1
$\begingroup$

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 #$2$.

$\endgroup$
2
$\begingroup$

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 $\wedge$.

$\endgroup$
  • $\begingroup$ Why does the very last equality hold? $\endgroup$ – wilsonw Jun 27 at 12:24
  • $\begingroup$ I ask because $A$ is not a number but a matrix...and can this equality be written in components? $\endgroup$ – wilsonw Jun 27 at 12:27
  • $\begingroup$ Another reason I asked is because $\text{Tr }( A\wedge A\wedge A) \neq 0$, which leaves me puzzled. Had the anticommutativity worked wouldn't it be zero in general? $\endgroup$ – wilsonw Jun 27 at 12:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.