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Alain Valette
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Assume that $G$ acts on $H$ through a unitary irreducible representation. Then by Schur's lemma, $GL_G(H)$ is $\mathbb{C}^\times$, which is of course not contractible.

For $G=S^1$: consider the left regular representation on $L^2(S^1)$. Then $GL_G(H)$ is the multiplicative group of bounded sequences with values in $\mathbb{C}^\times$, which is not contractible.

Assume that $G$ acts on $H$ through a unitary irreducible representation. Then by Schur's lemma, $GL_G(H)$ is $\mathbb{C}^\times$, which is of course not contractible.

Assume that $G$ acts on $H$ through a unitary irreducible representation. Then by Schur's lemma, $GL_G(H)$ is $\mathbb{C}^\times$, which is of course not contractible.

For $G=S^1$: consider the left regular representation on $L^2(S^1)$. Then $GL_G(H)$ is the multiplicative group of bounded sequences with values in $\mathbb{C}^\times$, which is not contractible.

Source Link
Alain Valette
  • 11.1k
  • 44
  • 62

Assume that $G$ acts on $H$ through a unitary irreducible representation. Then by Schur's lemma, $GL_G(H)$ is $\mathbb{C}^\times$, which is of course not contractible.