I've been having a little trouble proving that [Page 138 of "Introduction to Topological Quantum Computation" by Jiannis K. Pachos]:

$S_{CS} = \dfrac{k}{4 \pi} \int_{M} d^{3}x \, \epsilon^{\mu \nu \rho} tr \left( A_{\mu}\partial_{\nu}A_{\rho} + i\dfrac{2}{3} A_{\mu} A_{\nu} A_{\rho} \right)$

(with $A_{\mu} = A_{\mu}^{a} T_{a}$ and $T_{a}$ a generator of the gauge group's Lie alegebra) can be written as:

$S_{CS} = \dfrac{k}{8 \pi} \int_{M} d^{3}x \, \epsilon^{\mu \nu \rho} \left( A_{\mu}^{a}\partial_{\nu}A_{\rho}^{a} - \dfrac{1}{3} f_{abc} A_{\mu}^{a} A_{\nu}^{b} A_{\rho}^{c} \right)$

where$f^{abc}$ is antisymmetric.

My attempt thus far:

$tr \left( A_{\mu}\partial_{\nu}A_{\rho} + i\dfrac{2}{3} A_{\mu} A_{\nu} A_{\rho} \right)$

$= tr \left( A_{\mu}^{a} T_{a} \, \partial_{\nu} \, A_{\rho}^{b} T_{b} + i\dfrac{2}{3} A_{\mu}^{a} T_{a} \, A_{\nu}^{b} T_{b} \, A_{\rho}^{c} T_{c} \right)$

$= tr \left( A_{\mu}^{a} \partial_{\nu} A_{\rho}^{b} (T_{a} T_{b}) + i\dfrac{2}{3} A_{\mu}^{a} A_{\nu}^{b} A_{\rho}^{c} (T_{a}T_{b}T_{c}) \right)$

Using $tr(T_{a} T_{b}) = \dfrac{1}{2}\delta_{ab}$ and $(T_{a}T_{b}T_{c}) = \dfrac{1}{2}[T_{a}, T_{b}]T_{c} + \dfrac{1}{2}\{ T_{a}, T_{b} \}T_{c}$ we get:

$= \left( \dfrac{1}{2}A_{\mu}^{a} \partial_{\nu} A_{\rho}^{a} + i\dfrac{1}{3} A_{\mu}^{a} A_{\nu}^{b} A_{\rho}^{c} \, tr \left( [T_{a},T_{b}]T_{c} + \{ T_{a},T_{b} \}T_{c} \right) \right)$

Using

$$[T_{a}, T_{b}] = if_{abd}T_{d}$$

and $\{T_{a}, T_{b}\} = \dfrac{1}{N} \delta_{ab} + d_{abd}T_{d}$, where $d_{abc}$ is symmetric, we get:

$= \left( \dfrac{1}{2}A_{\mu}^{a} \partial_{\nu} A_{\rho}^{a} + i\dfrac{1}{3} A_{\mu}^{a} A_{\nu}^{b} A_{\rho}^{c} \, tr \left( if_{abd}T_{d}T_{c} + \dfrac{1}{N} \delta_{ab} T_{c} + d_{abd}T_{d}T_{c} \right) \right)$

Using $tr(T_{A}) = 0$ and $tr(T_{a} T_{b}) = \dfrac{1}{2}\delta_{ab}$, we get:

$$= \left( \dfrac{1}{2}A_{\mu}^{a} \partial_{\nu} A_{\rho}^{a} + i\dfrac{1}{3} A_{\mu}^{a} A_{\nu}^{b} A_{\rho}^{c} \, \dfrac{1}{2} \left( if_{abc} + d_{abc}\right) \right)$$

Now this would be precisely what we need if the $d_{abc}$ disappeared somehow but I'm not sure how to make that happen. If $A_{\mu}^{a} A_{\nu}^{b} A_{\rho}^{c}$ were anti-symmetric under permutation of the indices (abc) then if would cancel with the symmetric $d_{abc}$ but as far as I'm aware the $A_{\mu}^{a}$'s don't behave that way.

Any help would be greatly appreciated.

Cheers in advance.

This post imported from StackExchange Physics at 2014-06-21 08:58 (UCT), posted by SE-user Siraj R Khan