I'm currently following a course in representation theory for physicists, and I'm rather confused about irreps, and how they relate to states in Hilbert spaces.
First what I think I know:
If a representation $D: G \rightarrow L(V)$ of a group $G$ is reducible, then that means there exists a proper subspace of $V$, such that that for all $g$ in $G$ the action of $D(g)$ on any vector in the subspace is still in that subspace: that subspace is invariant under transformations induced by $D$.
Irreducible means not reducible: my interpretation is that an irrep is a representation restricted to its invariant subspace. In other words, an irrep $R$ only works the subspace that it leaves invariant. Is this a correct view?
Now, my confusion is the following:
Say we have a system invariant under the symmetries of a group $G$. If this group is finite then any rep $D$ of $G$ can be fully decomposed into irreps $R_i$. We could write any $D(g)$ as the following block diagonal matrix:
$D(g) = \left( \begin{array}{cccc} R_1(g) & & & \\ & R_2(g) & & \\ & & \ddots & \\ & & & R_n(g) \end{array} \right)$
I suppose the basis of this matrix is formed by vectors in the respective subspaces that are left invariant by $R_i(g), \forall g \in G$, but here is where I'm not clear on the meaning of it all. How does such a matrix transform states in the Hilbert space, when Hilbert space is infinite dimensional, and this rep $D$ isn't?
I've found a book that gives an example of parity symmetry, using $Z_2 = \{ e,p \}$.
The Hilbert space of any parity invariant system can be decomposed into states that behave like irreducible representations.
So we can choose a basis of Hilbert space consisting of such states, which I suppose would be the basis of the matrix $D(g)$ above? Then the Hilbert space is the union of all these invariant subspaces? In the case of parity there exist two irreps: the trivial one (symmetric) and the one that maps $p$ to $-1$ (anti-symmetric). I suppose this is also a choice of basis, but in this basis $D(g)$ is $2$-dimensional, so I don't understand how this could possibly work on the entire Hilbert space.
I apologize for the barrage of questions, but I honestly can't see the forest for the trees anymore.