Non-abelian Chern-Simons from fermion effective action (1) Starting from the fermion effective action
$$S_\text{eff}[A,m] = \log \det(i\gamma^\mu{\partial_\mu} + \gamma^\mu A_\mu + m)\tag{223}$$
once can do a loop expansion following https://arxiv.org/abs/hep-th/9902115, eq. 223. How does one derive the non-abelian CS action eq. 233:
$$S_\text{CS} \propto \frac{1}{8\pi}\frac{m}{|m|}\epsilon^{\delta\mu\nu}\int d^3 x\ \mathrm{tr}(A_{\delta}\partial_{\mu}A_{\nu} + \frac{2}{3}A_{\delta}A_{\mu}A_{\nu})\tag{233}$$
in the same article? I wish to understand how the 2/3 comes.
(2) Also, how does the variation of the above action lead to eq.56:
$$\delta L_\text{CS} \propto \epsilon^{\nu\mu\delta}\mathrm{tr}(F_{\nu\mu}\delta A_{\delta})\tag{56}$$
from the same article?
 A:  Conventions 
Equations (223) and (233) aren't quite consistent with each other, because (223) assumes a convention in which $A_\mu$ is hermitian and (233) assumes a convention in which it's antihermitian. The two conventions are related by a factor of $i$. I'll use the convention in which $A_\mu$ is antihermitian, consistent with (233), so that the Dirac operator is
$$
\newcommand{\Tr}{\text{Tr}}
\newcommand{\pl}{\partial}
\newcommand{\sp}{{\gamma p}}
m+i\gamma^\mu\partial_\mu+i\gamma^\mu A_\mu.
\tag{1}
$$
Also:

*

*I'll use the mostly-minus convention for the metric, but the result (the Chern-Simons effective action) doesn't depend on the metric.


*I'll write $\Tr_A(\cdots)$ and $\Tr_\gamma(\cdots)$ for the traces over gauge and Dirac indices, respectively, and I'll write $\int$ for integration over spacetime.
 Question #2 
I'll address question #2 first because it's shorter. Use the antisymmetry of $\epsilon^{abc}$ and the cyclicity of the trace to get
\begin{gather}
\delta\epsilon^{abc}\int\Tr_{A,\gamma}\left(A_a\pl_b A_c+\frac{2}{3}A_aA_bA_c\right)
\hspace{2cm}
\\
\hspace{2cm}
=\epsilon^{abc}\int\Tr_{A,\gamma}\Big((\delta A_a)\pl_b A_c+ A_b\pl_c\delta A_a + 2(\delta A_a)A_b A_c\Big).
\tag{2}
\end{gather}
To finish, use integration-by-parts to move the derivative in the second term, and remember that $F_{bc}\propto [\pl_b - A_b,\,\pl_c -A_c]$.
 Question #1, indirect approach 
Question #1 asks how the relative factor of $2/3$ between the $O(A^2)$ and $O(A^3)$ terms comes about. An indirect way to answer this is to use invariance under small gauge transformations (gauge transformations that are continuously connected to the do-nothing transformation). The eigenvalues of the Dirac operator are invariant under these transformations, and the determinant is the (regulated) product of eigenvalues, so (223) is also invariant. The relative coefficient $2/3$ is the only possible value for which the result is invariant. Therefore, once we've computed the first term in (233), invariance under small gauge transformations dictates the coefficient of the second term.
By the way, the result (233) is not quite invariant under large gauge transformations. For more detail, see Gauge Invariance of the Non-abelian Chern-Simons Term, and also this Math Overflow post.
 Question #1, direct approach 
The direct approach is to expand (223) in powers of $A$. Write (1) as $X+Y$ with
$$
X \equiv  m+i\gamma^\mu\partial_\mu
\hspace{2cm}
Y \equiv i\gamma^\nu A_\nu,
\tag{3}
$$
and expand
\begin{align}
\log\det(X+Y) 
&=\int\Tr_{A,\gamma}\log(X+Y)
\\
&= c + \int\Tr_{A,\gamma}\log(1+X^{-1}Y)
\\
&=c  + \int\Tr_{A,\gamma}\left(
 X^{-1}Y
 -\frac{(X^{-1}Y)^2}{2}
 +\frac{(X^{-1}Y)^3}{3}
 -\cdots
\right)
\tag{4}
\end{align}
with an $A$-independent term $c\equiv \int\Tr\log(X)$. I'm using matrix notation in which each "index" includes a Dirac-matrix index, a gauge-matrix index, and the coordinates of a point in spacetime. The definition (3) of $X$ implies
$$
X^{-1}(x,y) = (m-i\gamma^\mu\partial_\mu)G(x-y)
\tag{5}
$$
with
$$
G(x)=\int_p\,\frac{e^{ip\cdot(x-y)}}{m^2-p^2}
\hspace{1cm}
\int_p\cdots \equiv \int\frac{d^3p}{(2\pi)^3}\cdots
\tag{6}
$$
Notice that the relative coefficient between the $O(Y^2)$ and $O(Y^3)$ terms in (4) already has magnitude $2/3$, so the goal is to show that the remaining steps in the calculation don't change the relative coefficient.
We're only interested in the terms for which the trace over Dirac matrices gives $\epsilon^{abc}$, because this is where (233) comes from. We're also interested only in the long-wavelength limit, which is the same as the large-$m$ limit. Using these conditions, we find that the relevant parts of the $O(A^2)$ and $O(A^3)$ terms in (4) have magnitudes
$$
\frac{1}{2}
\left|\Tr_A\int A_a(x)\pl_b A_c(x)\,\Gamma_2^{abc}\right|
\tag{7}
$$
and
$$
\frac{1}{3}
\left|\Tr_A
\int A_a(x)A_b(x)A_c(x)\,\Gamma_3^{abc}
\right|
\tag{8}
$$
with
\begin{align}
\Gamma_2^{abc} &=\Tr_\gamma\int_p 
\frac{m\,\gamma^a\gamma^b\gamma^c}{(m^2-p^2)^2}
\\
\Gamma_3^{abc} &=\left.\Tr_\gamma\int_p
 \frac{(m-\sp)\gamma^a(m-\sp)\gamma^b(m-\sp)\gamma^c}{(m^2-p^2)^3}
\right|_\epsilon
\tag{9}
\end{align}
with $\sp\equiv \gamma^\mu p_\mu$, and where the subscript $\epsilon$
indicates that we only keep terms that contribute to $\epsilon^{abc}$, which means we can discard terms with odd powers of $p$ in the numerator. Use the identity
$$
 (\gamma p)\gamma^a(\gamma p) = 2p^a \gamma p - p^2\gamma^a
\tag{10}
$$
to get
\begin{align}
\Gamma_3^{abc} &=\Tr_\gamma\int_p
 \frac{m\,\gamma^a\gamma^b\gamma^c}{(m^2-p^2)^2}.
\tag{11}
\end{align}
Notice that one of the original factors of $m^2-p^2$ in the denominator has been cancelled by the numerator. This shows that $\Gamma_3$ has the same magnitude as $\Gamma_2$. Combined with equations (7)-(8), this gives the desired result: the relative coefficient between the $O(A^2)$ and $O(A^3)$ terms has magnitude $2/3$. Notes:

*

*Before using the long-wavelength approximation, the integrand of $\Gamma_3$ in (9) includes terms with $p_1$ and $p_2$, the arguments of the Fourier transforms of the factors of $A$. In the long-wavelength approximation, those terms are negligible compared to $m$.


*I didn't bother writing the "Feynman $i\varepsilon$" offsets in the denominators of the integrals (9), but they're implied. Or we can use Wick rotation from lorentzian to euclidean signature so that the factors $m^2-p^2$ become $m^2+p^2$. Wick rotation doens't affect the Chern-Simons action at all, not even the overall factor of $i$.


*Since our goal was only to determine the relative coefficient, we didn't need to use an explicit regulator to define (223). The terms contributed by the Pauli-Villars regulator have the same relative coefficient.


*Since our goal was only to determine the relative coefficient, we didn't need to evaluate the momentum integrals. We only needed to evaluate the Dirac traces to show that the (relevant parts of the) integrals are equal to each other.
