MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For 2D (topological) conformal field theory the corresponding moduli space is the space of Riemann surface with boundary, right? What does it mean by extending the theory to Deligne-Mumford space? How can we do that in terms of the methods we have by now? Thanks for any expository answers!

share|cite|improve this question

Here is my rough understanding. In a TCFT we have chains on moduli spaces $\widetilde{M}$ of Riemann surfaces with parameterized boundary acting (in the sense of operads, PROPs, whatever) on a chain complex $V$. The action should be compatible with the various gluing maps.*

(For simplicity, let's just deal with closed TCFTs, and not open or open-closed TCFTs.)

I think that to extend to DM space means something like the following:

First, we would like to have an action of moduli spaces $M$ of Riemann surfaces with unparameterized boundary -- which is the same up to homotopy as moduli spaces of Riemann surfaces with marked points and no boundary** -- which is compatible with the action of $\widetilde{M}$. Compatible here means, I think, compatible with the map $\widetilde{M} \to M$ which forgets the parameterization. Moreover we want the action of chains on $M$ to be compatible with the various gluing maps, which are well-defined up to homotopy.***

Finally, we would like to extend this action from chains on $M$ to chains on $\overline{M}$, the Deligne-Mumford spaces of stable Riemann surfaces with marked points and no boundary. The actions should be compatible with $M \hookrightarrow \overline{M}$. Moreover the action of $\overline{M}$ should be compatible with the various gluing maps.****

See pages 62 and 63 of Kontsevich-Soibelman for some details. The statement is that, if the Calabi-Yau category corresponding to your (open) TCFT satisfies the "degeneration conjecture" (an analogue of Hodge-de Rham degeneration), then there exists an extension to DM space. Recall that Costello proved that TCFTs correspond to Calabi-Yau categories.

Also take a look at the first two sections of this paper of Teleman.

For an alternative approach to these sorts of questions, see this paper of Costello.

* The gluing maps here are the obvious gluing maps.

** The homotopy equivalence induces a quasi-isomorphism of the chain complexes. We're working with chain complexes everywhere, and it's OK to replace things with quasi-isomorphic things.

*** I think that we can define maps, for instance, $M_{g,n} \times M_{g',n'} \to M_{g+g',n+n'-2}$ by taking two Riemann surfaces with marked points and gluing them together at some specified marked points to obtain a nodal Riemann surface, and then smoothing out that node. This doesn't really make sense on the level of topological spaces because there's no canonical way to smooth out the node. However, again since we're working not with spaces but chains on spaces, I think this is OK because the different ways of smoothing out the node are the same, up to homotopy; in other words, there are many different possible maps of topological spaces $M_{g,n} \times M_{g',n'} \to M_{g+g',n+n'-2}$, but they're all homotopic, so on the level of chains the different maps are also (chain) homotopic.

**** The gluing maps here are the standard maps $\overline{M}\_{g,n} \times \overline{M}\_{g',n'} \to \overline{M}\_{g+g',n+n'-2}$. This is essentially the standard notion of "cohomological field theory" as defined by Kontsevich-Manin.

share|cite|improve this answer
@Kevin, is $\overline{M}$ the moduli space of stable Riemann surface with boundary (because for $M$ it is so)? Thanks! – Hao Nov 27 '10 at 23:03
@Hao: I have edited my answer, maybe this will clarify things. – Kevin H. Lin Nov 27 '10 at 23:49
@Kevin, I don't understand why everything is defined "up to homotopy",is TCFT only "up to homotopy"?(it is not obvious according to the definition) – Hao Nov 28 '10 at 11:07
@Hao: I have added some footnotes. – Kevin H. Lin Nov 29 '10 at 4:23

Your Answer


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.