Strongly positive representations of $GSpin_{2n+1}$ and the Jacquet module method

Abstract

We explicitly construct the structure of Jacquet modules of parabolically induced representations of $GSpin_{2n+1}$ over a $p$-adic field $F$ of any characteristic. Using this construction of the Jacquet module, we obtain a classification of strongly positive representations of $GSpin_{2n+1}$ over $F$ and describe the general discrete series representations of $GSpin_{2n+1}$ over $F$, assuming the half-integer conjecture. One application of this paper is the proof of the equality of $L$-functions from the Langlands–Shahidi method and Artin $L$-functions through the local Langlands correspondence (Kim in Langlands–Shahidi $L$-functions for $GSpin$ groups and the generic Arthur packet conjecture, preprint).

Appendix: Strongly positive representations in an exceptional rank-one reducibility case

The purpose of this appendix is to provide a proper treatment of an exceptional case which appears in the investigation of strongly positive discrete series in [17]. As in [17], we choose to work with metaplectic groups, but the same arguments can be used in the classical or $GSpin$ group case.

Let $\sigma \in D(\rho , \sigma _{cusp})$ denote a strongly positive discrete series of a metaplectic group over a non-Archimedean local field $F$ of a characteristic different than two, $\sigma \ne \sigma _{cusp}$. Also, we suppose that $\rho$ is a self-dual irreducible genuine cuspidal representation of $\widetilde{GL(l, F)}$, a two-fold cover of the general linear group. Further, let $\sigma = \delta (\varDelta _{1}, \varDelta _{2}, \ldots , \varDelta _{k}; \sigma _{cusp})$, i.e., we realize $\sigma$ as a unique irreducible subrepresentation of the induced representation of the form

$$\begin{aligned} \delta (\varDelta _{1}) \times \delta (\varDelta _{2}) \times \cdots \times \delta (\varDelta _{k}) \rtimes \sigma _{cusp}, \end{aligned}$$

where $\varDelta _{1}, \varDelta _{2}, \ldots , \varDelta _{k}$ is a sequence of strongly positive genuine segments satisfying $0 < e(\varDelta _{1}) \le e(\varDelta _{2}) \le \cdots \le e(\varDelta _{k})$. Let us write $\varDelta _{i} = [\nu ^{a_{i}} \rho , \nu ^{b_{i}} \rho ], i = 1, 2, \ldots , k$. The following result complements Theorem 4.4 of [17].

Theorem 7.1

Suppose that $\nu ^{s} \rho \rtimes \sigma _{cusp}$ reduces for $s = \frac{1}{2}$. Then $k = 1$ and $a_{1} = \frac{1}{2}$.

Proof

Strong positivity of $\sigma$ and assumption of the theorem immediately give $a_{k} = \frac{1}{2}$.

Observe that $\sigma$ is a subrepresentation of $\delta (\varDelta _{1}) \times \cdots \times \delta (\varDelta _{k-2}) \rtimes \delta (\varDelta _{k-1}, \varDelta _{k}; \sigma _{cusp})$. Thus, it is enough to prove that $\delta (\varDelta _{k-1}, \varDelta _{k}; \sigma _{cusp})$ is not strongly positive, i.e., to prove $k \ne 2$.

Suppose, on the contrary, $k = 2$. Then we have an embedding $\sigma \hookrightarrow \delta ([ \nu ^{a_{1}} \rho , \nu ^{b_{1}} \rho ]) \times \delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{b_{2}} \rho ]) \rtimes \sigma _{cusp}$. If $a_{1} > \frac{1}{2}$ we get $b_{1} < b_{2}$ and in the same way as in the proof of Theorem 4.4 from [17] we obtain a contradiction with the strong positivity of $\sigma$.

It remains to consider the case $a_{1} = \frac{1}{2}$.

We will first show that there are no strongly positive irreducible subquotients of $\nu ^{\frac{1}{2}} \rho \rtimes \delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{b_{2}} \rho ]; \sigma _{cusp})$, using induction over $b_{2} - \frac{1}{2}$.

First, it can be seen in the same way as in discussion preceding Proposition 3.12 of [9] that the representation $\nu ^{\frac{1}{2}} \rho \rtimes \delta (\nu ^{\frac{1}{2}} \rho ; \sigma _{cusp})$ does not have a strongly positive irreducible subquotient (it contains two irreducible subquotients, the Langlands quotient and a tempered representation).

Further, the representation $\pi = \nu ^{\frac{1}{2}} \rho \rtimes \delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{\frac{3}{2}} \rho ]; \sigma _{cusp})$ does not have a strongly positive irreducible subquotient since we have:

$$\begin{aligned} r_{GL}(\pi )&= \nu ^{\frac{1}{2}} \rho \times \delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{\frac{3}{2}} \rho ]) \otimes \sigma _{cusp} + \nu ^{-\frac{1}{2}} \rho \times \delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{\frac{3}{2}} \rho ]) \otimes \sigma _{cusp} \\ r_{(l)}(\pi )&= \nu ^{\frac{1}{2}} \rho \otimes \delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{\frac{3}{2}} \rho ]; \sigma _{cusp}) + \nu ^{-\frac{1}{2}} \rho \otimes \delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{\frac{3}{2}} \rho ]; \sigma _{cusp})\\&\quad +\,\nu ^{\frac{3}{2}} \rho \otimes \nu ^{\frac{1}{2}} \rtimes \delta (\nu ^{\frac{1}{2}} \rho ; \sigma _{cusp}). \end{aligned}$$

If $\sigma '$ is some strongly positive irreducible subquotient of $\pi$, then obviously $r_{GL}(\sigma ') \ge \nu ^{\frac{1}{2}} \rho \times \delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{\frac{3}{2}} \rho ]) \otimes \sigma _{cusp}$, but this implies that $r_{(l)}(\sigma ')$ contains some irreducible subquotient of $\nu ^{\frac{3}{2}} \rho \otimes \nu ^{\frac{1}{2}} \rtimes \delta (\nu ^{\frac{1}{2}} \rho ; \sigma _{cusp})$, contradicting strong positivity of $\sigma '$.

Let us now suppose that the representation $\nu ^{\frac{1}{2}} \rho \rtimes \delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{m} \rho ]; \sigma _{cusp})$ does not have a strongly positive irreducible subquotient for $m < b_{k}$. We study the induced representation $\pi = \nu ^{\frac{1}{2}} \rho \rtimes \delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{b_{k}} \rho ]; \sigma _{cusp})$. Similarly as in the previously considered case we have:

$$\begin{aligned} r_{GL}(\pi )&= \nu ^{\frac{1}{2}} \rho \times \delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{b_{k}} \rho ]) \otimes \sigma _{cusp} + \nu ^{-\frac{1}{2}} \rho \times \delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{b_{k}} \rho ]) \otimes \sigma _{cusp} \\ r_{(l)}(\pi )&= \nu ^{\frac{1}{2}} \rho \otimes \delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{b_{k}} \rho ]; \sigma _{cusp}) + \nu ^{-\frac{1}{2}} \rho \otimes \delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{b_{k}} \rho ]; \sigma _{cusp}) \\&\quad +\,\nu ^{b_{k}} \rho \otimes \nu ^{\frac{1}{2}} \rtimes \delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{b_{k}-1} \rho ]; \sigma _{cusp}). \end{aligned}$$

Using the inductive assumption, in completely same way as in the case $b_{k} = \frac{3}{2}$ we deduce that $\pi$ does not contain a strongly positive irreducible subquotient.

Now, suppose that strongly positive discrete series $\sigma$ is a subrepresentation of $\delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{b_{1}} \rho ]) \rtimes \delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{b_{2}} \rho ]; \sigma _{cusp})$, where $b_{1} \le b_{2}$. We have

$$\begin{aligned} \sigma&\hookrightarrow \delta ([ \nu ^{\frac{3}{2}} \rho , \nu ^{b_{1}} \rho ]) \times \nu ^{\frac{1}{2}} \rho \rtimes \delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{b_{2}} \rho ]; \sigma _{cusp}). \end{aligned}$$

There is some irreducible representation $\pi$ such that $\sigma \hookrightarrow \delta ([ \nu ^{\frac{3}{2}} \rho , \nu ^{b_{1}} \rho ]) \rtimes \pi$. Since $\sigma$ is strongly positive, $\pi$ also has to be strongly positive. Also, Frobenius reciprocity gives $\mu ^{*}(\sigma ) \ge \delta ([ \nu ^{\frac{3}{2}} \rho , \nu ^{b_{1}} \rho ]) \otimes \pi$. Using the structural formula for $\mu ^{*}$, we get $\pi \le \nu ^{\frac{1}{2}} \rho \times \delta ([ \nu ^{\frac{3}{2}} \rho , \nu ^{b_{2}} \rho ]) \times \nu ^{\frac{1}{2}} \rho \rtimes \sigma _{cusp}$. Since $\pi$ is strongly positive, looking at Jacquet modules with respect to Siegel parabolic subgroup first we obtain $\pi \le \nu ^{\frac{1}{2}} \rho \times \delta ([\nu ^{\frac{3}{2}} \rho , \nu ^{b_{2}} \rho ]) \rtimes \delta (\nu ^{\frac{1}{2}} \rho ; \sigma _{cusp})$. Since $\delta ([\nu ^{\frac{1}{2}} \rho , \nu ^{b_{2}} \rho ]; \sigma _{cusp})$ is the only irreducible subquotient of $\delta ([\nu ^{\frac{3}{2}} \rho , \nu ^{b_{2}} \rho ]) \rtimes \delta (\nu ^{\frac{1}{2}} \rho ; \sigma _{cusp})$ whose Jacquet module with respect to Siegel parabolic subgroup contains only representations of the form $\pi ' \otimes \sigma _{cusp}$ with no $\nu ^{x} \rho , x \le 0$, appearing in the cuspidal support of $\pi '$, it follows that $\pi$ is a subquotient of $\nu ^{\frac{1}{2}} \rho \rtimes \delta ([ \nu ^{\frac{1}{2}} \rho , \nu ^{b_{2}} \rho ]; \sigma _{cusp})$, a contradiction.

Therefore, $k = 1$ and the proof is complete. $\square$