# Problem with the proof of a corollary of Schur's lemma

Hello,

I'm reading the book 'A course in Modern Mathematical Physics' by 'Szekeres' and encountered a problem in interpreting the proof of the following corollary of Schur's lemma.

The corollary and the proof in the book is as follows (I mark my problem area's in the proof with (<1>) and (<2>) :

Corollary 4.2. Let $T : G \rightarrow {GL} (V)$ be a irreducible representation of a finite (<1>) group G on a complex vector space $V$ and $A : V \rightarrow V$ an operator such that $\forall g \in G$ we have $T (g) \circ A = A \circ T (g)$ then $A = \alpha {id}_V$ for some complex scalar $\alpha$.

Proof: As $\alpha {id}_V$ commutes with $T (g)$ for every $g$ and every complex $\alpha$ we have $\forall g \in G$ that $T (g) \circ (A - \alpha {id}_V) = (A - \alpha {id}_V) \circ T (g)$ and from Shur's lemma it follows then that either $A - \alpha {id}_V = 0$ or $A - \alpha {id}_V$ is a isomorphism (and thus invertible). Now if $V$ is complex then $A$ must have a eigenvalue $\alpha$ (<2>) so that $A - \alpha {id}_V$ is not invertible and thus we must have $A = \alpha {id}_V$

The problems I have with this proof are as follows :

1. For (<2>) to be valid (that $A$ has a eigenvalue) must we not have that $V$ is finite dimensional. The corollary tells nothing about finite dimensionality of $V$.

2. For what do we need (<1>) ($G$ is finite). In the same book at the formulation of Shur's lemma $G$ does not need to be finite, so I think the fact that $G$ is finite should be used somewhere in the proof of the corollary but I fail to see where (unless from $G$ finite it follows that $V$ is finite which would solve my previous problem but I fail to see how this could be true). This is my biggest problem, the previous problem is in my opinion a typo in the text of the book, but here I have the impression that I'm missing something in the proof, in my opinion none of the steps in the proof do not need finiteness of G so can we omit the requirement that G must be finite in the corollary.

Marc Mertens

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It seems to me that you can get away with milder assumptions: $G$ can be any group so long as $V$ is a Banach space and $A$ is a bounded linear operator. Then in the proof you take $\alpha$ to be the Banach space analogue of an eigenvalue for $A$, i.e., an element of the spectrum of $A$.
Let $v \in V$ be non-zero. Then the span of the elements $g v$ ($g \in G$) is finite-dimensional, because $G$ is finite, and is a $G$-invariant subspace of $V$. Since $V$ is irred., it is just equal to this span, and so is itself finite-dimensional. This should deal with both your points 1 and 2.