The relation between Chern-Simons Theory and Yang-Mills Theory So from this page, I know that there is a relation between Chern-Simons Theory and Yang-Mills Theory, but I have difficulty proving the identities in the document.

I was going to prove $$\partial_\mu(\epsilon^{\mu\alpha\beta\gamma}(A_\alpha^a\partial_\beta A^a_\gamma+\dfrac13f^{abc}A^a_\alpha A^b_\beta A^c_\gamma)) = 4\epsilon^{\mu\alpha\beta\gamma}F^a_{\mu\alpha}F^a_{\beta\gamma}$$
where
$F^a_{\mu\alpha}=\partial_\mu A_\alpha^a - \partial_\alpha A_\mu^a+f^{abc}A^b_\mu A^c_\alpha$.

My attempt:
$$\partial_\mu(\epsilon^{\mu\alpha\beta\gamma}A_\alpha^a\partial_\beta A^a_\gamma) = \epsilon^{\mu\alpha\beta\gamma}\partial_\mu(A_\alpha^a\partial_\beta A^a_\gamma) 
= \epsilon^{\mu\alpha\beta\gamma}(\partial_\mu A_\alpha^a)(\partial_\beta A^a_\gamma)+\epsilon^{\mu\alpha\beta\gamma} A_\alpha^a(\partial_\mu\partial_\beta A^a_\gamma)$$
but note that $\partial_\beta\partial_\mu = \partial_\mu\partial_\beta$, so we have
$$\epsilon^{\mu\alpha\beta\gamma} A_\alpha^a(\partial_\mu\partial_\beta A^a_\gamma) = \epsilon^{\mu\alpha\beta\gamma} A_\alpha^a(\partial_\beta\partial_\mu A^a_\gamma) =  \epsilon^{\beta\alpha\mu\gamma} A_\alpha^a(\partial_\beta\partial_\mu A^a_\gamma) = -\epsilon^{\mu\alpha\beta\gamma} A_\alpha^a(\partial_\beta\partial_\mu A^a_\gamma) = 0$$
so we are left with the product of two derivatives. We can rewrite that term as
$$\epsilon^{\mu\alpha\beta\gamma}(\partial_\mu A_\alpha^a)(\partial_\beta A^a_\gamma) = \dfrac 14 (\epsilon^{\mu\alpha\beta\gamma}(\partial_\mu A_\alpha^a)(\partial_\beta A^a_\gamma) - (\partial_\alpha A_\mu^a)(\partial_\beta A^a_\gamma)-(\partial_\mu A_\alpha^a)(\partial_\gamma A^a_\beta)+(\partial_\alpha A_\mu^a)(\partial_\gamma A^a_\beta))=\dfrac 14 \epsilon^{\mu\alpha\beta\gamma}(\partial_\mu A_\alpha^a - \partial_\alpha A_\mu^a)(\partial_\beta A^a_\gamma-\partial_\gamma A^a_\beta)$$
Now, onto the product of three $A$'s.
$$\partial_\mu(\epsilon^{\mu\alpha\beta\gamma}\dfrac13f^{abc}A^a_\alpha A^b_\beta A^c_\gamma)) = \epsilon^{\mu\alpha\beta\gamma}\dfrac13 f^{abc} \partial_\mu(A^a_\alpha A^b_\beta A^c_\gamma)= \epsilon^{\mu\alpha\beta\gamma}\dfrac13 f^{abc}((\partial_\mu A^a_\alpha)A^b_\beta A^c_\gamma + A^a_\alpha(\partial_\mu A^b_\beta) A^c_\gamma + A^a_\alpha A^b_\beta(\partial_\mu A^c_\gamma))
= \dfrac13(\partial_\mu A^a_\alpha)A^b_\beta A^c_\gamma (\epsilon^{\mu\alpha\beta\gamma}  f^{abc}+ \epsilon^{\mu\beta\alpha\gamma} f^{bac}+\epsilon^{\mu\gamma\alpha\beta} f^{cab}) $$
Since the structure constants are completely antisymmetric, we have
$$\partial_\mu(\epsilon^{\mu\alpha\beta\gamma}\dfrac13f^{abc}A^a_\alpha A^b_\beta A^c_\gamma))=(\partial_\mu A^a_\alpha)A^b_\beta A^c_\gamma \epsilon^{\mu\alpha\beta\gamma}  f^{abc}=\dfrac 12(\partial_\mu A^a_\alpha-\partial_\alpha A^a_\mu)A^b_\beta A^c_\gamma \epsilon^{\mu\alpha\beta\gamma}  f^{abc}$$
That's difficulty #$1$: the prefactor is $\dfrac 12$ not $\dfrac 14$.
There is another difficulty #$2$: we are missing the product of $4$ $A$'s, i.e. the term
$$\epsilon^{\mu\alpha\beta\gamma}f^{abc}A^b_\mu A^c_\alpha f^{ade}A^d_\beta A^e_\gamma$$
which has to equal $0$, but I am not able to show it is.
Edit (1): On second thought I can solve difficulty #$1$, it's a mistake in which the term $$ (\partial_\beta A^c_\gamma-\partial_\gamma A^c_\beta)A^a_\mu A^b_\alpha \epsilon^{\mu\alpha\beta\gamma}  f^{abc}$$ is left out, which is the same as the term with the prefactor $\dfrac 12$ left out after index rearrangement, so the prefactor becomes $\dfrac 14$. So I am left with difficulty #$2$ only.
Edit (2): Using Jacobi identity, we have
$$f^{bac}f^{dea} = -f^{abc}f^{ade}= - f^{dac}f^{eba}-f^{eac}f^{bda}= - f^{acd}f^{aeb}-f^{ace}f^{abd}$$
The term becomes
$$\epsilon^{\mu\alpha\beta\gamma}(f^{acd}f^{aeb}A^b_\mu A^c_\alpha A^d_\beta A^e_\gamma + f^{ace}f^{abd}A^b_\mu A^c_\alpha A^d_\beta A^e_\gamma)
=\epsilon^{\gamma\mu\alpha\beta}f^{acd}f^{aeb}A^b_\gamma A^c_\mu A^d_\alpha A^e_\beta + \epsilon^{\beta\mu\gamma\alpha}f^{ace}f^{abd}A^b_\beta A^c_\mu A^d_\gamma A^e_\alpha
=f^{abc}f^{ade}A^b_\mu A^c_\alpha A^d_\beta A^e_\gamma(-\epsilon^{\mu\alpha\beta\gamma}-\epsilon^{\mu\alpha\beta\gamma})$$
The quantity is the same as $-2$ times itself, i.e. $X = -2X$, so we have either $3=0$ or $X=0$. Since the field $\mathbb R$ does not have characteristic $3$, we are left with the product of the $4$ $A$'s being $0$, so we have solved difficulty #$2$.
 A: This is one of those situations where performing calculations in a more covariant way is a great idea.
In particular, if we treat the gauge potential $A$ as a Lie algebra-valued differential form, then the statement you wish to prove is
$$\mathrm{d}\left(\text{Tr}\left[A\wedge\mathrm{d}A+\frac{2}{3}A\wedge A\wedge A\right]\right)=\text{Tr}\left[F\wedge F\right],$$
where $F=\mathrm{d}A+A\wedge A$, and $\mathrm{d}$ is the exterior derivative. Noting that the exterior derivative is nilpotent (i.e. $\mathrm{d}^2=0$), we have
$$
\begin{split}
\mathrm{d}\left(\text{Tr}\left[A\wedge\mathrm{d}A+\frac{2}{3}A\wedge A\wedge A\right]\right)&=\text{Tr}\left[\mathrm{d}A\wedge\mathrm{d}A+\frac{2}{3}\mathrm{d}A\wedge A\wedge A-\frac{2}{3}A\wedge\mathrm{d}A\wedge A+\frac{2}{3}A\wedge A\wedge\mathrm{d}A\right]\\
&=\text{Tr}\left[\mathrm{d}A\wedge\mathrm{d}A+\mathrm{d}A\wedge A\wedge A+A\wedge A\wedge\mathrm{d}A\right]\\
&=\text{Tr}\left[\left(\mathrm{d}A+A\wedge A\right)\left(\mathrm{d}A+A\wedge A\right)\right]=\text{Tr}\left[F\wedge F\right],
\end{split}
$$
where we used the fact that
$$\text{Tr}\left[\mathrm{d}A\wedge A\wedge A\right]=-\text{Tr}\left[A\wedge\mathrm{d}A\wedge A\right]=\text{Tr}\left[A\wedge A\wedge\mathrm{d}A\right],$$
and
$$\text{Tr}\left[A\wedge A\wedge A\wedge A\right]=0,$$
both of which are consequences of the cyclicity of the trace and the antisymmetry of $\wedge$.
