When I first starting studying differential geometry, I asked my lecturer a question about smooth manifolds that didn't admit a partition of unity. He promptly told not to worry about such objects as they were only studied by the extremely eccentric. I would like to know if this is true, ie, does anyone study manifolds that don't admit a partition of unity (not whether such people are eccentric).
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2$\begingroup$ If your manifolds are second countable, Hausdorff and locally Euclidian, then they are paracompact, and hence any open cover admits a partition of unity subordinate to it, at least in the smooth category. In the analytic category, you will not have partitions of unity, but that is because there are no bump functions. $\endgroup$– Charlie FrohmanCommented Mar 24, 2010 at 18:12
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$\begingroup$ I mean manifolds that aren't second countable but are locally euclidean and Hausdorff .... do we need to specify Hausdorff here, surely it should from the manifold being locally euclidean. $\endgroup$– Aston SmytheCommented Mar 24, 2010 at 18:29
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3$\begingroup$ Yes, you need to specify Hausdorff. Consider the real line with an extra zero grafted onto it. It's locally Euclidean, but not Hausdorff. $\endgroup$– Harald Hanche-OlsenCommented Mar 24, 2010 at 18:33
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3$\begingroup$ Hausdorff does not follow from any local condition. Consider the "line with a double point": two copies of $\mathbb{R}$ glued to each other along $\mathbb{R} \setminus \{ 0 \}$. This is locally as nice as you could want in any sense. $\endgroup$– David E SpeyerCommented Mar 24, 2010 at 18:34
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1$\begingroup$ More generally, the etale space of any sheaf over a manifold is always locally Euclidean but rarely Hausdorff. $\endgroup$– David CarchediCommented Mar 25, 2010 at 0:17
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2 Answers
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The answer to your stated question ("Does anyone study non-paracompact manifolds?") is certainly yes. Here are a few papers which do just this:
Gauld, David. Manifolds at and beyond the limit of metrisability. (English summary) Proceedings of the Kirbyfest (Berkeley, CA, 1998), 125--133 (electronic), Geom. Topol. Monogr., 2, Geom. Topol. Publ., Coventry, 1999.
Among other things, Gauld references that there are two paracompact and two nonparacompact 1-manifolds, and $\aleph_0$ paracompact and $2^{\aleph_1}$ non-paracompact 2-manifolds. (That's a lot!)
Foliations and foliated vector bundles, 1969 MIT lecture notes
http://www.foliations.org/surveys/FoliationLectNotes_Milnor.pdf
Milnor entertains non-paracompact manifolds. In particular he constructs a (necessarily non-paracompact) surface with uncountable fundamental group. Milnor also says: "The main object of this exercise is to imbue the reader with suitable respect for non-paracompact manifolds."
Balogh, Zoltan; Gruenhage, Gary. Two more perfectly normal non-metrizable manifolds. (English summary) Topology Appl. 151 (2005), no. 1-3, 260--272.
The existence of perfectly normal, non-metrizable (hence non-paracompact) manifolds is shown to depend upon one's set-theoretic assumptions.
And so forth. I could find 10 more papers without much effort. I'm not sure I could find 100. (A MathSciNet search with "manifold" and ("nonmetrizable" or "non-paracompact") in the anywhere fields doesn't return many hits.) So some serious mathematicians take non-paracompact manifolds seriously enough to write some papers about them. On the other hand, although one could use more complimentary language than "extremely eccentric", your lecturer's take on non-paracompact manifolds seems to be an accurate reflection of how most geometric topologists feel: they seem mostly to be used as a source of counterexamples and to be of interest to general and set-theoretic topologists.
answered Mar 24, 2010 at 21:35
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$\begingroup$ Wow! These are some cool results. I wonder what the other non-paracompact 1-manifold is? $\endgroup$ Commented Mar 24, 2010 at 22:36
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$\begingroup$ @Chris: don't get too excited: assuming that the long line is "long on both sides", the other one is the long ray. :) $\endgroup$ Commented Mar 24, 2010 at 23:01
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1$\begingroup$ Ahh, I see. It is open on both ends, but only one end is long... $\endgroup$ Commented Mar 24, 2010 at 23:39
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In my experience the answer is not really.
My favorite example of a Hausdorff manifold which is not paracompact (which incidentally is also my all-time favorite counter example to most point-set topology questions) is the Long Line.
Since it is not paracompact, it does not admit partitions of unity. All of it's homotopy groups vanish, yet it is not contractible.
If I remember correctly, its tangent bundle is non-trivial.
Even if people don't study these, then it is still important to know about the type of things which don't satisfy the usual hypotheses. You never know when you might run into one or need such a counter example.
answered Mar 24, 2010 at 18:43