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$M$ is a smooth manifold. It's known that if $M$ is compact, then the space of smooth Riemannian metrics has a Frechet manifold structure. For the space of $C^k$($k<\infty$) Riemannian metrics, does it have a Banach manifold structure?

If $M$ is not compact, does the space of $C^k$($k<\infty$) Riemannian metrics have a Banach manifold structure?

I need some references about those problems.

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    $\begingroup$ The question does not make sense unless you specify (a) what topology you put on the space of $C^k$-metrics (b) what you mean by a Banach manifold. If you give the space of metrics the uniform $C^k$ topology and by a Banach manifold you mean a separable metrizable space locally homeomorphic to a Banach space, then the space of metrics is a Banach manifold because it is an open subspace in the Banach space of all symmetric $C^k$ $2$-tensors equipped with the $C^k$ topology. $\endgroup$ Nov 27, 2015 at 12:53
  • $\begingroup$ Well,I'm asking if there exist such a topology to make it a Banach manifold.By a Banach manifold I mean a Hausdorff topological space that is locally homeomorphic to a Banach space, but may not be metrizable or connected. Your comment is right for the compact case. For the noncompact case, some other assumptions on the metrics may be needed, but I don't know. $\endgroup$ Nov 28, 2015 at 1:24
  • $\begingroup$ The question still makes no sense because with your definition the discrete topology makes any set into a Banach manifold. You need to decide what topology to use. One reasonable choice is the topology of $C^k$ uniform convergence on compact sets. With this topology the space of all symmetric $C^k$ 2-tensors is Frechet, and the subset of $C^k$ Riemannian metrics is a convex $G_\delta$ subset. I think with a little work one can show that this subset is homeomorphic to a separable Hilbert space. Very similar arguments can be found in section 6 of arxiv.org/abs/1510.07269. $\endgroup$ Nov 28, 2015 at 4:19

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