Parity transformation on fermionic bilinears

Question

In the Fermi weak theory we have the fermion bilinears which look like

$$ V_\mu = \bar{\psi} \gamma_\mu\psi $$ $$ A_\mu = \bar{\psi} \gamma_\mu \gamma_5 \psi $$

Under a parity transformation

$$ x = (x_0, \vec{x}) \rightarrow \tilde{x} = ( x_0, - \vec{x}) $$

The fields transform like

$$ V^\mu(x) \rightarrow V_\mu(\tilde{x}) $$ $$ A^\mu(x) \rightarrow - A_\mu(\tilde{x}) $$

Why do the contravariant indices transform to covariant indices as well as a coordinate transformation? I thought it would have something to do with the fact that you also have to transform the actual vector components under a coordinate/parity transformation, but I don´t know how to formalize it starting from the explicit form of the bilinears.

Answer 1

A four vector $V^\mu$ has components that we can regard as $(\rho, {\bf j})$. Under parity the charge $\rho$ does not change, but the current changes sign: $P:{\bf j}\mapsto -{\bf j}$. If the metric signature is $(+,-,-,-)$ one way to say this is $V^\mu \to V_\mu$. If the metric were $(-,+,+,+)$ we would have to say $V^\mu\to -V_\mu$.