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  Modular invariance for higher genus

+ 4 like - 0 dislike
897 views

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 has been migrated from (A51.SE)
asked Jan 14, 2012 in Theoretical Physics by Squark (1,725 points) [ no revision ]

2 Answers

+ 1 like - 0 dislike

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 has been migrated from (A51.SE)
answered Jan 15, 2012 by Luboš Motl (10,278 points) [ no revision ]
I know about the higher genus moduli group, but how do you show h=1 modular invariance implies h>1 modular invariance?

This post has been migrated from (A51.SE)
+ 0 like - 0 dislike

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 has been migrated from (A51.SE)
answered Jan 15, 2012 by crackjack (110 points) [ no revision ]
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 has been migrated from (A51.SE)

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