# Endomorphism algebras of restricted representations

Let $G$ be a group, and $$\rho:G\to \mathrm{GL}(V)$$ be an absolutely irreducible, finite-dimensional representation over a characteristic $0$ field $k$. For each finite index subgroup $H\le G$, let $$\mathrm{End}_H(V) = \{\phi\in\mathrm{End}(V) :g\cdot\phi(v) = \phi(g\cdot v)\;\;\forall v\in V, \ g\in H\}$$ be the endomorphism ring of $\rho|_H$.

Fix some finite index normal subgroup $N\lhd G$. Then there is a natural map \begin{align}\begin{Bmatrix}\text{Subgroups H with }\\N\le H\le G\end{Bmatrix}&\to\begin{Bmatrix}\text{Subalgebras A with } \\\mathrm{End}_G(V)\subseteq A\subseteq\mathrm{End}_N(V)\end{Bmatrix}\\ \\H&\mapsto \mathrm{End}_H(V).\end{align}

Can we determine which subalgebras $A$ are in the image of this map?

I'm particularly interested in the case where $\dim V = 4$ and $\mathrm{End}_N(V) = M_2(k)$. In this case, is there a sugbroup $H$ such that $\mathrm{End}_H(V) = k\times k$?

Edit:

My motivation is as follows. Suppose I have a representation (in my case, a Galois representation) $$\rho:G\to\mathrm{Aut}(V)\cong\mathrm{GL}_4(k)$$ and I know that for some normal subgroup $N$, $$\rho|_N = \sigma\oplus\sigma$$ for some representation $\sigma:N\to\mathrm{GL}_2(k)$. A priori, I have no information about $N$. Under what circumstances can I find an $H$ such that $$\rho|_H=\sigma_1\oplus\sigma_2,$$ where the $\sigma_i:H\to\mathrm{GL}_2(k)$ are distinct? Since I know nothing about $N$, I'm hoping for a condition that is intrinsic to the representation in some way.

Further edit:

Let $G$ act on $\mathrm{End}_N(V)$ by $$g\cdot \phi = \rho(g)\circ\phi\circ\rho(g^{-1})$$ for $g\in G$, $\phi\in \mathrm{End}_N(V)$. Enlarging $N$ if necessary, we can assume that $N$ is the kernel of this action, and we get an injection $$G/N\hookrightarrow \mathrm{Aut}(M_2(k))=\mathrm{PGL}_2(k).$$ Since $G/N$ is finite, it is either cyclic, dihedral, $A_4$, $S_4$ or $A_5$. As Johannes points out in his answer, it is only the $A_5$ case which presents a difficulty.

Are there any facts that I could use about $\rho$, without knowing what $N$ is, that would enable me to rule out this case?

• Btw do you assume $G$ to be finite? I'm afraid infinite groups could be arbitrarily bad behaved unless we restrict ourselves to some form of "good" groups and "good" representations. – Johannes Hahn Jul 27 '17 at 16:45
• @JohannesHahn I really care about Galois representations, so $G = \mathrm{Gal}(\overline K/K)$ is profinite, and representations are continuous. – Ariel Weiss Jul 27 '17 at 16:51
• In particular $\rho$ factors through a finite group, right? Do I remember that correctly? In that case we can assume wlog that $G$ is a(n irreducible) subgroup of $U_4(\mathbb{C})$. – Johannes Hahn Jul 27 '17 at 17:06
• That's only true if $k =\mathbb C$ or a finite field. Typically, $k$ will be something like $\overline{\mathbb Q}_p$ for me, and the image won't be finite. – Ariel Weiss Jul 27 '17 at 17:07

## 1 Answer

EDIT: Finiteness isn't necessary for this argument. Instead I use that $V$ is semisimple over $N$, $|G:N|<\infty$ and $char(k)=0$.

$Res_N^G(V)$ is semisimple because it is the restriction of a simple module to a normal subgroup. And it having $k^{2\times 2}$ as endomorphism ring is equivalent to it being the sum of two isomorphic two-dimensional, absolutely simple $N$-modules by the Wedderburn theorem. In particular, every $v\in V\setminus\{0\}$ generates a two-dimensional, simple $N$-module and all of those are isomorphic.

Now note that $Res_H^G(V)$ is also semisimple if $N\leq H\leq G$. This is because every $H$-invariant $U\leq V$ is also $N$-invariant and therefore has a $N$-invariant complement. Averaging over $H/N$ gives a a $H$-invariant complement.

Therefore $Res_H^G(V)$ can have the following endomorphism rings: $k$ (iff the restriction is still absolutely simple), a quadratic extension of $k$ (iff the restriction is simple, but not absolutely simple), $k\times k$ (iff the restriction decomposes into two different irreducibles) and $k^{2\times 2}$ (iff the restriction decomposes into two isomorphic irreducibles).

Theorem: Let $G$ is a group and $V$ a absolutely simple $k[G]$-module, $N\unlhd G$ a normal subgroup such that $End(Res_N^G(V))=k^{2\times 2}$. Then the following are equivalent:

1. There exists a subgroup $N\leq H\leq G$ such that $End_k(Res_H^G(V)) = k\times k$.
2. There exists a $g\in G$ such that $Res_{\langle g\rangle N}^G(\chi_V)$ is not divisible by two in the character ring $Ch(\langle g\rangle N)$.

Proof: If $g\in G$ is arbitrary and $v\in V$ an eigenvector of $g$, then $U:=span\{nv \mid n\in N\}$ is a $g$-invariant subspace. Therefore we get a decomposition $V=U\oplus U'$ into $\langle g,N\rangle$-invariant subspaces. It follows that $\chi_V(g) = \chi_U(g) + \chi_{U'}(g)$.

$1.\implies 2.$ Now $Res_H^G(V) = U_1 \oplus U_2$ for two irreducible, non-isomorphic $H$-modules and $Res_{H'}^H(U_i)$ is still irreducible for all $N\leq H'\leq H$. If we choose $g\in H$ such that $\chi_{U_1}(g) \neq \chi_{U_2}(g)$ and an eigenvector $v\in U_1$, then our construction gives $U=U_1$ and $U' \cong U_2$, $Res_{\langle g\rangle N}(\chi_V) = Res_{\langle g\rangle N}^H(\chi_{U_1}) + Res_{\langle g\rangle N}^H(\chi_{U_2})$ and these two summands are different irreducible characters. Therefore $Res_{\langle g\rangle N}(\chi_V)$ can not be divisible by two.

$2.\implies 1.$ Conversely if such a $g$ exists, then $H:=\langle g\rangle N$ satisfies the conditions.

Corollary: If there is a $g\in G$ such that $2 \nmid \chi_V(g)$, then $H=\langle g,N\rangle$ satisfies $End(Res_H^G(V)) = k\times k$.

Here is an example where the Corollary is actually applicable: Let $G=S_3\times S_3$ and $\rho=\sigma\otimes\sigma$ where $\sigma$ is the two-dimensional character of $S_3$. Over $N:=S_3\times 1$ this representation restricts to $\sigma+\sigma$. The element $g:=(c,c)$ where $c\in S_3$ satisfies $\chi_V(g)=1$.

Corollary 2: If $G/N$ is supersolvable, $H$ exists with the desired properties.

Being supersolvable means that there exists a normal series $N=M_0 < M_1 < \ldots < M_k = G$ such that every quotient is cyclic. We have just demonstrated that $\sigma$ can be extended to a (automatically irreducible) character $\widehat{\sigma}$ of $\langle g,N\rangle$ which is also a constituent of the restriction of $\rho$, no matter what $g\in G$ is. In particular it can be extended to $\sigma_1\in Irr(M_1)$. Let $\sigma_2 := Res_{M_1}^G(\rho) -\sigma_1$. If $\sigma_1\neq\sigma_2$, we are done. If not, we're back in the original situation: We have a normal subgroup $M_1\unlhd G$ such that $Res_{M_1} ^G(\rho)$ is the sum of two isomorphic $M_1$-modules so that we can proceed by induction. QED

• Thanks for your answer! I'm not sure the corollary helps: if $\mathrm{End}(\mathrm{Res}^G_H(V)) = k\times k$, then enlarging $H$ if necessary, $V=\mathrm{Ind}^G_H(U)$, where $U$ is one of the two distinct subreps. For any $g\in G$, wouldn't $\chi_V(g)$ usually (always?) be even? – Ariel Weiss Jul 27 '17 at 15:39
• My motivation behind the question is this: I have a four dimensional representation $\rho$ which decomposes as $\sigma + \sigma$ after restriction to $N$. I only know that $N$ is a finite index subgroup, but nothing else. I'd like to know under what circumstances there exists an $H$ and a rep $\sigma'$ of $H$ such that $\rho =\mathrm{Ind}_H^G\sigma'$. – Ariel Weiss Jul 27 '17 at 15:42
• At least for finite $G$ that is a much simpler question! Of course $(H,\sigma')=(G,\rho)$ is always possible; I assume you want $H$ to be a proper subgroup, i.e. you would like to know if $\rho$ can be imprimitive. $(H,\sigma')$ can only exist if $|G:H|=\sigma'(1)=2$ because $Ind_H^G(\sigma')$ has degree $|G:H|\sigma'(1)$ and $\sigma'$ cannot be linear. So first you have to determine if an $H$ with index 2 exists. In fact if you have an $H$ above $N$ with $|G:H|=2$, then every $\sigma' | Res_H^G(\rho)$ of degree 2 does the job... – Johannes Hahn Jul 27 '17 at 17:14
• ... In fact there are only finitely many $(H,\sigma')$ at all, because we now all finite subgroups of $U_2(\mathbb{C})$. $H$ has to be a subdirect product of two copies of such a group and $\sigma'$ is one of the two components in that subdirect product. Furthermore you want a $H$ with a normal subgroup $N$ such that $\sigma=Res_N^H(\sigma')$ is still irreducible. Again, this is a finite problem. – Johannes Hahn Jul 27 '17 at 17:14
• As you say, the situation is indeed simple in the finite case! In the infinite case, it comes down to whether $G/N$ has an index two subgroup. Is there any information that is intrinsic to the representation that I can make use of? I know nothing about $N$ a priori. – Ariel Weiss Jul 27 '17 at 17:18