# Critical Ising T transformation

How can the following be consistent?

The $T$ transformation of a Virasoro character $\chi(q)$ of central charge $c$ and weight $h$ is given by $$\chi(q+1) =e^{2 \pi i (h - c/24)}\chi(q)$$ for instance see eqn (125) in Jean-Bernard Zuber's Introduction to Conformal Field Theory.

On the other hand, the $c = 1/2$ character $\chi_{1,2}(q)$ which is supposed to be one chiral half the Ising spin operator has $h = 1/16$ and $$\chi_{1,2}(\tau + 1) = e^{i \pi/8}\chi_{1,2}(\tau)$$ according to the big yellow book. This appears to be missing the factor of $e^{-2\pi i c/24}$.... what happened?

There must be something obvious I'm missing here.

• Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Sep 12 '18 at 19:08
• Did you mean $\tau\to\tau+1$? It follows straightforwardly from the fact that all weights in the module differ by integers, and doesn't depend on the module being short. – Peter Kravchuk Sep 13 '18 at 0:48
• Thanks @PeterKravchuk my bad. And I agree. It's just a mistake in the book. – Ryan Thorngren Sep 13 '18 at 1:53

I've decided this must be a typo in the big yellow book (di Fracesco et al). The proper transformation rule should be $$\chi_{1,2}(q+1) = e^{2\pi i/24}\chi_{1,2}(q).$$ I was worried since $\chi_{1,2}$ is not quite the same as the Verma module $V_{2,1}$ (which is reducible), the transformation rule I quoted might become modified. Actually it doesn't, as one can see looking at table 8.1 in the big yellow book: One can compute $h_{1,2} = 1/16$ and derive the correct transformation rule with exponent $1/16 - 1/48 = 1/24$.
Edit: note in this table they've exchanged $p$ with $p'$ and $r$ with $s$ so my $\chi_{1,2}$ is $\chi_{2,1}$ in the table, annoyingly.