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A smooth structure on a manifold $M$ can be given in the form of a sheaf of functions $\mathcal{F}$ such that there is an open cover $\mathcal{U}$ of $M$ with every $U\in \mathcal{U}$ isomorphic (along with $\mathcal{F}|_U$) to an open subset $V$ of $\mathbb{R}^n$ (along with $\mathcal{O}|_V$, where $\mathcal{O}$ is the sheaf of smooth functions on $\mathbb{R}^n$). I think we might also need to say that this satisfies a smooth-coordinate-change axiom, although maybe that's already tied up in the definition of a sheaf. In any case, here is my question:

Given two smooth manifolds $(M,\mathcal{F})$ and $(N,\mathcal{G})$, is there an easy way to write the sheaf of functions on $M\times N$ without reference to coordinate neighborhoods?

I'm wondering this because in one of my classes we defined smooth manifolds in this way (and we defined analytic and holomorphic manifolds similarly). It seems like some people are very fond of this alternative definition because it doesn't refer to an atlas, which at first seems like it's an inherent part of the structure of the manifold. So okay fine, everyone loves a canonical definition. However, this is only going to be useful as long as we can tell our whole story in this canonical language. In class, the only way the professor was able to give the sheaf on the product was by breaking down and using coordinates. (Admittedly, he was on the spot and presumably unprepared for the question.)

This also suggests the broader, more open-ended question:

Are there longer-run advantages to the above definition (compared to the usual definition involving an atlas and perhaps a maximal atlas)?

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This way of defining manifolds is the same one as the one which consists of giving an atlas, up to very minor differences: if you are given a sheaf satisfying the conditions you mention in your first paragraph, the atlas can be immediately be found by looking at all subsets $U$ involved there together with the isomorphisms to open subsets of $\mathbb R^n$; going the other way requires as little effort, too. –  Mariano Suárez-Alvarez Sep 1 '10 at 21:00
I think over open sets which are of the form V= $U_1$ x $U_2$ where U_i is an open set in M_i you can define F(V) to be the "projective tensor product" of $F_i(U_i)$.Then this should determine the sheaf of smooth functions on the product by the concept of "sheaves on a base". I don't know if this "sheafy" approach to smooth manifolds buys you much... but a different theorem that the functor which assigns M---> $C^{\infty}(M)$, the algebra of global sections is fully faithful is really cool and might have some applications. –  Daniel Pomerleano Sep 1 '10 at 23:33
Yes, $C^{\infty}(M \times N)$ is some completed tensor product $C^{\infty}(M) \widehat{\otimes} C^{\infty}(N)$. –  Martin Brandenburg Sep 2 '10 at 0:19
This is subject to one's taste, but a lot of things (e.g. tangent sheaf, cotangent sheaf) are "easier" if you use the sheaf definition. It also generalizes better -- c.f. Hartshorne Chapter II. –  Kevin H. Lin Sep 2 '10 at 21:11
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4 Answers

up vote 5 down vote accepted

Smooth manifolds are affine, thus the sheaf of smooth functions is determined by its global sections. Now C^∞(M×N)=C^∞(M)⊗C^∞(N). The tensor product here is the projective tensor product of complete locally convex Hausdorff topological algebras.

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Projective space is a non-affine smooth manifold... –  Daniel Loughran Sep 2 '10 at 10:59
@Daniel: Projective space is an affine smooth manifold. –  Dmitri Pavlov Sep 2 '10 at 11:04
Ahh I see sorry and thanks for the clarification. I have only come across the word "affine" in algebraic geometry before, but it does indeed make sense in this context. –  Daniel Loughran Sep 2 '10 at 11:21
I see. This is probably the answer I'm looking for, but I don't think I know enough to be sure one way or the other. Could you either explain what this means in other words, or say a little bit about "projective tensor product" and "complete locally convex Hausdorff topological algebras"? Thanks! –  Aaron Mazel-Gee Sep 3 '10 at 6:01
Dimitri: I think it goes back to Grothendieck's book Produits tensoriels topologiques et espaces nucléaires but I don't have that in front of me at the moment so I'll quote Schaefer's Topological Vector Spaces instead, chapter III section 6. He doesn't go in to much detail on the other topologies on tensor products, but there are plenty. At the other end is the inductive topology where one asks simply for the bilinear maps to be separately continuous. Nuclear spaces, such as smooth functions, can be characterised in terms of their behaviour with tensor products. –  Andrew Stacey Sep 3 '10 at 13:15
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Let $Mfd$ be the category of smooth manifolds (over $\mathbb{R},\mathbb{C}$) and $LRS$ be the category of locally ringed spaces (over $\mathbb{R},\mathbb{C}$). Then the functor $Mfd \to LRS$ is full and faithful and indeed, you may define $Mfd$ to be the full subcategory of $LRS$, which consists of those locally ringed spaces, which are locally isomorphic to $\mathbb{R}^n$ together with the sheaf of smooth functions. Unfortunately, the functor $Mfd \to LRS$ does not preserve products; see below.

It should be remarked that products in $LRS$ do exist (even infinite ones); take the obvious product in $RS$ and "make it local" by introducing new points, namely prime ideals in the stalks, and take the localizations of the stalks as the new stalks. Now if we take the product of two manifolds $M,N$ in $LRS$, we get as a topological space the usual product $M \times N$; however, the structure sheaf consists only of those functions, which are locally of the form $(u,v) \mapsto f(u) * g(v)$ for smooth functions $f,g$ defined locally on $M,N$, or sums and also quotients of these functions. These functions are also smooth with respect to the usual smooth structure on $M \times N$, but the reverse is not true: For example it seems to be true that $\mathbb{R}^2 \to \mathbb{R}, (u,v) \mapsto exp(uv)$ is not such a functions (however, I don't know how to prove this).

However, I think that Stone-Weierstraß implies that these simple functions are dense within all smooth functions. Thus, we may regard $M \times N$ in $Mfd$ as the "completion" of $M \times N$ in $LRS$.

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If I may be forgiven for climbing on my soap box ...

This kind of thing highlights one of the problems with these "one sided" approaches to the theory of smooth spaces. If one takes smooth functions out, then the resulting functor, Manifolds → Functions, does not preserve limits (as Martin says). But if one takes smooth functions in then the resulting functor does not preserve colimits. The resolution is to take both smooth functions in and out, whereupon one does get a functor that preserves both colimits and limits. The resulting category is Hausdorff Frölicher spaces.

Taking functions both in and out also follows in the footsteps of the study of manifolds. I asked a bunch of topologists not long ago whether a chart was a map to a manifold or from it. It was about a 50-50 split! At the least, this shows that the direction of charts is ambiguous. At best, it shows that it's useful to have access to both directions.

So the answer to the question is as follows: the smooth functions $M \times N \to \mathbb{R}$ are those functions $f \colon M \times N \to \mathbb{R}$ with the property that $f \circ \gamma \in C^\infty(\mathbb{R},\mathbb{R})$ whenever $\gamma \colon \mathbb{R} \to M \times N$ is such that $(g,h) \circ \gamma \in C^\infty(\mathbb{R},\mathbb{R}^2)$ for all $g \in C^\infty(M)$ and $h \in C^\infty(N)$. This can probably be "sheafified" but I'm not a farmer.

Obligatory nlab link: for more on $C^\infty$-rings and the like: take a look at smooth algebra.

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LOL. I'm not much of a farmer either. –  Aaron Mazel-Gee Sep 2 '10 at 6:53
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If one denotes $\pi_1:M\times N\to M$ and $\pi_2:M\times N\to N$ the two projections, then I think that the Sheaf of functions on $M\times N$ is simply $\pi_1^*\mathcal{F}\otimes\pi_2^*\mathcal G$.

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I don't think this is right. Taking $M=N=\mathbb{R}$, this would say that the smooth functions on $\mathbb{R}^2$ are exactly those that are sums of products of smooth functions of $x$ and $y$. This doesn't include e.g. $\sin(xy)$... –  Aaron Mazel-Gee Sep 2 '10 at 6:57
@Aaron, you have to use the correct tensor product and then it works. –  Mariano Suárez-Alvarez Sep 2 '10 at 12:22
@Aaron: The statement is true, because the tensor product used here is the projective tensor product, not the algebraic tensor product. See also my answer below. –  Dmitri Pavlov Sep 2 '10 at 12:24
I prefer to take the tensor product of sheaves. First consider the naive tensor product (without completion). It's only a preasheaf... then sheafify it. –  DamienC Apr 25 '11 at 21:25
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