quantum field theory - What does Weinberg–Witten theorem want to express?

Weinberg-Witten theorem states that massless particles (either composite or elementary) with spin $j > 1/2$ cannot carry a Lorentz-covariant current, while massless particles with spin $j > 1$ cannot carry a Lorentz-covariant stress-energy. The theorem is usually interpreted to mean that the graviton ( $j = 2$ ) cannot be a composite particle in a relativistic quantum field theory.

Before I read its proof, I've not been able to understand this result. Because I can directly come up a counterexample, massless spin-2 field have a Lorentz covariant stress-energy tensor. For example the Lagrangian of massless spin-2 is massless Fierz-Pauli action:

$$S=\int d^4 x (-\frac{1}{2}\partial_a h_{bc}\partial^{a}h^{bc}+\partial_a h_{bc}\partial^b h^{ac}-\partial_a h^{ab}\partial_b h+\frac{1}{2}\partial_a h \partial^a h)$$

We can calculate its energy-stress tensor by $T_{ab}=\frac{-2}{\sqrt{-g}} \frac{\delta S}{\delta g^{ab}}$, so we get

$$T_{ab}=-\frac{1}{2}\partial_ah_{cd}\partial_bh^{cd}+\partial_a h_{cd}\partial^ch_b^d-\frac{1}{2}\partial_ah\partial^ch_{bc}-\frac{1}{2}\partial^ch\partial_ah_{bc}+\frac{1}{2}\partial_ah\partial_bh+\eta_{ab}\mathcal{L}$$ which is obviously a non-zero Lorentz covariant stress-energy tensor.

And for U(1) massless spin-1 field, we can also have the energy-stress tensor

$$T^{ab}=F^{ac} F^{b}_{\ \ \ c}-\frac{1}{4}\eta^{ab}F^{cd}F_{cd}$$ so we can construct a Lorentz covariant current $J^a=\int d^3x T^{a 0}$ which is a Lorentz covariant current.

Therefore above two examples are seeming counterexamples of this theorem. I believe this theorem must be correct and I want to know why my above argument is wrong.

Answer

1. 
   

   The stress tensor for $h_{ab}$ is not Lorentz covariant, despite the fact that it looks like it is. This is because $h_{ab}$ itself is not a Lorentz tensor. Rather under Lorentz transformations

   $$h_{ab} \to \Lambda_a{}^c \Lambda_b{}^d h_{cd} + \partial_a \zeta_b + \partial_b \zeta_a ~.$$ The extra term is present to make up for the fact that $h_{ab}$ is not a tensor of the Lorentz group. Plug this into the stress tensor and you will find that the stress tensor also transforms with a inhomogeneous piece thereby making it non-covariant.
   
   


2. 
   

   The photon is not charged under the $U(1)$ gauge symmetry. Thus, its $U(1)$ current is zero. The current you have defined is not the $U(1)$ current. Rather it is the current corresponding to translations. Weinberg-Witten theorem has nothing to say about this current.