parity invariance of Einstein, Maxwell and Dirac Lagrangians How can we show that Einstein, Maxwell and Dirac Lagrangians are parity invariant?
 A: Parity is a symmetry that we impose to the Dirac and Maxwell Lagrangian because we know experimentally that electromagnetism conserves parity. Demanding parity invariance will then tell you how Dirac spinors and vectors transform under such a symmetry. The correct transformation properties are:
$\bullet$ Spinors: $$\mathcal{P}\thinspace\psi(t,x)\mathcal{P}^\dagger\ =\ \gamma^0\ \psi\thinspace(t,-x)\ ,$$
$$\mathcal{P}\thinspace\bar{\psi}\thinspace(t,x)\mathcal{P}^\dagger\ =\   \psi^\dagger\thinspace(t,-x)\ ,$$
$\bullet$ Vectors:
$$\mathcal{P}\thinspace V^\mu\thinspace(t,x)\mathcal{P}^\dagger\ =\  \eta_{\mu\nu}\thinspace V^\nu\thinspace(t,-x)\ =\ V_\mu(t,-x)$$
Here by vectors I mean quantities such as the potential $A^\mu$, the partial derivative $\partial^\mu$ or the current $\bar{\psi}\gamma^\mu\psi$.
You can check that these are the transformation properties you want. The Dirac Lagrangian is
$$\mathcal{L}_D=\bar{\psi}(i\gamma_\mu\partial^\mu-m)\psi.$$
It is Parity invariant since both the mass and kinetic term are Parity invariant:
$$\mathcal{P}\bar{\psi}\psi\mathcal{P}^\dagger\ =\ \mathcal{P}\bar{\psi}\mathcal{P}^\dagger\thinspace\mathcal{P}\psi\mathcal{P}^\dagger\ =\ \psi^\dagger\gamma^0\psi\ =\ \bar{\psi}\psi.$$
$$\mathcal{P}\thinspace\bar{\psi}\gamma_\mu\partial^\mu\psi\mathcal{P}^\dagger\ =\ \mathcal{P}\bar{\psi}\mathcal{P}^\dagger\thinspace\gamma_\mu\mathcal{P}\partial^\mu\mathcal{P}^\dagger\mathcal{P}\psi\mathcal{P}^\dagger\ =\ \psi^\dagger\gamma_\mu\gamma^0\partial_\mu\psi\ =\ \bar{\psi}\gamma^\mu\partial_\mu\psi$$
where in the last equality we used the property $\gamma^0\gamma^\mu=\gamma_\mu\gamma^0$.
The Maxwell Lagrangian is
$$\mathcal{L}_M\ = \ -\frac{1}{4} F_{\mu\nu}F^{\mu\nu}$$
where the Maxwell tensor is defined as $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$. From this definition it is clear that each index in $F^{\mu\nu}$ transforms as a vector, so
$$\mathcal{P} F^{\mu\nu}\mathcal{P}^\dagger=F_{\mu\nu}.$$
Consequently Maxwell's lagrangian is parity invaiant.
By Einstein Lagrangian, you mean the Einstein-Hilbert action? or Einsteins equations?
