A question in a Representation Theory midterm got me thinking, and made me realise I didn't really understand irreps. The question was on the subject of reps of $S_4$, and went:
An obvious representation of $S_4$ is the explicit geometric symmetry group of the regular tetrahedron in 3-space. You can choose the vertices of the tetrahedron at the positions $(1,1,1)$, $(-1,-1,1)$, $(-1,1,-1)$ and $(1,-1,-1)$. Is this rep irreducible?
I started reasoning as follows: a representation is a mapping from group elements to linear operators on some linear vector space. If it is an irrep, then there exists no proper subspace of that vector space which is invariant under the action of the representation. So I just need to figure out whether such a subspace exists. Then I realised I had no idea how to visualise that in terms of this question. What is the appropriate vector space? How do I know what the action of the representation on the vectors in this space is? What basis do I choose? This seems quite a gaping hole in my understanding, so any help would be greatly appreciated.
Continuing this line of thought, I know that any rep of a finite group can be decomposed into its irreps. In matrix form:
$D(g) = \left( \begin{array}{cccc} R_1(g) & & & \\ & R_2(g) & & \\ & & \ddots & \\ & & & R_n(g) \end{array} \right)$
Where I suppose the basis is that of the vectors that transform according to the irrep. I was under the impression that this decomposition essentially partitions the space into subspaces that transform according to the different irreps. However, there is always the trivial irrep which leaves any space invariant, so I'd say that the subspace that this irrep works on is the whole space, which leaves no room in the decomposition for any other irreps, as it were.
There is most likely a fatal flaw in my thinking somewhere.