Let $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ be a short exact sequence of groups. What can be said about group representations of $B$ if we assume a complete classification of the representations of $A$ and $C$? Precisely, assume we have determined the categories $Rep_R(A)$ and $Rep_R(C)$ of $R$-linear representations for a commutative ring $R$, up to equivalence.

If the sequence splits, then by identifying a representation of $B$ with its "classifying map" $B\rightarrow Aut(M)$ for an $R$-module $M$, we classify a representation of $B$ with its component representations of $A$ and $C$, so it seems like we have an equivalence between $Rep_R(B)$ and the pullback of the diagram $$ Rep_R(A) \rightarrow Mod_R \leftarrow Rep_R(C) $$ where both maps are the canonical forgetful functors taking a map $A \rightarrow Aut(M)$, say, to $M$.

What can be said if the sequence does not split?

I am most interested in the case when $B$ is a semidirect product of $A$ and $C$. I was inspired to ask this by this question, which asks about the specific case of the sequence $$0\rightarrow SO(n) \rightarrow O(n) \rightarrow \mathbb{Z}_2 \rightarrow 0$$

  • $\begingroup$ You have the inclusion $SO(n)\rightarrow O(n)$, which is the kernel of the determinant $O(n)\rightarrow \mathbb{Z}_2$. The first map is injective, the last one is surjective, so it is an exact sequence. What is the map $O(n)\rightarrow SO(n)$ you had in mind? $\endgroup$ – Espen Nielsen May 7 '14 at 19:07
  • $\begingroup$ Oh silly me, I was thinking of modding out by the center, which is projectivizing. My mistake! $\endgroup$ – Viktor Vaughn May 7 '14 at 19:08

Even in the simplest case of a direct product, so that $B=A\times C$, this is complicated. The pullback of $$Rep_R(A) \rightarrow Mod_R \leftarrow Rep_R(C)$$ will give $R$-modules with actions of $A$ and $C$ that don't necessarily commute, so representations of the free product $A*C$ rather than the direct product $A\times C$.

If $R$ is a field of characteristic $p>2$ and $A$ and $C$ are cyclic groups of order $p$ then there's a fairly simple classification of modules for $RA$ and $RC$, but classifying $R[A\times C]$-modules is a "wild" problem that there's no hope of solving.

  • $\begingroup$ Thank you very much. I assume it is safe to assume that the characteristic zero, even algebraically closed case is not any simpler? $\endgroup$ – Espen Nielsen May 7 '14 at 22:30
  • $\begingroup$ For direct products of finite groups over an algebraically closed field of characteristic zero it is simpler if you work one irreducible representation at a time. The irreducible representations of $A\times C$ are just tensor products $U\otimes V$, where $U$ and $V$ are irreducible representations of $A$ and $C$ respectively. $\endgroup$ – Jeremy Rickard May 7 '14 at 22:51
  • $\begingroup$ Right. But, say, semidirect products are still very difficult to handle? $\endgroup$ – Espen Nielsen May 7 '14 at 23:06

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