Suppose that $X$ is a compact, finite dimensional manifold and $Y$ is an infinite dimensional, second countable ($C^\infty$-)Banach manifold. Let $\nu \in \mathbb{N}$.

Question: Is the space $C^\nu(X,Y)$ a ($C^\infty$ or $C^k$ for some $k \in \mathbb{N}$) Banach manifold which is Lindelöf (i.e. every open cover has a countable subcover)?

Any help, as well as references are very much appreciated.

Some motivation:
My goal is to apply a parametric transversality theorem, where my parameter space is $C^\nu(X,Y)$. In order to so it is required that the parameter space is Lindelöf and a ($C^k$-) Banach manifold. I don't have much experience with function spaces and their topology, especially when the target is infinite dimensional. So I apologize if it turns out that my question is "well known".


See the paper

  • MR0226681 (37 #2268) Elĭasson, Halldór I. Geometry of manifolds of maps. J. Differential Geometry 1 1967 169–194.

for a description of the manifold structure on $C^\nu(X,Y)$. But you need somthing like a Riemannian exponential mapping on $Y$ which is a diffeomorphism from a neighborhood of the 0-section in $TY$ to a neighborhood of the diagonal in $Y\times Y$; you can forget some properties of the exponential mapping, then this is called a local addition (see here). If $Y$ is nice enough (smoothly paracompact or at least $C^1$-paracompact allows to construct a linear connection on $TY$; or you may be given a linear connection on $TY$) then you have this.


If $X$ is compact and $Y$ is a separable Banach space (added in second edit), then $C^\nu(X,Y)$ is separable; In fact it is $C^\nu(X)\hat{\hat\otimes} Y$ for the completed inductive tensor product. (see here, e.g) Then it is obvious that the tensor product is separable.

Second edit: About $C^k$-mapping manifolds.

In "Kriegl, Michor: The convenient setting ..." (see my homepage) 14.11.1 states that there does not exist any Frechet differentiable bump function. You can embed an interval in any manifold, thus $C^0(M)$ does not admit differentiable bump functions. Furthermore, $C^0$ sits at the top of $C^k(M)$, so also $C^k(M)$ does not admit differentiable bump functions. On the other hand, $C^\infty(M)$ is nuclear Frechet, so has smooth partitions of unity. This is the reason why I vastly prefer $C^\infty$, and Sobolev completions, if I have to solve equations.

  • $\begingroup$ Thank you very much for your answer and the references! From your answer I gather that the existence of such a local addition is not a very restrictive condition on $Y$, is this correct? Do you know if $C^\nu(X,Y)$ is a Lindelöf space (or second countable)? $\endgroup$
    – Dave
    Jun 5 '13 at 15:12
  • $\begingroup$ @Peter Michor: Thank you for the edit. Although I don't understand why it is true, I'm very glad to hear that $C^\nu(X,Y)$ is separable if $X$ is compact and $Y$ is separable! I'd appreciate it very much if you could explain in some more detail why this is in fact true (or point me to a reference). Unfortunately I don't know what the "completed inductive tensor product" is (and I didn't find much about it on the internet...). Thank you for your help. $\endgroup$
    – Dave
    Jun 7 '13 at 15:45
  • $\begingroup$ @Dave Hartman: What is your Banach manifold $Y$. Is it an open set in a Banach space? $\endgroup$ Jun 12 '13 at 9:41
  • $\begingroup$ @Peter Michor: Thank you for the additional reference. Yes, $Y$ is an open set in a Banach space. In fact, $Y=\mathbb{R}^2\times F_2(Gvect(f))$ which is an open set in $\mathbb{R}^2\times G_1^2$. $G_1$ is the space from your answer here: mathoverflow.net/questions/127843/… . But I consider $G_1$ with the $C^k$ (instead of $C^\infty$) topology so that it is a Banach space (instead of a Fréchet space). $\endgroup$
    – Dave
    Jun 12 '13 at 21:21
  • $\begingroup$ @Peter Michor: Thank you for elaborating. Are smooth bump functions necessary in order to get a local addition? Does my space $Y$ admit a local addition? $\endgroup$
    – Dave
    Jun 17 '13 at 5:53

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