While studying representations of finite groups I got confused by the the statement that any irreducible representation is at the same time a completely reducible representation.
This doesn't seem to make any sense to me, since an irrep has per definition no (non-trivial) invariant subspace and therefore the carrier space can't be a direct sum of the invariant subspaces.
Furthermore I am puzzled by the statement that any representation of a finite group is equivalent to a completely reducable representation.
Let's consider for instance the symmetric group $S_3$ and it's 1-dim. representation $D_1: S_3 \rightarrow \mathbb{R}$ , $g_i \mapsto 1$. According to the just mentioned statement $D_1$ has to be equivalent to a completely reducable representation and therefore the direct sum of it's invariant subspaces (which is the set $\{1\}$) should make up the carrier space. Which is obviously not the case.
Any ideas?
I might add in view of the latest discussion about the question if not any arbitrary rep is completely reducible and hence the concept being useless:
A representation $(D,V)$ is completely reducible if:
$V$ is the direct sum of invariant subspace (true for any representation, if one considers $V$ and $0$ as invariant subspaces) and
the projection of $D$ on the invariant subspaces $D|_{V_i}$ is irreducible (which is not true in general if one just considers $V$ and $0$ as the only invariant subspaces).
Therefore the statement, that any representation is completely reducible is not true due to the second criterion.
Thanks
Philipp
Best Answer
Subspaces do not have to be proper, meaning the "carrier" space, as you call it, is a subspace of itself which may or may not be invariant. Also when you sum together spaces, you are allowed to say that you're just summing one space, and the result of that sum is that one space.
So an irreducible representation (like your $1$-dimensional $S_3$ rep) is completely reducible because the carrier space itself is irreducible and so the carrier space is indeed the sum of it's irreducible subspaces because it is the sum of a single subspace: itself.