Quadratic Twists for Representations of Finite Symplectic Groups

I started mentioning descent in this post. During the course of writing my masters results in article form, I found a cute descent-like result that can be proven with similar ideas.

Theorem. Let $\sigma$ be a representation of $Sp_{2m}(\mathbb{F}_q)$ in general position. Then ${\textup{Ind}_{P_{1,2m,1}}^{Sp_{2m+2}(\mathbb{F}_q)}(1_{\mathbb{F}_q^\times}\times\sigma)}$ is reducible and has a unique semisimple component, call it $\pi(\sigma)$. Then

$\tilde{\sigma}:=(\textup{Res}_{Sp_{2m}\ltimes H_{2m}}(\pi(\sigma))\otimes\omega_\psi)^{H_{2m}}$

is the “quadratic twist” of $\sigma$, i.e. the characters in its Deligne-Lusztig parameter are the same, but multiplied by the quadratic character. A representation is said to be in general position if its character is plus or minus a Deligne-Lusztig character in general position.

Here $\psi$ is an additive character of $\mathbb{F}_q$, $\omega_\psi$ is the Weil representation of $Sp_{2m}\ltimes H_{2m}$, and $P_{1,2m,1}$ is a rational parabolic subgroup with Levi subgroup isomorphic to $\mathbb{F}_q^\times\times Sp_{2m}(\mathbb{F}_q)$.

The strategy of proof is as follows:

1. Express $\pi(\sigma)$ as a combination of Deligne-Lusztig characters.
2. Compute the degree of $\pi(\sigma)$.
3. Deduce a bound on the degree of $\tilde{\sigma}$
4. Compute the character value of $latex \tilde{\sigma}$ at all non-unipotent elements.
5. Deduce the dimension of $\tilde\sigma$.
6. Deduce character values at unipotent elements.

The first four points are pretty straightforward computations with Deligne-Lusztig characters. The fifth point is an application of the following crucial lemma:

Lemma 1. Let $\textup{\textbf{G}}$ be a connected reductive algebraic group over $\mathbb{F}_q$ with Frobenius morphism $F$. Then, as $n\rightarrow\infty$, for any pair of representations $\rho_1,\rho_2$ of $\textup{\textbf{G}}^{F^n}$, such that ${\textbf{dim}\ \rho_1\ne \emph{dim}\ \rho_2}$ and ${\chi_{\rho_1}-\chi_{\rho_2}}$ is zero on all non-unipotent elements, we have:

$\emph{dim}\ \rho_1\oplus\rho_2 \gg \sqrt{|\textup{\textbf{G}}^{F^n}|(q^n)^{r(\textup{\textbf{G}})}}$

where the implied constant depends only on $\textup{\textbf{G}}$.

In our case, $\rho_1$ is $\tilde\sigma$, and $\rho_2$ is the quadratic twist of $\sigma$. Assume that $q$ is large, and that ${\emph{dim}\ \rho_1\ne \emph{dim}\ \rho_2}$. Then the hypotheses of the lemma are satisfied. But then the conclusion of the lemma contradicts the bound on the dimension of $\tilde\sigma$. So, for all large enough $q$, we must have ${\emph{dim}\ \rho_1= \emph{dim}\ \rho_2}$. But the dimension of $\tilde\sigma$, as well as the dimension of the quadratic twist of $\sigma$, are polynomials in $q$ ($\sigma$ going over a family of Deligne-Lusztig characters in general position for the same torus). Hence, for all $q$ we have equality of dimensions.

We get the desired result by applying the following lemma:

Lemma 2. Let $\rho_1,\rho_2$ be two representations of a finite group $G$ with equal dimensions, such that their characters agree on a subset $H\subset G$, and that $\rho_2$ is irreducible. Assume there exists a $z\in Z(G)\cap H$, such that $z\cdot(G\backslash H)\subset H$. Then $\chi_{\rho_1}=\chi_{\rho_2}$ on all of $G$.

Our result says that we can twist representations of $Sp_{2m}(\mathbb{F}_q)$ with an explicit representation theoretic construction. I am not aware of such a result over local and global fields. All proofs that I have seen for twisting general automorphic forms use converse theorems. I will note that Shimura gave a geometric meaning to twisting modular forms by a general Dirichlet character, but being geometric in nature precludes it from working for Maass forms.

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Proofs for lemmas. I’ll start with the easier one.

Proof of Lemma 2: Since $\rho_2$ is irreducible, and $z\in Z(G)$, $z$ acts as a scalar on $\rho_2$, say $\lambda_z$. So

$\chi_{\rho_1}(z) = \chi_{\rho_2}(z) = \lambda_z\chi_{\rho_2}(1)=\lambda_z\chi_{\rho_1}(1)$

which implies that $z$ also acts as $\lambda_z$ on $\rho_1$.

Let $g\in G\backslash H$. Then we have

$\chi_{\rho_1}(g)=\chi_{\rho_1}(z^{-1}zg)=\lambda_z^{-1}\chi_{\rho_1}(zg)=\lambda_z^{-1}\chi_{\rho_2}(zg)=\chi_{\rho_2}(g).$
$\blacksquare$

Proof of Lemma 1: Let $n$ be a positive integer, and $\rho_1$,$\rho_2$ be as in the statement of the lemma. The assumption ${\textup{dim} \rho_1\ne \textup{dim} \rho_2}$ implies that ${(\chi_{\rho_1}-\chi_{\rho_2},\textup{reg}_{\textup{\textbf{G}}})\ne 0}$. Using Corollary 7.7 in [DL], there exists a pair $({\textup{\textbf{T}}},\theta)$, $\theta\in\widehat{{\textup{\textbf{T}}}^{F^n}}$, such that

$(\chi_{\rho_1}-\chi_{\rho_2}, R_{{\textup{\textbf{T}}}}^{\textup{\textbf{G}}}(\theta)) \ne 0.$

Since ${\chi_{\rho_1}-\chi_{\rho_2}}$ is supported only at unipotent elements, this character pairing depends only on the values at unipotent elements. But this in turn is independent of $\theta$ by the character formula, Theorem 4.2 in [DL], so

$(\chi_{\rho_1}-\chi_{\rho_2}, \varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{\textup{\textbf{T}}}^{\textup{\textbf{G}}}(\theta')) \ne 0.$

for all $\theta'\in\widehat{{\textup{\textbf{T}}}^{F^n}}$.

For any $\theta'\in\widehat{{\textup{\textbf{T}}}^{F^n}}$ in general position ${\varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{\textup{\textbf{T}}}^{\textup{\textbf{G}}}(\theta')}$ is the character of an irreducible representation by Theorem 6.8 and Theorem 7.1 in [DL], thus the above shows that it is a constituent of at least one of $\rho_1,\rho_2$. The number of such $\theta'$ is

$\gg (q^n)^{r({\textup{\textbf{G}}})}$

and each $\varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{{\textup{\textbf{T}}}}^{\textup{\textbf{G}}}(\theta')$ has dimension

$\varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{\textup{\textbf{T}}}^{\textup{\textbf{G}}}(\theta')(1) = |{\textup{\textbf{G}}}^{F^n}/{\textup{\textbf{T}}}^{F^n}|_{p'}\gg \sqrt{|{\textup{\textbf{G}}}^{F^n}|(q^n)^{-r({\textup{\textbf{G}}})}}$

by Theorem 7.1 in [DL]. This proves the lemma.
$\blacksquare$

After finishing with my masters results, I’ll write the first four points as well. Stay tuned!

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[DL] P. Deligne, G. Lusztig, Representations of reductive groups over finite fields, Ann. of Math. (2) 103, no. 1, 103?161. (1976).