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Short question:

Let $M$ and $N$ be smooth manifold, with appropriate smooth function algebras $C^\infty(M,\mathbb{R})$ and $C^\infty(N,\mathbb{R})$.

Can we express the smooth function algebra of the cartesian product manifold in terms of $C^\infty(M,\mathbb{R})$ and $C^\infty(N,\mathbb{R})$?

I know it is neither (equivalent to) $C^\infty(M,\mathbb{R}) \oplus C^\infty(N,\mathbb{R})$ nor $C^\infty(M,\mathbb{R})\otimes_{\mathbb{R}}C^\infty(N,\mathbb{R})$.

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A more general question is,if there is a general rule to get from categorical constructions on manifolds to constructions on the appropriate smooth function algebra.

Maybe this boils don to the question whether or not the functor $C^\infty(\cdot,\mathbb{R})$ from smooth manifolds to ass. comm. unitary $\mathbb{R}$-algebras preserves (co)limits.

That's indeed the deeper question.

...

P.S.: I tagged it in particular as algebraic-geometry related, do to the category theory related part...

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    $\begingroup$ Assume $M$, $N$ compact. Isn't then $C^\infty(M\times N, \mathbb{R})$ the completion of $C^\infty(M, \mathbb{R}) \otimes C^\infty(N, \mathbb{R})$ with respect to the supremum norm? $\endgroup$
    – Piotr Achinger
    Mar 19, 2013 at 18:35
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    $\begingroup$ @Piotr: That is true for continuous functions. Spaces of smooth functions are not normable or Banach spaces. $\endgroup$
    – Peter Michor
    Mar 19, 2013 at 18:58
  • $\begingroup$ @Peter: Piotr's comment would be fine for functions of class $C^k$, $k finite$, right? $\endgroup$
    – Claudio Gorodski
    Mar 19, 2013 at 19:45
  • $\begingroup$ I mean, wrt $C^k$-norm (uniform convergence of functions and their first $k$ derivatives). $\endgroup$
    – Claudio Gorodski
    Mar 20, 2013 at 15:52

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If $M$ and $N$ are compact, then $C^\infty(M\times N)=C^\infty(M)\bar\otimes_{i}C^\infty(N)$, the completed injective tensor product which coincides with the completed projective tensor product, since the locally convex spaces involved are nuclear.

Edit: This also holds for for non-compact $M,N$; see [Treves: Topological Vector Spaces, Distributions, and Kernels, Page 530].

If the manifolds are not compact, the same holds for the space of smooth functions with compact support.

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  • $\begingroup$ Just a little remark: That $\otimes_i$ coincides with $\otimes_pi$ in this case is not only because of nuclearity (your remark about the spaces of smooth functions with compact support shows this since $\mathscr D(\mathbb R) \tilde{\otimes}_\pi \mathscr D(\mathbb R) \neq \mathscr D(\mathbb R^2)$).You use first that $\otimes_i = \otimes_\varepsilon$ for Frechet spaces and then nuclearity. $\endgroup$
    – Jochen Wengenroth
    Mar 21, 2013 at 12:18
  • $\begingroup$ I know this question is super old, but: Jochen's comment and Peter's answer seem to contradict each other, no? Jochen says that the projective tensor product of compactly supported functions on $\mathbb{R}$ does not equal the compactly supported functions on the product space, but the last sentence in Peter's answer appears to say that it does. The cited Treves page seems not to comment on this situation (he only considers the product of the compactly supported function on compact spaces). If one of you sees this, could you clarify? $\endgroup$
    – Lukas Miaskiwskyi
    Nov 12, 2020 at 18:28
  • $\begingroup$ @JochenWengenroth Sorry for my confusion about your notations. What is $\otimes_i$? I guess that $\otimes_\epsilon$ is with $\epsilon$-topology, and $\otimes_\pi$ is with $\pi$-topology (in Treves' terms)? $\endgroup$
    – Z. M
    Apr 23, 2022 at 20:58
  • $\begingroup$ 9 years ago I probably meant by $\otimes_i$ the inductive tensor product of Grothendieck which differs for the injective tensor product $\otimes_\varepsilon$. $\endgroup$
    – Jochen Wengenroth
    Apr 24, 2022 at 13:34

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