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Let $M,N,X$ be smooth manifolds. Equip the space of smooth functions between two manifolds with the (strong) Whitney- $\mathcal{C}^\infty$-topology.

The evaluation map $$ev\colon N\times\mathcal{C}^\infty(N,X)\rightarrow X$$ is continuous. (This can e.g. be found in "Margalef-Roig/Dominguez - Differential Topology", Proposition 9.6.7)

Therefore one gets a map of sets $$\mathcal{C}(M,\mathcal{C}^\infty(N,X))\rightarrow \mathcal{C}(M\times N,X).$$

I searched the literature for a reference to the "other direction", but did not find anything. I.e. starting with a smooth map $M\times N\rightarrow X$ I want to know whether the adjoint $M\rightarrow \mathcal{C}^\infty(N,X)$ is continuous or if not, in which cases.

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  • $\begingroup$ Ha! My attempt to look this up in an obvious place quickly ran into an obstacle. In Hirsch's Differential topology, Exr.2.4.2 (a starred exercise) is essentially identical to your question, without suggesting an answer: "Under what conditions is the natural map $C_S(X,C_S(Y,Z)) \to C_S(X\times Y, Z)$ a homeomorphism?" $\endgroup$ – Igor Khavkine Jun 15 '15 at 12:38
  • $\begingroup$ The exercice deals with the continuous case. Hirsch denotes with $C_S$ the topology on the space of all continuous maps coming from considering graphs. $\endgroup$ – Kathrin L. Jun 15 '15 at 12:57
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    $\begingroup$ Time to create an "Ask-Michor-tag". More seriously, Peter Michor, Andreas Kriegl, and collaborators did a lot on such questions for a big variety of function spaces. Look for "convenient calculus". $\endgroup$ – Jochen Wengenroth Jun 15 '15 at 20:59
  • $\begingroup$ @KathrinL. Igor's comment is still relevant. The graph topology on the space of continuous maps is the Whitney $C^0$ topology, and Peter Michor's argument works in the continuous case as well. $\endgroup$ – Pedro Lauridsen Ribeiro Jun 17 '15 at 5:02
  • $\begingroup$ @JochenWengenroth or perhaps a "convenient-calculus" tag... $\endgroup$ – Pedro Lauridsen Ribeiro Jun 17 '15 at 5:05
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The other direction is not true. The map $$\mathcal{C}^\infty(M,\mathcal{C}^\infty(N,X))\rightarrow \mathcal{C}^\infty(M\times N,X)$$ is not surjective if $N$ is not compact. The image consists of functions $f$ such that

  • for each compact $K\subset M$ there is a compact $L\subset N$ such that $f(x,y)$ is constant in $x\in K$ for each $y\in N\setminus L$.

This is, because smooth (and continuous) curves in $C^\infty (N,X)$ can move only within a compact of $N$ in finite time. See 4.4.4 (page 34) of

  • Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp. (pdf)

for the very easy proof. There even the continuous case that you ask for is treated.

Added:

The following might give you feeling for the reason:

Consider $f\in C^\infty(N,\mathbb R)$, equip $C^\infty(N,\mathbb R)$ with the strong Whitney topology. If $t\mapsto t.f$ is continuous $\mathbb R\to C^\infty(N,\mathbb R)$, then $f$ has compact support. Thus $C^\infty_c(N,\mathbb R)$ is the maximal topological vector space in ($C^\infty(N,\mathbb R)$, Whitney topology).

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  • $\begingroup$ Motivated by your last paragraph I would like to ask you the following: Is the continuous case true if one restricts to $\mathcal{C}^\infty_c(N,\mathbb{R})$? I.e. if I have a smooth map $M\times N\rightarrow\mathbb{R}$ such that the adjoint $M\rightarrow \mathcal{C}^\infty(N,\mathbb{R})$ has image in $\mathcal{C}^\infty_c(N,\mathbb{R})$, will it be continuous? Thank you very much. $\endgroup$ – Kathrin L. Jun 23 '15 at 9:41
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If you take the compact-open ($C^\infty$) topology (instead of the Whitney one), then you have continuity/smoothness of the evaluation map and the following exponential laws $$ C(M \times N, P) = C(M, C(N, P)) $$ $$ C^\infty(M \times N, R) = C^\infty(M, C^\infty(N, R)) $$ if $N$ is a (locally compact) manifold without boundary, $M$ and $P$ are manifolds and $R$ is a vector space (possibly infinite-dimensional). In particular, for a compact manifold $N$ the above exponential laws also hold for the Whitney topology (which of course coincides with the compact-open topology in this case).

More details can be found for example in some lecture notes by H. Glöckner. Also Michor's "Convienient Calculus" book contains a discussion of these matters but there a different topology on the space of smooth maps is used.

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  • $\begingroup$ Just to add a citable reference to Tobias comment on the compact-open $C^\infty$ topology: You can find this exponential law as a special version of the general exponential law in arxiv.org/abs/1208.6510 . $\endgroup$ – Alexander Schmeding Feb 5 '18 at 22:40

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