Gluing of manifolds and the Hausdorff condition. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:41:22Z http://mathoverflow.net/feeds/question/65684 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65684/gluing-of-manifolds-and-the-hausdorff-condition Gluing of manifolds and the Hausdorff condition. Eivind Dahl 2011-05-21T22:55:41Z 2011-08-18T13:23:11Z <p>Hi!</p> <p>I apologize in advance if this question is better fit for <a href="http://math.stackexchange.com/" rel="nofollow">http://math.stackexchange.com/</a>.</p> <p>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 $\mathbb{R}^n$.</p> <p>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").</p> <p>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.</p> <p>Sincerely,</p> <p>Eivind</p> http://mathoverflow.net/questions/65684/gluing-of-manifolds-and-the-hausdorff-condition/65713#65713 Answer by David Carchedi for Gluing of manifolds and the Hausdorff condition. David Carchedi 2011-05-22T14:21:04Z 2011-05-22T14:21:04Z <p>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?</p> <p>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.</p> <p>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.</p> <p>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. <code>$$\left(x \in U_\alpha \cap U_\beta\right) \mapsto \left(x \in U_\alpha, x \in U_\beta\right),$$</code> is a proper map. If you would like, you can reinterpret this result in terms of the simplicial nerve.</p> http://mathoverflow.net/questions/65684/gluing-of-manifolds-and-the-hausdorff-condition/73147#73147 Answer by Alexander Alldridge for Gluing of manifolds and the Hausdorff condition. Alexander Alldridge 2011-08-18T13:23:11Z 2011-08-18T13:23:11Z <p>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!</p> <p>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. </p> <p>Notice that the condition on the map <code>$\coprod U_{\alpha\beta}\to\coprod U_\alpha\times\coprod U_\beta$</code> 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 <code>$X':=\coprod U_\alpha$</code> such that $X'/R$ is the glued space $X$. </p> <p>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.</p>