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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asked Dec 3, 2013 at 18:44
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$\begingroup$ 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. $\endgroup$– Peter MuellerDec 3, 2013 at 19:00
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$\begingroup$ @PeterMueller what does this actually mean in practical terms? $\endgroup$– Igor RivinDec 3, 2013 at 19:14
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2$\begingroup$ 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. $\endgroup$– Peter MuellerDec 3, 2013 at 19:29
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4$\begingroup$ 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 $R=n^2$. $\endgroup$– YCorDec 3, 2013 at 19:35
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1$\begingroup$ 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. $\endgroup$– Peter MuellerDec 3, 2013 at 19:38
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1 Answer
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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.
answered Dec 22, 2013 at 20:45