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I was reading David Carchedi's answer for a question on Grothendieck topology for a non-small category. It "reads" like people "choose" if they allow manifolds to be Hausdorff and/or second countable. When I came across the notion of smooth manifolds for the first time, by definition, smooth manifolds are Hausdorff and second countable.

When (why) did we allow manifolds to be non-Hausdorff and/or non-second countable?

I have observed this when I am reading about stacks.

Is it because of the constructions when we do in the set up of manifolds (stacks) which resulted in spaces that are "same as" manifolds but are not Hausdorff or not second countable?

Is it because of the influence of algebraic geometry?

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    $\begingroup$ Is it because of the influence of algebraic geometry (where the very first topological space (Zariski topology) we come across is non-Hausdorff)? $\endgroup$ – Praphulla Koushik Jun 4 at 5:57
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    $\begingroup$ The Zariski topology on the maximal spectrum of a C^∞-ring is Hausdorff, so this example is not relevant. $\endgroup$ – Dmitri Pavlov Jun 4 at 7:37
  • $\begingroup$ @DmitriPavlov oh, I did not knew (do not know yet) about C^\infty rings.. $\endgroup$ – Praphulla Koushik Jun 4 at 8:35
  • $\begingroup$ @DmitriPavlov It is easy to blame anything on Algebraic geometry :P It has all kind of surprising properties/set up/constructions.. :D $\endgroup$ – Praphulla Koushik Jun 4 at 12:23
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    $\begingroup$ One elementary reason for contemplating non-second countable manifolds is that it allows you to view any set, not just countable ones, as a discrete manifold. If one replaces second countable with paracompact, any connected component of a manifold is still second countable. $\endgroup$ – Adrian Clough Jun 5 at 7:14
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The étale space construction produces non-Hausdorff and nonparacompact spaces (e.g., smooth manifolds) in many practical examples that have nothing to do with algebraic geometry.

The étale space is often non-Hausdorff because two germs can coincide on some nontrivial part of the domain without being equal. For example, germs of continuous real-valued functions around 0 can be equal on (−ε,0), but different on (0,ε) for arbitrarily small ε>0.

Nonparacompactness arises for the same reason. For connected manifolds, paracompactness coincides with second countability. The étale space, even if connected, is often not second countable because there are uncountably many disjoint open subsets, e.g., germs of constant real-valued functions.

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  • $\begingroup$ By etale space construction do you mean the one from sheaves on a topological space $X$? This is the only etale space construction I know. $\endgroup$ – Praphulla Koushik Jun 4 at 7:14
  • $\begingroup$ "non-Hausdorff and nonparacompact spaces (e.g., smooth manifolds)" so, for you smooth manifolds are not necessarily Hausdorff and paracompact?? $\endgroup$ – Praphulla Koushik Jun 4 at 7:15
  • $\begingroup$ @PraphullaKoushik: The étale space construction works for sheaves of sets on any site. In particular, for sheaves of sets on manifolds it produces a (typically) non-Hausdorff non-paracompact manifold. $\endgroup$ – Dmitri Pavlov Jun 4 at 7:29
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    $\begingroup$ @PraphullaKoushik: No, and you cannot make such inferences from such a use of language. Just like when somebody says “nonunital ring”, it does not mean that rings are not assumed to be unital. See ncatlab.org/nlab/show/red+herring+principle. $\endgroup$ – Dmitri Pavlov Jun 4 at 7:31
  • $\begingroup$ :) I understand what you mean.. Yes, Yes, sheaves of sets on a site.. can you suggest me to look at some paper which (one of which) started using this possibility of non-Hausdorff non-paracompact manifolds.. $\endgroup$ – Praphulla Koushik Jun 4 at 8:26
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Hmm, I don't know anything about stacks and all that, but below is a situation where one cares about non-Hausdorff 1-manifolds.

If one considers a codimension-1 foliation of a manifold, a very useful thing to do in dimension 3, it is often fruitful to study the leaf space of the universal cover of this manifold. That is to lift the foliation to the universal cover and examine the quotient space you get by collapsing every leaf to a point.

The generic case of this space is a non-Hausdorff, second countable, locally Euclidean topological space. The leaf space is homeomorphic to $\mathbb{R}$ if and only if it is Hausdorff. When the leaf space is $\mathbb{R}$ the foliation is called $\mathbb{R}$-covered.

The leaf space contains a remarkable amount of information about the manifold, the original foliation, and its fundamental group. This comes by way of studying how the fundamental group naturally acts on the leaf space, which is by deck transformations before quotienting. Non-Hausdorff 1-manifolds are in the literature often referred to as $\mathbb{R}$-trees, and perhaps less commonly as order trees.

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  • $\begingroup$ Thank you.. I heard many times that one would come across non-Hausdorff spaces when studying foliation groupoids (manifolds).. I just need some time to respond.. $\endgroup$ – Praphulla Koushik Jun 4 at 9:57
  • $\begingroup$ @PraphullaKoushik Not only the leaf space but some times the groupoid associated to foliation is sometimes non Haussdorff. But this can not occured in the real analyytic case. $\endgroup$ – Ali Taghavi Jun 4 at 21:56
  • $\begingroup$ Search for example Ann. Maria Torpe: K theory for foliation with Reeb components $\endgroup$ – Ali Taghavi Jun 4 at 22:02
  • $\begingroup$ Right, I meant to say something about $C^*$ algebras, K-theory, and foliations, but I think it's better that I forgot, as I'm not really qualified to address that aspect of the theory. $\endgroup$ – guest Jun 4 at 22:12
  • $\begingroup$ @guest but non Haussdorfness of leaf space is not a surprising fact however the graph of a foliation and non Haussdorfness of it is more interesting. In 2 master thesis I tried to understand these objects with collaboration of my students. The graph of a foliation is also an interesting paper. $\endgroup$ – Ali Taghavi Jun 4 at 22:19
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Non-Hausdorffness shows up in several contexts when dealing with Lie groupoids: the integration (Lie's 3rd Theorem) for Lie algebroids to Lie groupoids will typically produce a non-Hausdorff one, if it works at all. So there seems to be a large part of oid-geometry involved with non-Hausdorff manifolds.

However, to abandon second countability is a more serious step, at least in my eyes. It is not so much the existence of partition of unities (which requires paracompactness) but the second countability itself which is extremely useful. Consider the following statement:

A bijective immersion is a diffeomorphism

This is a theorem in differential geometry which you definitely want to be true and it relies directly (well, a bit hidden) on second countability. To see that it fails right on the nose if you drop second countability, consider a manifold $M$ in the usual sense of positive dimension and $M_{\mathrm{discrete}}$ as a zero-dimensional manifold with uncountably many connected components, each of which is paracompact (it's a point...) Then $\mathrm{id}\colon M_{\mathrm{discrete}} \longrightarrow M$...

Now why is this theorem important: Lie theory depends strongly on it. Any group would be a zero dimensional Lie group otherwise. In particular, the manifold structure of a Lie group would not be determined by the group structure. This would also imply that a transitive smooth group action of a Lie group on a manifold is not the same thing as a homogeneous space $G/H$ and many more problems...

So before asking why one should abandon a property, it might be good to understand what it is good for. For Hausdorffness the situation seems to be very different to second countability.

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  • $\begingroup$ Thanks for the answer.. I wanted to read about integration of Lie algebroids.. "before asking why one should abandon a property, it might be good to understand what it is good for" +1 $\endgroup$ – Praphulla Koushik Jun 4 at 10:37
  • $\begingroup$ "In particular, the manifold structure of a Lie group would not be determined by the group structure.": The manifold structure is never determined by the group structure, unless the group is countable. For instance, the discrete group R of real numbers can be equipped with uncountably many manifold structures that turn it into a Lie group because there are uncountably many automorphisms of R. $\endgroup$ – Dmitri Pavlov Jun 4 at 17:07
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If you're asking a historical question, it is probably because of Whitney, who set out a clear account of differentiable manifolds and proved the embedding theorem, that the Hausdorff and second countability assumptions have been regarded as "standard." Nevertheless, violations of these assumptions had been contemplated prior to Whitney's work, notably the non-second-countable Prüfer surface, which dates back to the 1920s.

In addition to the examples mentioned by others, note that people studying general relativity have sometimes considered non-Hausdorff spacetime manifolds (see here for a non-paywalled version). The consensus, however, seems to be that these are mathematical curiosities that are not physically significant. Luc and Placek amusingly quote Penrose as saying, "I must … return firmly to sanity by repeating three times: spacetime is a Hausdorff differentiable manifold; spacetime is a Hausdorff …"

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    $\begingroup$ Names of references, in case of linkrot: Prüfer surface -> Gabard - A separable manifold failing to have the homotopy type of a CW-complex; non-Hausdorff spacetime manifolds -> Luc and Placek - Interpreting non-Hausdorff (generalized) manifolds in general relativity. $\endgroup$ – LSpice Jun 4 at 18:20
  • $\begingroup$ I once posted this question (but not on MathOverflow). It was pointed out to me that non-Hausdorff manifolds arise in the study of dynamical systems. Unfortunately, I know nothing beyond that. $\endgroup$ – Deane Yang Jun 5 at 0:23
  • $\begingroup$ I am not sure if I am reading it correctly.. You are saying because of Whitehead’s theorem (which required space to be second countable and Hausdorff), people start taking Hausdorff and second countable as “standard” assumptions in definition of a manifold.. Before that, manifold was not assumed to be Hausdorff and second countable? $\endgroup$ – Praphulla Koushik Jun 5 at 2:29
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    $\begingroup$ @PraphullaKoushik: It's Whitney's theorem, not Whitehead's. And there is no “before”: the 1931 paper by Veblen and Whitehead defined abstract smooth manifolds for the first time, and Whitney's paper followed shortly. Veblen and Whitehead assume smooth manifolds to be Hausdorff, but not necessarily paracompact. $\endgroup$ – Dmitri Pavlov Jun 5 at 3:36
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    $\begingroup$ @PraphullaKoushik : Often, mathematicians study important examples, then gradually sense that it would be useful to make an abstract definition. There may be several "false starts" as people fumble around for the "right" definition. The "right" definition is often the one that yields a good theorem. People certainly had some vague idea of what a manifold was, or ought to be, before the 1930s. But then Whitney came along and laid out everything cleanly and proved a wonderful theorem. So it was natural to think of an abstract manifold as "the thing that Whitney's theorem was about." $\endgroup$ – Timothy Chow Jun 5 at 4:06

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