# Modular invariance for higher genus

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As far as I understand, there are roughly 2 "common" kinds of 2D conformal field theories:

1. Theories that are defined only on the plane, more precisely, on any surface of vanishing genus. Such a theory can be mathematically described by a vertex operator algebra, or by "usual" axiomatic QFT in 2D with the added requirement of conformal invariance
2. Theories defined on arbitrary surfaces. Such a theory can be described e.g. by Segal's axioms

In order for a theory of the 1st category to become a theory of the 2nd category, it must pass the test of modular invariance. In my experience, this term usually means that the theory has to be defined in genus 1, i.e. that the would-be torus partition function

$$Tr(q^{L_0-\frac{c}{24}}\bar{q}^{\bar{L_0}-\frac{\bar{c}}{24}})$$

is invariant under the modular group $SL(2,\mathbb{Z})$

What about higher genus? Is the torus condition sufficient to make the theory well defined there? Or does it impose additional conditions? If so, can they be cast into an anologous elegant algebraic form? Have these conditions been proved for CFTs used in string theory?

This post imported from StackExchange Physics at 2014-04-05 17:28 (UCT), posted by SE-user Squark
This is a very good question. Put it in another way: Can representation of modular transformation for torus completely determine a (chiral) CFT?

This post imported from StackExchange Physics at 2014-04-05 17:28 (UCT), posted by SE-user Xiao-Gang Wen

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The modular group $SL(2,{\mathbb Z})$ is generated to $Sp(2h,{\mathbb Z})$, for genus $h$. That's the group exchanging the 1-cycles of the Riemann surface while preserving the intersection numbers (an antisymmetric tensor). Recall that the moduli space of Riemann surfaces is higher-dimensional, namely $(6h-6)$-dimensional (real dimensions) for $h>1$.

Quite generally, the modular invariance at $h=1$ guarantees the modular invariance at all finite $h$.

This post imported from StackExchange Physics at 2014-04-05 17:28 (UCT), posted by SE-user Luboš Motl
answered Jan 15, 2012 by (10,248 points)
I know about the higher genus moduli group, but how do you show h=1 modular invariance implies h>1 modular invariance?

This post imported from StackExchange Physics at 2014-04-05 17:28 (UCT), posted by SE-user Squark
Hi Squark, see e.g. sciencedirect.com/science/article/pii/0370269387909464 - scholar.google.com/… - sorry for not reviewing the papers here

This post imported from StackExchange Physics at 2014-04-05 17:28 (UCT), posted by SE-user Luboš Motl
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A higher genus surface can be considered as a connected sum of tori, and the linking cylinder in the connected sum can be replaced by a sum over all the propagating particles. If the theory is modular invariant on the torus, you know that you can do modular transformations on each torus separately, and everything is consistent. It is intuitive that every large diffeomorphism of a high genus surface can be generated using generators each in the individual SL(2,Z)s of the different tori of the connected sum. So by knowing torus modular invariance of the theory, correct for all tori and insertions on the tori, you learn that the large diffeomorphisms are ok.

A sketch of a proof: you cut up the torus along loops into pieces homeomorphic to triangles, then consider the image of the loops under diffeomorphism. Restricting to each torus, there is a nonzero intersection number of the image of the loops with the old loops, and you perform a large diffeomorphism whenever this can reduce the intersection number. continue on each torus until you can't go reduce the intersection numbers any further. At this point, the intersection number must be as low as possible, and this means that the curves are isotopic to their original position, so that the diffeomorphism is now continuously connected to the identity.

I did not check this in detail (I should), but the main lemma you need--- that you can always reduce a non-minimal intersection number by isotoping the curves to go near one genus, and doing an SL(2,Z) is very plausible. There might also be an easier proof.

This post imported from StackExchange Physics at 2014-04-05 17:28 (UCT), posted by SE-user Ron Maimon
answered May 31, 2012 by (7,535 points)
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A review reference for the case of superstrings in NSR formalism: http://arxiv.org/abs/0804.3167

I have been told that pure-spinor formalism has gone slightly further on the genus count.

This post imported from StackExchange Physics at 2014-04-05 17:28 (UCT), posted by SE-user crackjack
answered Jan 15, 2012 by (110 points)
Thx, but I'm not sure we are talking about the same thing. You are talking about performing integration over the moduli space in higher genus while I am talking about having a CFT well defined in higher genus which is a weaker requirement

This post imported from StackExchange Physics at 2014-04-05 17:28 (UCT), posted by SE-user Squark

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