E-writing Action on a different slice of space-time, Quantum Hall Effect

[binbagsss]

Hi, I'm looking at QHE notes D.Tong and wondering how he gets from equation 5.46 to 5.48 ( http://www.damtp.cam.ac.uk/user/tong/qhe/five.pdf )

$S_{CS}=\frac{k}{4\pi}\int d^3 x \epsilon^{\mu \nu \rho} tr(a_{\mu}\partial_{\nu}a_{\rho} -\frac{2i}{3}a_{\mu}a_{\nu}a_{\rho})$.
manifold $\bf{R} \times \Sigma$ where $\bf{R}$ is time and $\Sigma$ is a spatial compact manifold.

$S_{CS}=\frac{k}{4\pi}\int dt \int\limits_{\sigma} d^2x tr(\epsilon^{ij} a_i \frac{\partial}{\partial t} a_j + a_0 f_{12)}$

I'm very stuck , I'm not sure where to begin. Any hint or explanation very much appreciated.

For example how have we gone from levi - civita tensor in '3-d to 2-d', how have we gone to only a time derivative in the first term- all spatial derivatives vanish? why would this be?

also very confused about the last term. I can't see no reason why spatial derivatives would vanish so I guess instead it's made use of compactness combined with the levi-civita symbol causing something to vanish? I guess such vanishing might also be the reason we able to write in terms of the '2-d' levi-civita symbol instead, but I'm pretty clueless.

Many thanks in advance.

 

[king vitamin]

how have we gone to only a time derivative in the first term- all spatial derivatives vanish?

Perhaps you have forgotten the definition of $f_{\mu \nu}$? It is
$$
f_{\mu \nu} = \partial_{\mu} a_{\nu} - \partial_{\nu} a_{\mu} - i [a_{\mu},a_{\nu}],
$$
so the $f_{12}$ term contains spatial derivatives.

Let's just write everything out. Eq (5.46) is
$$
S_{CS} = \frac{k}{4 \pi} \int d^3 x \, \epsilon^{\mu \nu \rho} \mathrm{tr} \left( a_{\mu} \partial_{\nu} a_{\rho} - \frac{2 i}{3} a_{\mu} a_{\nu} a_{\rho} \right).
$$
First of all,
$$
\epsilon^{\mu \nu \rho} a_{\mu} a_{\nu} a_{\rho} = a_0 a_1 a_2 + a_1 a_2 a_0 + a_2 a_0 a_1 - a_0 a_2 a_1 - a_2 a_1 a_0 - a_1 a_0 a_2.
$$
Now, if this expression is inside of a trace, we can use the cyclicity of the trace to combine terms which are the same up to a cyclic permutation. So
$$
\epsilon^{\mu \nu \rho} \mathrm{tr} \left( a_{\mu} a_{\nu} a_{\rho} \right) = 3 \mathrm{tr} \left( a_0 [a_1,a_2] \right).
$$

A similar computation on the other term finds
$$
\epsilon^{\mu \nu \rho} a_{\mu} \partial_{\nu} a_{\rho} = a_0 \partial_1 a_2 - a_0 \partial_2 a_1 + a_1 \partial_2 a_0 - a_1 \partial_t a_2 + a_2 \partial_t a_1 - a_2 \partial_1 a_1.
$$
At this point, we use the fact that everything sits inside of a spatial integral over a compact manifold $\Sigma$ (see the text prior to (5.48)), so we can freely integrate by parts on each term without any "boundary term" appearing. So clearly
$$
\epsilon^{\mu \nu \rho} \int dt \int_{\Sigma} d^2 x \, a_{\mu} \partial_{\nu} a_{\rho} = \int dt \int_{\Sigma} d^2 x \, \left( 2 a_0 \left( \partial_1 a_2 - \partial_2 a_1 \right) + a_2 \partial_t a_1 - a_2 \partial_1 a_0 \right).
$$
Combining the terms, we then find
$$
S_{CS} = \frac{k}{4 \pi} \int dt \int_{\Sigma} d^2 x \, \mathrm{tr}\left( 2 a_0 f_{12} - \epsilon^{ij} a_i \frac{\partial}{\partial t} a_j \right)
$$
So I actually get a different expression than Tong, unless perhaps there is some notational different which is important (e.g. the Levi-Civita symbol has implicit minus signs in Minkowski signature?). I believe that my expression is equivalent to Equations (21) and (64) in this review article.