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 $\mathrm{End}_H(V) = k\times k$?


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?

  • $\begingroup$ 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. $\endgroup$ – Johannes Hahn Jul 27 '17 at 16:45
  • $\begingroup$ @JohannesHahn I really care about Galois representations, so $G = \mathrm{Gal}(\overline K/K)$ is profinite, and representations are continuous. $\endgroup$ – Ariel Weiss Jul 27 '17 at 16:51
  • $\begingroup$ 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 $U_4(\mathbb{C})$. $\endgroup$ – Johannes Hahn Jul 27 '17 at 17:06
  • $\begingroup$ 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. $\endgroup$ – Ariel Weiss Jul 27 '17 at 17:07

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

  • $\begingroup$ 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? $\endgroup$ – Ariel Weiss Jul 27 '17 at 15:39
  • $\begingroup$ 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 $\rho =\mathrm{Ind}_H^G\sigma'$. $\endgroup$ – Ariel Weiss Jul 27 '17 at 15:42
  • $\begingroup$ 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... $\endgroup$ – Johannes Hahn Jul 27 '17 at 17:14
  • $\begingroup$ ... 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. $\endgroup$ – Johannes Hahn Jul 27 '17 at 17:14
  • $\begingroup$ 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. $\endgroup$ – Ariel Weiss Jul 27 '17 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.