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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).

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  • $\begingroup$ Yes the $1/2$ is present just for this $\endgroup$ Commented Oct 28, 2020 at 19:42
  • $\begingroup$ Related: Matrix Representation of Lorentz Group Generators $\endgroup$ Commented Apr 2, 2021 at 5:18

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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. $$ \mathbb{G}^{\alpha\gamma} {\mathbb{I}^\delta}_\beta - \mathbb{G}^{\alpha\delta} {\mathbb{I}^\gamma}_\beta \doteq {\mathbb{J}^{\gamma\delta\alpha}}_\beta $$

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  • $\begingroup$ I am impressed by the amount of effort it must have taken to type all this out. $\endgroup$ Commented Aug 1, 2022 at 23:31
  • $\begingroup$ @RC_23 Not so much actually, I adapted my own notes ahah $\endgroup$ Commented Aug 2, 2022 at 8:00

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