Construction 2.7. Consider the vector space $U=\operatorname {\mathrm {M}}_2(F)$ , and let W be a two-dimensional symplectic vector space. We turn $U\otimes W$ into a symplectic vector space via

$$\begin{align*}\left<m\otimes v,m'\otimes v'\right>=(v\wedge v')\cdot\left<m,m'\right>. \end{align*}$$

In particular, we get a map

$$\begin{align*}G^{3,\det=1}=({\mathrm{GL}}_2(F)\times{\mathrm{GL}}_2(F)\times{\mathrm{GL}}_2(F))^1 \rightarrow \operatorname{\mathrm{Sp}}(U\otimes W) \end{align*}$$

of the group

$$\begin{align*}G^{3,\det=1}=\left\{(g_1,g_2,g_3)\in G^3{\,|\,}\det(g_1 g_2 g_3)=1\right\} \end{align*}$$

into the group of symplectic automorphisms of $U\otimes W$ . This embedding is defined by

$$\begin{align*}(g_1,g_2,g_3)\cdot(m\otimes v)=g_1 m g_3^{T}\otimes g_2 v. \end{align*}$$

Construction 2.9. The map $G^{3,\det =1} \rightarrow \operatorname {\mathrm {Sp}}(U\otimes W)$ of Construction 2.7 can be extended to a map

$$\begin{align*}S_3\ltimes G^{3,\det=1} \rightarrow \operatorname{\mathrm{Sp}}(U\otimes W) \end{align*}$$

where $S_3$ acts by interchanging the order of the factors. In particular, we fix the convention that the transposition $(1,3)\in S_3$ acts by

$$\begin{align*}(1,3)\cdot(m\otimes v)=m^{T}\otimes v. \end{align*}$$

Recollection 2.10 (Weil representation of metaplectic group)

Let V be a symplectic vector space over F. The Heisenberg group $H(V)$ is defined as the underlying set $V\times F$ , with multiplication given by

$$\begin{align*}(v,x)\cdot(v',x')=(v+v',\frac{1}{2}\left<v,v'\right>+x+x'). \end{align*}$$

The group $H(V)$ has a unique (up to scalar automorphism) irreducible representation $(\pi ,X)$ such that $(0,x)$ acts via any given nontrivial additive character $e{\,:\,} F\to {\mathbb {C}}^\times $ . For any symplectomorphism $g\in \operatorname {Sp}(V)$ , one can define the representation $\pi ^g$ given by $\pi ^g(v,x)=\pi (g(v),x)$ . The group of pairs $(g,\gamma )$ , where $g\in \operatorname {Sp}(V)$ and $\gamma {\,:\,}\pi \to \pi ^g$ is an $H(V)$ -linear map is called the metaplectic group $\operatorname {\mathrm {Mp}}(V)$ . This group sits in an exact sequence

and it has a canonical representation on the space X, called the Weil representation.

For $L\subseteq V$ a Lagrangian subspace and a splitting

one can give an explicit model for this representation, $X=S(L)$ . Here, $S(L)$ is the space of smooth and compactly supported functions on L, with group action

$$\begin{align*}\pi(v,x)\cdot \Psi(\ell)=e\left(-\left<v,\ell+\frac{1}{2}p(v)\right>+x\right)\cdot\Psi(\ell+p(v)) \end{align*}$$

for $v\in V$ , $\ell \in L$ , $x\in F$ . This defines an action of $\operatorname {\mathrm {Mp}}(V)$ on $S(L)$ , called the Schrödinger model of the Weil representation.

Construction 2.11. Consider the Lagrangian subspace $\operatorname {\mathrm {M}}_2(F)=\operatorname {\mathrm {M}}_2(F)\otimes e_2\subseteq U\otimes W$ , where $e_1,e_2\in W$ is a standard basis. We define a splitting split $U\otimes W\to \operatorname {\mathrm {M}}_2(F)$ with kernel $\operatorname {\mathrm {M}}_2(F)\otimes e_1$ . Then by the Schrödinger model of the Weil representation of $\operatorname {\mathrm {Mp}}(U\otimes W)$ on $S(\operatorname {\mathrm {M}}_2(F))$ corresponding to the character $e{\,:\,} F \rightarrow {\mathbb {C}}^\times $ , we get an action of the metaplectic group

$$\begin{align*}\operatorname{\mathrm{Mp}}(U\otimes W) \end{align*}$$

on $S(\operatorname {\mathrm {M}}_2(F))$ .

Construction 2.13. We use induction with compact support to extend the action of $G^{3,\det =1}$ on $S(\operatorname {\mathrm {M}}_2(F))$ to an action of $G^3$ on Y. Specifically, we identify

$$\begin{align*}Y=S(\operatorname{\mathrm{M}}_2(F)\times F^\times)=\operatorname{\mathrm{cInd}}_{S_3\ltimes G^{3,\det=1}}^{S_3\ltimes G^3}S(\operatorname{\mathrm{M}}_2(F)) \end{align*}$$

via the section

$$\begin{align*}G^{3,\det=1}\backslash G^3=F^\times \rightarrow G^3 \end{align*}$$

given by $y\mapsto \left (1,\begin {pmatrix}y & \\ & 1\end {pmatrix},1\right )$ , together with a determinant twist.

Explicitly, if $\Psi (m,y)=\Phi (m)f(y)\in S(\operatorname {\mathrm {M}}_2(F)\times F^\times )$ for $\Phi \in S(\operatorname {\mathrm {M}}_2(F))$ and $f\in S(F^\times )$ , then we define

(2.4) $$ \begin{align} & ((g_1,g_2,g_3)\cdot\Psi)(m,y) \notag \\ &\quad ={\left|{\det(g_1 g_2 g_3)}\right|}\cdot\left(\left(g_1,\begin{pmatrix}y & \\ & 1\end{pmatrix}g_2\begin{pmatrix}y^{-1}\det(g_1 g_2 g_3)^{-1} & \\ & 1\end{pmatrix},g_3\right)\cdot\Phi\right)(m) f(y\cdot\det(g_1 g_2 g_3)), \end{align} $$

where the action of $\left (g_1,\begin {pmatrix}y & \\ & 1\end {pmatrix}g_2\begin {pmatrix}y^{-1}\det (g_1 g_2 g_3)^{-1} & \\ & 1\end {pmatrix},g_3\right )$ on $\Phi $ is the Weil representation action.

We refer to the resulting action of the middle copy of $G={\mathrm {GL}}_2(F)$ on Y as the middle action.

Proof of Proposition 2.12

Let us begin by showing that there is at most one metaplectic lift for the map $S_3\ltimes G^{3,\det =1} \rightarrow \operatorname {\mathrm {Sp}}(U\otimes W)$ compatible with the specified action of the metaplectic lift of $(1,3)$ . This follows because the abelianization of $S_3\ltimes G^{3,\det =1}$ is ${\mathbb {Z}}/2$ and is generated by the image of the transposition $(1,3)$ due to the fact that any two lifts to $\operatorname {\mathrm {Mp}}(U\otimes W)$ differ by a character. Also, recall that the Weil representation is faithful for the metaplectic group so that describing lifts of elements of $S_3\ltimes G^{3,\det =1}$ is the same as describing their actions on $S(\operatorname {\mathrm {M}}_2(F))$ .

Let us show the existence of this lift. We will do so in several steps. Let us first show that the subgroup $\operatorname {\mathrm {SL}}_2(F)^3\subseteq S_3\ltimes G^{3,\det =1}$ admits a unique lift. We will then handle the rest of the group. Indeed, the lift of each factor $\operatorname {\mathrm {SL}}_2(F)$ is unique because its abelianization is trivial. Note that the subgroup $\operatorname {\mathrm {SL}}_2(F)\times \{1\}\times \operatorname {\mathrm {SL}}_2(F)$ consisting of two of the copies of $\operatorname {\mathrm {SL}}_2(F)$ admits an explicit metaplectic lift by the following action on $S(\operatorname {\mathrm {M}}_2(F))$ :

(2.5) $$ \begin{align} (g_1,1,g_3)\cdot \Psi(m)=\Psi(g_1^{-1}m g_3^{-T}). \end{align} $$

Because the pairs $\operatorname {\mathrm {SL}}_2(F)\times \operatorname {\mathrm {SL}}_2(F)\times \{1\}$ and $\{1\}\times \operatorname {\mathrm {SL}}_2(F)\times \operatorname {\mathrm {SL}}_2(F)$ are conjugate to it, these pairs admit metaplectic lifts as well, and all of those lifts agree on each factor $\operatorname {\mathrm {SL}}_2(F)$ separately by uniqueness. This shows that there is a metaplectic lift for $\operatorname {\mathrm {SL}}_2(F)\times \operatorname {\mathrm {SL}}_2(F)\times \operatorname {\mathrm {SL}}_2(F)$ . For the sake of being explicit, we note that the action of the middle copy of $\operatorname {\mathrm {SL}}_2(F)$ is given by

(2.6) $$ \begin{align} \left(1,\begin{pmatrix}1 & b \\ & 1\end{pmatrix},1\right)\cdot \Psi(m) & =e\left(b\det(m)\right)\cdot\Psi(m) \end{align} $$

(2.7) $$ \begin{align} \left(1,\begin{pmatrix}t & \\ & t^{-1}\end{pmatrix},1\right)\cdot \Psi(m) & ={\left|{t}\right|}^2\cdot\Psi(t\cdot m) \end{align} $$

(2.8) $$ \begin{align} \left(1,\begin{pmatrix}& -1 \\ 1 &\end{pmatrix},1\right)\cdot \Psi(m) & =\int\limits_{\operatorname{\mathrm{M}}_2(F)}\Psi(n)e(-\left<m,n\right>){\,\mathrm{d}{n}}, \end{align} $$

where the measure ${\mathrm {d}{n}}$ on $\operatorname {\mathrm {M}}_2(F)$ is normalized such that $e(\left <\cdot ,\cdot \right>)$ is a self-dual character.

Now, we observe that there is a split short exact sequence

where $(F^\times )^{3,\Pi =1}=\left \{(a,b,c)\in F^\times {\,|\,} abc=1\right \}$ . We think of $(F^\times )^{3,\Pi =1}$ as embedded in $G^{3,\det =1}$ via the splitting:

$$\begin{align*}(a,b,c)\mapsto\left(\begin{pmatrix}a & \\ & 1\end{pmatrix},\begin{pmatrix}b & \\ & 1\end{pmatrix},\begin{pmatrix}c & \\ & 1\end{pmatrix}\right). \end{align*}$$

Thus, it remains to show that there is a metaplectic lift for $S_3\ltimes (F^\times )^{3,\Pi =1}$ as it will act by conjugation on $\operatorname {\mathrm {SL}}_2(F)^3$ correctly by uniqueness.

We construct the lift for $S_3\ltimes (F^\times )^{3,\Pi =1}$ in three steps. We will first construct a lift $S_2\ltimes (F^\times )^{3,\Pi =1}$ for the copy of $S_2\subseteq S_3$ generated by the permutation $\tau =(1,3)$ . Then, we will construct a lift for a permutation $\sigma \in S_3$ of order $3$ . Finally, we will show that all required relations between $\sigma $ and $S_2\ltimes (F^\times )^{3,\Pi =1}$ must hold.

We lift $S_2\ltimes (F^\times )^{3,\Pi =1}$ explicitly by defining

(2.9) $$ \begin{align} \left(\begin{pmatrix}a & \\ & 1\end{pmatrix},\begin{pmatrix}b & \\ & 1\end{pmatrix},\begin{pmatrix}c & \\ & 1\end{pmatrix}\right)\cdot\Psi(m) & ={\left|{b}\right|}\cdot\Psi\left(\begin{pmatrix}a^{-1} & \\ & 1\end{pmatrix}m\begin{pmatrix}c^{-1} & \\ & 1\end{pmatrix}\right) \end{align} $$

(2.10) $$ \begin{align} \tau\cdot\Psi(m) & =\Psi(m^{T}) \end{align} $$

for $\Psi \in S(\operatorname {\mathrm {M}}_2(F))$ , and $\tau =(1,3)\in S_3$ being the transposition of the first and third actions.

Let $\sigma \in S_3$ be the permutation $\sigma (1)=2,\sigma (2)=3,\sigma (3)=1$ of order $3$ . Observe that there is a unique lift for sigma in $\operatorname {\mathrm {Mp}}(U\otimes W)$ which respects the relation

$$\begin{align*}\tau\sigma\tau^{-1}=\sigma^2. \end{align*}$$

Note that because $\tau $ has order $2$ , such a lift automatically respects all relations of $S_3$ . For the sake of being explicit, we note that this lift is given by

(2.11) $$ \begin{align} \sigma\cdot\Psi\left(\begin{pmatrix}m_{00} & m_{01} \\ m_{10} & m_{11}\end{pmatrix}\right)=\int\limits_{F\times F}e(\beta\cdot m_{00}-\alpha\cdot m_{10})\Psi\left(\begin{pmatrix}\alpha & \beta \\ m_{01} & m_{11}\end{pmatrix}\right){\,\mathrm{d}{\alpha}}{\,\mathrm{d}{\beta}}. \end{align} $$

We have now chosen lifts for both $S_2\ltimes (F^\times )^{3,\Pi =1}$ and $\sigma $ . Thus, to show that $S_3\ltimes (F^\times )^{3,\Pi =1}$ admits a metaplectic lift and finish the proof, it suffices to check that the action of $S_3$ on the metaplectic lift of $(F^\times )^{3,\Pi =1}$ by conjugation is the correct one. This can be verified explicitly using equations (2.9) and (2.11)

Proof. Observe that the statement is true for the left and right translation actions of G on $S(\operatorname {\mathrm {M}}_2(F))$ (recall Remark 2.16) because the $G\times G$ -equivariant pairing

$$ \begin{align*} S(\operatorname{\mathrm{M}}_2(F))\otimes S(G) & \rightarrow {\mathbb{C}} \\ \Psi\otimes f & \mapsto \int_G \Psi(g)f(g){\,\mathrm{d}^\times\!{g}} \end{align*} $$

is nondegenerate and because the Bernstein center of G acts on $S(G)$ the same way from both sides. As a result, the left and right actions of the center of G on Y coincide. By symmetry (i.e., by conjugation by a permutation from $S_3$ ), this also applies to the middle action.

Proof. By equation (2.6) and the definition in equation (2.4), the middle action of the matrix $\begin {pmatrix} a & b \\ & 1\end {pmatrix}$ on the element $\Psi \in Y$ is

(2.12) $$ \begin{align} \begin{pmatrix} a & b \\ & 1\end{pmatrix}\cdot\Psi(g,y)={\left|{a}\right|}\cdot e\left(b\det(g)y\right)\cdot\Psi(g,ay). \end{align} $$

By the exact sequence of part (e) of Proposition 3.2 of [Reference Bernstein and Zelevinsky3], it follows that the map is injective and its image consists of the functions supported away from the set $\{\det (g)y=0\}$ .

Proof of Theorem 2.2

We consider the middle action of G on Y, given by Construction 2.13. That it commutes with the left and right G-actions is by construction. The existence of the isomorphism $\nu $ of Theorem 2.2 is proven as follows.

First, we suppose that $(\pi ,V)$ is either an irreducible principal series representation or a supercuspidal representation. By Lemma 2.17, the space

$$\begin{align*}\tilde{V}\otimes_G Y\otimes_G V \end{align*}$$

is V-isotypic, because such irreducible representations are determined by the characters they define on the Bernstein center (also known as their central character, or infinitesimal character). Our goal is to count the number of copies of V appearing in it and show that it is $1$ . By the assumptions on V, we have that $\Phi ^-V$ is one-dimensional. Therefore, it is enough to test the dimension of

$$\begin{align*}\Phi^-\left(\tilde{V}\otimes_G Y\otimes_G V\right). \end{align*}$$

However, this is the same as

$$\begin{align*}\Phi^-\Phi^+\Phi^-\left(\tilde{V}\otimes_G Y\otimes_G V\right), \end{align*}$$

which by Lemma 2.19 is the same as evaluating

$$\begin{align*}\Phi^-\left(\tilde{V}\otimes_G Y^\circ\otimes_G V\right)\xrightarrow{\sim}\Phi^- S(F^\times)={\mathbb{C}}. \end{align*}$$

Note that the first isomorphism is given by $\Phi ^-(\mu )$ for $\mu $ as in equation (2.2), and it follows from equation (2.12) that $\mu $ is an isomorphism of P-modules. This shows that the number of copies of V appearing in the space above is indeed $1$ , and shows that the diagram in Item 3 is commutative.

The above argument does not quite work in the case that $(\pi ,V)$ is a Steinberg representation because the Bernstein center is not strong enough to distinguish V from the trivial representation. However, in this case, it is sufficient to verify that the image of the Jacquet functor on $\tilde {V}\otimes _G Y\otimes _G V$ is one dimensional (because we already know that $\Phi ^-\left (\tilde {V}\otimes _G Y\otimes _G V\right )$ is one dimensional by the argument above). By equation (2.12), it follows that the Jacquet functor on $\tilde {V}\otimes _G Y\otimes _G V$ is given by

$$\begin{align*}\tilde{V}\otimes_G S(\operatorname{\mathrm{M}}_2^{\det=0}(F)\times F^\times)\otimes_G V. \end{align*}$$

Hence, we need to prove that this space is one dimensional. This follows immediately from Lemma 2.20 below (note that V is a Steinberg representation and therefore generic, meaning that it admits no $\operatorname {\mathrm {SL}}_2(F)$ -invariant linear functionals, as the lemma requires).

Proof. We use the decomposition of $S(\operatorname {\mathrm {M}}_2^{\det =0}(F))$ via an exact sequence

where

is the trivial representation. This uses an identification of the quotient of

$$\begin{align*}P\backslash G\times P\backslash G \end{align*}$$

by the diagonal left action of $F^\times $ with the space of rank-one matrices.

If we tensor this sequence by $S(F^\times )$ , we obtain the decomposition

where $U\subseteq G$ is the unipotent radical. Tensoring by $\tilde {V}$ and V from the left and right, we get a morphism:

(2.13)

and we want to show that it is an isomorphism.

The fact that equation (2.13) is an isomorphism will follow because our assumption about $\operatorname {\mathrm {SL}}_2(F)$ -invariant linear functionals implies that

(2.14) $$ \begin{align} S(F^\times)\otimes_G V=S(F^\times)\otimes_G \tilde{V}=0. \end{align} $$

Indeed, surjectivity immediately follows from $\tilde {V}\otimes _G S(F^\times )\otimes _G V=0$ and because relative tensor products are right exact. Injectivity is trickier but can be proven as follows. Essentially, it remains to prove that

$$\begin{align*}\operatorname{\mathrm{Tor}}_1^{G\times G}(\tilde{V}\otimes V,S(F^\times))=0. \end{align*}$$

Plugging equation (2.14) in to the spectral sequence

$$\begin{align*}\operatorname{\mathrm{Tor}}_i^G(\tilde{V},\operatorname{\mathrm{Tor}}_j^G(V,S(F^\times))) \implies \operatorname{\mathrm{Tor}}_{i+j}^{G\times G}(\tilde{V}\otimes V,S(F^\times)), \end{align*}$$

we see that the objects at $(i,j)=(0,0),(1,0),(0,1)$ are all $0$ .

Construction 3.3. We have a functor

given by

with the $G\times G$ action on Y being given by the left and right translation actions, respectively, and where the left G-action on

is given by the middle action of G on Y.

Figure 1

String diagram for the bifunctor .

Warning 3.25. Note that, while most generic irreducible representations are degeneracy-free, this is not always the case. Specifically, let ${\mathrm {St}}$ be the Steinberg representation with trivial central character. While ${\mathrm {St}}$ is generic, it is not degeneracy-free. Instead, let us denote by the unique nontrivial extension of the trivial representation ${\mathrm {St}}$ by . Then is degeneracy-free.

Construction 3.44. Let $\nu (e)$ be the largest integer such that $e(\pi ^{-\nu (e)}{\mathcal {O}})=1$ , with $\pi \in {\mathcal {O}}$ a uniformizer. We let the map

be given as follows.

Let f be the distribution on G given by the delta distribution

$$\begin{align*}f(g)=e(b)\delta_1(\det(k)) \end{align*}$$

if g has the form

$$\begin{align*}\begin{pmatrix}1 & b \\ & 1\end{pmatrix}\begin{pmatrix}1 & \\ & -\pi^{\nu(e)}\end{pmatrix}k \end{align*}$$

with $k\in {\mathrm {GL}}_2({\mathcal {O}})$ , and we set $f(g)=0$ otherwise. It is clear that f is compactly supported modulo U, left- $\operatorname {\mathrm {SL}}_2({\mathcal {O}})$ -invariant, and right- $\theta $ -equivariant.

Therefore, f defines an $\operatorname {\mathrm {SL}}_2({\mathcal {O}})$ -invariant element of the roughening $\operatorname {\mathrm {Hom}}(S(G),\operatorname {\mathrm {cInd}}_U^G(\theta ))$ given explicitly by

$$\begin{align*}\phi(g)\mapsto\int_U\int_{\operatorname{\mathrm{SL}}_2({\mathcal{O}})}e(-b)\phi\left(k^{-1}\begin{pmatrix}1 & \\ & -\pi^{-\nu(e)}\end{pmatrix}\begin{pmatrix}1 & b \\ & 1\end{pmatrix}g\right){\,\mathrm{d}^\times\!{k}}{\,\mathrm{d}{b}}. \end{align*}$$

By the universal property in Remark 3.39, we conclude that the distribution f defines a map $\varepsilon {\,:\,} E \rightarrow \operatorname {\mathrm {cInd}}_U^G(\theta )$ . Note that the resulting map is independent of the choice of measure on U.

Construction 3.45. Let ${Y}^{\operatorname {\mathrm {SL}}_2({\mathcal {O}})\times \operatorname {\mathrm {SL}}_2({\mathcal {O}})}$ be the space of vectors in Y which are invariant under the left and right actions of $\operatorname {\mathrm {SL}}_2({\mathcal {O}})$ . We let $\mu $ be the map

given as in Remark 3.39 by the $S_3\ltimes \operatorname {\mathrm {SL}}_2({\mathcal {O}})^3$ -invariant distribution (more precisely, $\operatorname {\mathrm {SL}}_2({\mathcal {O}})$ -invariant element of the roughening of $Y^{\operatorname {\mathrm {SL}}_2({\mathcal {O}})\times \operatorname {\mathrm {SL}}_2({\mathcal {O}})}$ )

for $(g,y)\in \operatorname {\mathrm {M}}_2(F)\times F^\times $ . This can be more explicitly described as the $\operatorname {\mathrm {SL}}_2({\mathcal {O}})$ -invariant element of $\operatorname {\mathrm {Hom}}(S(G),Y^{\operatorname {\mathrm {SL}}_2(O)\times \operatorname {\mathrm {SL}}_2({\mathcal {O}})})$ given by

where $\pi $ denotes the middle action on Y.

Proof of Claim 3.49

In order to prove that diagram (3.7) commutes, it suffices to evaluate it for the roughenings of the respective objects. Specifically, one may evaluate the two compositions on the object of the roughening of E given informally by the uniform distribution of volume $1$ on $\operatorname {\mathrm {SL}}_2({\mathcal {O}})$ as this generates all of E. Under $\mu $ , this distribution maps into

We wish to apply the map $\varepsilon $ , followed by the isomorphism of Construction 3.16, and then compare the result with the delta distribution on $\operatorname {\mathrm {SL}}_2({\mathcal {O}})$ , from which we started. We do so via the formula in equation (3.3); plugging $\Psi $ and $\varepsilon $ in, we must show that

$$\begin{align*}\int_{\operatorname{\mathrm{SL}}_2({\mathcal{O}})}\left(w\begin{pmatrix}1 & \\ & -\pi^{\nu(e)}\end{pmatrix}^{-T}k^{-T}\cdot\Psi\right)\left(g^{-1},-\det(g)\right){\,\mathrm{d}^\times\!{k}} \end{align*}$$

is the uniform distribution of volume $1$ on $\operatorname {\mathrm {SL}}_2({\mathcal {O}})$ .

However, because $\Psi $ is $\operatorname {\mathrm {SL}}_2({\mathcal {O}})$ -invariant, the above simplifies to

and we deduce that this is indeed the desired distribution from

$$\begin{align*}\operatorname{\mathrm{M}}_2({\mathcal{O}})\cap\operatorname{\mathrm{SL}}_2(F)=\operatorname{\mathrm{SL}}_2({\mathcal{O}}). \end{align*}$$

It remains to show that $\mu $ is an isomorphism. Let us start by noting that applying the functor $\Phi ^-_{\mathrm {enh}}$ to $\mu $ , we obtain a retract of ${\mathcal {Z}}_{\mathrm {sph}}$ -modules

Since $I_{RL}(E)\otimes _G E\cong {\mathcal {Z}}_{\mathrm {sph}}$ has no nontrivial retracts when restricted to any component, we deduce that $\Phi ^-(\mu )$ is an isomorphism.

Having shown that $\Phi ^-$ takes the map $\mu $ into an isomorphism, we see that the cokernel of $\mu $ is degenerate. However, this cokernel must be split (diagram (3.7) splits it), and it is easy to see that Y has no degenerate subrepresentations. Indeed, because Y consists of smooth and compactly supported functions on $\operatorname {\mathrm {M}}_2(F)\times F^\times $ , it is clear that it has no vector invariant under the action of U.

Proof. Consider any extension of to a symmetric monoidal structure (that is, suppose that there is some associator for satisfying the axioms of a symmetric monoidal structure).

We want to show that any two such extensions are equal. Because

respects colimits, it is enough to compare their associators

on objects $A,B,C$ that are either of the two generators

. Note that if any of the objects $A,B,C$ is equal to

, then the associator is uniquely determined by unitality.

Thus, it remains to show that the only possible choice for the associator

is the one coming from the functor

$$\begin{align*}\operatorname{\mathrm{Mod}}({\mathcal{Z}})\times\operatorname{\mathrm{Mod}}({\mathcal{Z}}_{\mathrm{sph}}) \rightarrow \operatorname{\mathrm{Mod}}(G). \end{align*}$$

That is, it remains to show that the only possible choice for the associator

is the identification

coming from repeated applications of Claim 3.49.

Indeed, in this case the pentagonal axiom

fully determines the associator; any such associator must be given by a map $a{\,:\,} E\xrightarrow {\sim } E$ (after using Claim 3.49 to identify

), which must then satisfy

$$\begin{align*}a^2=a^3. \end{align*}$$

Note that this is the same argument as the idempotent theorem (Theorem 3.1.2) of [Reference Gurevich and Hadani9].

Proof of Theorem 3.52

The functor

$$\begin{align*}\iota{\,:\,}\operatorname{\mathrm{Mod}}({\mathcal{Z}})\times\operatorname{\mathrm{Mod}}({\mathcal{Z}}_{\mathrm{sph}}) \rightarrow \operatorname{\mathrm{Mod}}(G) \end{align*}$$

defines associator isomorphisms

for $A,B,C\in \operatorname {\mathrm {Mod}}({\mathcal {Z}})\times \operatorname {\mathrm {Mod}}({\mathcal {Z}}_{\mathrm {sph}})$ , by considering the images of the associators in its domain. Note that, because $\iota $ is not fully faithful, this family of isomorphisms does not automatically descend to its essential image.

Let ${\mathcal {C}}_0\subseteq \operatorname {\mathrm {Mod}}(G)$ be the fully faithful subcategory given by finite direct sums of

and E. Because

is a ${\mathcal {Z}}$ -linear functor that respects colimits, it is uniquely determined by its restriction to ${\mathcal {C}}_0$ . Indeed, $\operatorname {\mathrm {Mod}}(G)$ is generated as a ${\mathcal {Z}}$ -linear category by

and E. In fact,

is given by the left Kan extension

Thus, it is sufficient to show that the associators $a_{A,B,C}$ commute with the morphisms of $\operatorname {\mathrm {Mod}}(G)$ on the finite direct sums of ${\mathcal {Z}}$ and ${\mathcal {Z}}_{\mathrm {sph}}$ in $\operatorname {\mathrm {Mod}}({\mathcal {Z}})\times \operatorname {\mathrm {Mod}}({\mathcal {Z}}_{\mathrm {sph}})$ , and therefore induce a natural morphism on ${\mathcal {C}}_0$ . The required compatibilities with the rest of the symmetric monoidal structure (and the pentagonal axiom) will follow immediately from those of $\operatorname {\mathrm {Mod}}({\mathcal {Z}})\times \operatorname {\mathrm {Mod}}({\mathcal {Z}}_{\mathrm {sph}})$ .

Because $\operatorname {\mathrm {Mod}}({\mathcal {Z}})\times \operatorname {\mathrm {Mod}}({\mathcal {Z}}_{\mathrm {sph}})$ is generated under colimits by

and E, it is sufficient to prove that for all

and all maps $f{\,:\,} C \rightarrow D$ , the following diagram commutes:

The claim follows immediately if either A or B is . Additionally, because respects the action of the center and because both and ${\mathrm {End}}(E)$ are covered by the center, it is enough to consider the case $C\neq D$ , so either or .

Since the proof in either case is identical, it is enough to consider the case

. Letting

, we must show that the following diagram commutes:

We can rewrite this as the following diagram commuting:

which is equivalent to asking that the two maps

given by

and

be equal.

In other words, it is sufficient to prove that the map

$$\begin{align*}\operatorname{\mathrm{Hom}}(E,E) \rightarrow \operatorname{\mathrm{Hom}}(E,E) \end{align*}$$

given by

is the identity map of $\operatorname {\mathrm {Hom}}(E,E)$ . But the center of $\operatorname {\mathrm {Mod}}(G)$ covers $\operatorname {\mathrm {Hom}}(E,E)$ . Therefore, it is enough to test this property on the identity map of E, and verify that

However, this holds by functoriality of

.

Proof. Recall that for every place v, the space is a cocommutative, counital, coalgebra by Remark 3.56. For places v where $e_v$ is unramified, denote by $f_v$ the vector .

We observe that under the counit of Construction 3.44, the vector $f_v$ maps to the distinguished vector of . Moreover, observe that under the comultiplication map of Construction 3.45, the vector $f_v$ maps to $f_v\otimes y_v\otimes f_v\in E_v\otimes _{G_v}Y_v\otimes _{G_v} E_v$ .

Therefore, the counitality of $E_v$ implies Item 1, the cocommutativity of $E_v$ implies Item 2, and the coassociativity of $E_v$ implies Item 3.

Construction 4.13. We think of as the space of smooth functions on the automorphic quotient ${\mathrm {GL}}_2(F)\backslash {\mathrm {GL}}_2({\mathbb {A}})$ , and of as the space of $\theta ^{-1}$ -equivariant functions on ${\mathrm {GL}}_2({\mathbb {A}})$ . We define the map by sending an automorphic function to its corresponding Whittaker function:

$$\begin{align*}\phi(g)\mapsto W_\phi(g)=\int_{U({\mathbb{A}})/U(F)}\phi(ug)\theta(u){\,\mathrm{d}{u}}. \end{align*}$$

Here, we use a normalized measure on $U({\mathbb {A}})/U(F)$ such that the total volume is $1$ .

Construction 4.15. Define the map

as the composition of

$$ \begin{align*} S({\mathrm{GL}}_2({\mathbb{A}}))_{/U_2({\mathbb{A}}),\theta} & \rightarrow S(U_2(F)\backslash{\mathrm{GL}}_2({\mathbb{A}})) \\ f(g) & \mapsto\int_{U_2({\mathbb{A}})}f(ug)\theta(u^{-1}){\,\mathrm{d}{u}} \end{align*} $$

with the projection

. In other words, using the equivalence of Remark 4.2, a function

is mapped to the Poincaré series

It is a standard fact that this Poincaré series is an element of $S({\mathrm {GL}}_2(F)\backslash {\mathrm {GL}}_2({\mathbb {A}}))$ .

Proof of Proposition 4.16

We want to show that the two maps

coincide. This is a straightforward verification. We will show that the diagram

commutes by separately evaluating the clockwise and counter-clockwise compositions.

We begin with the counterclockwise composition. Using equation (3.3), it is given by

$$ \begin{align*} f(h)\otimes\Psi(g,y)\mapsto\sum_{\xi\in{\mathrm{GL}}_2(F)}\int_{{\mathrm{GL}}_2({\mathbb{A}})}(wh^{-T}\cdot\Psi)(\xi^{-1},-\det(\xi))\cdot f(h){\,\mathrm{d}^\times\!{h}}. \end{align*} $$

We should compare this with the clockwise composition

$$ \begin{align*} f(h)\otimes\Psi(g,y) &\mapsto \sum_{(\xi,q)\in\operatorname{\mathrm{M}}_2(F)\times F^\times}\int_{U_2(F)\backslash{\mathrm{GL}}_2({\mathbb{A}})}\int_{U_2({\mathbb{A}})}(h^{-T}\cdot\Psi)(\xi,q)\cdot f(uh)\cdot\theta(u^{-1}){\,\mathrm{d}{u}}{\,\mathrm{d}^\times\!{h}} \\ &=\sum_{(\xi,q)\in\operatorname{\mathrm{M}}_2(F)\times F^\times}\int_{U_2(F)\backslash{\mathrm{GL}}_2({\mathbb{A}})}\int_{U_2({\mathbb{A}})}(wh^{-T}\cdot\Psi)(\xi,-q)\cdot f(uh)\cdot\theta(u^{-1}){\,\mathrm{d}{u}}{\,\mathrm{d}^\times\!{h}} \\ &=\sum_{(\xi,q)\in\operatorname{\mathrm{M}}_2(F)\times F^\times}\int_{{\mathrm{GL}}_2({\mathbb{A}})}(wh^{-T}\cdot\Psi)(\xi,-q)\cdot f(h)\cdot \int_{{\mathbb{A}}/F}e(x(q\det{\xi}-1)){\,\mathrm{d}{x}} {\,\mathrm{d}^\times\!{h}}. \end{align*} $$

Thus, the two maps coincide.

Proof. Consider the commutative diagram

The left vertical map is surjective, while the right vertical map is injective. This means that the diagram can be completed as above.

Proof. Follows immediately from the surjectivity of

and the axioms of symmetric monoidal categories. By Corollaries 4.18 and 4.21, it is sufficient to show that the multiplication

is associative. Indeed, we have two maps

and we want to show that they are the same. By Proposition 4.19, we can test this by precomposing with the unit map

at one of the coordinates. However, the two maps

are the same by the unitality result of Corollary 4.18.

Proof. Identical to Corollary 4.21.