Why do we need Hausdorff-ness in definition of topological manifold?

Question

Suppose $M^n$ is a topological manifold, then $M^n$ locally looks like $\mathbb{R}^n$. $M^n$ is locally Hausdorff, since $\mathbb{R}^n$ is Hausdorff and Hausdorff-ness is a topological invariant.

All I want to understand is:

1) Does locally Hausdorff-ness imply Hausdorff-ness? (I can not imagine a locally Hausdorff topological space that is not globally Hausdorff)

2) Why do we need Hausdorff-ness in definition of the topological manifold? Is locally Hausdorff-ness not sufficient? If not, why?

Can anyone say anything that might be helpful?

Answer 1

There are non-Hausdorff spaces that are locally Euclidean; some people include them in the class of manifolds, and some prefer to exclude them by requiring a manifold to be Hausdorff. The line with two origins is a simple example of a non-Hausdorff manifold: it is locally Euclidean, since it has a base of open sets homeomorphic to $(0,1)$, but the two origins cannot be separated by disjoint open sets.

Answer 2

I know three main reasons we require manifolds to be Hausdorff (and 2nd countable):

1. Make classification of 1-dimensional manifolds possible. Without such classification, classifying (or even understanding) manifolds in higher dimensions is pretty hopeless.

2. One would like to be able to embed manifolds in some higher-dimensional Euclidean spaces. Again, it would be impossible without requiring both conditions.

3. Theory of manifolds did not come from nowhere, its origins are in analysis (primarily complex analysis) and differential geometry. Riemann first defined Riemann surfaces using multivalued complex-analytic functions, since he was interested in making sense of those. Later, about 100 years ago, Weyl gave the first rigorous definition of an abstract manifold, again in the context of Riemann surfaces. Since Riemann surfaces were the primary motivation, he required them to be both 2nd countable and Hausdorff. Then, I think, differential geometers jumped onto the bandwagon, since they realized that Weyl gave a definition which served them perfectly well. Much later, they realized that there are natural settings where one should weaken "Hausdorfness", such generalizations were primarily motivated by the theory of foliations (and, maybe quantum mechanics, but I am unsure about this).

Answer 3

One author who explicitly avoids assuming the Hausdorff property is Serge Lang, "Fundamentals of differential geometry", Springer 1999, 2001. In Chapters II and II, he does not assume Hausdorff. He introduces the condition at the beginning of Chapter IV. He says:

We see no reason to assume that $X$ is Hausdorff. If we wanted $X$ to be Hausdorff, we would have to place a separation condition on the covering. This plays no role in the formal development in Chapters II and III.

It's fairly clear from the literature that the reason for assuming Hausdorff is because it is always true for embedded and immersed manifolds, and that's what differential geometry was about in its first few decades. If you don't have this condition, you make possible not just the "line with two origins". You can also have two closed unit intervals $[0,1]$, for example, and a whole Hilbert spaces of unit intervals if you like, and you can have any number of "foliations" from individual lines going off to create vast complicated networks of lines. And then in $n$-dimensional spaces, you can get an astonishing variety of spaces with topologically closed regions containing "bubbles", and in space-time, you can get bubbles opening and closing in astonishingly complicated ways over time.

You can regard non-Hausdorff manifolds as either a huge opportunity to construct physical models which have mixed states in the quantum sense, or you can regard it as a nightmare to be avoided if all you want to do is geometry of manifolds embedded/immersed in Euclidean spaces.

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PS. 2015-8-23.
Just one more little point about this issue.
It's fairly obvious that you can't metrize a non-Hausdorff manifold. What's not quite so obvious is that it's difficult to even pseudometrize it. I was thinking that you can put a pseudometric on the "real line with two origins" by letting the distance between the two origins equal zero. However, there is a very easy theorem which says that any pseudometric (i.e. a metric which does not require equality of two points if their distance equals zero) must be a metric if the topology is $T_0$. But every locally Cartesian space is $T_1$, which implies $T_0$. Therefore every non-Hausdorff locally Cartesian space is non-metrizable and also non-pseudometrizable. (I'm assuming that the set contains more than one point!)

This makes these spaces somewhat useless for Riemannian geometry. You can put differentiable structures on them, and maybe you can put affine connections on them, but the familiar Riemannian metric can't work. (Any affine connection would not be a metric connection for any metric.) So there's another reason to reject non-Hausdorff manifolds.

Answer 4

I'll add to the answers already given by giving yet another reason for Hausdorfficity (and paracompactness, incidentally).

A topological space is a topological manifold if and only if it is locally Euclidean and admits partitions of unity subordinate to any open cover.

A proof of this is added at the end of this answer.

The use of these partitions of unity is to pass from local geometry to global geometry. For example, to get a Riemannian metric, you take an open cover consisting of Euclidean spaces, you put metrics on each of these (this is easy, they're Euclidean), and then glue them together using a partition of unity subordinate to this cover. You'll rarely see the Hausdorffness or paracompactness invoked directly, but partitions of unity pop up in a whole lot of proofs.

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The alternative definition of manifolds follows immediately from the following topological lemma (I take "locally $T_1$" to mean that every point has a $T_1$ open neighborhood).

Let $X$ be a topological space. The following are equivalent:

$\bullet$ $X$ is locally $T_1$ and has partitions of unity subordinate to any open cover;

$\bullet$ $X$ is $T_2$ and paracompact.

The implication from the second bullet to the first can be found in Lee's book on topological manifolds. To go from the first to the second, one uses that partitions of unity imply paracompactness (this is exercise 4-33 in Lee's book, which I will not spoil, partly because professor Lee doesn't like it when you do that, partly because I haven't done the exercise). To get global $T_1$-ness, take an open cover $\{U_i\}_{i \in I}$ of $X$ consisting of $T_1$ sets, and let $\{\rho_i\}_{i \in I}$ be a partition of unity subordinate to this cover. Let $x \in X$, and $I_1 = \{i \in I \mid x \in \text{Supp}(\rho_i)\}$. Then $\bigcap_{i \in I_1} \text{Supp}(\rho_i)$ is a closed $T_1$ neighborhood of $x$, proving that $x$ is closed in $X$.

Now if $x$ and $y$ are distinct points of $X$, $\{X \setminus x, Y \setminus y\}$ is an open cover of $X$. Using the existence of partitions of unity we find $\rho{:}X \rightarrow [0,1]$ such that $\rho(x) = 1$ and $\rho(y) = 0$. The open sets $\rho^{-1}([0,{1}/{2}[)$ and $\rho^{-1}(]{1}/{2},1])$ now separate $x$ and $y$.

Answer 5

I think the main reason behind the hausdorffness condition in the definition of topological manifold is that one needs to be able to do calculus on manifolds. More to the point, one needs to deal with convergent sequences on manifolds and here comes the necessity of including the hausdorffness condition.

Answer 6

We know that the limit of sequences in a Hassdorf space is unique. This is one of the most important reasons why we assume such condition. For example Lie derivative or other derivations are defined by a limit.