In Weinberg's Gravitation and Cosmology, the author mentions that an infinitesimal Lorentz transformation (in the four-vector representation of the Lorentz group) has the form $$\Lambda^{\alpha}_{\phantom{\alpha}\beta}=\delta^{\alpha}_{\phantom{\alpha}\beta}+\omega^{\alpha}_{\phantom{\alpha}\beta}.\tag{$\dagger$}$$ It is then straightforward to verify that the $\omega$-matrix must satisfy $$\omega_{\gamma\delta}=-\omega_{\delta\gamma}.\tag{$*$}$$ I'm okay with that. Then, Weinberg says that the matrix representation $D(\Lambda)$ of such a transformation (now a general, say $n\times n$ representation) must satisfy $$D(1+\omega)=1+\frac{1}{2}\omega^{\alpha\beta}\sigma_{\alpha\beta}\tag{$**$}$$ where $\sigma_{\alpha\beta}$ are a set of matrices which may be chosen to be antisymmetric, by virtue of (*).
I have no idea where the exact form ( ** ) comes from. What's a little confusing to me is that in (†), I believe that $\omega$ may be any any linear combination of the generators of the four-vector representation of the Lorentz group and $\omega^\alpha_{\phantom{\alpha}\beta}$ are its matrix elements, while in (**) $\omega$ seems to be a set of infinitesimal parameters, whilst $\sigma_{\alpha\beta}$ are now the generators (I think I know the $1/2$ prevents counting each generator twice).