I wonder why one requires that the base manifold of a Lie groupoid is second-countable?

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    $\begingroup$ Isn't second-countability in the definition of manifold? en.wikipedia.org/wiki/Differentiable_manifold#Definition $\endgroup$ – Zev Chonoles May 15 '11 at 20:41
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    $\begingroup$ Usually manifolds tout court are required to be second countable. $\endgroup$ – Mariano Suárez-Álvarez May 15 '11 at 20:42
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    $\begingroup$ Sorry, I should state this question more carefully. Of course, Zev Chonoles and Mariano Suarez-Alvarez are right: the usual definition of a manifold requires second-countability and Hausdorff and locally euclidean. My question should merely be: At which point in the theory of Lie groupoids does one really need that the base manifold is second-countable? When constructing a Lie groupoid from a foliation one actually has to be a bit careful at this point. If one takes uncountably many charts the base manifold of the Lie groupoid won't be second-countable. $\endgroup$ – Dave Lewis May 15 '11 at 20:57
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    $\begingroup$ @Dave Lewis: Can I request that you edit your question to include your comments above? (Mark the edit as an edit, so that @Zev and @Mariano 's comments still make sense.) It sounds like you have a more specific direction that you're thinking about, and in any case clearly recognize that "When constructing a Lie groupoid from a foliation one actually has to be a bit careful at this point", for example. I do not know of a good reason to have questions on MO that are only one sentence long, and there are many good reasons for including a few paragraphs. $\endgroup$ – Theo Johnson-Freyd May 15 '11 at 22:50
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    $\begingroup$ @Zev, Mariano, and Dave: If you require manifolds to be second countable, then a disjoint union of manifolds is not always a manifold. Replacing second countability by paracompactness allows you to keep all good properties of second countable manifolds and makes the category of manifolds closed under coproducts, which seems like a good property to have. $\endgroup$ – Dmitri Pavlov May 16 '11 at 4:19

Answer #1:
There is no real reason for imposing that the base manifold of a groupoid be second countable.

Answer #2:
You lose some desirable properties if you don't impose second countability: For example, without it, the homotopy type of the geometric realisation of the nerve will no longer be an invariant of the Morita equivalence class of the groupoid.

  • $\begingroup$ Re Answer #2: Weird! I would have expected that the homotopy type of the nerve was a well-defined invariant for any topological groupoid, and that the construction should factor through forgetting from Manifolds to Homotopy Types. Could you either explain more, or include a reference? $\endgroup$ – Theo Johnson-Freyd May 15 '11 at 22:52
  • $\begingroup$ @Theo: I take my favourite non-second countable manifold: the long line $L$, and I look at the cover consisting of all of its bounded connected open subsets. The corresponding Cech groupoid is Morita equivalent to $L$. There is an obvious projection from the geometric realization of the Cech groupoid back to $L$. But there is no section of that map: that's because the cover does not admit partitions of unity. More generally, you can show that the projection does not admit a homotopy inverse. $\endgroup$ – André Henriques May 15 '11 at 23:12
  • $\begingroup$ Ah, Andre - that is why you redefine Morita equivalence not to use 'local sections', but 'local sections wrt a numerable cover'. This class of weak equivalences of topological/Lie groupoids (take your pick) is closer to what people think of when they restrict to paracompact spaces. $\endgroup$ – theHigherGeometer May 16 '11 at 0:35

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