# 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).

• Yes the $1/2$ is present just for this Oct 28, 2020 at 19:42
• – Sean
Apr 2, 2021 at 5:18

## 1 Answer

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

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