# 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. Aug 2, 2016 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. Aug 2, 2016 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? Aug 2, 2016 at 21:14

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"? Aug 25, 2018 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? Aug 25, 2018 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? Aug 26, 2018 at 0:48

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.

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.

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

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

This proof does not require the dimensionality of the representation space to be finite. So Schur's lemma is valid in both finite and infinite dimensions, countable or uncountable.

• The inverse of a bijection is a bijection. Apr 8, 2019 at 14:17