Proving the Maxwell action is conformally invariant

Question

I want to show that the Maxwell action $$S = -\frac{1}{4}\int d^4 x F_{\mu\nu} F^{\mu\nu}$$ is invariant under conformal transformations in $d=4$.

For this I considered the proof given in Zee's book on GR, section IX.9 page 621. However, I think he is making an error there. His proof goes as follows. For a conformal transformation $x'^\mu = x^\mu - \xi^\mu,$ the gauge boson transforms as the Lie derivative $\delta A_\mu = \mathcal{L}_\xi A_\mu$. From here he generalizes this to the field strength, i.e. $\delta F_{\mu\nu} = \mathcal{L}_\xi F_{\mu\nu}$. But then when computing the action on the Lagrangian density, he writes $$\delta(F_{\mu\nu}F^{\mu\nu}) = 2 F^{\mu\nu}\delta F_{\mu\nu}.\tag{IX.9.16}$$ This seems incorrect to me, because the Lie derivative transforms differently depending on whether the indices are upstairs or downstairs. I would expect to see $$\delta(F_{\mu\nu}F^{\mu\nu}) = F^{\mu\nu}\delta F_{\mu\nu} +F_{\mu\nu}\delta F^{\mu\nu}.$$

Now I want to use the fact that these variations are Lie derivatives found by

\begin{align} \delta F^{\mu\nu} &= -(\partial_\rho \xi^\mu) F^{\rho\nu} - (\partial_\rho \xi^\nu) F^{\mu\rho} + \xi^\alpha \partial_\alpha F^{\mu\nu}\\ \delta F_{\mu\nu} &= (\partial_\mu \xi^\rho) F_{\rho\nu} + (\partial_\nu \xi^\rho) F_{\mu\rho} + \xi^\alpha \partial_\alpha F_{\mu\nu}.\tag{IX.9.16} \end{align}

However, this results in

$$\delta(F_{\mu\nu}F^{\mu\nu}) = \partial_\alpha(\xi^\alpha F_{\mu\nu}F^{\mu\nu}) - (\partial \cdot \xi)F_{\mu\nu}F^{\mu\nu}.$$

This is decisively not what we should obtain.

How can I fix this proof?

Answer 1

Conformal transformations are NOT just diffeomorphisms. Rather, they are a special type of diffeomorphism (conformal diffs) followed by a particular Weyl transformation. Let's see this precisely,

1. Under infinitesimal diffeomorphisms $$ \delta^D_\zeta g_{\mu\nu} = {\cal L}_\zeta g_{\mu\nu} , \qquad \delta^D_\zeta \Phi = {\cal L}_\zeta \Phi . $$

2. Under infinitesimal Weyl transformations, $$ \delta^W_\omega g_{\mu\nu} = 2 \omega g_{\mu\nu} , \qquad \delta^W_\omega \Phi = - (\Delta-s) \omega \Phi . $$$s$ is the spin of $\Phi$. If $\Phi$ is a tensor field, then $s = \# \text{covariant indices} - \# \text{contravariant indices}$. Fields $\Phi$ transforming this way are known as primary fields. Non-primary fields have more complicated Weyl transformations.

3. An infinitesimal conformal transformation is an infinitesimal diffeomorphism generated by a conformal Killing vector field $\xi$ + an infinitesimal Weyl transformation with $\omega(\xi) = - \frac{1}{d} \nabla_\alpha \xi^\alpha$. In particular, \begin{align} \delta_\xi g_{\mu\nu} &= \delta^D_\xi g_{\mu\nu} + \delta^W_{\omega(\xi)} g_{\mu\nu} = {\cal L}_\xi g_{\mu\nu} - \frac{2}{d} \nabla_\alpha \xi^\alpha g_{\mu\nu} , \\ \delta_\xi \Phi &= \delta^D_\xi \Phi + \delta^W_{\omega(\xi)} \Phi = {\cal L}_\xi \Phi + \frac{\Delta-s}{d} \nabla_\alpha \xi^\alpha \Phi . \end{align} In particular, because $\xi$ is a conformal Killing vector, $\delta_\xi g_{\mu\nu} = 0$.

---

Let us now apply all of this to your problem. In this case, $d=4$. $F_{\mu\nu}$ has $\Delta=2, s=2$ and $F^{\mu\nu}$ has $\Delta=2, s=-2$. This implies $$ \delta_\xi F_{\mu\nu} = {\cal L}_\xi F_{\mu\nu} , \qquad \delta_\xi F^{\mu\nu} = {\cal L}_\xi F^{\mu\nu} + \nabla_\alpha \xi^\alpha F^{\mu\nu} . $$

We can evaluate $\delta_\xi ( F^{\mu\nu} F_{\mu\nu} )$ in three ways.

1. We first follow the method you propose in the question and write \begin{align} \delta_\xi ( F^{\mu\nu} F_{\mu\nu} ) &= \delta_\xi F^{\mu\nu} F_{\mu\nu} + F^{\mu\nu} \delta_\xi F_{\mu\nu} \\ &= ( {\cal L}_\xi F^{\mu\nu} + \nabla_\alpha \xi^\alpha F^{\mu\nu} ) F_{\mu\nu} + F^{\mu\nu} {\cal L}_\xi F_{\mu\nu} \\ &= \nabla_\alpha ( \xi^\alpha F^{\mu\nu} F_{\mu\nu} ) . \end{align}

2. We use the fact that the metric $g$ is invariant under conformal transformations (by definition), so we have \begin{align} \delta_\xi ( F^{\mu\nu} F_{\mu\nu} ) &= 2 F^{\mu\nu} \delta_\xi F_{\mu\nu} \\ &= 2 F^{\mu\nu} {\cal L}_\xi F_{\mu\nu} \\ &= \nabla_\alpha ( \xi^\alpha F^{\mu\nu} F_{\mu\nu} ) . \end{align}

3. Following the same argument as before, we can obviously also write \begin{align} \delta_\xi ( F^{\mu\nu} F_{\mu\nu} ) &= 2 \delta_\xi F^{\mu\nu} F_{\mu\nu} \\ &= 2 ( {\cal L}_\xi F^{\mu\nu} + \nabla_\alpha \xi^\alpha F^{\mu\nu} ) F_{\mu\nu} \\ &= \nabla_\alpha ( \xi^\alpha F^{\mu\nu} F_{\mu\nu} ) . \end{align}

Of course, all three answers are the same, as they should be.