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This is related to this question. Suppose I have an $n$-dimensional representation of a finitely generated group, and I want to know whether it is absolutely irreducible. This can, of course, be done if you are willing to compute the eigenspaces of the generators, but that involves working over the splitting field of the characteristic polynomial (it should, presumably, be the same for all the generators if the representation is reducible), but that is not what I would describe as pleasant. Is there anything more efficient? (note that I can, of course, diagonalize the generators numerically, but I am not sure what this tells me).

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If you know for some reason that your representation is completely reducible, then the double centralizer theorem tells you that the representation is absolutely irreducible if and only if the scalar maps are the only endomorphisms which commute with the representation. –  Peter Mueller Dec 3 '13 at 19:00
@PeterMueller what does this actually mean in practical terms? –  Igor Rivin Dec 3 '13 at 19:14
Well, you have finitely many invertible matrices $a_i$ in $M_n(K)$. So the common centralizer $C$ of all the $a_i$ (which is the same as the centralizer of the group generated by the $a_i$) is the intersection of the subspaces $C_i$, where $C_i$ is the space of matrices commuting with $a_i$. So $C$ can be computed by simple linear algebra. Of course the scalar matrices are in $C$, and your representation is absolutely irreducible if and only if $C$ isn't bigger. –  Peter Mueller Dec 3 '13 at 19:29
The representation $\rho$ of $G$ is absolutely irreducible iff $\rho(G)$ linearly generates the algebra of matrices. So an algorithm is to compute the rank $r_k$ (in the sense of linear algebra) of the subspace spanned by $\rho(B_k)$, where $B_k$ is the $k$-ball, until it stabilizes i.e. $r_{k}=r_{k+1}=R$. Then the representation is absolutely irreducible iff $R=n^2$. –  YCor Dec 3 '13 at 19:35
What I use here, in the irreducible case: The $K$-algebra $S$ generated by the $a_i$ is absolutely irreducible if and only if the group is. By Jacobson's density theorem, $S$ is a simple algebra, so $dim S\cdot\dim C=n^2$ by the double centralizer theorem. By Schur's Lemma, $C$ is a divison field over $K$. So if $C$ is bigger than the scalars, then there is a field $K<F\le C$. But then $S$ is reducible over $F$, hence not absolutely irreducible. –  Peter Mueller Dec 3 '13 at 19:38

1 Answer 1

There are very complicated groups $G$ such that $G = \langle x,y : x^{2} = y^{3} = 1 \rangle,$ possibly with further relations. In a sense, the most complicated such group is ${\rm PSL}(2,\mathbb{Z}),$ which is the free product of a group of order $2$ and a group of order $3$.

For any given representation of such a group, it is easy to calculate the invariant subspaces of the generators, but calculating the invariant subspaces for the whole group seems hard to me.

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