Representation theory is painfully new to me, but I am trying to understand simple examples of Schur's (second) lemma. My understanding of Schur's lemma is that, for a given complex and irreducible matrix representation $D$, if a matrix $Q$ commutes with $D(g)$ for every group element $g$, then the matrix $Q$ is proportional to the identity.
In what follows, I think my confusion may be because I am over-simplifying Schur's lemma. Consider the group $\mathbb{Z}_2$, with the irreducible representation $D: \mathbb{Z}_2 \to GL(2, \mathbb{C})$:
$$D(e) = \begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}, \,\, D(p) = \begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$$
It feels like I can construct a counterexample to Schur's lemma: the matrix $D(p)$ commutes with $D(e)$ and $D(p)$, but it is not proportional to the identity.
Why isn't the above a counterexample? What am I misunderstanding? Heightening my confusion, this website (in example 1) even claims that that the only matrix commuting with $D(e)$ and $D(p)$ is proportional to the identity, but that seems false to me given that $D(p)$ commutes with both $D(e)$ and $D(p)$.
Best Answer
Your representation $D$ is not irreducible! Indeed, you can see this by just carrying out the proof of Schur's lemma for it: any eigenspace of $D(p)$ will be a subrepresentation of $D$. Explicitly, $D(p)$ has eigenvalues $1$ and $-1$ with eigenvectors $(1,1)$ and $(1,-1)$, respectively, so the spans of each of these eigenvectors are subrepresentations.