# Schur's lemma for infinite dimensional group representations

Why does Schur's lemma (that there are no intertwining operators between two non-isomorphic, finite-dimensional, and irreducible representations of a group) require that the reps be finite-dimensional? Is this true in general for any group and any pair of irreps, or is there a counter-example when the representations are infinite-dimensional?

EDIT: Thank you for the responses so far! The reason I was asking was that I've been reading Terrance Tao's blogpost on the proof of the Peter-Weyl theorem. He uses Schur's lemma to prove it, and uses functional analysis to prove Schur's lemma. But from what I can tell, his proof only seems to use the version stated above (so not the fact that a $G$-stable map $T: V\to V$ must be a scalar multiple of the identity). Am I mistaken in thinking that, or did T. Tao just prove more than was necessary?

• There are many versions of Schur's Lemma, some of which only apply in the finite-dimensional case. The version you've stated is general. – Dustan Levenstein Aug 2 '16 at 20:43
• This version of Schur's Lemma holds generally. The version which states: if $T:V\to V$ is an intertwiner, then $T = \lambda I$, requires the underlying field to be algebraically closed, and $V$ to be finite dimensional. The proof relies on the fact that every endomorphism of a finite dimensional vector space over an algebraically closed field has an eigenvalue. This is false in general. – Mathmo123 Aug 2 '16 at 20:54
• Thank you! I understood that the version of the lemma as I've stated it can be general. Could you answer the new question I added on the post? – Steven Aug 2 '16 at 21:14

## 3 Answers

For the record, there is a version of Schur's lemma due to Dixmier initially and then to Quillen, which applies to infinite dimensional modules. You can find Quillen's paper here.

Let $G$ be a group and $U,V$ irreducible representations of $G$, consider $f:U\rightarrow V$ a non trivial equivariant $G$-morphism, $ker f$ is a submodule of $U$, so $ker f=0$ since $U$ is irreducible, $Im f$ is a submodule of $V$ so $Im f=V$ since $V$ is irreducible. We deduce that $f$ is bijective.It is easy to check that $f^{-1}$ is a $G$-morphism. Henceforth the Schur' lemma is true if the dimension $U,V$ is infinite.

The differences between finite and infinite dimensions appear when one compute the group of automorphisms of an irreducible representation.

https://mathoverflow.net/questions/23502/when-does-schurs-lemma-break

• Dear Tsemo, (1) How to show that $\,f^{-1}\,$ is bijective? (2) Is there a special reason why you said "bijection" and not "isomorphism"? – Michael_1812 Aug 25 '18 at 18:44
• Also, could you please comment more extensively on your statement that the differences between finite and infinite dimensions appear when one computes the group of automorphisms of an irreducible representation? – Michael_1812 Aug 25 '18 at 18:54
• Finally, when you say that the morphism is nontrivial, do you mean that it is not nonzero ($\,f\,\neq\,0\,$) ? Or that it is not isomorphism? – Michael_1812 Aug 26 '18 at 0:48

Using the notation introduced above by Tsemo Aristide, $\,f:\;U\longrightarrow V\,$, let me try to sum up:

(1) $\,\mbox{Ker}\,f\,$ is a subspace of $U$, and is invariant under the action of the representation of G.

This agrees with our assumption of irreducibility in two cases: either $\,\mbox{Ker}\,f = U\,$ and the rep. is trivial, or $\,\mbox{Ker}\,f = 0\,$.

(2) $\,\mbox{Im}\,f\,$ is a subspace of $V$, and is invariant.

This agrees with our assumption of irreducibility either when $\,\mbox{Im}\,f\,=\,0\,$ or when $\,\mbox{Im}\,f\,=\,V\,$.

(3) Setting aside the trivial case, we see that $\,\mbox{Ker}\,f = 0\,$ and $\,\mbox{Im}\,f\,=\,V\,$, wherefrom $\,f\,$ is bijective.

Now, if $\,f^{-1}\,$ is bijective too, then the representations $U$ and $V$ are isomorphic.

So, there is no nontrivial (= nonzero) intertwiner between nonisomorphic irreducible reps.

Is my understanding correct that this is the infinite-dim. version of Schur's lemma?

Can someone please tell me why $\,f^{-1}\,$ is bijective?

• The inverse of a bijection is a bijection. – punctured dusk Apr 8 at 14:17