F^\times $ such that $i^2=-1$ . Then

$$ \begin{align*}r_\tau \cdot f = \tau(\omega_{\psi,X}^-(m_i,1)f)\end{align*} $$

is a $\tau $ -linear $\text {Mp}(W)$ -automorphism of $\omega _{\psi ,X}^-$ . It satisfies $r_\tau ^2 = \omega _{\psi ,X}^-(m_{-1},1) = - \text {id}_{S_{\psi ,X}^-}$ .

Remark that any $\lambda r_\tau $ with $\lambda \in \mathcal {L}$ satisfies

$$ \begin{align*}(\lambda r_\tau)^2 = - \lambda \tau(\lambda) \text{id}_{S_{\psi,X}^-}.\end{align*} $$

But $\lambda \tau (\lambda ) \in \mathbb {R}_+$ since $\mathcal {L}$ is a CM-field, so there is no $\tau $ -linear $\text {Mp}(W)$ -automorphism of order $2$ and $\omega _{\psi ,X}^-$ does not descend to any strict subextension of $\mathcal {L}$ .

Let $\sigma $ be a generator of $G_2$ . Then $G_2 \cap \mathcal {O}_F^{\times 2} = \langle \sigma ^a \rangle ,$ where $a=1$ if q is a square and $a=2$ otherwise. We set $\tau = \sigma ^a$ and choose $\alpha \in \mathcal {O}_F^\times $ such that $\tau =\alpha ^2$ . The element $\tau $ has order $2^{k_a}$ , where $k_a=k-a+1$ is k or $k-1$ . Define

$$ \begin{align*}r_\tau \cdot f = \tau(\omega_{\psi,X}^-(m_\alpha,1)f).\end{align*} $$

Then $r_\tau ^{2^{k_a}} = \omega _{\psi ,X}^-(m_{-1},1) = - \text {id}_{S_{\psi ,X}^-}$ .

To simplify notations, we set R for the character field, which is also the fixed field of $G_2 \cap \mathcal {O}_F^{\times 2}$ in $\mathcal {L}$ . Let $A = \text {End}_{R[\text {Mp}(W)]}(\omega _{\psi ,X}^-|_R)$ . On the one hand, A contains $\mathcal {L}$ and $r_\tau $ , so it contains the twisted R-algebra generated by $\mathcal {L}$ and $r_\tau ,$ which is of the form

$$ \begin{align*}A_\tau = \mathcal{L}'[X_\tau]/(X_\tau^{2^{k_a}}+1).\end{align*} $$

On the other hand, $\omega _{\psi ,X}^-|_R \otimes _R \mathcal {L} \simeq 2^{k_a} \cdot \omega _{\psi ,X}^-$ , therefore

$$ \begin{align*}\text{End}_{\mathcal{L}[\text{Mp}(W)]}(\omega_{\psi,X}^-|_R \otimes_R \mathcal{L}) \simeq \mathcal{M}_{2^{k_a}}(\mathcal{L}).\end{align*} $$

Comparing the latter with the dimension of $A_\tau \otimes _R \mathcal {L}$ , we deduce that $A_\tau \otimes _R \mathcal {L} \simeq \mathcal {M}_{2^{k_a}}(\mathcal {L})$ . Therefore, $A_\tau = A$ is a central simple R-algebra.

Proof We know that $A = A_\tau = \mathcal {L}'[X_\tau ]/(X_\tau ^{2^{k_a}}+1)$ . We show that $A \otimes _R R_v$ is split if the place v does not lie above p or $\infty $ . Since $\mathcal {L}/R$ is not ramified at v, we have $\mathcal {L} \otimes _R R_v \simeq \mathcal {L}_v^{2^i}$ , where $\mathcal {L}_v/R_v$ is an unramified field extension. The group $G_2 \cap \mathcal {O}_F^{\times 2}$ acts on $\mathcal {L} \otimes _R R_v$ in the following way. There exists a generator $\tau _v$ of $\text {Gal}(\mathcal {L}_v/R_v)$ and an ordering on $\mathcal {L}_v^{2i}$ such that $\tau $ acts via

$$ \begin{align*}\tau \cdot (\lambda_1,\ldots,\lambda_{2^i}) = (\tau_v(\lambda_{2^i}),\lambda_1, \ldots, \lambda_{2^i-1}).\end{align*} $$

By local class field theory, there exists $\lambda _v \in \mathcal {L}_v$ such that $N_{\mathcal {L}_v/R_v}(\lambda _v) = -1$ and therefore the element $\lambda = (\lambda _v,1,\ldots ,1) \in \mathcal {L}_v^{2^i}$ satisfies

$$ \begin{align*}N_{\mathcal{L}_v^{2^i}/R_v}(\lambda) = \prod_{k=0}^{2^i-1} \tau^k(\lambda) = (-1,\ldots,-1) = -1.\end{align*} $$

As a result, $\lambda r_\tau $ has order $2^{k_a}$ and we obtain a semilinear action of $\langle \tau \rangle = G_2 \cap \mathcal {O}_F^{\times 2}$ on the Weil representation, or in other words, there exists $\mathcal {V} \in \text {Irr}_{R_v}(\text {Mp}(W))$ such that

$$ \begin{align*}(\omega_{\psi,X}^- |_R) \otimes_R R_v \simeq 2^{k_a} \mathcal{V}.\end{align*} $$

This implies that $A \otimes _R R_v$ is split.

We claim that $A \simeq A^{\text {op}}$ . Indeed, since $-1$ is a square in $\mathcal {O}_F^\times $ , we have $\omega _{\psi ^{-1},X}^- \simeq \omega _{\psi ,X}^-$ . Furthermore, we have the well-known result

$$ \begin{align*}(\omega_{\psi,X}^-)^\vee \simeq \omega_{\psi^{-1},X}^-.\end{align*} $$

When V is a representation, there is a canonical identification $\text {End}(V^\vee ) \cong \text {End}(V)^{\text {op}}$ . Therefore, A is isomorphic to $A^{\text {op}}$ so the order of A in the Brauer group is either $1$ or $2$ .

Since R is a number field, the $2$ -torsion in the Brauer group of R is generated by quaternion algebras. Therefore, there exists a quaternion algebra D over R such that $A \simeq \mathcal {M}_{2^{k_a-1}}(D)$ . Note that A can’t be split because of Lemma 9.5; otherwise, we would be able to descend the Weil representation to a strict subextension of $\mathcal {L}$ . This implies that D is a quaternion division algebra, which only ramifies at an even number of places.

If $R= \mathbb {Q}$ , we proved that A splits at all places different from p and $\infty $ . So D is the unique quaternion algebra ramified at p and $\infty $ . When $R=\mathbb {Q}[\sqrt {p}]$ , recall that $k_a = k-1$ and $\tau $ has order $2^{k-1}$ , where $k=\text {val}_2(p-1)$ . Let v be the place above p. Then $A \otimes _R R_v$ is split. Indeed, $\mathbb {Q}_p$ contains a primitive $2^k$ -root of unity $\lambda $ , therefore $\lambda r_\tau $ had order $2^{k-1}$ so defines a descent data. We conclude as in the first paragraph. So D ramifies at $\infty $ ’s.

Since $\omega _{\psi ,X}^-|_R$ is semisimple and its endomorphism ring A is isomorphic to $\mathcal {M}_{2^{k_a-1}}(D)$ , there exists $\mathcal {V} \in \text {Irr}_R(\text {Mp}(W))$ such that

$$ \begin{align*}\omega_{\psi,X}^-|_R \simeq 2^{k_a-1} \mathcal{V}.\end{align*} $$

Moreover, $\text {End}_{R[\text {Mp}(W)]}(\mathcal {V}) \simeq D$ and $\mathcal {V} \otimes _R \mathcal {L} \simeq 2 \omega _{\psi ,X}^-$ .

Note that D splits over the quadratic extension $R[\sqrt {-p}]$ of R, therefore there exists $\mathcal {V}' \in \text {Irr}_{R[\sqrt {-p}]}(\text {Mp}(W))$ such that $\mathcal {V} \otimes _R R[\sqrt {-p}] \simeq 2 \mathcal {V}'$ . Going to the composite $\mathcal {L}[\sqrt {-p}]$ of $R[\sqrt {-p}]$ and $\mathcal {L}$ , we get that $\mathcal {V}'$ is a realization of $\omega _{\psi ,X}^- \otimes _{\mathcal {L}} \mathcal {L}[\sqrt {-p}]$ over $R[\sqrt {-p}]$ and the Schur index $m(\omega _{\psi ,X}^-)$ must be $2$ as $\omega _{\psi ,X}^-$ can’t be realized over R.

Proof The situation is rather similar to the case $p \neq 2$ , but there are a few technical complications. In order to define our descent argument, we first want to be able to extract roots from $G_{2'}$ to $\mathcal {O}_F^\times $ , that is, to define a group morphism $\sigma \in G_{2'} \mapsto \lambda \in \mathcal {O}_F^\times $ subject to the relation $\sigma = \lambda ^2$ . To do so, we can embed $G_{2'}$ in $1+ 2 \delta + 8\mathbb {Z}_2$ – note that $1+4 \delta + 8 \mathbb {Z}_2$ works equally well – and extract roots in this subgroup of $\mathbb {Z}_2^\times \subseteq \mathcal {O}_F^\times $ , that is, for all $\sigma \in G_{2'}$ , there exists a unique $\lambda \in 1+ 2 \delta + 8\mathbb {Z}_2$ such that $\sigma = \lambda ^2$ . We obtain a group isomorphism $\sigma \in G_{2'} \mapsto \lambda \in \mathcal {O}_F^\times $ and we denote by $G'$ its image in $\mathcal {O}_F^\times $ , which contains $G_{2'}$ as an index $2$ subgroup.

As opposed to the case $p \neq 2$ , the quadratic Hilbert symbol for $\lambda , \lambda ' \in \mathcal {O}_F^\times $ in the action of $M(X)$ is nontrivial, that is, $\omega _{\psi ,X}(m_\lambda ,1) \omega _{\psi ,X}(m_{\lambda '},1) = (\lambda ,\lambda ')_F \omega _{\psi ,X}(m_{\lambda \lambda '},1)$ . We can consider the restriction of $(-,-)_F$ to $G'$ , which is a subgroup of $\mathcal {O}_F^\times $ . We define an action of $G_{2'}$ on the Weil representation via

$$ \begin{align*}r_\sigma \cdot f = \sigma(\omega_{\psi,X}(m_\lambda,\gamma_\lambda) f),\end{align*} $$

where $\partial \gamma $ trivializes the $2$ -cocycle $(- , - )_F$ on $G'$ , for example, we can take

• • $\gamma _u = 1$ for $u \in G'$ when $(-,-)_F$ is trivial on $G'$ ;

• • $\gamma _u = 1$ for $u \in G_{2'}$ and $\gamma _u = i$ for $u \in G' \backslash G_{2'}$ otherwise.

This defines an effective descent data as in the proof of Theorem 9.1 and we can realize the Weil representation over the fixed field of $G_{2'}$ , that is, over $\mathcal {L}$ .