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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
    Commented Oct 18, 2016 at 4:06

2 Answers 2

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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
    Commented Mar 18, 2013 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
    Commented Mar 18, 2013 at 22:15
  • $\begingroup$ @user20327 There are nonsemisimple representations too (in particular, over vector spaces in positive characteristic). $\endgroup$
    – anon
    Commented Mar 18, 2013 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
    Commented Mar 18, 2013 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
    Commented Mar 18, 2013 at 23:03
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A reducible group representation

$\Phi: G \rightarrow GL(n,V)$

is a representation that contains invariant subspaces different from the zero vector space and the complete vector space $V$.

Now, a representation is irreducible if it the only invariant subspaces that it contains are $0$ and $V$

So, a completely irreducible representation is a a representation that can be expressed as the direct sum of irreducible representations. If you'd like to see a more visual mathematical way of seeing this:

$\Phi$ is completely irreducible $\leftrightarrow \Phi = \oplus \Phi_{\lambda}$ with $\Phi_{\lambda}$ irreducible

I hope this was clear.

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