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I have a very basic question on Hilbert manifolds.

Consider the Hilbert space $$ \mathcal{H}:= L^2(S^1) $$ with $S^1$ the unit circle. On $\mathcal{H}$ let us introduce the equivalence relation $$ f\sim g : \Leftrightarrow f(\cdot ) = g(\cdot + \alpha)\quad \mbox{for some }\alpha \in S^1. $$ Now define the factor space $$ \overline{\mathcal{H}}:= \mathcal{H}/\sim. $$ What is the structure of $\overline{\mathcal{H}}$? Is it a Hilbert manifold? If so, how to construct the smooth structure? I am particularly interested in computing a (Riemannian) distance between two elements of $\overline{\mathcal{H}}$.

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  • $\begingroup$ Sorry, I has an answer that was wrong, using Fourier transform. If it can be corrected, I will repost. $\endgroup$
    – BS.
    Jun 21, 2012 at 13:40
  • $\begingroup$ I would be equally interested in the answer for $L^2(G)/G$ if $G$ is a (compact) topological group. $\endgroup$
    – Spice the Bird
    Jun 21, 2012 at 23:58
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    $\begingroup$ See Andrew Stacy's answer to the question mathoverflow.net/questions/10200/… $\endgroup$
    – Spice the Bird
    Jun 22, 2012 at 0:10
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    $\begingroup$ Not a Hilbert manifold. As the answer(s) have pointed out, you get problems at fixed points of the circle action (indeed, at any point where the circle action has a non-trivial stabiliser). There's another problem which is that the circle does not act continuously on the Hilbert space - how much of a problem this is will depend on how you want to fix the first problem (for example, you could go for a stratified space). $\endgroup$
    – Andrew Stacey
    Jun 25, 2012 at 9:28

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It probably is NOT a smooth manifold. I think finding a chart around the point corresponding to constants, namely, the fixed points of the action of the group of rotation, is problematic.More precisely, at a fixed point, there is not a well-defined tangent space.

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