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

Hi!

I apologize in advance if this question is better fit for http://math.stackexchange.com/.

Out of curiosity I'm interested in a particular case of the problem of what properties of a manifold is given by combinatorial information associated to the gluing of the manifold from pieces of $\mathbb{R}^n$.

Let $X$ be a manifold and $K$ be a covering of $X$ by subsets of $X$ homeomorphic to $\mathbb{R}^n$. We may produce a simplicial object $N_\cdot K$ called the Čech nerve of $K$. If the Hausdorff condition on $X$ is a constraint on the allowable gluings of $X$ from pieces of $\mathbb{R}^n$, we might expect to find this as a constraint on the set of nerves $N_\cdot K$ arising from manifolds (contra non-Hausdorff "premanifolds").

So my question becomes: Does the Hausdorff condition affect the set of possible nerves arising from such coverings? If so by what criterion? Is this criterion both necessary and sufficient? I would be interested in a characterization of such "separatedness" of simplicial objects.

Sincerely,

Eivind

share|improve this question
2  
The line with two origins can be covered by two subsets homeomorphic to $\mathbb{R}$ which intersect nontrivially. Similarly, the line can be covered by two subsets homeomorphic to $\mathbb{R}$ which intersect nontrivially, e.g. $(-\infty, 1)$ and $(0,\infty)$. Does this give a negative answer to your question? –  Tom Church May 22 '11 at 1:00
1  
@Tom : It would make more sense for you to require all intersections of your nerves to be contractible since then the nerve of the covering is weakly h.e. to your original space. This is possible for any Hausdorff manifold, but I'm fairly certain that you can't do it for the line with two origins. –  Andy Putman May 22 '11 at 2:53
    
@Andy, the nerve is always weakly h.e. to the original manifold since its fat geometric realization is. Stated another way, weak homotopy types of classifying spaces of Lie groupoids are invariant under Morita equivalence. –  David Carchedi May 22 '11 at 14:06
1  
@David : You need some condition on intersections of elements of your cover. Otherwise, you could cover the circle by two charts which intersect and get that the circle was weakly h.e. to the interval. –  Andy Putman May 22 '11 at 14:35
    
@Andy: I suppose you are right, if you are considering it as a simplicial set. I was considering it as a simplicial space. –  David Carchedi May 22 '11 at 15:12
add comment

2 Answers 2

up vote 3 down vote accepted

First a comment. You don't need the full Cech nerve of the cover. All the information in it is encoded in the Cech Lie groupoid. So the question boils down to: When does a Lie groupoid have a Hausdorff quotient?

Secondly, if $M_K$ denotes this Lie groupoid, it is etale, meaning all of its structure maps are local diffeomorphisms. If $M$ were a Hausdorff manifold, then in particular, it would be an orbifold, and since $M_K$ is Morita equivalent to $M,$ $M_K$ would be a proper Lie groupoid. Recall that a proper Lie groupoid $\mathcal{G}$ is one such that source and target map put together $$\left(s,t\right):\mathcal{G}_1 \to \mathcal{G}_0 \times \mathcal{G}_0$$ is a proper map, i.e. the pre-image of compact subsets are compact.

Conversely, if $M_K$ is a proper etale Lie groupoid, it is an orbifold groupoid and it follows that the quotient, which is diffeomorphic to $M,$ must be Hausdorff.

So, the quotient is Hausdorff if and only if the map $$\coprod U_\alpha \cap U_\beta \to \coprod U_\alpha \times \coprod U_\alpha$$ which sends a point $x$ in the intersection of $U_\alpha$ and $U_\beta$ to its copy in $U_\alpha$ and its copy in $U_\beta$, i.e. $$\left(x \in U_\alpha \cap U_\beta\right) \mapsto \left(x \in U_\alpha, x \in U_\beta\right),$$ is a proper map. If you would like, you can reinterpret this result in terms of the simplicial nerve.

share|improve this answer
    
This is actually a condition on the simplicial nerve when considered as a simplicial space, not as a simplicial set. I am not sure what to do without the added topological information. –  David Carchedi May 22 '11 at 15:13
    
I like this! Do you have a recommendation for literature about this kind of stuff? –  Eivind Dahl May 22 '11 at 16:41
1  
"Introduction to Foliation Theory and Lie Groupoids" by Moerdijk and Mrcun is a good place to start probably. –  David Carchedi May 22 '11 at 18:09
1  
Nice answer! Hence a Lie groupoid represents a Hausdorff manifold if it is proper, etale, and an equivalence relation ($X_1 \to X_0\times X_0$ is injective) –  David Roberts May 22 '11 at 18:49
add comment

I am leaving this as an answer since I am new here and therefore can't comment on answers. It's not an answer, since I am not saying anything about the nerve of the covering. Apologies to the regulars here!

Here's what I wanted to say: David Carchedi's answer to the question is rather nice. However, it can be stated in much simpler terms, as follows.

Notice that the condition on the map $\coprod U_{\alpha\beta}\to\coprod U_\alpha\times\coprod U_\beta$ to be proper is simply that it be closed, since it is injective and its domain is Hausdorff. Moreover, it's open, so it's a homeomorphism onto its image, so it is closed if and only if its range is closed. The range of the map is just the graph of the equivalence relation $R$ on $X':=\coprod U_\alpha$ such that $X'/R$ is the glued space $X$.

Thus the statement is that $X$ is Hausdorff if and only $R$ has a closed graph in $X'\times X'$. This is true for any equivalence relation $R$ such that the canonical projection $X'\to X'/R=X$ is open. You may find this in Chapter 1, § 8.3 of Bourbaki, General Topology, vol. 1. The type of equivalence relation $R$ coming from such gluings as considered always has this property. This is also in Bourbaki, somewhere in § 4. Anyway, these things are not difficult to check by hand.

share|improve this answer
add comment

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.