Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Modular invariance for higher genus

+ 4 like - 0 dislike
528 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)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar\varnothing$sicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...