Generic representations of metaplectic groups and their theta lifts

Abstract

In this paper we give the description of generic representations of metaplectic groups over p-adic fields in terms of their Langlands parameters and calculate their theta lifts on all levels for any tower of odd orthogonal groups. We also describe precisely all the occurrences of the failure of the standard module conjecture for metaplectic groups.

Appendix A: Proof of Theorem 3.10.

Before we begin the proof we remind the reader of our notation: if $\pi$ is an irreducible representation, we let $\phi _\pi$ denote its L-parameter.

Proof

We first prove reducibility. Recall that $a_0= \min \{a: a \text { is even and }1\otimes S_a\hookrightarrow \phi _\sigma \}$. If $4s_i < a_0$, then $\phi _{\Theta _0(\sigma )}$ (which is the same as $\phi _\sigma$) does not contain $1\otimes S_{4s_i}$. Therefore, $\delta ([\nu ^{-(2s_i-\frac{1}{2})},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \Theta _0(\sigma )$ is reducible, and so is $\gamma _{\psi }^{-1}\delta ([\nu ^{-(2s_i-\frac{1}{2})},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \sigma$, [13]. We now imitate the proof of Proposition 4.2.

We thus assume that $\sigma$ is square-integrable, so that the long intertwining operator

$$\begin{aligned} \begin{aligned}&A(w_0,s): \gamma _{\psi }^{-1}\delta ([\nu ^{-(2s_i-\frac{1}{2})}, \nu ^{2s_i-\frac{1}{2}}])\nu ^s \rtimes \sigma \\&\quad \rightarrow \gamma _{\psi }^{-1}\delta ([\nu ^{-(2s_i-\frac{1}{2})},\nu ^{2s_i-\frac{1}{2}}])\nu ^{-s} \rtimes \sigma \end{aligned} \end{aligned}$$

is holomorphic at $s=0$. As before, $A(w_0,s)$ is the restriction of an intertwining operator $B(w_0,s)$ which is composed of the following intertwining operators:

$$\begin{aligned} \begin{aligned}&\gamma _{\psi }^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{(2s_i-\frac{1}{2})}])\nu ^s \times \gamma _{\psi }^{-1}\delta ([\nu ^{-(2s_i-\frac{1}{2})},\nu ^{-\frac{1}{2}}])\nu ^s \rtimes \sigma \\&\quad \xrightarrow {T_1(s)} \gamma _{\psi }^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}])\nu ^s \times \gamma _{\psi }^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}])\nu ^{-s} \rtimes \sigma \\&\quad \xrightarrow {T_2(s)} \gamma _{\psi }^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}])\nu ^{-s} \times \gamma _{\psi }^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}])\nu ^s \rtimes \sigma \\&\quad \xrightarrow {T_3(s)} \gamma _{\psi }^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}])\nu ^{-s} \times \gamma _{\psi }^{-1}\delta ([\nu ^{-(2s_i-\frac{1}{2})},\nu ^{-\frac{1}{2}}])\nu ^{-s} \rtimes \sigma . \end{aligned} \end{aligned}$$

Now assume $\gamma _{\psi }^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}])\rtimes \sigma$ is irreducible. This implies that (after multiplying by a certain positive power of s, if necessary) the operators $T_1(s)$, $T_2(s)$ and $T_3(s)$ are all holomorphic isomorphisms. We thus have a $k \geqslant 0$ such that $\lim _{s \rightarrow 0} s^k B(w_0,s)$ is a holomorphic isomorphism. Furthermore, we know that k is strictly positive since $T_2(s)$ has a pole at $s=0$. This implies that $A(w_0,s)$ has a pole for $s=0$, which contradicts the fact that $A(w_0,s)$ is holomorphic at $s=0$. Therefore, $\gamma _{\psi }^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}])\rtimes \sigma$ must be reducible.

If $\sigma$ is tempered, but not in discrete series, we complete the proof just like in Prop. 4.2.

We now turn to the proof of irreducibility when $4s_i=a_0$. First we need the following

Claim

Apart from the Langlands quotient, all the irreducible subquotients of $\gamma _\psi ^{-1} St_{2s_i}\nu ^{s_i} \rtimes \sigma$ are tempered.

Proof

If $\sigma$ is in discrete series, this is Corollary 2.1 of [28], which transfers easily to the metaplectic case. When $\sigma$ is tempered, we can embed it in its tempered support:

$$\begin{aligned} \sigma \hookrightarrow \gamma _\psi ^{-1}\delta _1' \times \cdots \times \gamma _\psi ^{-1}\delta _k' \rtimes \sigma _0. \end{aligned}$$

We then have, with $\delta =St_{2s_i}\nu ^{s_i}$

$$\begin{aligned} \gamma _\psi ^{-1}\delta \times \gamma _\psi ^{-1}(\delta _1 \times \cdots \times \delta _k) \rtimes \sigma _0&\xrightarrow {T_1} \gamma _\psi ^{-1}(\delta _1 \times \cdots \times \delta _k) \times \gamma _\psi ^{-1}\delta \rtimes \sigma _0\\&\xrightarrow {T_2} \gamma _\psi ^{-1}(\delta _1 \times \cdots \times \delta _k) \times \gamma _\psi ^{-1}\delta ^\vee \rtimes \sigma _0 \\&\xrightarrow {T_3} \gamma _\psi ^{-1}\delta ^\vee \times \gamma _\psi ^{-1}(\delta _1 \times \cdots \times \delta _k) \rtimes \sigma _0. \end{aligned}$$

Restricting $T_3 \circ T_2 \circ T_1$ to the subrepresentation $\gamma _\psi ^{-1}\delta \rtimes \sigma$ we get

$$\begin{aligned} T: \gamma _\psi ^{-1}\delta \rtimes \sigma \rightarrow \gamma _\psi ^{-1}\delta ^\vee \rtimes \sigma . \end{aligned}$$

The image of T is exactly the Langlands quotient of $\gamma _\psi ^{-1}\delta \rtimes \sigma$. We want to show that the kernel consists only of tempered subquotients. We first note that $T_1$ and $T_3$ are isomorphisms: since $4s_i$ is minimal among a such that $1\otimes S_a \hookrightarrow \phi$, none of the segments which define $\delta _1, \dotsc , \delta _k$ are linked to the segment which defines $\delta$ (or $\delta ^\vee$). On the other hand, $T_2$ need not be an isomorphism, but it is induced from $T_2': \gamma _\psi ^{-1}\delta \rtimes \sigma _0 \rightarrow \gamma _\psi ^{-1}\delta ^\vee \rtimes \sigma _0$. By the part of this claim which concerns discrete series representations, $\ker T_2'$ only has tempered subquotients. From here we easily deduce the same for $T_2$. Thus, any subquotient of $\ker T$ is in fact contained in $\ker T_2$ and is therefore tempered. This proves the claim. $\square$

We are now ready to prove irreducibility. The $l=0$ lift of $\sigma$ to the split tower is $\Theta _0(\sigma )$. This is a generic representation of the orthogonal group, so we can use the results of [12]. By Propositions 4.7 and 5.1 of [12], we know that there is an irreducible tempered generic representation $\sigma _1$ such that $\Theta _0(\sigma ) \hookrightarrow \delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \sigma _1$. Therefore, we have $\delta ([\nu ^{-(2s_i-\frac{1}{2})},\nu ^{-\frac{1}{2}}]) \rtimes \sigma _1 \twoheadrightarrow \Theta _0(\sigma )$. We may now use a slight generalization of Lemma 3.8 (cf. Corollary 3.7 of [5]) to obtain either

$$\begin{aligned} \gamma _\psi ^{-1}\delta ([\nu ^{-(2s_i-\frac{1}{2})},\nu ^{-\frac{3}{2}}]) \rtimes \Theta _{-2}(\sigma _1) \twoheadrightarrow \sigma \end{aligned}$$

or

$$\begin{aligned} \gamma _\psi ^{-1}\delta ([\nu ^{-(2s_i-\frac{1}{2})},\nu ^{-\frac{1}{2}}]) \rtimes \Theta _0(\sigma _1) \twoheadrightarrow \sigma . \end{aligned}$$

We show that the latter cannot be true. To this end, we use Theorem 4.1 (2) of [4]. It shows that we have $m^{\text {down}}(\sigma _{1{|SO}}) = m^\alpha (\sigma _{1{|SO}})$, with $\alpha = \eta (z_{\phi _{\sigma _1}})\cdot \epsilon ({\phi _{\sigma _1}}) = \epsilon ({\phi _{\sigma _1}})$ (recall that $\eta$ is trivial since $\sigma _1$ is generic). Similarly, $m^{\text {down}}(\Theta _0(\sigma )_{{|SO}}) = m^{\alpha '}(\Theta _0(\sigma )_{{|SO}})$, with $\alpha ' = \epsilon ({\phi _{\Theta _0(\sigma )_{{|SO}}}})$. Now it is easy to see that $\sigma _1$ and $\Theta _0(\sigma )$ have the same central character. On the other hand, one shows that $\epsilon ({\phi _{\sigma _1}}) = -\epsilon ({\phi _{\Theta _0(\sigma )_{{|SO}}}})$. Since $\Theta _0(\sigma )$ obviously has a non-zero lift on level $l=0$ (it is equal to $\sigma$), this implies that $\Theta _0(\sigma _1) = 0$ (and $\Theta _0(\sigma _1 \otimes \det ) \ne 0$).

We have thus shown that

$$\begin{aligned} \gamma _\psi ^{-1}\delta ([\nu ^{-(2s_i-\frac{1}{2})},\nu ^{-\frac{3}{2}}]) \rtimes \Theta _{-2}(\sigma _1) \twoheadrightarrow \sigma . \end{aligned}$$

Moreover, we now know that $\Theta _{-2}(\sigma _1)$ is the first non-zero lift of $\sigma _1$, so it is tempered by Theorem 4.5 (1) of [4]. It is also generic, as shown by the above map (and the hereditary property).

Now assume $\gamma _\psi ^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \sigma$ reduces. We have shown that any irreducible subquotient of $\gamma _\psi ^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \sigma$ which is not the Langlands quotient is necessarily tempered. Thus let T be an irreducible tempered representation such that

$$\begin{aligned} T \hookrightarrow \gamma _\psi ^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \sigma . \end{aligned}$$

By the above discussion, this implies

$$\begin{aligned}&T \hookrightarrow \gamma _\psi ^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}]) \times \gamma _\psi ^{-1}\delta ([\nu ^{\frac{3}{2}},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \Theta _{-2}(\sigma _1)\\&\quad \Rightarrow \quad T \hookrightarrow \gamma _\psi ^{-1}\delta ([\nu ^{\frac{3}{2}},\nu ^{2s_i-\frac{1}{2}}]) \times \gamma _\psi ^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \Theta _{-2}(\sigma _1)\\&\quad \Rightarrow \quad T \hookrightarrow \gamma _\psi ^{-1}\delta ([\nu ^{\frac{3}{2}},\nu ^{2s_i-\frac{1}{2}}]) \times \gamma _\psi ^{-1}\delta ([\nu ^{\frac{3}{2}},\nu ^{2s_i-\frac{1}{2}}]) \times \gamma _\psi ^{-1}\nu ^\frac{1}{2}\rtimes \Theta _{-2}(\sigma _1). \end{aligned}$$

Now $\gamma _\psi ^{-1}\nu ^\frac{1}{2}\rtimes \Theta _{-2}(\sigma _1) = \gamma _\psi ^{-1}\nu ^{-\frac{1}{2}}\rtimes \Theta _{-2}(\sigma _1)$ by Proposition 4.2, so that

$$\begin{aligned}&T \hookrightarrow \gamma _\psi ^{-1}\delta ([\nu ^{\frac{3}{2}},\nu ^{2s_i-\frac{1}{2}}]) \times \gamma _\psi ^{-1}\delta ([\nu ^{\frac{3}{2}},\nu ^{2s_i-\frac{1}{2}}]) \times \gamma _\psi ^{-1}\nu ^{-\frac{1}{2}}\rtimes \Theta _{-2}(\sigma _1)\\ \text {i.e.} \quad&T \hookrightarrow \gamma _\psi ^{-1}\nu ^{-\frac{1}{2}} \times \gamma _\psi ^{-1}\delta ([\nu ^{\frac{3}{2}},\nu ^{2s_i-\frac{1}{2}}]) \times \gamma _\psi ^{-1}\delta ([\nu ^{\frac{3}{2}},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \Theta _{-2}(\sigma _1). \end{aligned}$$

However, this contradicts Casselman’s criterion. We have thus shown that it is impossible for $\gamma _\psi ^{-1}\delta ([\nu ^{\frac{1}{2}},\nu ^{2s_i-\frac{1}{2}}]) \rtimes \sigma$ to reduce. This completes our proof. $\square$