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 be a representation of in general position. Then is reducible and has a unique semisimple component, call it . Then*

*is the “quadratic twist” of , 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 is an additive character of , is the Weil representation of , and is a rational parabolic subgroup with Levi subgroup isomorphic to .

The strategy of proof is as follows:

- Express as a combination of Deligne-Lusztig characters.
- Compute the degree of .
- Deduce a bound on the degree of
- Compute the character value of $latex \tilde{\sigma}$ at all non-unipotent elements.
- Deduce the dimension of .
- 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 be a connected reductive algebraic group over with Frobenius morphism . Then, as , for any pair of representations of , such that and is zero on all non-unipotent elements, we have:*

*where the implied constant depends only on .*

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

We get the desired result by applying the following lemma:

**Lemma 2.** *Let be two representations of a finite group with equal dimensions, such that their characters agree on a subset , and that is irreducible. Assume there exists a , such that . Then on all of .*

Our result says that we can twist representations of 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 is irreducible, and , acts as a scalar on , say . So

which implies that also acts as on .

Let . Then we have

* Proof of Lemma 1:* Let be a positive integer, and , be as in the statement of the lemma. The assumption implies that . Using Corollary 7.7 in [DL], there exists a pair , , such that

Since is supported only at unipotent elements, this character pairing depends only on the values at unipotent elements. But this in turn is independent of by the character formula, Theorem 4.2 in [DL], so

for all .

For any in general position 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 . The number of such is

and each has dimension

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

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).