Invariance of Yang-Mills Lagrangian under charge conjugation

Question

The Yang-Mills Lagrangian gauge invariant under an $SU(N)$ tranformation can be written as

$${\cal L} = -\frac{1}{4}F_{\mu\nu}^i F^{i\ \mu\nu} \tag1$$

(Sum over $i$ implicit)

This Lagrangian contains a term of the form

$${\cal L'} = -g·f_{ijk}A_\mu^i A_\nu^j\partial^\mu A^{k\ \nu} \tag2$$

$f_{ijk}$ are the structure constants.

Under charge conjugation, the self-adjoint gauge field transforms as

$$A_\mu \rightarrow -A_\mu \tag3$$

And therefore Eq. (2) isn't charge conjugation (${\cal C}$) invariant while the other terms in the complete Lagrangian (Eq. (1)) are invariant. This implies that the QCD Lagrangian isn't ${\cal C}$-invariant.

But is this correct or what am I misunderstanding?

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Also, you can't just pick $A_\mu \rightarrow A_\mu$ because in that case the couplings to matter, i.e., $J_\mu A^\mu = \bar{\psi}\gamma_\mu \psi A^\mu$ wouldn't be ${\cal C}$-invariant since $J_\mu \rightarrow -J_\mu$; as you can check in Invariance of the QED Lagrangian under charge conjugation

Answer 1

Just like the asker's transformation in the question you link, your transformations are wrong. Charge conjugation is literal conjugation - you are replacing all fields that transform in a non-trivial representation of the gauge group by fields that transform in its conjugate representation. It just so happens that this transformation is the same as "inverting the sign" for a $\mathrm{U}(1)$ gauge field, but this is not true for a more general non-Abelian field. Conjugation means taking the representation $\rho(g)$ of the group $T^a$ and replacing it by $\overline{\rho(g)}$, where $\bar{}$ is the transpose in the representation space. Given that a group element is written as an exponential $\exp(\mathrm{i}A^a T^a)$ for $T^a$ the representations of generators of the algebra, you can see that this sends $\rho(T^a)$ to $-\overline{T^a}$.

For a $\mathrm{U}(1)$-theory, the $T^a$ is just equal to the identity on the one-dimensional representation space, meaning you can express the charge conjugation map as $A^a \mapsto -A^a$ map on the coefficients $A^a$, explaining the origin of the minus sign in the Abelian case.

Answer 2

Let me try to bring some physical intuition to what happens under charge conjugation. In the case of Yang-Mills with SU(N) the charge conjugation on the gauge fields acts explicitly in the following manner $$ \mathcal{C} A_\mu^i \mathcal{C} \,T^i = - A_\mu^j (T^j)^T , $$ where $T^j$ are the generators of $SU(N)$, more specifically you can compute the transformation by expressing $T^i$ in the fundamental representation $$ \mathcal{C} A_\mu^i \mathcal{C} = - 2 \text{tr}(T^i (T^j)^T) A_\mu^j = -M^{ij}A_\mu^j, $$ since $M$ is a symmetric matrix it can always be diagonalized. Specifically in the case of $U(1)$ the only generator is the identity, and that is why the photon gets transformed into the negative of itself. In the case of SU(2) then $M^{ij}$ will look like $$ M = \begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. $$ You can imagine that there is a basis where two of the gauge bosons are exchanged under charge conjugation, where the matrix will be $$ M = \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}, $$ which is what the intuition would tell us about how the $Z$ boson only flips sign like the photon under a charge conjugation, but the $W^\pm$ bosons will get exchanged. Finally, to address more specifically your question, this transformation induces a flip sing in the structure constants $f_{ijk}$, so that the cross term of the Lagrangian that you were worried about is charge conjugation invariant.

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The flip in sign of the structure constants arises since $$ \mathcal{L}^\prime=\text{tr}(\partial_\mu A^i_\nu T^i[A^{j\mu}T^j,A^{k\mu}T^k])\to \text{tr}(-\partial_\mu A^i_\nu (T^i)^T[-A^{j\mu}(T^j)^T,-A^{k\mu}(T^k)^T])\\ =-\partial_\mu A^i_\nu A^{j\mu}A^{k\mu}\text{tr}\{(T^i)^T[(T^j)^T,(T^k)^T]\}=-\partial_\mu A^i_\nu A^{j\mu}A^{k\mu}(-f_{jki})/2\,. $$