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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 $\rho$.

(Asked first on MSE: https://math.stackexchange.com/questions/3096704/failure-of-schurs-lemma-for-topological-group-representations)

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.

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    $\begingroup$ There are several different statements that get referred to as Schur's lemma, what is the version you are referring to? $\endgroup$ – Nate Feb 11 at 22:13
  • $\begingroup$ Keep in mind that Schur's lemma originated in the setting of finite dimensjonal representations, so what Nate points out is highly relevant here. $\endgroup$ – Jim Humphreys Feb 11 at 22:40
  • $\begingroup$ 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. $\endgroup$ – YCor Feb 11 at 23:04
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    $\begingroup$ 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. $\endgroup$ – YCor Feb 11 at 23:06
  • $\begingroup$ @Nate I expanded the question with a statement of Schur's lemma. (Also, I added that $\rho$ is irreducible.) $\endgroup$ – safety stegosaurus Feb 11 at 23:13

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