# Is there a sheaf theoretical characterization of a differentiable manifold?

I'm going through the crisis of being unhappy with the textbook definition of a differentiable manifold. I'm wondering whether there is a sheaf-theoretic approach which will make me happier. In a nutshell, is there a natural condition to impose on a structure sheaf $\mathcal{C}^k_M$ of a topological space $M$ that can stand-in for the requirement that $M$ be second-countable Hausdorff?

### Background

Most textbooks introduce differentiable manifolds via atlases and charts. This has the advantage of being concrete, and the disadvantage of involving an arbitrary choice of atlas, which obscures the basic property that a differential manifold "looks the same at all points" (for $M$ connected, without boundary: diffeomorphism group acts transitively). And isn't introducing local coordinates "an act of violence"?

I saw a much nicer definition of differentiable manifolds on Wikipedia, which I don't know a good textbook reference for. This definition proceeds via sheaves of local rings. The Wikipedia definition stated:

A differentiable manifold (of class $C_k$) consists of a pair $(M, \mathcal{O}_M)$ where $M$ is a topological space, and $\mathcal{O}_M$ is a sheaf of local $\mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,\mathcal{O}_M)$ is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$.
[$\mathcal{O}(U)=C^k(U,\mathbb{R})$ is the structure sheaf on $\mathbb{R}^n$.]

Beautiful, really! Entirely coordinate free. But isn't there a General Topology condition missing?

I confirmed on math.SE (to make sure that I wasn't hallucinating) that this definition is indeed missing the condition that $M$ be second-countable Hausdorff. That indeed turned out to be the case, so I edited the Wikipedia definition to require $M$ to be second-countable Hausdorff.

### Why am I still not happy?

The deep reason that we require a differentiable manifold to be paracompact, as per Georges Elencwajg's extremely informative answer, is that paracompactness makes sheaves of $C_M^k$-modules (maybe $k=\infty$) acyclic. This is a purely sheaf-theoretic property (a condition on the structure sheaf of $M$ rather than on $M$ itself), which quickly implies good things like that every subbundle of a vector bundle on $M$ be a direct summand. Is this in fact enough?

If it were enough to require that $\mathcal{O}_M$ be acyclic, or maybe fine, then the nicest, most flexible (and, in a strange sense, most enlightening) definition of differentiable manifold, would be:

Definition: A differentiable manifold (of class $C_k$) consists of a pair $(M, \mathcal{O}_M)$ where $M$ is a topological space, and $\mathcal{O}_M$ is an acyclic sheaf of local $\mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,\mathcal{O}_M)$ is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$.

Maybe the word acyclic should be fine. Maybe soft and acyclic. Maybe a bit more, but still something that can be stated in terms of the structure sheaf.

Question: Can I put a natural sheaf-theoretic condition on $\mathcal{O}_M$ (acyclic? fine?) which ensures that $M$ (a topological space) must be a second-countable Hausdorff space? If not, would such a condition at least ensure that $M$ be a generalized differentiable manifold in some kind of useful sense?

Update: This question really bothers me, so I've started a bounty. I'd like to narrow it down a little in order to make it easier to answer:

Does acyclicity (or a slightly stronger condition) on $\mathcal{O}_M$ of a topological (Hausdorff?) space imply paracompactness (or a slightly weaker but still useful condition)?

Hausdorff bothers me as well, of course; but a sheafy characterization of paracompactness somehow seems like it has the potential to be lovely and really enlightening.

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I'm all in favor of sheafy points of view, but: You can avoid the unpleasantness of an arbitrary choice of atlas by defining a smooth structure to be a maximal atlas. (Of course in practice you will use the fact that an atlas determines a maximal atlas.) Also, is "looks the same at all points" really the same as "diffeomorphism group acts transitively"? What if $M$ is not connected? – Tom Goodwillie Feb 10 '12 at 1:24
@Paul Siegel: I don't think that statement is true, since either origin has a neighborhood excluding the other origin. Hausdorff is not a local condition, and everything that's locally Euclidean is locally Hausdorff. – Will Sawin Feb 10 '12 at 2:20
@Daniel: Separatedness, in the sense of being a Hausdorff topological space, is a property of topological spaces with points, but not a property of locales (pointless topological spaces). The assignment $X \mapsto \textbf{Sh}(X) : \textbf{Top} \to \mathfrak{Topos}$ factors through the category of locales $\textbf{Loc}$. As such there is no hope of finding a purely sheaf-theoretic characterisation without referring to stalks or points somewhere. – Zhen Lin Feb 10 '12 at 18:10
It's not clear to me there's any advantage in this formalism for manifolds. From a historical perspective, demanding someone to know what a sheaf is before a manifold seems kind of backwards. And the end result is, you've got a definition that pre-supposes the student is comfortable with a higher-order level of baggage and formalism than the manifold concept, moreover, you haven't really made the subject any easier, you've just brought in an extra layer of definitions to wade through. Cute? Yes. Nice? Maybe not. – Ryan Budney Feb 10 '12 at 23:01
@Daniel: yes, manifolds are a special case of schemes, but this viewpoint is not so useful for the study of manifolds, because the scheme formalism is taylor-made to globalize results from commutative algebra. But from the viewpoint of commutative algebra , $C^{\infty}(\mathbb{R})$ is rather pathological. This is why commutative algebra is not the right local tool for differential geometry. – Johannes Ebert Feb 12 '12 at 11:48

Definition: ''A smooth manifold is a locally ringed space $(M;C^{\infty})$ which satisfies the conditions:

1. Each $x \in M$ admits a neighborhood $U$, such that $(U,C^{\infty})$ is isomorphic to $(\mathbb{R}^n,C^{\infty})$ as a locally ringed space.
2. The global sections of $C^{\infty}(M)$ separate points.
3. The structure sheaf $C^{\infty}$ is fine as a sheaf of modules over itself.
4. $C^{\infty}(M)$ has at most countably many indecoposable idempotents.''

Explanations:

1) is evident.

2) means that for $x \neq y \in M$, there exists $f \in C^{\infty}(M)$ with $f(x)=0$, $f(y) \neq 0$. This ensures the Hausdorff condition once we know that elements of $C^{\infty}(M)$ give rise to continuous maps $M \to \mathbb{R}$. This is as follows: $f \in C^{\infty}(M)$ given, $x \in M$. Pick a chart $h:U \to \mathbb{R}^n$; under this chart, $f|_U$ corresponds to a smooth function on $\mathbb{R}^n$, whose value at $h(x)$ does not depend on the choice of the chart. Call this value $f(x)$. Checking the continuity of $x \mapsto f(x)$ can be done in charts.

3.) By this I mean that for each open cover $(U_i)$, there is a partition of unity $\lambda_i$ with the usual properties and that $\lambda_i$ is a map of $C^{\infty} (M)$-modules. A standard argument shows that $\lambda_i$ is given by multiplication with a smooth function. Therefore, the underlying space $M$ is paracompact.

4.) An idempotent $p\neq 0$ in a commutative ring $A$ is called indecomposable if

$$p=q +r; r^2 =r; q^2 =q , q \neq 0 \Rightarrow p = q$$

holds. Indecomposable idempotents in $C^{\infty}(M)$ correspond to connected components. Therefore condition 4 means that $M$ has only countably many connected components.

These conditions together imply that $M$ is Hausdorff and second countable, because a locally euclidean, connected and paracompact Hausdorff space is second countable, see Gauld, "Topological properties of manifolds", Theorem 7 (see http://www.jstor.org/stable/2319220 ). Paracompactness alone does not guarantee second countability, see $\mathbb{R}$ with the discrete topology.

However, I think that sheaf theory and locally ringed spaces are the wrong software for differential geometry and differential topology.

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This was my first idea, but if we are staying purely within the abstract framework of sheaves does it really make sense to evaluate a section at a point? If it does then this should definitely be the right condition. – Paul Siegel Feb 10 '12 at 12:36
It does indeed make sense--the stalks of the structure sheaf are local rings, and evaluation at a point is quotienting by the maximal ideal. Completely intrinsic. – Daniel Litt Feb 10 '12 at 15:46
And you can state the separation property by saying: $x\neq y$, then there is an $f$ with $f(x)=0$ and $f(y)\neq 0$, so one does not need an isomorphism of differerent stalks. – Johannes Ebert Feb 10 '12 at 22:21
You can restate Johannes' criterion as follows: The canonical morphism $X \to \mathrm{Spec}(\Gamma(X,\mathcal{O}_X))$ (which is defined for every locally ringed spaace, see EGA I, 1.6.3) is injective. – Martin Brandenburg Feb 12 '12 at 11:45
One thing is "easiest" for the beginning of the theory, and another is what a power may be gained when going to subtleties and extensions of the theory. By no means introduction of sheaves means one should leave other tools like coordinates. One often needs to extend some constructions from manifolds to categories closed under interesting operations like quotients etc. It is easier to pass to e.g. supermanifolds, stacks and derived geometry, understanding the sheaf aspect of the story first, not only about sheaves on spaces, but also understanding of spaces as sheaves on site of local models. – Zoran Skoda Feb 14 '12 at 14:32

For another perspective, think about definitions of "complex manifold" or "real analytic manifold". Normally you use atlases for this, imposing the Hausdorff and paracompactness conditions separately. You can restate the atlas part of the definition in sheaf language, but you can't hope to get the rest of it that way, can you? I mean, you can't get the paracompactness from properties like acyclicity or flasqueness, because you don't have those properties.

Of course, you can always fall back on "a complex manifold is a real manifold plus the following extra structure", but that seems to go against the spirit of the question.

On a related note, of course the "basic" property of transitivity of the automorphism group fails in the complex-analytic case. (Query: what about the real-analytic case?)

EDIT From the comments it is clear that I haven't expressed myself clearly. Let me try again.

The usual definition of "smooth manifold" says (1) the space is equipped with an atlas in which all the charts are pairwise smoothly compatible, or rather an equivalence class of such atlases, or if you prefer a maximal such atlas, (2) the space is paracompact, (3) the space is Hausdorff.

Daniel likes the option of replacing (1) by the logically equivalent:

(1)' the space is equipped with a sheaf of algebras such that every point has a nbhd such that ...

He is wondering if he can simultaneously replace (2) by some extra requirement on the sheaf (something implying acyclicity) to get either an equivalent definition or a slightly weaker but still useful notion, and he offers as evidence the idea that the usefulness of (2) can be expressed in sheaf language. I am not offering an opinion about that.

I am just pointing out the following: If you did succeed in reworking the definition of smooth manifold in this esthetically pleasing way, then you might want to do the same for the definition of complex analytic manifold. But you'd get stuck on the fact that the structure sheaf is not acyclic.

So maybe you would fall back on describing a complex manifold as a smooth manifold plus extra structure.

But wait, what about topological manifolds? Don't we like them to be paracompact, too? For similar reasons as in the real case. Does it seem esthetically right to emphasize the ring of continuous functions when dealing with topological manifolds?

By the way, why do we want complex manifolds to be paracompact? Probably to allow the occasional use of $C^\infty$ methods.

Oh, and what about piecewise linear manifolds? Here (and also, by the way, for $C^k$ manifolds for $k\ge 1$ finite) there is no sheaf of rings involved. But there is the option of using charts, where compatibility of charts depends on the notion of PL homeomorphism between open subsets of $\mathbb R^n$). If you don't also assume paracompactness and Hausdorff then you won't have triangulations. Is there some sheafy way to discuss this?

OK, I've rambled, and probably replaced one unclear thing by several other unclear things.

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So, taking the extreme case, what you're saying is that there exists a Hausdorff topological space $M$ with fine, soft, acyclic sheaf $\mathcal{O}_M$, such that $(M,\mathcal{O}_M)$ is locally isomorphic as a locally ringed space to $(\mathbb{R}^n,\mathcal{O})$, yet $M$ is not paracompact? – Daniel Moskovich Feb 10 '12 at 5:23
Dear Tom, I don't understand your sentence "you can't get the paracompactness from properties like acyclicity or flasqueness, because you don't have those properties". Like Daniel, I wonder if you are claiming that acyclicity of the structure sheaf implies or doesn't imply paracompactness. Can you please clarify? – Georges Elencwajg Feb 10 '12 at 11:35
Thanks for the clarification in your edit: indeed the structural sheaf has no reason to be acyclic in the holomorphic or algebraic category, and usually isn't (except for Stein or affine manifolds). – Georges Elencwajg Feb 10 '12 at 14:26
This is interesting! What about leaving (3) alone, and trying to replace (2) by a sheaf condition? (these are actually different questions, so I should edit the question to make that clear) – Daniel Moskovich Feb 11 '12 at 23:47
I actually meant (2) when I said (3). I've edited again to correct that now. – Tom Goodwillie Feb 14 '12 at 14:18

Dear Daniel, you should take a look into the nice book "Global Calculus" of S. Ramanan.

From the Review of the book by John Miller: "This is a decidedly individual course on analysis and geometry on manifolds. The book begins by introducing smooth manifolds as spaces equipped with sheaves of differentiable functions. Although this approach is unusual in a textbook, it works rather smoothly. ...."

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How does that answer the question though? I don't think Ramanan addresses questions like this at all- his manifolds are explicitly Hausdorff second-countable, and the sheafs are explicitly sheafs of functions. – Daniel Moskovich Feb 10 '12 at 7:59
@Sebastian: This point of view on manifolds was already mentioned in the "Background" section of the question. It does not answer the question. – Martin Brandenburg Feb 12 '12 at 14:28

This is more a long comment than an answer, but the comments concerning the question Hausdorff or non-Hausdorff triggered this...

The second countable condition is certainly desirable for many reasons (embedding theorems etc) but there are prominent examples of non-Hausdorff manifolds in differential geometry. If you have a Lie algebroid, i.e. a vector bundle $E \longrightarrow M$ such that the sections of $E$ are equipped with a Lie bracket which satisfies a Leibniz rule along a bundle map $E \longrightarrow TM$ (the anchor) then there is a notion of a corresponding Lie groupoid integrating this. The obstructions for existence have been described in a beautiful paper by Crainic and Fernandes. Nevertheless, the resulting Lie groupoid is typically not Hausdorff but the non-Hausdorffness (funny word) is not so bad. The fibers of the source/target as well as the base of the Lie groupoid are all Hausdorff, it is only the way they are glued together which makes it non-Hausdorff.

So this does not answer your question at all, but I think that an answer should also take care of these kind of situation as this is really important in many areas of differential geometry (Lie groupoids are everywhere...).

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Is there a nice way to formulate "mild non-Hausdorfness"? Maybe again, in terms of the structure sheaf? – Daniel Moskovich Feb 11 '12 at 23:44
The arrow space is foliated with Hausdorff leaves and with Hausdorff leaf space, but I don't know how to say it with sheaves. – David Roberts Feb 12 '12 at 0:16

Why do you want manifolds to be second-countable Hausdorff? You exclude long lines and the line with double origin. You can develop most of the theory without these conditions. You won't have metrizability, though.

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I want to be able to do differential topology. In particular, I want approximation theorems which approximate smooth functions on a closed subset extended by continuous functions, by smooth functions. I want to spline diffeomorphisms together. I want to glue along boundaries, I want inverse and implicit function theorems. Maybe I even want Morse Theory. – Daniel Moskovich Feb 10 '12 at 0:45
Also, I'm not "married" to the textbook definition of a differentiable manifold, but if one were to define anything strictly more general, the usefulness of this extra generality (including extra objects) would need to be strongly justified. – Daniel Moskovich Feb 10 '12 at 1:13
It seems to me that hardly any of the theory works without these conditions. Can you really think of a nontrivial theorem about manifolds which uses neither partitions of unity nor a metric? – Paul Siegel Feb 10 '12 at 2:04
The answers to this question include some more examples of why it is important that your manifolds be Hausdorff: mathoverflow.net/questions/13072/… – Tom Church Feb 10 '12 at 4:04
I don't dispute that non-Hausdorff manifolds come up naturally - the quotient of any manifold by a bad group action comes to mind. I dispute that "you can develop most of the theory without these conditions". – Paul Siegel Feb 10 '12 at 12:38