# What's wrong with compact-open topology on the space of maps?

Given a smooth vector bundle $E$ with non-compact base, let $\Gamma(E)$ be the space of $C^\infty$ sections equipped with compact-open $C^\infty$-topology.

1. I have heard that $\Gamma(E)$ is not locally-contractible. Why not?

2. Is $\Gamma(E)$ contractible? Visibly any section can be joined to the zero section by "straight line", doesn't this prove that $\Gamma(E)$ is contractible?

3. Is it true that every convex subset of $\Gamma(E)$ is contractible? The argument of 2 seems to apply, but then it seems plausible that each section has an arbitrary small convex neighborhood, contradicting 1.

CLARIFICATION: One source of "rumor 1" is the book "The Convenient Setting of Global Analysis" freely available here. On page 429 one reads: "Unfortunately, for non-compact $M$, the space $C^\infty(M, N)$ is not locally contractible in the compact-open $C^\infty$-topology". Another source is the discussion in Hirsch's book in the beginning of Chapter 2, which says "It can be shown that $C^\infty(M, N)$ has very nice features, e.g. it has a complete metric, and a countable base; if $M$ is compact, it is locally contractible and $C^r(M, \mathbb R^n)$ is a Banach space for $2\le r<\infty$".

Thus I assumeed that in general, if $M$ is non-compact, then the space $\Gamma(E)$ is not (or maybe just need not be?) locally contractible. Also I am uncertain whether $C^\infty(M, \mathbb R^n)$ or $\Gamma(E)$ is a topological vector space, is it really? There seems to be a sequence of semi-norms giving these spaces a structure of Frechet spaces, but then they must be locally convex, hence locally contractible. Obviously, I am missing something.

Question. Let $T_{r,s}(M)$ denote the space of $C^\infty$-smooth $(r,s)$-tensors on a connected non-compact $C^\infty$ manifold $M$. For $k$ with $2\le k\le \infty$, give $T_{r,s}(M)$ the weak $C^k$-topology as in Hirsch's book (roughly for $k$ finite we require that given $\epsilon>0$ and compact subset $K$ all derivatives up to $k$ are $\epsilon$-close over $K$, and for $k=\infty$ we take the union of all $C^k$-topologies for all finite $k$ under the inclusions $C^\infty\to C^k$). Now I ask

Is $T_{r,s}(M)$ a Fréchet space with respect to the weak $C^k$-topology?

In particular, I want to conclude that $T_{r,s}(M)$ is locally contractible, and any convex subset of $T_{r,s}(M)$ is contractible; I think Fréchet spaces must have this property.

• This is not an answer, but Hirsch's Differential Topology book has a nice chapter on function space topologies. – user1835 May 14 '10 at 4:07
• Andrew Stacey has started to write up some stuff on the nLab about the "differential topology of mapping spaces" based on a series of lectures he gave at NTNU. It seems like it might be relevant: ncatlab.org/nlab/show/… – Harry Gindi May 14 '10 at 4:29
• And an arXiv paper giving an overview/introduction: arxiv.org/abs/math/0510097 . I hope it's relevant. – Harry Gindi May 14 '10 at 4:33
• As stated, the rumours that you have heard are false. The simplest case is C^\infty(R,R) - this is sections of the trivial R-bundle over R. As Sergei hints in his "answer", this is a locally convex topological vector space, and hence contractible. I suggest that you track down the source of the rumour and get some clarification and then ask a more focussed question. – Andrew Stacey May 14 '10 at 9:17
• Comments on the added quotes: The first is taken from the introduction to a chapter and so should be taken as such. You should read the whole chapter to discover the meaning behind those words. I don't have the source for the second quote to hand, but there's still a mistake there: C^\infty(M,R^n) is not a Banach space. Generally, "compact-open" is a bit misused and so one should always ask "what exactly do you mean?". In particular, the KM book has quite an extensive discussion on the different topologies and I recommend that you read that and then ask a more focussed question. – Andrew Stacey May 14 '10 at 11:43

(For the more specific question)

Yes for $k = \infty$ if $M$ can be exhausted by a countable number of compact sets, no otherwise. However, the failure is due to a lack of completeness (for $k \ne \infty$) or size issues (if $M$ can't be exhausted) rather than anything else and thus the local contractibility still holds. Indeed, contractibility holds simply by contracting the vector bundle itself down to the image of the zero section.

The reason is due to the fact that the topology can be described by a family of semi-norms, as Sergei indicates in his comment to his earlier answer, so you get a locally convex topological vector space. I recommend that you read about these spaces; Schaefer's book is a good place to start (as in Jarchow's but that doesn't seem to be available any more).

(In particular, be wary of saying "union of all $C^k$-topologies"; actually you are taking a projective limit here which means that the space has a $0$-neighbourhood base which is a union of the $0$-neigbourhood bases from each of the $C^k$ topologies, but that doesn't mean that the final topology is the union of all of the $C^k$ topologies. Simply take a set of point $x_n$ that are "far apart" and put a set $U_n$ about each one so that $U_n$ is open in the $C^n$-topology but not in $C^{n-1}$. Then $\bigcup U_n$ is open in the $C^\infty$ topology but not in any of the $C^n$-topolgies.)

• Thanks! When you say ''M can be exhausted by a countable number of compact sets'' you must be thinking of non-metrizable manifolds. Any "usual" (i.e. finite dimensional smooth metrizable Hausdorff) manifold certainly has such exhaustion, e.g. the exhaustion by metric balls of any complete Riemannian metric on the manifold; the balls are compact by Hopf-Rinow – Igor Belegradek May 14 '10 at 14:44
• Yes, I was just being extra careful in stating the conditions. Partly because at the moment I'm thinking about how results like this generalise to things that aren't manifolds. – Andrew Stacey May 14 '10 at 15:08

Are you sure it is about vector bundles?

Any topological vector space is contractible. Any locally convex topological vector space is locally contractible. This is straightforward from the definitions. The compact-open $C^\infty$ topology is defined by a family of semi-norms and hence locally convex.

In the case of arbitrary fiber bundles, the space of sections can indeed fail to be locally contractible. For example, consider the trivial bundle $\mathbb N\times\{0,1\}$ over $\mathbb N$ (they are 0-dimensional manifolds). The space of sections is not locally contractible (in fact, homeomorphic to the Cantor set). You can build connected examples as well, e.g. consider the trivial circle bundle over a surface with infinitely many handles.

• I am interested in sections of vector bundles because they form a vector space, and hence, $\Gamma(E)$ is more likely to be locally contractible. As you say the space of maps $C^\infty(M, N)$ need not be locally contractible. One specific space I care about is the space of $(r,s)$-tensors on a non-compact manifold. – Igor Belegradek May 14 '10 at 13:04
• For every local trivialization of $E$, every compact subset $K$ of the respective region in $M$ and every positive integer $r$ you define a semi-norm on $\Gamma(E)$ as the coordinate-wise $C^r$ norm of the restriction of a section to $K$. The compact-open $C^\infty$ topology is the topology defined by this family of semi-norms. Thus $\Gamma(E)$ is locally convex topological vector space and hence locally contractible. – Sergei Ivanov May 14 '10 at 13:36
• Looks like you are confirming that $\Gamma(E)$ is a Fréchet space. I thought this should be the case, but then I got confused. Thanks! – Igor Belegradek May 14 '10 at 13:59
• Not quite so, you need to be careful if you want a countable family of semi-norms, as explained in Andrew's answer. For local contractibility, this does not matter. – Sergei Ivanov May 14 '10 at 18:29
• Not sure I understand the last comment. Andrew, does say that with $C^\infty$ topology $\gamma(E)$ is Fréchet. Indeed, consider semi-norms corresponding to an exhaustion by a countable family of compact sets. – Igor Belegradek May 14 '10 at 19:51