Irreducible representations and Hilbert spaces I am reading Howard Georgi's book "Lie Algebras in Particle Physics" where he writes the following (chapter 1.14:eigenstates):

"... if some irreducible representation appears only once
in the Hilbert space, then the states in that representation must be eigenstates of $H$ (and any other invariant operator)."

The irreducible representation here, as far as I can tell, is meant to be part of a representation $D(g)$ on the full Hilbert space and we assume $H$ to commute with $D(g): [H, D(g)] = 0.$
My question is: what is meant by "appearing only once" in the Hilbert space?
Does it mean that, when I write the full representation D(g) as a direct sum of irreps, it appears only once in this direct sum?
To motivate why I think this is the case:
in this work explaining Schur's Lemma it is stated that, if the Hamiltonian commutes with $D(g) = \begin{pmatrix}\pi(g) & 0 \\ 0 & \pi(g) \end{pmatrix}$ where $\pi(g)$ is an irrep, then Schur's lemma does not apply but we can say that $H = \begin{pmatrix}A \mathbb{I} & B \mathbb{I} \\C\mathbb{I} & D\mathbb{I} \end{pmatrix}$.
So my questions are: 1)  is my assumption correct? and 2) can you point me to an example for the two different cases (an irrep appearing once and more than once) that may potentially clarify my confusion?
 A: If we can decompose
$$
{\mathcal H}=\bigoplus_{{\rm irreps}\, J} {\mathcal H}_J
$$
into $\hat H$-invariant irreps  of $G$ then Schur's lemma tells us that in each ${\mathcal H}_J$  the hamiltonian $\hat H$ will act as a multiple of  the identity operator. In other words every state in  ${\mathcal H}_J$ will be an eigenstate of $\hat H$ with a common  energy $E_J$.
If an irrep $J$ occurs only once in the decomposition of ${\mathcal H}$ then it is automatically an $H$ invariant subspace and we can find the eigenstates directly  by    applying  projection  to vectors in  the total Hilbert space ${\mathcal H}$. If the irrep occurs $n_J$ times in the decomposition, then we can project onto the reducible subspace
$$
 \underbrace{{\mathcal H}_J\oplus {\mathcal H}_J\oplus\cdots {\mathcal H}_J}_{n_J\, {\rm copies}}={\mathcal M}\otimes {\mathcal H}_J.
 $$
Here ${\mathcal M}$ is an $n_J$ dimensional multiplicity space.
The hamiltonian $\hat H$ will act in ${\mathcal M}$  as an $n_J$-by-$n_J$ matrix.  In other words, if the vectors
$$
 |n,i\rangle \equiv |n\rangle \otimes |i \rangle \in {\mathcal M}\otimes {\mathcal H}_J
 $$
form  a basis for ${\mathcal M}\otimes {\mathcal H}_J$, with $n$ labelling which copy of ${\mathcal H}_J$ the vector $|n,i\rangle $ lies  in, then the hamiltonian and the symmetry group act as
$$
 \hat H |n,i\rangle = |m,i\rangle H^J_{mn},\nonumber\\
 D(g)|n,i\rangle = |n,j\rangle D^J_{ji}(g),
 $$
where $D^J_{ji}(g)$ is the representation matrix in representation $J$.
Diagonalizing $H^J_{nm}$ provides us with $n_j$ $\hat H$-invariant copies of ${\mathcal H}_J$ and  gives us the energy eigenstates.
