The space underlying the projective unitary group of a separable, infinite-dimensional Hilbert space has a number of topologies, so for the purposes of this question, pick you favourite and answer for that one.


I've read that $PU(H)$ is a Fréchet manifold, but that was without saying which topology. There are two ways I can see to think about this. First is that if we know what sort of manifold $U(H)$ is, then we know what $PU(H)$ is, as the former looks locally like a chart of $PU(H)$ times $U(1)$. So we can consider the closed (I think!) subspace $U(H) \subset End(H)$.


Alternately we can consider $PU(H)$ as sitting inside the Hilbert-Schmidt operators on $H$ as the projective unitaries act freely on the latter. (EDIT: this is not right, as pointed out by Andrew. I was thinking of the inclusion $PU(H) \hookrightarrow U(HS(H))$, which doesn't really tell us much in hindsight.)


I'm not familiar enough with the analysis to turn the above observations into results, and I may be interested in other topologies.


So the question is, is there the structure of a Banach or even Hilbert manifold on $PU(H)$?



What sort of manifold is PU(H)?
*Edit: Theo's comments are spot-on - that'll teach me to try to post a quick answer without thinking too much*


There are two sensible topologies on $U(H)$, each leads to a topology on $P U(H)$ by quotienting:


1. The norm topology. In this topology, $U(H)$ is viewed as a sub-Lie group of the Banach Lie group $Gl(H)$. Here, $Gl(H)$ is open in $B(H)$ and the exponential map is a diffeomorphism onto a neighbourhood of the identity and this takes skew-Hermitian operators onto $U(H)$, thus defining charts. As the quotient $U(H) \to P U(H)$ admits slices, this makes $P U(H)$ a Banach manifold.
2. The weak topology. In this topology, we start with $Gl(H)$ as a subspace of $B(H)$ where this has the weak topology in which $A\_\lambda \to A$ if $A\_\lambda v \to A v$ for each $v \in H$. However, this isn't good enough for $Gl(H)$ as taking the inverse isn't continuous. So we put on it the topology so that both the inclusion $Gl(H) \to B(H)$ and the inclusion after the inversion map $A \to A^{-1}$ are continuous. However, then $Gl(H)$ is not an open subspace of $B(H)$ so it isn't obviously a manifold. Indeed, I don't think that with this topology then it is a manifold at all (I don't think it is an ANR). As Theo says, the induced topology on $U(H)$ doesn't need the inversion fix, but even so it still is not a manifold.


The map to Hilbert-Schmidt operators is not useful here because the action is not free, it is only faithful. It acts by conjugation whereupon the stabiliser group of a particular element are all the unitary elements that commute with it and this can be quite large.


