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

Question

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

Comments

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

Answer 1

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.

Answer 2

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)

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