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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 \begin{equation} \rho: G\longrightarrow GL_2(\mathbb{F}_q) \end{equation} 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 \begin{equation} \rho': G\longrightarrow GL_2(\mathbb{F}_p) \end{equation} 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.

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  • $\begingroup$ sorry, what does $\rho'\otimes\mathbb{F}_q$ mean? $\endgroup$ – vidyarthi Aug 21 at 8:44
  • $\begingroup$ maybe, this question is somewhat related $\endgroup$ – vidyarthi Aug 21 at 8:53
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    $\begingroup$ The subgroups of $GL_2(q)$ are known, so you can check this with, say, maximal subgroups. $\endgroup$ – Dima Pasechnik Aug 21 at 9:11
  • $\begingroup$ 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 $\rho'$? $\endgroup$ – kneidell Aug 21 at 9:16
  • $\begingroup$ @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 $\rho'(g)\in GL_2(\mathbb{F}_p)\hookrightarrow GL_2(\mathbb{F}_p)$. $\endgroup$ – Leo D Aug 21 at 15:04
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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 $G\xrightarrow{\chi_1}\mathbb{F}_{p^2}^{\times}\to GL_2(\mathbb{F}_p)$

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  • $\begingroup$ Great! Thanks for the elegant proof. (Sorry for the late respnse). $\endgroup$ – Leo D Aug 27 at 18:22

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