Timeline for Laplacians on graphs vs. Laplacians on Riemannian manifolds: $\lambda_2$?
Current License: CC BY-SA 3.0
16 events
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| Aug 7, 2012 at 8:31 | comment | added | Camilo Sarmiento | Juergen Jost and coworkers have been looking at this relation for some time. See for instance arxiv.org/pdf/0910.3118 , where they present some "dual" Cheeger bound for their last Laplacian eigenvalue (last since they work with some normalized version). The "primal" bound relates the Cheeger constant to the "first" Laplacian eigenvalue, and from the expression for the Cheeger constant one can read easily the "connection" to connectedness both in the discrete and continuous cases. See also math.ucsd.edu/~fan/wp/cheeger.pdf | |
| Aug 6, 2012 at 23:58 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 110 characters in body
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| Aug 6, 2012 at 20:04 | comment | added | Richard Montgomery |
It is apparently an open (and fun) question to compute explicitly the fundamental solution to the Laplacian on the integer lattice. Google lattice Green's functions' to get some of the literature. Maybe it is not even possible. Kleiner recently began a new branch of research by reproving Gromov's main theorem from Groups of Polynomial Growth' by using harmonic maps on the Cayley graphs of these groups.
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| Apr 2, 2012 at 21:47 | answer | added | Highwind | timeline score: 1 | |
| Jan 23, 2012 at 0:25 | comment | added | Joseph O'Rourke | Thanks Asaf, Valerio, and Paul--All very useful comments! | |
| Jan 22, 2012 at 23:42 | comment | added | Paul Siegel | This probably won't answer your question directly, but I got a lot of useful references in the responses to this question: mathoverflow.net/questions/66892/… | |
| Jan 22, 2012 at 23:09 | answer | added | Alain Valette | timeline score: 7 | |
| Jan 22, 2012 at 13:23 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Removed "connected."
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| Jan 22, 2012 at 13:12 | comment | added | Valerio Capraro | For instance, if the manifold is not connected, then the graph on the any triangulation $T_n$ is not connected and maybe you can get the analogue that you are looking for. | |
| Jan 22, 2012 at 13:10 | comment | added | Valerio Capraro | This is a general comment about the relation between the laplacian graph and the one on Riemannian manifold. Suppose $M$ compact and let $T$ be a triangulation of $M$. Given $n\in\mathbb N$ denote by $T_n$ the triangulation of $M$ obtained by $T$ doing $n$-times a barycentric subdivision of $T$. Denote by $\Delta_n$ the discrete Laplacian of the natural connected graph constructed on the 1-skeleton of $T_n$. It is possible that $\Delta_n$ converges in some suitable sense to $\Delta$ and this should give some transference principle. | |
| Jan 22, 2012 at 12:43 | answer | added | Liviu Nicolaescu | timeline score: 12 | |
| Jan 22, 2012 at 7:45 | comment | added | Asaf | If the manifold has finitely many connected components, then L^2(M) decomposes as direct sum over the L^2 spaces of the separate connected components. Every connected component contribute the constant space $$\mathbb{C}\cdot 1$$ to be an eigenvector of eigenvalue $0$, hence the multiplicity of the the eigenvalue $0$ equals to the number of connected components. The connection between graphs and manifolds is more explicit by approximating the manifold with its 1 skeleton structure see in the book by Lubotzky - ma.huji.ac.il/~alexlub/BOOKS/On%20property/On%20property.pdf , p.34 | |
| Jan 22, 2012 at 3:12 | comment | added | Qiaochu Yuan | (Above I need to assume that the graph has finitely many connected components.) | |
| Jan 22, 2012 at 2:49 | comment | added | Joseph O'Rourke | @Qiaochu: Well, maybe that is precisely what I am missing, I have overdefined $M$... | |
| Jan 22, 2012 at 2:15 | comment | added | Qiaochu Yuan | I don't understand why you require that $M$ is connected. Isn't it true in both cases that the space of harmonic functions is precisely the space of locally constant functions, so its dimension is the number of connected components? | |
| Jan 22, 2012 at 2:01 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |