Is the $U(1)_A$ axial vector current even under charge conjugation? The axial current of a Dirac spinor is given by $j_A^\mu = \bar{\psi} \gamma^5 \gamma^\mu \psi$. In this book, in the paragraph under equation (2.18) it is stated that the current is even under charge conjugation. I am trying to show this. Charge conjugation of a spinor $\psi$ is defined as $\psi^{(c)} = C \psi^*$, where $C$ is the unitary charge conjugation matrix which satisfies $C^\dagger \gamma^\mu C = -(\gamma^\mu)^*$ for all gamma matrices and $\bar{\psi}=\psi^\dagger \gamma^0$.
We have 
$$ (j^\mu_A)^{(c)} = \overline{\psi^{(c)}}\gamma^5 \gamma^\mu \psi^{(c)} \\
=(C\psi^*)^\dagger \gamma^0 \gamma^5\gamma^\mu C \psi^* \\
= (\psi^\dagger)^* (C^\dagger\gamma^0 C)(C^\dagger \gamma^5 C)(C^\dagger \gamma^\mu C) \psi^* \\
= - (\psi^\dagger)^*(\gamma^0)^*(\gamma^5)^* (\gamma^\mu)^*\psi^* \\
=(\psi^\dagger \gamma^0\gamma^5 \gamma^\mu \psi)^* \\
= (\bar{\psi}\gamma^5 \gamma^\mu \psi)^*$$
where going from the 4th to 5th line, pulling out the complex conjugate picks up a minus sign as the components of the spinors are anti-commuting Grassman numbers. Now that book is telling me that the current is even under charge conjugation so I would expect $\bar{\psi}\gamma^5 \gamma^\mu \psi$ is real, but I can't show this.
Any help would be greatly appreciated.
 A: You're almost there. We know that the components of the current are complex numbers, and so we can rewrite your second last line as
\begin{equation}
(\psi^\dagger \gamma^0 \gamma^5 \gamma^\mu \psi)^* = (\psi^\dagger \gamma^0 \gamma^5 \gamma^\mu \psi)^\dagger.
\end{equation}
We then perform the Hermitian conjugate, and use following the properties of the gamma matrices 
\begin{align}
(\gamma^0)^\dagger &= \gamma^0, \\
(\gamma^5)^\dagger &= \gamma^5, \\
(\gamma^\mu)^\dagger &= \gamma^0 \gamma^\mu \gamma^0, \\
\{\gamma^5,\gamma^\mu\} &= 0, \\
(\gamma^0)^2 &= \mathbb{I}_{4},
\end{align}
where $\mathbb{I}_{4}$ is the identity to proceed as follows:
\begin{equation}
\begin{split}
(\psi^\dagger \gamma^0 \gamma^5 \gamma^\mu \psi)^\dagger &= \psi^\dagger (\gamma^\mu)^\dagger (\gamma^5)^\dagger (\gamma^0)^\dagger \psi \\
&= \psi^\dagger \gamma^0 \gamma^\mu \gamma^0 \gamma^5 \gamma^0 \psi\\
&= -\bar{\psi} \gamma^\mu \gamma^5 (\gamma^0)^2 \psi \\
&= \bar{\psi} \gamma^5 \gamma^\mu \psi, 
\end{split}
\end{equation}
as required.
