# Completing the trace in the Non-abelian Chern-Simons Term

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.

• It's been a while since I did this calculation but I expect you can use the anti-symmetry of the Levi-Civita symbol $\epsilon^{\mu\nu\rho}$ to show that the unwanted symmetric combination of the gauge fields vanishes. – Mark Mitchison Jun 19 '14 at 16:37

## 1 Answer

I think is quite simple if you split the product of two $A_\mu$'s in symmetric and anti-symmetric terms as $$A_\mu A_\nu=\frac{1}{2}[A_\mu,A_\nu]+\frac{1}{2}\{A_{\mu},A_\nu\}\,.$$ Thus, the action of $\epsilon^{\mu\nu\rho}$ on $A_\mu A_\nu$ just kills the symmetric piece $$\epsilon^{\mu\nu\rho}A_\mu A_\nu=\epsilon^{\mu\nu\rho}\frac{1}{2}[A_\mu,A_\nu]\,.$$ Now, with the normalization $\mathrm{Tr}(T^a T^b)=\delta^{ab}/2$, is trivial to show that $$\mathrm{Tr}\left([A_\mu,A_\nu]A_\rho\right)=A_\mu^a A_\nu^b A^c_\rho \mathrm{Tr}\left([T^a,T^b]T^c\right)=iA_\mu^a A_\nu^b A^c_\rho f^{abd}\mathrm{Tr}(T^d T^c)=i\frac{f^{abc}}{2}A_\mu^a A_\nu^b A^c_\rho$$ and $$\mathrm{Tr}(A_\mu\partial_\nu A_\rho)=\frac{1}{2}A_\mu^a \partial_\nu A^a_\rho$$. Putting everything together we arrive at the result, $$S_{CS}=\frac{k}{8\pi}\int_M d^3 x\, \epsilon^{\mu\nu\rho}\left(A_\mu^a \partial_\nu A^a_\rho-\frac{1}{3}f^{abc}A_\mu^a A_\nu^b A^c_\rho\right)$$

• Thank you for the extremely speedy response, this really helped me out. – Siraj R Khan Jun 19 '14 at 16:55