Lorentz transformation of an antisymmetric tensor I'm trying to find the infinitesimal Lorentz transformation of a rank 2 antisymmetric tensor. Looking through Peskin, all I can see is the transformation of a vector, and even there it is simply given. I thought about developing it by writing it as a tensor product of two rank-1 tensors. This gives me:
$$
C_{\mu\nu} = A_\mu B_\nu - A_\nu B_\mu\\
\delta C_{\mu\nu} = \delta A_\mu B_\nu + A_\mu\delta B_\nu - \delta A_\nu B_\mu - A_\nu\delta B_\mu
$$
where
$$
\delta A_\mu = \epsilon_\mu^{\phantom\mu\nu}A_\nu - \epsilon^\nu_{\phantom\nu\mu}A_\nu
$$
Can this also be generalized for any rank-n tensor? Is this the way to find the spin tensor for every integer Lorentz representation?
 A: Sort of, except that you can't generally decompose a rank-2 tensor into a product of rank-1 tensors.
Let $\Lambda^{\mu}{}_{\nu}$ be an arbitrary Lorentz transformation. As you probably saw in Peskin, this transformation acts on vectors as
$$x^{\mu} \mapsto \Lambda^{\mu}{}_{\nu} x^{\nu}.$$
We can extend this principle to a tensor with an arbitrary number of up-indices. For example, for a rank-2 tensor $T^{\mu \nu},$ we have
$$T^{\mu \nu} \mapsto \Lambda^{\mu}{}_{\rho} \Lambda^{\nu}{}_{\sigma} T^{\rho \sigma}.$$
So, for example, since the statement that $\Lambda$ is a Lorentz transformation is equivalent to the statement that it leaves the Minkowski metric $\eta^{\mu \nu}$ invariant, $\Lambda$ must satisfy $\eta^{\mu \nu} \mapsto \eta^{\mu \nu},$ or
$$\Lambda^{\mu}{}_{\rho} \Lambda^{\nu}{}_{\sigma} \eta^{\rho \sigma} = \eta^{\mu \nu}.$$
Now, how should $\Lambda$ act on down-indices? Well, we can obtain a down-index from an up-index by lowering using the metric. So, starting with $x_{\mu} = \eta_{\mu \nu} x^{\nu}$ and using the fact that $\Lambda$ leaves the metric invariant, we have
$$x_{\mu} = \eta_{\mu \nu} x^{\nu} \mapsto \eta_{\mu \nu} \Lambda^{\nu}{}_{\rho} x^{\rho} = \Lambda_{\mu}{}^{\rho} x_{\rho}.$$
This tells us how $\Lambda$ should act on down-indices. However, in the particular case of Lorentz transformations, the tensor $\Lambda_{\mu}{}^{\rho}$ has a particular property. If we multiply it by the tensor $\Lambda^{\tau}{}_{\rho},$ we find
$$\Lambda^{\tau}{}_{\rho} \Lambda_{\mu}{}^{\rho}
= \Lambda^{\tau}{}_{\rho} (\eta_{\mu \nu} \Lambda^{\nu}{}_{\sigma} \eta^{\sigma \rho}) = \eta_{\mu \nu} (\Lambda^{\nu}{}_{\sigma} \Lambda^{\tau}{}_{\rho} \eta^{\sigma \rho}) = \eta_{\mu \nu} \eta^{\nu \tau} = \delta_{\mu}{}^{\tau}.$$
So we have $\Lambda^{\tau}{}_{\rho} \Lambda_{\mu}{}^{\rho} = \delta_{\mu}{}^{\tau} = \Lambda^{\tau}{}_{\rho} (\Lambda^{-1})^{\rho}{}_{\mu}.$
So, generally, we conclude $\Lambda_{\mu}{}^{\rho} = (\Lambda^{-1})^{\rho}{}_{\mu}.$ So while up-indices transform naturally under $\Lambda$, down-indices transform naturally under $\Lambda^{-1}.$ That is,
$$x_{\mu} \mapsto \Lambda_{\mu}{}^{\rho} x_{\rho} = (\Lambda^{-1})^{\rho}{}_{\mu} x_{\rho}.$$
Now it's easy to answer your original question of "how does a rank-2 tensor $C_{\mu \nu}$ transform under a Lorentz transformation?" As in the case of multiple up-indices, we can just extend our principle to see
$$
C_{\mu \nu} \mapsto (\Lambda^{-1})^{\rho}{}_{\mu} (\Lambda^{-1})^{\sigma}{}_{\nu} C_{\rho \sigma}.
$$
Or, equivalently, 
$$
C_{\mu \nu} \mapsto \Lambda_{\mu}{}^{\rho} \Lambda_{\nu}{}^{\sigma} C_{\rho \sigma}.
$$
