A Understanding that QCD is not CP invariant

[Gaussian97]

Where the first term drops because $\varepsilon^{\mu\nu\rho\sigma}$ is antisymmetric under $\mu \leftrightarrow \nu$ while $\partial_{\mu} A_{\nu}$ is symmetric (not completely sure if this is OK).

Mmm... How would you argue that $\partial_{\mu}A_{\nu}$ is symmetric?? And more important, if the first term vanishes, how in the world the second does not vanish?

Next I thought of product rule + divergence theorem

\begin{equation*}
\underbrace{\partial_\mu\left(\varepsilon^{\mu\nu\rho\sigma}A_\nu^i F_{\rho\sigma}^i\right)}_{=0?} =\underbrace{\partial_\mu(\varepsilon^{\mu\nu\rho\sigma})A_\nu^i F_{\rho\sigma}^i}_{=0} + \varepsilon^{\mu\nu\rho\sigma}\partial_\mu A_\nu^i F_{\rho\sigma}^i + \varepsilon^{\mu\nu\rho\sigma}A_\nu^i \partial_\mu F_{\rho\sigma}^i
\end{equation*}

You shouldn't use the divergence theorem here. First of all, because we are told that the boundary terms don't vanish. And furthermore, we want to express the whole expression as a total divergence, if you use the DT to get rid of such terms...

 

[JD_PM]

And more important, if the first term vanishes, how in the world the second does not vanish?

Absolutely right.

Hence I have

\begin{align*}
\varepsilon_{\mu\nu\rho\sigma}F_i^{\mu\nu}F^{\rho\sigma}_i &= \varepsilon^{\mu\nu\rho\sigma}F^i_{\mu\nu}F_{\rho\sigma}^i \\
&= \varepsilon^{\mu\nu\rho\sigma}(\partial_{\nu} A_{\mu}^i - \partial_{\mu} A_{\nu}^i)F_{\rho\sigma}^i \\
&= -2\varepsilon^{\mu\nu\rho\sigma}\partial_{\mu} A_{\nu}^i F_{\rho\sigma}^i \\
&= -\frac 1 2 \partial_\mu\left(\varepsilon^{\mu\nu\rho\sigma}A_\nu^i F_{\rho\sigma}^i\right) + \frac 1 2 \varepsilon^{\mu\nu\rho\sigma}A_\nu^i \partial_\mu F_{\rho\sigma}^i
\end{align*}

Mmm how to deal with the $\frac 1 2 \varepsilon^{\mu\nu\rho\sigma}A_\nu^i \partial_\mu F_{\rho\sigma}^i$ term though? I thought of Bianchi identity for $F_{\mu\nu}$ but I do not think it is going to lead us to the desired result.

 

[JD_PM]

\begin{align*}
G \cdot \tilde G &= \varepsilon_{\mu \nu \rho \sigma} G_i^{\mu \nu}G_i^{\rho \sigma}\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu} + g_s f_{ijk} A_j^{\mu} A_k^{\nu})(F_i^{\rho \sigma} + g_s f_{ilm} A_l^{\rho} A_m^{\sigma})\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu}F_i^{\rho \sigma} + g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} + g_s f_{ijk} A_j^{\mu} A_k^{\nu} F_i^{\rho \sigma} + g^2_sf_{ijk}f_{ilm}A_j^{\mu} A_k^{\nu}A_l^{\rho} A_m^{\sigma})\\
&= ?\\
&= \partial_{\mu}(\varepsilon^{\mu \nu \rho \sigma} A_{\nu}^i [G_{\rho \sigma}^i - \frac{g_s}{3}f_{ijk} A_{j\rho} A_{k\sigma}])
\end{align*}

I advanced just a bit: $A^4$ term is symmetric so

\begin{align*}
G \cdot \tilde G &= \varepsilon_{\mu \nu \rho \sigma} G_i^{\mu \nu}G_i^{\rho \sigma}\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu} + g_s f_{ijk} A_j^{\mu} A_k^{\nu})(F_i^{\rho \sigma} + g_s f_{ilm} A_l^{\rho} A_m^{\sigma})\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu}F_i^{\rho \sigma} + g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} + g_s f_{ijk} A_j^{\mu} A_k^{\nu} F_i^{\rho \sigma} + g^2_sf_{ijk}f_{ilm}A_j^{\mu} A_k^{\nu}A_l^{\rho} A_m^{\sigma})\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu}F_i^{\rho \sigma} + g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} + g_s f_{ijk} A_j^{\mu} A_k^{\nu} F_i^{\rho \sigma})\\
&= ?\\
&= \partial_{\mu}(\varepsilon^{\mu \nu \rho \sigma} A_{\nu}^i [G_{\rho \sigma}^i - \frac{g_s}{3}f_{ijk} A_{j\rho} A_{k\sigma}])
\end{align*}

 

[Gaussian97]

Absolutely right.

Hence I have

\begin{align*}
\varepsilon_{\mu\nu\rho\sigma}F_i^{\mu\nu}F^{\rho\sigma}_i &= \varepsilon^{\mu\nu\rho\sigma}F^i_{\mu\nu}F_{\rho\sigma}^i \\
&= \varepsilon^{\mu\nu\rho\sigma}(\partial_{\nu} A_{\mu}^i - \partial_{\mu} A_{\nu}^i)F_{\rho\sigma}^i \\
&= -2\varepsilon^{\mu\nu\rho\sigma}\partial_{\mu} A_{\nu}^i F_{\rho\sigma}^i \\
&= -\frac 1 2 \partial_\mu\left(\varepsilon^{\mu\nu\rho\sigma}A_\nu^i F_{\rho\sigma}^i\right) + \frac 1 2 \varepsilon^{\mu\nu\rho\sigma}A_\nu^i \partial_\mu F_{\rho\sigma}^i
\end{align*}

Mmm how to deal with the $\frac 1 2 \varepsilon^{\mu\nu\rho\sigma}A_\nu^i \partial_\mu F_{\rho\sigma}^i$ term though? I thought of Bianchi identity for $F_{\mu\nu}$ but I do not think it is going to lead us to the desired result.

Not sure how you convert the 2 to a 1/2, but the idea is OK. Using the definition of $F$ and symmetry is not difficult to show that
$$\varepsilon^{\mu\nu\rho\sigma}\partial_\mu F_{\rho\sigma}^i=0$$

I advanced just a bit: $A^4$ term is symmetric so

Yes... But I think is not completely trivial to show how the symmetry of $A^4$ cancels this term.

 

[JD_PM]

The authors seem to be using the definition $F_i^{\mu \nu} := \partial^{\mu} A^{\nu}_i - \partial^{\nu} A^{\mu}_i$ so let us use it as well from here on.

Not sure how you convert the 2 to a 1/2

Oops indeed, we have

\begin{align*}
&\partial_\mu\left(\varepsilon^{\mu\nu\rho\sigma}A_\nu^i F_{\rho\sigma}^i\right) =\underbrace{\partial_\mu(\varepsilon^{\mu\nu\rho\sigma})A_\nu^i F_{\rho\sigma}^i}_{=0} + \varepsilon^{\mu\nu\rho\sigma}\partial_\mu A_\nu^i F_{\rho\sigma}^i + \varepsilon^{\mu\nu\rho\sigma}A_\nu^i \partial_\mu F_{\rho\sigma}^i \\
&\Rightarrow 2\varepsilon^{\mu\nu\rho\sigma}\partial_\mu A_\nu^i F_{\rho\sigma}^i = 2\partial_\mu\left(\varepsilon^{\mu\nu\rho\sigma}A_\nu^i F_{\rho\sigma}^i\right) - 2\varepsilon^{\mu\nu\rho\sigma}A_\nu^i \partial_\mu F_{\rho\sigma}^i
\end{align*}

Then

\begin{align*}
\varepsilon_{\mu\nu\rho\sigma}F_i^{\mu\nu}F^{\rho\sigma}_i &= \varepsilon^{\mu\nu\rho\sigma}F^i_{\mu\nu}F_{\rho\sigma}^i \\
&= \varepsilon^{\mu\nu\rho\sigma}(\partial_{\mu} A_{\nu}^i - \partial_{\nu} A_{\mu}^i)F_{\rho\sigma}^i \\
&= 2\varepsilon^{\mu\nu\rho\sigma}\partial_{\mu} A_{\nu}^i F_{\rho\sigma}^i \\
&= 2 \partial_\mu\left(\varepsilon^{\mu\nu\rho\sigma}A_\nu^i F_{\rho\sigma}^i\right) - 2 \varepsilon^{\mu\nu\rho\sigma}A_\nu^i \partial_\mu F_{\rho\sigma}^i
\end{align*}

Mmm so there is a 2 factor mismatch between your result and mine.

Using the definition of $F$ and symmetry is not difficult to show that
$$\varepsilon^{\mu\nu\rho\sigma}\partial_\mu F_{\rho\sigma}^i=0$$

\begin{align*}
\varepsilon^{\mu\nu\rho\sigma}\partial_\mu F_{\rho\sigma}^i &= \varepsilon^{\mu\nu\rho\sigma}\partial_\mu(\partial_\rho A_\sigma^i - \partial_\sigma A_\rho^i)\\
&= \varepsilon^{\mu\nu\rho\sigma}\partial_\mu \partial_\rho A_\sigma^i - \varepsilon^{\mu\nu\rho\sigma}\partial_\mu \partial_\sigma A_\rho^i \\
&= 0
\end{align*}

Where we noticed that the object $\partial_\mu \partial_\rho A_\sigma^i$ is symmetric under $\mu \leftrightarrow \rho$ and is contracted with the object $\varepsilon^{\mu\nu\rho\sigma}$, which is antisymmetric under $\mu \leftrightarrow \rho$. Hence $\varepsilon^{\mu\nu\rho\sigma}\partial_\mu \partial_\rho A_\sigma^i = 0$. Analogously, $\varepsilon^{\mu\nu\rho\sigma}\partial_\mu \partial_\sigma A_\rho^i=0$

 

[JD_PM]

So let us now draw our attention back to the main prove. We first see that $\varepsilon_{\mu\nu\rho\sigma} g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} + \varepsilon_{\mu\nu\rho\sigma} g_s f_{ijk} F_i^{\rho \sigma} A_j^{\mu} A_k^{\nu} = 2 \varepsilon_{\mu\nu\rho\sigma}g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma}$, given that $\varepsilon_{\mu\nu\rho\sigma} = \varepsilon_{\rho\sigma\mu\nu}$.

All is left is to work out the term (let me drop the 2 factor for the explicit computation) $\varepsilon_{\mu\nu\rho\sigma}g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma}$

\begin{align*}
\varepsilon_{\mu\nu\rho\sigma}g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} &= \varepsilon_{\mu\nu\rho\sigma}g_s f_{ilm}(\partial^\mu A^\nu_i - \partial^\nu A^\mu_i)A_l^{\rho} A_m^{\sigma} \\
&= 2 \varepsilon_{\mu\nu\rho\sigma} g_s f_{ilm} \partial^\mu A^\nu_i A_l^{\rho} A_m^{\sigma}\\
&= \frac 2 3 \partial^\mu (\varepsilon_{\mu\nu\rho\sigma} g_s f_{ilm} A^\nu_i A_l^{\rho} A_m^{\sigma})\\
&= \partial^\mu (\varepsilon_{\mu\nu\rho\sigma} A^\nu_i A_l^{\rho} A_m^{\sigma}) -\frac 1 3 \partial^\mu (\varepsilon_{\mu\nu\rho\sigma} A^\nu_i A_l^{\rho} A_m^{\sigma})
\end{align*}

This smells really good! (at least to me!)

We finally have

\begin{align*}
G \cdot \tilde G &= \varepsilon_{\mu \nu \rho \sigma} G_i^{\mu \nu}G_i^{\rho \sigma}\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu} + g_s f_{ijk} A_j^{\mu} A_k^{\nu})(F_i^{\rho \sigma} + g_s f_{ilm} A_l^{\rho} A_m^{\sigma})\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu}F_i^{\rho \sigma} + g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} + g_s f_{ijk} A_j^{\mu} A_k^{\nu} F_i^{\rho \sigma} + g^2_sf_{ijk}f_{ilm}A_j^{\mu} A_k^{\nu}A_l^{\rho} A_m^{\sigma})\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu}F_i^{\rho \sigma} + g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} + g_s f_{ijk} A_j^{\mu} A_k^{\nu} F_i^{\rho \sigma})\\
&= \varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu}F_i^{\rho \sigma} + 2g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma})\\
&= 2\partial_{\mu}(\varepsilon^{\mu \nu \rho \sigma} A_{\nu}^i [G_{\rho \sigma}^i - \frac{g_s}{3}f_{ijk} A_{j\rho} A_{k\sigma}])
\end{align*}

Mmm so my result has a 2 extra factor...

This issue is simply solved by redefining $G \cdot \tilde G := \frac 1 2 \varepsilon_{\mu \nu \rho \sigma} G_i^{\mu \nu}G_i^{\rho \sigma}$ but I might have missed some detail in the computation and that is why I get the 2 extra factor

What do you think?

 

[Gaussian97]

Ok, I think the factor 2 is due to the definition of $\tilde{G}$, but anyway I don't think it is really important.
Regarding your proof, there are some technical details missing, but I think that you have the general idea.

 

[JD_PM]

Ok, I think the factor 2 is due to the definition of $\tilde{G}$

Oh indeed! The Hodge dual tensor is actually defined as $\tilde{G}_{i\mu \nu} := \frac 1 2 \varepsilon_{\mu \nu \rho\sigma} G_i^{\rho \sigma}$ (I made the analogy from QED from here) so the 2 factor is taken care by the $\frac 1 2$ included in the dual tensor, so the guess was right!

Vamos! (this is a mother-tongue word I love to use when celebrating something; you could translate it as "let's go!" :) ).

I have recently started to learn more about gauge theories in the context of particle physics and it is a truly fascinating subject. I am currently reading these fantastic notes written by Australian physicist Rod Crewther.

Another discussion in which I have learned A LOT. Thank you very much @Gaussian97, I hope to come across you very very soon.

 

[vanhees71]

A nice older review is by Abers and Lee:

E. Abers and B. Lee, Gauge Theories, Phys. Rept. 9, 1
(1973), https://doi.org/10.1016/0370-1573(73)90027-6