Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
1  
Isn't second-countability in the definition of manifold? en.wikipedia.org/wiki/Differentiable_manifold#Definition –  Zev Chonoles May 15 '11 at 20:41
2  
Usually manifolds tout court are required to be second countable. –  Mariano Suárez-Alvarez May 15 '11 at 20:42
2  
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. –  Dave Lewis May 15 '11 at 20:57
2  
@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. –  Theo Johnson-Freyd May 15 '11 at 22:50
2  
@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. –  Dmitri Pavlov May 16 '11 at 4:19

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
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? –  Theo Johnson-Freyd May 15 '11 at 22:52
    
@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. –  André Henriques May 15 '11 at 23:12
    
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. –  David Roberts May 16 '11 at 0:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.