Timeline for Algorithm to check is representation irreducible ? Algorithm to decompose the reducible one ?
Current License: CC BY-SA 3.0
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| Nov 4, 2012 at 16:08 | history | edited | Andy B | CC BY-SA 3.0 |
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| Nov 4, 2012 at 16:01 | comment | added | Andy B | @daniel litt. good point. in haste i'd thought that by taking a random isomorphic copy of the representation you could get rid of this problem. i guess it makes the problem worse(!), since the rep is now not even visibly reducible. as derek holt points out, you could factorize the characteristic polynomial to see that this is reducible. | |
| Nov 4, 2012 at 10:44 | comment | added | Derek Holt | You can improve the probability by choosing a "random" element $x$ of the algebra generated by the matrices of the group elements, and factorizing its characteristic polynomial $f$. (This is what the Holt-Rees variation of the Meataxe does.) Unfortunately, in characteristic 0, the probability that $f$ is irreducible is high (like 1). But if $f$ has a factor $g$, then choosing your random vector in the nullspace of $g(x)$ is likely to be effective in finding submodules or proving irreducibility. | |
| Nov 4, 2012 at 9:11 | comment | added | Daniel Litt | This ``randomized algorithm" simply won't work. Even for a representation which is not irreducible, a random unit vector (say, chosen uniformly) will with probability one not lie in any proper subrepresentation. Think about the case of e.g $\mathbb{Z}/2\mathbb{Z}$ acting on $\mathbb{C}^2$ via the matrix $(1~0;0~-1)$. Then unless the chosen vector is a scalar multiple of $(1~0)$ or $(0~1)$, the span of its orbit will be all of $\mathbb{C}^2$. | |
| Nov 4, 2012 at 1:44 | history | answered | Andy B | CC BY-SA 3.0 |