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.

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.

Is there an explanation for this coincidence?

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?

  • 3
    $\begingroup$ One can "explain" why such graphs show up in path-integral QFT in the following way: they exactly parametrize the $\hbar \to 0$ asymptotics of finite-dimensional integrals of the form $\int\exp\bigl((i\hbar)^{-1}f(x)\bigr)dx$, where $f(x)\in C^\infty(\mathbb R^n)[[\hbar]]$. (More generally, the asymptotics only depend on the Taylor coefficients of $f$ near critical points of $[f]\in C^\infty(\mathbb R^n)=C^\infty(\mathbb R^n)[[\hbar]]/\hbar$.) But this isn't the kind of explanation you are looking for, since it does not explain the coincidence with geometry. And I don't know any geometry. $\endgroup$ – Theo Johnson-Freyd Apr 23 '10 at 5:34

From the sound of it, you are reading Costello's book.

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.

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.

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.

  • $\begingroup$ Yep, I'm reading Costello's book. Yes, I am familiar with the string theory/Gromov-Witten theory story you describe. But I am more wondering about QFTs outside of string theory/GW theory. $\endgroup$ – Kevin H. Lin Apr 23 '10 at 8:39
  • 1
    $\begingroup$ I don't mean to be contrary, but I am not sure you understand the string theory case very well. One can think of point particle theory as string theory in the limit of infinite string tension. This fact, along with my comment above, should give the relation you seek. $\endgroup$ – Kelly Davis Apr 23 '10 at 17:42
  • 1
    $\begingroup$ Kelly, IIRC, not all QFTs are obtainable as limits of string theories. However, any perturbative expansion in $\hbar$ can be organized using stable graphs. (This is a quibble, of course. The string case did inspire the QFT construction.) $\endgroup$ – userN Apr 26 '10 at 18:17
  • $\begingroup$ @A.J. I did not say all point-particle QFT are obtainable as limits of string theories, and hope I didn't imply such is the case. Also, yes, any perturbative expansion in $\hbar$ can be organized using stable graphs. However, the problem with saying "any perturbative...stable graphs" and only this to a mathematician is that they then have no idea why this is the case, the fountainhead of this question. The example I gave from string theory, I think we both agree, motivates why stable graphs play a role. $\endgroup$ – Kelly Davis Apr 28 '10 at 20:24

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.

  • $\begingroup$ From a physicist point-of-view this labeling is non-standard. However, for the context in which this appears, Kevin Costello's book on perturbative renormalization for mathematicians, it makes sense as a way of easing mathematicians in to Feynman digrams. $\endgroup$ – Kelly Davis Apr 25 '10 at 19:23
  • $\begingroup$ I don't think we're disagreeing with each other... $\endgroup$ – Dylan Thurston Apr 25 '10 at 23:20
  • $\begingroup$ What Grassmannian? $\endgroup$ – Kevin H. Lin Apr 26 '10 at 0:00
  • $\begingroup$ @Dylan I agree that we agree :-) $\endgroup$ – Kelly Davis Apr 26 '10 at 6:28
  • $\begingroup$ @Kevin: Sorry, typo for "Lagrangian". Fixed now. $\endgroup$ – Dylan Thurston Apr 26 '10 at 7:56

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 tadpole 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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