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The Lorentz invariant term $\epsilon^{\mu\nu\sigma\rho}F_{\mu\nu}F_{\sigma\rho}$ is not parity invariant. To show this one needs to find the parity transformation property of $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. Under parity i.e., $x^0\to x^0, x^i\to -x^i$, $$\partial_0\to\partial_0,~~\partial_i\to -\partial_i~~\text{and}~~ A_0\to A_0,~~A_i\to -A_i.\tag{1}$$

$\bullet$ To find the transformation property of $F_{\mu\nu}$, do I need to check each component of $F_{\mu\nu}$ separately? That would seem rather inelegant. I also do not want to change it into $\textbf{E}$ and $\textbf{B}$ and use their parity transformation properties.

$\bullet$ How do I show that $\epsilon^{\mu\nu\sigma\rho}$ change sign under parity? I have no clue why should it even change sign under parity.

$\bullet$ What about the parity property of gluon field strength $G_{\mu\nu}^a=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+g f^{abc}A_\mu^b A_\nu^c$? To find the relations in Eq.(1), I used the behaviour of $\textbf{E}$ and $\textbf{B}$ from classical electrodynamics. Where do we get the behaviour of gluon fields $A_\mu^\nu$ under parity? I derived (1) from $\textbf{E}=-\nabla A^0-\frac{\partial \textbf{A}}{\partial t}$ and $\textbf{B}=\nabla\times\textbf{A}$, and the parity properties of $\textbf{E}$ and $\textbf{B}$ where $A_\mu$ is the electromagnetic four-potential. On the other hand, $A_\mu^a$ are the gluons fields , and it's not obvious that they would behave the same way as in (1) under parity. I know that color index will not change under parity.

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    $\begingroup$ Under a general transformation $A$, we have $\epsilon^{\mu \nu \rho \sigma } \mapsto A^\mu{}_\alpha A^\nu{}_\beta A^\rho{}_\gamma A^\sigma{}_\delta \epsilon^{\alpha \beta \gamma \delta} = \mathrm{det}(A) \epsilon^{\mu \nu \rho \sigma}$. If $A$ is a parity transformation then its determinant is $-1$. $\endgroup$ – gj255 Jun 14 '18 at 14:41
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Since under parity only the spatial component of a 4-vector change, you can see the effect of a parity transformation as such

$$V^\mu \to V_\mu\quad\text{and}\quad V_\mu\to V^\mu$$

So you have that $F^{\mu\nu}\to F_{\mu\nu}$ and thus $F_{\mu\nu} F^{\mu\nu}\to F^{\mu\nu} F_{\mu\nu}$ is invariant.

For the second case you have

$$\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}\to\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$$

and you can see that you get a minus sign since $\epsilon^{\mu\nu\rho\sigma}=-\epsilon_{\mu\nu\rho\sigma}$

The reason why the $\epsilon$ changes under parity is that when you do a Lorentz transformation of this tensor you get the determinant of the Lorentz transformation matrix and a parity transformation is a Lorentz transformation with determinant $-1$.

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  • $\begingroup$ Great! I liked your argument. What do you have to say about the gluon fields? How do we know how $A_\mu^a$ changes under parity? @FrodCube $\endgroup$ – SRS Jun 14 '18 at 14:56
  • $\begingroup$ @SRS it is the same thing since the color index $a$ is unaffected by Lorentz transformations $\endgroup$ – FrodCube Jun 14 '18 at 15:24
  • $\begingroup$ I derived (1) from $\textbf{E}=-\nabla A^0-\frac{\partial \textbf{A}}{\partial t}$ and $\textbf{B}=\nabla\times\textbf{A}$, and the parity properties of $\textbf{E}$ and $\textbf{B}$ where $A_\mu$ is the electromagnetic four-potential. On the other hand, $A_\mu^a$ are the gluons fields , and it's not obvious that they would behave the same way as in (1) under parity. I know that color index will not change under parity. @FrodCube $\endgroup$ – SRS Jun 15 '18 at 5:12

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