# Failure of Schur's lemma for topological group representations

Is there an example of $$G$$, $$\rho$$ as below?

• $$G$$ is a locally compact group.

• $$\rho$$ is an irreducible continuous representation of $$G$$ on a complex Hilbert space $$V$$. This means that we have a continuous homomorphism from $$G$$ to the group of bounded linear operators on $$V$$ with bounded inverse, such that $$G \times V \rightarrow V$$ is continuous. And there are no nontrivial closed invariant subspaces.

• Schur's lemma fails for $$\rho$$.

Added later: For concreteness, I will record below one version of Schur's lemma that I have in mind. (But I'm not too picky about this.)

Schur's lemma: Every bounded operator commuting with $$\rho(G)$$ is a scalar.

• There are several different statements that get referred to as Schur's lemma, what is the version you are referring to? – Nate Feb 11 at 22:13
• Keep in mind that Schur's lemma originated in the setting of finite dimensjonal representations, so what Nate points out is highly relevant here. – Jim Humphreys Feb 11 at 22:40
• Probably the topology on $G$ is irrelevant, because if Schur's lemma (I guess, if the representation is topologically irreducible then its commutant is reduced to scalars- you probably mean complex Hilbert space) fails for some representation of a topological group $G$, it fails for the same representation with $G$ being endowed with the discrete topology. – YCor Feb 11 at 23:04
• If it's the correct interpretation of the question, then the question is equivalent to: find a non-scalar bounded operator of a (complex) Hilbert space whose centralizer acts irreducibly on the Hilbert space. – YCor Feb 11 at 23:06
• @Nate I expanded the question with a statement of Schur's lemma. (Also, I added that $\rho$ is irreducible.) – safety stegosaurus Feb 11 at 23:13