Descent a representation over finite fieldter1.0ht4xpaf1 viss T mkо ќx ddci
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.
-
$\\begingroup$ sorry, what does $\\rho'\\otimes\\mathbb{F}_q$ mean? $\\endgroup$ – vidyarthi 8 hours ago
-
$\\begingroup$ maybe, this question is somewhat related $\\endgroup$ – vidyarthi 7 hours ago
-
$\\begingroup$ The subgroups of $GL_2(q)$ are known, so you can check this with, say, maximal subgroups. $\\endgroup$ – Dima Pasechnik 7 hours ago
-
$\\begingroup$ 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 7 hours ago
-
$\\begingroup$ @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 1 hour ago
1 Answer
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 $G\\xrightarrow{\\chi_1}\\mathbb{F}_{p^2}^{\\times}\\to GL_2(\\mathbb{F}_p)$
