# How do the Lorentz generators act on a rank 2 tensor?

Let $M_{ab}$ be the Lorentz generators in 4 space-time dimensions in an arbitrary representation of the Lorentz group. In the $s=(\frac{1}{2},0)$, $\bar{s}=(0,\frac{1}{2})$ and $v=(\frac{1}{2},\frac{1}{2})$ representations the $M_{ab}$ act on spinors and vectors respectively as \begin{align} ~^sM_{ab}(\psi_{\alpha})&=(\sigma_{ab})_{\alpha}^{~~\beta}\psi_{\beta}\\ ~^{\bar{s}}M_{ab}(\bar{\psi}^{\dot{\alpha}})&=(\tilde{\sigma}_{ab})^{\dot{\alpha}}_{~~\dot{\beta}}\bar{\psi}^{\dot{\beta}}\\ ~^vM_{ab}(V^c)&=(\delta_a^cV_b-\delta_b^cV_a). \end{align} The vector representation $v=(\frac{1}{2},\frac{1}{2})$ is the tensor product $v=s\otimes\bar{s}$. Similarly, the Lorentz generators which act on rank 2 tensors would be in the $v\otimes v$ representation, but how do I write down its action like the above? $$~^{v\otimes v}M_{ab}(T^{cd})=\cdots?$$ Is there a systematic method to find how they act on arbitrary rank tensors, spinors and spin-tensors?

If I was to take a guess I would say $$~^{v\otimes v}M_{ab}(T_{cd})=(\eta_{ac}T_{bd}-\eta_{bc}T_{ad}+\eta_{ad}T_{cb}-\eta_{bd}T_{ca}).$$ But now I am confused if their action on an arbitrary rank 2 tensor is even defined, since the representations are realised on totally symmetric spinors (and hence totally antisymmetric tensors).

The space of rank-2 tensors is not an irreducible representation. The representation on all rank-2 tensors decomposes as $$(\frac{1}{2},\frac{1}{2})\otimes(\frac{1}{2},\frac{1}{2}) = \underbrace{\mathbf{1}}_\text{trace} + \underbrace{(1,0)}_\text{self-dual antisymmetric} + \underbrace{(0,1)}_\text{anti-self-dual antisymmetric} + \underbrace{(1,1)}_\text{traceless symmetric}.$$