2
$\begingroup$

The QCD Lagrangian has the term $\alpha G^{a}_{\mu\nu}\widetilde{G}^{a}_{\mu\nu}$, where $$G^{a}_{\mu\nu} = \partial_{\mu}G^a_{\nu} - \partial_{\nu} G^a_{\mu} -gf_{bca}G_{\mu}^bG_{\nu}^c$$ Is the field strength tensor and this: $$\widetilde{G}^a_{\mu\nu} = \frac{1}{2}\epsilon_{\mu\nu\lambda\rho}G_a^{\lambda\rho}$$ Is the dual of the field strength tensor.

Apparently this can be written as a total derivative. But with my (admittedly very limited) knowledge of tensors, I can't see how multiplying those two quantities together gives you something you can simplify at all: $$G^{a}_{\mu\nu}\widetilde{G}^{a}_{\mu\nu} = \frac{1}{2}\epsilon_{\mu\nu\lambda\rho}G_a^{\lambda\rho}(\partial_{\mu}G^a_{\nu} - \partial_{\nu} G^a_{\mu} -gf_{bca}G_{\mu}^bG_{\nu}^c)$$ Because of the derivatives you can't expand the brackets to multiply the G terms together and contract the $a$. The $\epsilon$ is the four component Levi-Civita symbol which is zero if two or more indices are the same, so if I try to expand the Einstein summation plenty of those terms will be zero and the remaining ones don't look like a total derivative.

Based on my question it's probably clear I'm missing a lot of important background, but I'd massively appreciate it if someone could sketch out how to approach this! The calculation is part of understanding the strong CP problem, which I'd love to know more about, but textbooks I've seen don't include the steps.

$\endgroup$
2
$\begingroup$

The "important background" that keeps this sort of calculation from being a super pain in the butt is the calculus of Lie algebra–valued differential forms. Let

$$A=A_{\mu a}\tau^a\mathrm{d}x^{\mu} $$ be the Lie algebra-valued one-form representing the gauge field (this is often off from physicists' conventions by a factors $\pm\mathrm{i}$ and $g$). Let $$F=\tfrac{1}{2}F_{a\mu\nu}\tau^a\,\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}=\mathrm{d}A+A\wedge A$$ be the Lie algebra–valued two form representing the associated field strength. Then the index-free version of the desired expression as a total derivative is

$$\mathrm{Tr}\,F\wedge F =\mathrm{d}\,\mathrm{Tr}(A\wedge\mathrm{d}A+\tfrac{2}{3}A\wedge A\wedge A)$$

See these MathOverflow questions for some further details on the calculus and its underlying geometry:

| cite | improve this answer | |
$\endgroup$

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.