Infinitesimal Lorentz Transformations 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).
 A: This is what's happening: you have $\vartheta^{\gamma\delta}$ parameters for describing the $\text{SO}^+(1,3)$ group; they constitute an antisimmetric matrix in their $\gamma,\delta$ indexes, so that the actual free parameters are the usual $6$ for the proper Lorentz transformation.
That said consider that an infinitesimal proper Lorentz transformation will be
$$
{\Lambda^\alpha}_\beta
\approx
{\mathbb{I}^\alpha}_\beta
+
\frac{1}{2}
\vartheta_{\gamma\delta}
{\mathbb{J}^{\gamma\delta\alpha}}_\beta
$$
where you got ${\left(\mathbb{J}^{\gamma\delta}\right)^\alpha}_\beta=-{\left(\mathbb{J}^{\delta\gamma}\right)^\alpha}_\beta,\,\forall\,\alpha,\beta$, so that you can think it as an antisymmetric (but just on $\gamma,\delta$ indices) matrix of matrices
\begin{gather*}
 \begin{pmatrix}
 {\left(\mathbb{J}^{\gamma\delta}\right)^\alpha}_\beta
 \end{pmatrix}
=
 \begin{pmatrix}
 \mathbb{O}&\mathbb{J}^{01}&\mathbb{J}^{02}&\mathbb{J}^{03}
 \\
 -\mathbb{J}^{01}&\mathbb{O}&\mathbb{J}^{12}&\mathbb{J}^{13}
 \\
 -\mathbb{J}^{02}&-\mathbb{J}^{12}&\mathbb{O}&\mathbb{J}^{23}
 \\
 -\mathbb{J}^{03}&-\mathbb{J}^{13}&-\mathbb{J}^{23}&\mathbb{O}
 \end{pmatrix}
\\
\mathbb{J}^{01}
=
 \begin{pmatrix}
 0&1&0&0
 \\
 1&0&0&0
 \\
 0&0&0&0
 \\
 0&0&0&0
 \end{pmatrix},
\,
\mathbb{J}^{02}
=
 \begin{pmatrix}
 0&0&1&0
 \\
 0&0&0&0
 \\
 1&0&0&0
 \\
 0&0&0&0
 \end{pmatrix},
\,
\mathbb{J}^{03}
=
 \begin{pmatrix}
 0&0&0&1
 \\
 0&0&0&0
 \\
 0&0&0&0
 \\
 1&0&0&0
 \end{pmatrix}
\\
\mathbb{J}^{12}
=
 \begin{pmatrix}
 0&0&0&0
 \\
 0&0&-1&0
 \\
 0&1&0&0
 \\
 0&0&0&0
 \end{pmatrix},
\,
\mathbb{J}^{13}
=
 \begin{pmatrix}
 0&0&0&0
 \\
 0&0&0&-1
 \\
 0&0&0&0
 \\
 0&1&0&0
 \end{pmatrix},
\,
\mathbb{J}^{23}
=
 \begin{pmatrix}
 0&0&0&0
 \\
 0&0&0&0
 \\
 0&0&0&-1
 \\
 0&0&1&0
 \end{pmatrix}
\end{gather*}
P.S.
$$
\mathbb{G}^{\alpha\gamma}
{\mathbb{I}^\delta}_\beta
-
\mathbb{G}^{\alpha\delta}
{\mathbb{I}^\gamma}_\beta
\doteq
{\mathbb{J}^{\gamma\delta\alpha}}_\beta
$$
