Schur's Lemma in Zee's Group Theory book for reducible representations

Question

Main question

Schur's lemma says: $$D(g) A = A D(g) \Rightarrow A = \lambda I\tag{1}$$ if $D$ is irreducible. How can I use this to show that if $D$ is reducible and if $SDS^{-1}$ is a direct sum of irreducible representations, then: $$SAS^{-1} = \lambda_1 I_{d_1} \oplus ... \oplus \lambda_n I_{d_n}?\tag{2}$$

What I understand

I want to understand a consequence of Schur's lemma as discussed in Anthony Zee's Group Theory book. In this book, the general theory of representations is avoided (rings, etc.), so answers that avoid this would be helpful.

On page 102, he discusses Schur's lemma. I'll provide the statement of the theorem (paraphrased) to show the sorts of technical terms that are avoided: If D(g) where $g \in G$ is a set of matrices representing group G, and furthermore D is an irreducible representation, then $D(g) A = A D(g)$ for all $g \in G$ implies $A = \lambda I$ for some number $\lambda$.

So far so good.

What I do not understand

At the end of this discussion, he says that if $D$ is reducible, so $D$ is block diagonal, and then in that basis $H$ is also block diagonal. Why?

I know that in some basis, $SDS^{-1}$ is block diagonal because it is a sum of reducible representations. However, we are loking at $WDW^\dagger$, where $W$ is unrelated to $S$ because $W$ diagonalizes $H$.

More details

The problem is that if I follow through the proof of the theorem where $D$ is a direct sum of irreducible representations:

1. We can take $A$ to be Hermitian, call it $H$.
2. We can diagonalize $H$ to get $H' = WHW^\dagger$.
3. Using the same basis, we get $D' = WDW^\dagger$.

Now we have $D'(g)A' = A'D'(g)$. And the rest of the argument shows that if $D'$ is block diagonal, then in that basis $H'$ is also block diagonal ($H'$ is also diagonal because of step 2). But why is $D'$ block diagonal?

Related questions:

1. A previous step of the same proof
2. Seems relevant but references Mashke's theorem and modules, which are not discussed in Zee's book

Answer 1

1. Here is a counterexample to OP's eq. (2): Let the decomposable representation $D=\begin{pmatrix} D_1 & 0 \cr 0 & D_1 \end{pmatrix}$ be 2 copies of the same representation $D_1$, and let $A=\begin{pmatrix} 0 & \lambda {\bf 1} \cr 0 & 0 \end{pmatrix}$. Then $D$ and $A$ commute, but $A$ is not on block-diagonal form.

2. What extra assumption should one make to secure OP's eq. (2)? Well, here's one approach. Given a decomposable representation $D:G\to GL(V,\mathbb{F})$ with vector space $V=V_1\oplus\ldots \oplus V_n$ such that $D$ and $A$ commute, then we can use Schur's lemma to conclude that $A|_{V_i}=\lambda_i {\bf 1}_{V_i}.$

   • Let us additionally assume that $A$ is diagonalizable $A=\oplus_{a=1}^m\mu_a {\bf 1}_{E_a}$ with eigenvalues $\mu_a$ and eigenspaces $E_a:={\rm Ker}(A-\mu_a{\bf 1}_V)$.

   The eigenspaces $E_a\subseteq V$ are $G$-invariant subspace, i.e. subrepresentations $D|_{E_a}$.

   • Let us additionally assume that all reducible representations are decomposable.

   Then we may assume (possibly after a similarity transformation of $A$) that $E_a=\oplus_{j=1}^{r_a}V_{i_j}$. By restricting to an irreducible $V_i\subseteq E_a$, we conclude that $\mu_a=\lambda_i$.

   Altogether, it follows that $A=\oplus_{i=1}^n\lambda_i {\bf 1}_{V_i}$. $\Box$