• 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.


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


(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 ...

  CFT calculation of the partition function of $2+1$ dimensional gravity

+ 7 like - 0 dislike

I want to reproduce formula (4.29) in http://arxiv.org/abs/0804.1773v1 given by:

$$ Z=Tr(q^{L_{0}}\bar q^{\bar L_{0}})=|q|^{-2k} \prod^{\infty}_{n=2}\frac{1}{|1-q^{n}|^{2}} $$

where the trace is over an irreducible representation of the Virasoro algebra with a ground state of weight $(h,\bar h)=(-k,-k)$.

When I'm correct one can use the formula for the character of the Verma module $$ \chi_{(c,h)}(\tau)=Tr(q^{L_{0}-c/24})= q^{h-c/24} \prod_{n=1}^{\infty}\frac{1}{1-q^{n}}. $$

With $$ \chi_{(0,-k)}(\tau)=Tr(q^{L_{0}})= q^{-k} \prod_{n=1}^{\infty}\frac{1}{1-q^{n}} $$ and $$ \bar\chi_{(0,-k)}(\bar\tau)=Tr(\bar q^{\bar L_{0}})= \bar q^{-k} \prod_{n=1}^{\infty}\frac{1}{1-\bar q^{n}} $$ I get $$ Z_{my}=\chi_{(0,-k)}(\tau) \bar\chi_{(0,-k)}(\bar\tau)=|q|^{-2k} \prod^{\infty}_{n=1}\frac{1}{|1-q^{n}|^{2}} $$

1.) My product starts at $n=1$ and I do not see why I could ignore the first contribution. Where is my mistake?

2.) Why do they set $c=0$?

This post imported from StackExchange Physics at 2014-10-12 11:42 (UTC), posted by SE-user ungerade

asked Oct 11, 2014 in Theoretical Physics by ungerade (125 points) [ revision history ]
edited Oct 12, 2014 by Arnold Neumaier

$c\neq0$, rather $c=24k$. I believe that $L_0$ includes the $c/24$ piece.

2 Answers

+ 4 like - 0 dislike

1) http://xxx.lanl.gov/pdf/0712.0155v1.pdf explain (p.15 bottom) that the product lacks the first term since $L_{-1}$ annihilates the CFT ground state.

2) They (the authors of the paper you cited) don't use your argument, so they don't set $c=0$. You set $c=0$ in order to simplify; but your argument would work for any $c$ since the additional exponents cancel on both sides of the trace fromula.

answered Oct 12, 2014 by Arnold Neumaier (15,787 points) [ no revision ]
+ 4 like - 0 dislike

The computation that is being discussed is in a CFT with central charge $c=24k$ and the trace is being computed for irrep associated  the vacuum state $|\Omega\rangle$, which has $h=\bar{h}=0$. It is easy to show that $L_{-1}|\Omega\rangle$ has zero norm with a similar statement for the anti-holomorphic part. In other word, we need to project out descendants of $L_{-1}|\Omega\rangle$ from the trace. One has

 \(\begin{align} \text{Tr}_{\Omega}\left(q^{L_0-\frac{c}{24}}\right ) &= q^{-k} \text{Tr}\left(q^{L_0} \right ) \\ &= q^{-k} \prod_{n=1}^{\infty} \frac1{(1-q^n)} - q^{-k+1} \prod_{n=1}^{\infty} \frac1{(1-q^n)} \\ &=q^{-k} \prod_{n=2}^{\infty} \frac1{(1-q^n)}\ . \end{align}\)

In the first line, I have put in $c=24k$. In the second line, I include a second term to subtract out the contribution of the null (zero-norm) state and its descendants. The second term is not present in a generic Verma module where there are no nulls at all levels and the Verma module is an irrep of the Virasoro algebra.  Please check the paper(s) to see if they include the $\frac{c}{24}$ factor in their definition of $L_0$ (it is the usual cylinder vs plane relation).

Remark: The vacuum character in the Ising model ($c=1/2$) is an interesting example. It has an additional null at level 2. However, the two nulls themselves are not irreducible and there are nulls over nulls and so on. This leads to an infinite sequence of addition and subtractions that lead to the character expansion (see Ginsparg, Applied Conformal Field Theory, for more details)

\(\chi_0 = \frac12 \left(\sqrt{\frac{\theta_3}{\eta}}+\sqrt{\frac{\theta_4}{\eta}}\right) = q^{-1/48} (1+q^2 +q^3 + 2q^4+\cdots)\ . \)

answered Oct 13, 2014 by suresh (1,545 points) [ revision history ]
edited Oct 13, 2014 by suresh

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:
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