Chern-Simons theory: Connection between Thermal and Quantum Partition Function

Question

I have been reading the Quantum Hall Effect from Prof. David Tong's notes. In the section on Chern-Simons theory, he describes the connection between the Thermal Partition Function and the Quantum partition function in section (5.1.2). He comments that the presence of $\epsilon_{\mu \nu \rho}$ symbol in (5.6) means that the action in Euclidean space picks up an extra i factor (in the last paragraph before section 5.1.3). I can't make sense of this statement. Why is there an extra factor of i because of epsilon symbol?

http://www.damtp.cam.ac.uk/user/tong/qhe.html

Answer 1

Let's Wick rotate the Chern-Simons action. The Lorentzian action is: $\newcommand{\d}{\mathrm{d}}\newcommand{\pd}{\partial}$ $$S_\text{CS}^{(\text{L})}[A] =\frac{k}{4\pi} \int_\Sigma \d{t}\;\d^2{x}\ \epsilon^{\mu\nu\rho} A_\mu\pd_\nu A_\rho,\label{1}\tag{1}$$ where I separated the temporal component, to fiddle with it in a second.

Now, upon Wick rotating, besides sending $$t\mapsto \tau := i\,t,$$ we should also send the temporal component of any vector to $-i\times$itself. That is, because we want the vector to transform like a derivative (or, in more high-brow language, $A_\mu\d{x}^\mu$ is a one-form). Therefore upon Wick rotating we also have the following transformation rules: $$A_0 \mapsto A_0^\text{E}:=-i A_0 \qquad\text{and}\qquad \pd_t \equiv \pd_0 \mapsto \pd_\tau \equiv \pd_0^\text{E} :=-i \pd_0.$$

The action \eqref{1} has two types of terms (up to boundary terms which we neglect either because of closedness of $\Sigma$ which is typical in Chern-Simons systems, or because of boundary conditions): $$\epsilon^{0mn}A_0 \pd_m A_n \qquad\text{and}\qquad \epsilon^{m0n}A_m\pd_0A_n,$$ where $m,n$ are spatial indices. Because of the above transformation rule, both terms simply pick up a factor of $i$, upon Wick rotating, so $$\epsilon^{\mu\nu\rho} A_\mu\pd_\nu A_\rho \mapsto i\,\epsilon^{\mu\nu\rho} A_\mu\pd_\nu A_\rho.$$ So now we're ready to see the Wick rotated, Euclidean action. It is: $$S_\text{CS}^{(\text{E})}\left[A^{(\text{E})}\right] =\frac{k}{4\pi} \int_\Sigma \d{(-i\,\tau)}\;\d^2{x}\ i\, \epsilon^{\mu\nu\rho} A_\mu^{(\text{E})}\pd_\nu^{(\text{E})} A_\rho^{(\text{E})} = \frac{k}{4\pi} \int_\Sigma \d{\tau}\;\d^2{x}\ \epsilon^{\mu\nu\rho} A_\mu^{(\text{E})}\pd_\nu^{(\text{E})} A_\rho^{(\text{E})}.\label{2}\tag{2}$$ This shows that $$ S_\text{CS}^{(\text{L})}[A] =S_\text{CS}^{(\text{E})}[A], $$ and the action does not pick up an extra $i$, as claimed.

This fact can be traced back to the $\epsilon$ symbol in the action, requiring that exactly one vectorial quantity (i.e. one of $A_\mu$ or $\pd_\mu$) will have its leg pointing at the time direction, so that it can pick up the correct factor of $i$ to cancel against the measure.