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Let $H$ be an infinite dimensional separable complex Hilbert space with Lie group action (I am mostly interested in the case $S^1$). Let $\text{Gl}_{G}(H)$ be the space of invertible, bounded and equivariant linear maps (from $H$ to $H$).

Now, in the non-equivariant case, Kuiper's theorem states that $\text{Gl}(H)$ is (weakly) contractible. Is this this also true for $\text{Gl}_{G}(H)$?

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If $G$ is compact so that $H$ splits up as a topological sum of isotypical components, then each isotypical component has to have infinite multiplicity. Under that assumption it seems to me that one can apply Kuiper's result to each of the components. – Torsten Ekedahl Oct 17 '11 at 15:59
up vote 7 down vote accepted

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.

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Thank you very much - is it at least true that $\text{GL}_{S^1}(H)$ is path-connected ($H$ a complex Hilbert space as above)? – J. Fabian Meier Oct 18 '11 at 9:36
@ J. Fabian Meier: assuming $G$ acts unitarily on $H$, the group $GL_G(H)$ is indeed path-connected. Indeed it is the invertible group of the commutant of $G$ in $B(H)$, which is a von Neumann algebra, and any von Neumann algebra $M$ has path-connected invertible group. [Recall the argument: by the polar decomposition, the invertible group retracts on the unitary group; and the unitary group is path-connected because, using a measurable branch of the logarithm on $S^1$, we may express any unitary as $e^{iH}$, for some self-adjoint $H\in M$.] – Alain Valette Oct 18 '11 at 12:09

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