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 $latex \sigma$ be a representation of $latex Sp_{2m}(\mathbb{F}_q)$ in general position. Then $latex {\textup{Ind}_{P_{1,2m,1}}^{Sp_{2m+2}(\mathbb{F}_q)}(1_{\mathbb{F}_q^\times}\times\sigma)}$ is reducible and has a unique semisimple component, call it $latex \pi(\sigma)$. Then
$latex \tilde{\sigma}:=(\textup{Res}_{Sp_{2m}\ltimes H_{2m}}(\pi(\sigma))\otimes\omega_\psi)^{H_{2m}}$
is the “quadratic twist” of $latex \sigma$, 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 $latex \psi$ is an additive character of $latex \mathbb{F}_q$, $latex \omega_\psi$ is the Weil representation of $latex Sp_{2m}\ltimes H_{2m}$, and $latex P_{1,2m,1}$ is a rational parabolic subgroup with Levi subgroup isomorphic to $latex \mathbb{F}_q^\times\times Sp_{2m}(\mathbb{F}_q)$.
The strategy of proof is as follows:
- Express $latex \pi(\sigma)$ as a combination of Deligne-Lusztig characters.
- Compute the degree of $latex \pi(\sigma)$.
- Deduce a bound on the degree of $latex \tilde{\sigma}$
- Compute the character value of $latex \tilde{\sigma}$ at all non-unipotent elements.
- Deduce the dimension of $latex \tilde\sigma$.
- 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 $latex \textup{\textbf{G}}$ be a connected reductive algebraic group over $latex \mathbb{F}_q$ with Frobenius morphism $latex F$. Then, as $latex n\rightarrow\infty$, for any pair of representations $latex \rho_1,\rho_2$ of $latex \textup{\textbf{G}}^{F^n}$, such that $latex {\textbf{dim}\ \rho_1\ne \emph{dim}\ \rho_2}$ and $latex {\chi_{\rho_1}-\chi_{\rho_2}}$ is zero on all non-unipotent elements, we have:
$latex \emph{dim}\ \rho_1\oplus\rho_2 \gg \sqrt{|\textup{\textbf{G}}^{F^n}|(q^n)^{r(\textup{\textbf{G}})}}$
where the implied constant depends only on $latex \textup{\textbf{G}}$.
In our case, $latex \rho_1$ is $latex \tilde\sigma$, and $latex \rho_2$ is the quadratic twist of $latex \sigma$. Assume that $latex q$ is large, and that $latex {\emph{dim}\ \rho_1\ne \emph{dim}\ \rho_2}$. Then the hypotheses of the lemma are satisfied. But then the conclusion of the lemma contradicts the bound on the dimension of $latex \tilde\sigma$. So, for all large enough $latex q$, we must have $latex {\emph{dim}\ \rho_1= \emph{dim}\ \rho_2}$. But the dimension of $latex \tilde\sigma$, as well as the dimension of the quadratic twist of $latex \sigma$, are polynomials in $latex q$ ($latex \sigma$ going over a family of Deligne-Lusztig characters in general position for the same torus). Hence, for all $latex q$ we have equality of dimensions.
We get the desired result by applying the following lemma:
Lemma 2. Let $latex \rho_1,\rho_2$ be two representations of a finite group $latex G$ with equal dimensions, such that their characters agree on a subset $latex H\subset G$, and that $latex \rho_2$ is irreducible. Assume there exists a $latex z\in Z(G)\cap H$, such that $latex z\cdot(G\backslash H)\subset H$. Then $latex \chi_{\rho_1}=\chi_{\rho_2}$ on all of $latex G$.
Our result says that we can twist representations of $latex Sp_{2m}(\mathbb{F}_q)$ 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 $latex \rho_2$ is irreducible, and $latex z\in Z(G)$, $latex z$ acts as a scalar on $latex \rho_2$, say $latex \lambda_z$. So
$latex \chi_{\rho_1}(z) = \chi_{\rho_2}(z) = \lambda_z\chi_{\rho_2}(1)=\lambda_z\chi_{\rho_1}(1)$
which implies that $latex z$ also acts as $latex \lambda_z$ on $latex \rho_1$.
Let $latex g\in G\backslash H$. Then we have
$latex \chi_{\rho_1}(g)=\chi_{\rho_1}(z^{-1}zg)=\lambda_z^{-1}\chi_{\rho_1}(zg)=\lambda_z^{-1}\chi_{\rho_2}(zg)=\chi_{\rho_2}(g).$
$latex \blacksquare$
Proof of Lemma 1: Let $latex n$ be a positive integer, and $latex \rho_1$,$latex \rho_2$ be as in the statement of the lemma. The assumption $latex {\textup{dim} \rho_1\ne \textup{dim} \rho_2}$ implies that $latex {(\chi_{\rho_1}-\chi_{\rho_2},\textup{reg}_{\textup{\textbf{G}}})\ne 0}$. Using Corollary 7.7 in [DL], there exists a pair $latex ({\textup{\textbf{T}}},\theta)$, $latex \theta\in\widehat{{\textup{\textbf{T}}}^{F^n}}$, such that
$latex (\chi_{\rho_1}-\chi_{\rho_2}, R_{{\textup{\textbf{T}}}}^{\textup{\textbf{G}}}(\theta)) \ne 0.$
Since $latex {\chi_{\rho_1}-\chi_{\rho_2}}$ is supported only at unipotent elements, this character pairing depends only on the values at unipotent elements. But this in turn is independent of $latex \theta$ by the character formula, Theorem 4.2 in [DL], so
$latex (\chi_{\rho_1}-\chi_{\rho_2}, \varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{\textup{\textbf{T}}}^{\textup{\textbf{G}}}(\theta’)) \ne 0.$
for all $latex \theta’\in\widehat{{\textup{\textbf{T}}}^{F^n}}$.
For any $latex \theta’\in\widehat{{\textup{\textbf{T}}}^{F^n}}$ in general position $latex {\varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{\textup{\textbf{T}}}^{\textup{\textbf{G}}}(\theta’)}$ 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 $latex \rho_1,\rho_2$. The number of such $latex \theta’$ is
$latex \gg (q^n)^{r({\textup{\textbf{G}}})}$
and each $latex \varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{{\textup{\textbf{T}}}}^{\textup{\textbf{G}}}(\theta’)$ has dimension
$latex \varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{\textup{\textbf{T}}}^{\textup{\textbf{G}}}(\theta’)(1) = |{\textup{\textbf{G}}}^{F^n}/{\textup{\textbf{T}}}^{F^n}|_{p’}\gg \sqrt{|{\textup{\textbf{G}}}^{F^n}|(q^n)^{-r({\textup{\textbf{G}}})}}$
by Theorem 7.1 in [DL]. This proves the lemma.
$latex \blacksquare$
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).