# Smooth function algebra on cartesian product and beyond

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.

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

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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? – Piotr Achinger Mar 19 '13 at 18:35
@Piotr: That is true for continuous functions. Spaces of smooth functions are not normable or Banach spaces. – Peter Michor Mar 19 '13 at 18:58
@Peter: Piotr's comment would be fine for functions of class $C^k$, $k finite$, right? – Claudio Gorodski Mar 19 '13 at 19:45
I mean, wrt $C^k$-norm (uniform convergence of functions and their first $k$ derivatives). – Claudio Gorodski Mar 20 '13 at 15:52

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].
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. – Jochen Wengenroth Mar 21 '13 at 12:18