Stable graphs: Feynman diagrams and Deligne-Mumford space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:55:45Z http://mathoverflow.net/feeds/question/22291 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22291/stable-graphs-feynman-diagrams-and-deligne-mumford-space Stable graphs: Feynman diagrams and Deligne-Mumford space Kevin Lin 2010-04-23T02:51:23Z 2010-04-26T07:53:52Z <p>I do not know very much about quantum field theory, but I have seen, in my reading, that stable graphs can appear in QFT in the form of, I think, Feynman diagrams. By stable graph I mean a "graph with tails", whose vertices are labelled by nonnegative integers, and such that each vertex with labeling 0 has valence at least 3, and each vertex with labeling 1 has valence at least 1.</p> <p>Algebraic geometers of course know that stable graphs also give a stratification of the Deligne-Mumford spaces $\overline{M}_{g,n}$: Vertices with label $g$ correspond to genus $g$ curves; edges correspond to nodes; tails correspond to marked points. Valency conditions correspond to finitude of automorphism group of the nodal curve.</p> <p>Is there an explanation for this coincidence?</p> <p>I guess there is probably some kind of explanation via Gromov-Witten theory. But I get the impression that stable graphs show up in QFTs more generally, and beyond Gromov-Witten theory. Do they? If so, how? And where?</p> http://mathoverflow.net/questions/22291/stable-graphs-feynman-diagrams-and-deligne-mumford-space/22310#22310 Answer by Kelly Davis for Stable graphs: Feynman diagrams and Deligne-Mumford space Kelly Davis 2010-04-23T07:08:17Z 2010-04-24T09:35:33Z <p>From the sound of it, you are reading Costello's book.</p> <p>In point particle QFT the stable graphs you are referring to can be thought of as the paths of particles through spacetime the vertices are where the particles interact.</p> <p>Now in the Deligne-Mumford case you can think of taking the stable graph and "thickening" it and replacing the particles with little loops of string to obtain a Riemann surface. The interactions then can be thought of as corresponding to the joining and splitting of these little loops of string.</p> <p>It happens to be the case that in bosonic string theory one is in this second "thickened" case and the symmetries of the theory are such that in doing the path integral one ends up integrating over the Deligne-Mumford spaces of such Riemann surfaces. Thus, the Deligne-Mumford stratification is of use there. Furthermore, you can relate Deligne-Mumford spaces of such Riemann surfaces to point particle QFT by simply taking the limit of infinite string tension.</p> http://mathoverflow.net/questions/22291/stable-graphs-feynman-diagrams-and-deligne-mumford-space/22348#22348 Answer by userN for Stable graphs: Feynman diagrams and Deligne-Mumford space userN 2010-04-23T13:38:19Z 2010-04-23T13:38:19Z <p>To amplify Theo's comment slightly: The graphs that show up in Feynman diagram perturbation theory are stable because the physicists use a different accounting system for the genus zero vertices with 0,1, or 2 edges. The diagrams with these graphs don't show up in the perturbation series because the physical effects they represent are the situation one is perturbing away from. A genus zero graph with two edges is an order $\mathcal{O}(\hbar^0)$ correction to the propagator. Likewise, one edge gives a <a href="http://en.wikipedia.org/wiki/Tadpole_%2528physics%2529" rel="nofollow">tadpole</a> correction to the expectation value of the field (usually gotten rid of by redefining the field), and zero edges a correction to the vacuum energy (usually set to zero by convention).</p> http://mathoverflow.net/questions/22291/stable-graphs-feynman-diagrams-and-deligne-mumford-space/22513#22513 Answer by Dylan Thurston for Stable graphs: Feynman diagrams and Deligne-Mumford space Dylan Thurston 2010-04-25T15:56:59Z 2010-04-26T07:53:52Z <p>I'm not sure exactly what the question is, but let me comment that lots of Feynman graphs with lots of different rules for labelling the vertices come up in QFT, as explained by Theo. From the QFT point of view, the labelling you're describing is not particularly common; you have infinitely many terms in your Lagrangian, for instance. So I would say that the answer is no, stable graphs do not come up much beyond the cases that are clearly related to Gromov-Witten theory or other string theories.</p>