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 $\text{Gl}\_{G}(H)$?



Theorem of Kuiper for Hilbert spaces with group action
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.


