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This should be a simple question, but i'm having a rather hard time though finding a explicit definition of a completely reducible group representation. Is it right to say that a presentation is completely reducible if the vector space on which the group $G$ is represented can be written as a direct sum of $G$-invariant subspaces?

Because if the answer is "yes", can't I say that every presentation is completely reducible, because it is the direct sum of the full vectorspace with the nullspace which are both $G$-invariant?

I wonder about this, because Mashke's theorem is formulated in my book as follows:

Every finite-dimensional representation of a finite group $G$ is completely reducible as the direct sum of irreducible representations.

But in the proof it's mentioned that $V$ itself can be irreducible (but they make no problem out of it), but by the theorem $V$ would then be irreducible and completely reducible at the same time. And if $V$ is completely irreducible, isn't any representation completely reducible then?

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  • $\begingroup$ As a direct sum of simpme G invariant subspaces. 0 is not simple $\endgroup$ – Amr Oct 18 '16 at 4:06
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Completely reducible is usually called semisimple, which is to say that it can be written as a direct sum of simples. What is a simple module $V$? One that has only the trivial module $\{0\}$ and itself $V$ as submodules. (It has no proper, non-trivial submodules).

This is analogous to how we exclude the integer 1 from the list of prime numbers, so that we don't write decompositions like $30 = 5 \times 3 \times 2 \times 1 \times 1 \times \cdots \times 1$.

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  • $\begingroup$ Thanks, but doesn't that imply that every representation is semisimple? Because I can always write it as the direct sum of V itself and {0} which are both simple modules? $\endgroup$ – yarnamc Mar 18 '13 at 22:09
  • $\begingroup$ I wonder about this, because Mashke's theorem is formulated in my book as follows: every finite-dimensional representation of a finite group G is completely reducible as the direct sum of irreducible representations. But in the proof it's mentioned that V itself can be irreducible (but they make no problem out of it), but by the theorem V would then be irreducible and completely reducible at the same time. And if V is completely irreducible, isn't any representation completely reducible then? $\endgroup$ – yarnamc Mar 18 '13 at 22:15
  • $\begingroup$ @user20327 There are nonsemisimple representations too (in particular, over vector spaces in positive characteristic). $\endgroup$ – anon Mar 18 '13 at 22:37
  • $\begingroup$ Yeah, i'm not really wondering about that. If in fact every representation would be semisimple, then the word "semisimple" would be meaningless. The problem is that every vectorspace can be written as the direct sum of the zero space with itself. Both of these are G-invariant. So the representation seems to be always fully reducible. $\endgroup$ – yarnamc Mar 18 '13 at 22:52
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    $\begingroup$ @user20327 Writing a representation as a direct sum of 0 and itself does not count as fully reducing it if the representation itself is not already irreducible. "Irreducible" and "fully reducible" do not exhaust all possibilities. For instance, the group algebra ${\bf F}_p[C_p]$ is not irreducible (the kernel of the trace map is a $C_p$-invariant space, which has no $C_p$-invariant complement) nor is it semisimple. $\endgroup$ – anon Mar 18 '13 at 23:03

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