# Descent a representation over finite field

Let $$p$$ be a prime integer, and $$q$$ a power of $$p$$. Let $$\mathbb{F}_p$$ and $$\mathbb{F}_q$$ be the corresponding finite fields. Suppose $$$$\rho: G\longrightarrow GL_2(\mathbb{F}_q)$$$$ is a representation of finite group $$G$$. Now if we know that for every element $$g\in G$$, the characteristic polynomial $$\rho(g)$$ is defined over $$\mathbb{F}_p$$. Can we conclude that we can descend $$\rho$$ as an $$\mathbb{F}_p$$-representation, i.e. can we find a representation $$$$\rho': G\longrightarrow GL_2(\mathbb{F}_p)$$$$ such that $$\rho'\otimes \mathbb{F}_q=\rho$$?

If the answer is no. How about the situation when $$\rho$$ is an abelian representation?

If the answer is still no, what kind of conditions we need to give a positive answer? Thanks.

• sorry, what does $\rho'\otimes\mathbb{F}_q$ mean? – vidyarthi Aug 21 at 8:44
• maybe, this question is somewhat related – vidyarthi Aug 21 at 8:53
• The subgroups of $GL_2(q)$ are known, so you can check this with, say, maximal subgroups. – Dima Pasechnik Aug 21 at 9:11
• Just to make sure I understand what you mean: Say $q=p^2$, $G=\mathbb F_q^\times$ and $\rho(x)=\mathrm{diag}(x,x^p)$. Then the characteristic polynomial of $\rho(x)$ is $t^2-\mathrm{Tr}(x) t+\mathrm{N}(x)\in\mathbb F_p[t]$. How would you define $\rho'$? – kneidell Aug 21 at 9:16
• @vidyarthi, I should be more careful when I was typing. In fact I should type $\rho'\otimes_{\mathbb{F}_p}{\mathbb{F}_q}$. To be more precise, it means for every $g\in G$, we consider $\rho'(g)\in GL_2(\mathbb{F}_p)\hookrightarrow GL_2(\mathbb{F}_p)$. – Leo D Aug 21 at 15:04

As stated, there is a non-semisimple counterexample: take $$G$$ to be the additive group $$\mathbb{F}_q$$ with a unipotent representation $$\rho(a)=\left(\begin{matrix}1 & a\\ 0 & 1\end{matrix}\right)$$. Its character is trivial but it is not defined over $$\mathbb{F}_p$$ because $$GL_2(\mathbb{F}_p)$$ does not have an order $$p^2$$ subgroup.
If we assume that $$\rho$$ is semi-simple then the answer is positive.
First, assume additionally that $$\rho$$ is absolutely irreducible. In general, let $$L/K$$ be a finite Galois extension and $$\rho:G\to GL_n(L)$$ be an irreducible representation with the characteristic polynomial of any element $$\rho(g)$$ defined over $$K$$. For any element $$\sigma\in Gal(L/K)$$ the representations $$\sigma(\rho)$$ and $$\rho$$ are semi-simple representations with equal characteristic polynomials, hence they are isomorphic(Bourbaki Algebra, Chapter 8, §20.6 Corollary 1 to thm 2) and the isomorphism is unique up to a scalar by the Schur's lemma. This gives an element $$A(\sigma)\in PGL_n(L)$$ for every $$\sigma\in Gal(L/K)$$ and by the unicity of the intertwining operators these matrices form a cocycle in $$H^1(Gal(L/K),PGL_n(L))$$. If this cocycle happens to be trivial, that is there is an element $$B\in PGL_n(L)$$ such that $$A(\sigma)=\sigma(B)B^{-1}$$, then $$B^{-1}\rho B$$ is a representation invariant under all the elements $$\sigma$$ and is the desired descent.
The group $$H^1(Gal(L/K),PGL_n(L))$$ injects into the Brauer group $$Br(K)$$. If $$K$$ is a finite field then $$Br(K)$$ vanishes and so does the group $$H^1(Gal(L/K),PGL_n(L))$$ which implies the vanishing of the obstruction to descending our representation.
If $$\rho$$ is not absolutely irreducible but is semi-simple(being semi-simple over $$\mathbb{F}_q$$ and over $$\overline{\mathbb{F}_q}$$ are equivalent conditions) then, after possibly enlarging $$q$$, it becomes isomorphic to a direct sum of two characters $$\chi_1\oplus\chi_2$$ such that $$\chi_1(g)$$ and $$\chi_2(g)$$ are roots of a quadratic polynomial with coefficients in $$\mathbb{F}_p$$. Thus theses characters must either be defined over $$\mathbb{F}_p$$ already or be defined over $$\mathbb{F}_{p^2}$$ such that $$\overline{\chi_1}=\chi_2$$ where $$\overline{\bullet}$$ is the non-trivial automorphism of this quadratic extension. The representation $$\rho$$ is then isomorphic to the base change of $$G\xrightarrow{\chi_1}\mathbb{F}_{p^2}^{\times}\to GL_2(\mathbb{F}_p)$$