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

  • $\begingroup$ It has another invariant subspace: The entire space. $\endgroup$ – Tobias Kildetoft Dec 30 '14 at 18:30
  • 2
    $\begingroup$ I know this sounds strange, but it is really just a matter of definition. If $V$ is the underlying space of an irreducible representation, then the only irreducible subspace of $V$ is $V$ itself, but $V$ is the direct sum of its subspaces $\{ V \}$, so it fulfils the definition of completely reducible. $\endgroup$ – Derek Holt Dec 30 '14 at 18:30
  • $\begingroup$ Thank you all for your answers. It seems that my notion of reducibility doesn't match up with yours. As far as I am concerned an unreducible representation can be defined as such a representation that has no non trivial invariant subspace. With saying "non trivial" I want to point out that $0$, $\emptyset$ and $V$ - the carrier space itself - is excluded in the definition of irreducibility. $\endgroup$ – Philipp Dec 31 '14 at 11:42
  • $\begingroup$ $0$ and $V$ are excluded from the consideration in the definition of irreducible. But $V$ (and $0$) is ok, when considering complete reducibility. $\endgroup$ – Jyrki Lahtonen Dec 31 '14 at 12:07
  • $\begingroup$ Pardon me if this sounds patronizing. You can think of it this way. An irreducible representation is an atom. A completely reducible element is something built from the atoms (in the most obvious way. Here "most obvious"="direct sum" - there are subtleties that I skip for now). By this definition the trivial rep corresponds to an empty set of atoms, and is thus completey reducible. An irreducible rep corresponds to a set of a single atom, an is thus completily reducible et cetera. The set of irreducible summands can have zero, $1$ or more parts. $\endgroup$ – Jyrki Lahtonen Dec 31 '14 at 12:25

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.

  • $\begingroup$ "which may or may not be invariant": the entire space of any representation would have a hard time not being invariant. $\endgroup$ – Marc van Leeuwen Jan 31 '15 at 9:37
  • $\begingroup$ The point is that you look at any subspace and you ask whether it's invariant or not. The entire space is in the list of subspaces to which you ask that question. $\endgroup$ – Jim Jan 31 '15 at 16:47
  • $\begingroup$ Yes, I got that. But your first sentence is really garbled. $\endgroup$ – Marc van Leeuwen Jan 31 '15 at 17:23

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