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This question is prompted by another one.

I want to motivate the definition of a scheme for people who know about manifolds(smooth, or complex analytic). So I define a manifold in the following way.

Defn: A smooth $n$-manifold is a pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and $\mathcal{O}_X$ is a sheaf of rings on it, such that, every point $x \in X$ has a neighborhood $U_x$ which is homeomorphic to an open set $V_x$ in $\mathbb{R}^n$ and $\mathcal{O}_X$ restricted to $U_x$ is isomorphic to the sheaf of ring of smooth functions on $V_x$ and its open subsets.

This agrees with the usual definition using charts and atlases, for all except the requirement that a manifold is a (separable) Hausdorff space. But indeed it seems that many things in differential topology can be proven without using the Hausdorff property. In a fleeting conversation in a brief encounter, a personal I shall refer to as O., informed me that even Stokes' theorem can be done this way. But I am unable to ask O. again about this, as he is not physically around.

If the above is true, then this is a really good point to mention when introducing schemes to a person who knows about differentiable manifolds.

So my main question:

Is it true that the proof of Stokes' theorem for smooth manifolds can be proven without the Hausdorff condition on the manifold? If so, is it done in any well-known reference?

Aside question:

What are some crucial propositions/theorems in differential topology that use Hausdorff condition, except those involving imbedding in some $\mathbb{R}^m$ for high enough $m$, for which of course Hausdorffness is a necessary condition(together with separable property)?

Tom Church answers below that partitions of unity does not work, for instance on the example of the line with the double point. However I believe that one can still make sense of integration of differential forms even with such pathologies, because by introducing a measure for instance, we can ignore sets of measure $0$.

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@Yemon: the same "line with a double origin" example given by Hartshorne works as an example of a non Hausdorff manifold. –  Mariano Suárez-Alvarez Jan 26 '10 at 22:31
I feel like removing the Hausdorff restriction is a pretty useless way to generalize manifolds. Instead, suppose that every point has a nbhd homeomorphic to a hilbert space, or a banach space, or even a frechet space. These are actually contexts where we can prove useful theorems cf. Lang's book on manifolds. One motivation I've heard for scheme theory is that we'd rather have a good category of bad spaces than a bad category of good spaces. Dropping the assumption of hausdorffness doesn't make our spaces category any nicer. –  Harry Gindi Jan 27 '10 at 1:43
Remember that hausdorff spaces are nicer categorically than general topological spaces because they are algebraic with respect to monadicity. Dropping the assumption of Hausdorffness actually makes things worse rather than better. –  Harry Gindi Jan 27 '10 at 1:44
Non-Hausdorff manifolds arise quite naturally in certain contexts. For example, if X is a foliated manifold the leaf space of X (e.g. the quotient of X defined by the equivalence relation x ~ y if and only if x and y lie on a common leaf) need not be Hausdorff. But it is otherwise a manifold (by the definition of a foliation). I have a colleague who at one point described their main field of study as 1-dimensional (non-Hausdorff) manifolds (these arising as the leaf spaces of certain codimension 1 foliations of three-manifolds). –  Emerton Jan 27 '10 at 2:44
I should modify my comment to be less categorical, and rather say that certain foliated manifolds have leaf space given by a non-Hausdorff manifold. –  Emerton Jan 27 '10 at 13:32

2 Answers 2

up vote 21 down vote accepted

The existence of flows in the direction of a vector field seems to require Hausdorff; indeed, consider the vector field $\frac{\partial}{\partial x}$ on the line-with-two-origins. We have no global existence of a flow for any positive t, even if we make our space compact (that is, considering the circle-with-one-point-doubled). If the nonexistence of the flow is not visibly clear, consider instead the real line with the interval [0,1] doubled.

Also, partitions of unity do not exist; for example, in the line with two origins, take the open cover by "the line plus the first origin" and "the line plus the second origin". There is no partition of unity subordinate to this cover (the values at each origin would have to be 1).

For me, a basic example of the beauty of this function-theoretic approach is the definition of a vector field as a derivation $D\colon C^\infty(M)\to C^\infty(M)$. The proof that such a derivation defines a vector field hinges upon the fact that $Df$ near a point p only depends on $f$ near the point p. To prove this fact you use the fineness of your sheaf $\mathcal{O}_X$, i.e. the existence of partitions of unity. (It is true though that the failure of fineness in the non-Hausdorff case is of a different sort and might not break this particular theorem.) I feel that the existence of partitions of unity, and the implications thereof, is one of the basic fundamentals of approaching smooth manifolds through their functions; more importantly, a good handle on how partitions of unity are used is important to understand the differences that arise when the same approach is extended to more rigid functions (holomorphic, algebraic, etc.).

Now that the question has been edited to ask specifically about Stokes' theorem, let me say a bit more. Stokes' theorem will be false for non-Hausdorff manifolds, because you can (loosely speaking) quotient out by part of your manifold, and thus part of its homology, without killing all of it.

For the simplest example, consider dimension 1, where Stokes' theorem is the fundamental theorem of calculus. Let $X$ be the forked line, the 1-dimensional (non-Hausdorff) manifold which is the real line with the half-ray $[0,\infty)$ doubled. For nonnegative $x$, denote the two copies of $x$ by $x^\bullet$ and $x_\bullet$, and consider the submanifold $M$ consisting of $[-1,0) \cup [0^\bullet,1^\bullet] \cup [0_\bullet,1_\bullet]$. The boundary of $M$ consists of the three points $[-1]$ (with negative orientation), $[1^\bullet]$ (with positive orientation), and $[1_\bullet]$ (with positive orientation); to see this, just note that every other point is a manifold point.

Consider the real-valued function on $X$ given by "$f(x)=x$" (by which I mean $f(x^\bullet)=f(x_\bullet)=x$). Its differential is the 1-form which we would naturally call $dx$. Now consider $\int_M dx$; it seems clear that this integral is 3, but I don't actually need this. Stokes' theorem would say that

$\int_M dx=\int_M df = \int_{\partial M}f=f(1^\bullet)+f(1_\bullet)-f(-1)=1+1-(-1)=3$.

This is all fine so far, but now consider the function given by $g(x)=x+10$. Since $dg=dx$, we should have

$\int_M dx=\int_M dg=\int_{\partial M}g=g(1^\bullet)+g(1_\bullet)-g(-1)=11+11-9=13$. Contradiction.

It's possible to explain this by the nonexistence of flows (instead of $df$, consider the flux of the flow by $\nabla f$). But also note that Stokes' theorem, i.e. homology theory, is founded on a well-defined boundary operation. However, without the Hausdorff condition, open submanifolds do not have unique boundaries, as for example $[-1,0)$ inside $X$, and so we can't break up our manifolds into smaller pieces. We can pass to the Hausdorff-ization as Andrew suggests by identifying $0^\bullet$ with $0_\bullet$, but now we lose additivity. Recall that $M$ was the disjoint union of $A=[-1,0)$ and $B=[0^\bullet,1^\bullet] \cup [0_\bullet,1_\bullet]$. So in the quotient $\partial [A] = [0]-[-1]$ and $\partial [B] = [1^\bullet]-[0]+[1_\bullet]-[0]=[1^\bullet]+[1_\bullet]-2[0]$, which shows that $\partial [M]\neq \partial [A]+\partial [B]$. This is inconsistent with any sort of Stokes formalism.

Finally, I'd like to point out that Stokes' theorem aside, even rather nice non-Hausdorff manifolds can be significantly more complicated than we might want to deal with. One nice example is the leaf-space of the foliation of the punctured plane by the level sets of the function $f(x,y)=xy$. The leaf-space looks like the union of the lines $y=x$ and $y=-x$, except that the intersection has been blown up to four points, each of which is dense in this subset. In general, any finite graph can be modeled as a non-Hausdorff 1-manifold by blowing up the vertices, and in higher dimensions the situation is even more confusing. So for any introductory explanation, I would strongly recommend requiring Hausdorff until the students have a lot more intuition about manifolds.

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I am a little unhappy about "if we make our space compact": if you accept the algebro-geometric point of view, compactness includes the Hausdorff property, just like complete schemes have to be separated. –  t3suji Jan 27 '10 at 0:19
Compactness has not included Hausdorffness for ages, just as no one says today preschéma. –  Mariano Suárez-Alvarez Jan 27 '10 at 0:57
I think that the theorem relating derivations to vector fields still goes through. As I recall, all one needs is the existence of a single bump function around every point, and you still get this as it's true locally in R^n. –  Jason DeVito Jan 27 '10 at 1:19
Lack of partition of unity is a serious problem. However I suppose we can make sense of integration of differential forms still, by transforming the situation to that of integrating a measure, wherein we can ignore sets of measure 0. –  Anweshi Jan 27 '10 at 11:35
I have edited my question a little bit, in order to focus on Stokes' theorem. I hope you don't mind. –  Anweshi Jan 27 '10 at 11:38

If you are a little bit careful as to what you mean by a "non-Hausdorff manifold" then I believe that Stokes' theorem will go through.

The crucial issue is that the smooth functions cannot detect the fact that the manifold is not Hausdorff. Thus when you try to integrate a smooth function over the manifold, the function that you are integrating does not know that the manifold was not Hausdorff. Thus the fact that partitions of unity don't work doesn't matter because they do work "up to Hausdorffness".

Thus if "non-Hausdorff manifold" essentially means "becomes a manifold upon Hausdorffification" then you ought to be alright. I can't think of a counterexample off the top of my head, but that doesn't mean that one doesn't exist (i.e. an example of a "locally Euclidean space" that does not quotient down to a "Hausdorff locally Euclidean space" - ignore paracompactness for this, that's cheating).

A slightly tangential point is that if you have a "locally Euclidean space" that is not Hausdorff then you probably have the wrong topology on it. Take, again, the double pointed line. You probably think that the correct topology on this is the one with basis either $(a,b)$ or $(-a,a)\setminus \{0\}\cup \{\ast\}$. Wrong! By doing so, you are artificially imposing the condition that $0$ and $\ast$ can be distinguished without justifying that assertion. If you try by differential topological means (i.e. not topological, but differential topological) to separate $0$ and $\ast$ you find that you cannot tell the difference between them. So the correct topology has basis $(a,b)$ with $a$ and $b$ of the same sign, and $(-a,a) \cup \{\ast\}$. That is, the induced topology on the subset $\{0,\ast\}$ is the indiscrete topology.

At this point I should come clean (especially in the light of Emerton's comment to the original question) and say that when thinking of things that are like-but-not-manifolds then I think of Froelicher spaces. So a non-Hausdorff manifold really means a non-Hausdorff Froelicher space with nice local properties. And for Froelicher spaces, the difference between Hausdorff and non-Hausdorff is extremely small.

But if what you are interested in is smooth functions, then that's the right view to take. And if you are integrating, then you are using smooth functions. Of course, in other contexts you may want to remember more structure than just what the smooth functions can detect, but that's a different story.

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