# Lorentz Group Generators: Two Methods

Members of the Lorentz group obey $$\eta=\Lambda^{T}\eta\Lambda$$ where $\eta=\textrm{diag}(1,-1,-1,-1)$ is the Minkowski metric.

First, in matrix form write $$\Lambda=I+T$$ where $T$ is an infinitesimal generator. The above condition implies to first order that $$T=-\eta T^{T}\eta$$ and so $T$ is anti-symmetric in the space-space and time-time components, and symmetric in the space-time components.

Next, in index form write $$\Lambda^{\mu}_{\,\,\nu}=\delta^{\mu}_{\,\,\nu}+\omega^{\mu}_{\,\,\nu}$$ The component version of the defining equation then implies that, to first order, $$\omega^{\mu}_{\,\,\nu}=-\omega^{\,\,\mu}_{\nu}$$ and so the generators are fully anti-symmetric.

As far as I can see, $\omega^{\mu}_{\,\,\nu}$ is just the index form of $T$. Therefore, I'm a little confused as to why they seem to obey different conditions. Could anybody please help?

• Hint: Think how the notation $\omega_{\nu}{}^{\mu}$ (as opposed to $\omega^{\mu}{}_{\nu}$) is defined in the last equation. – Qmechanic Mar 29 '17 at 19:15
• $T^\mu{}_\nu = - ( \eta T^T \eta)^\mu{}_\nu = - \eta^{\mu\alpha} (T^T)_{\alpha}{}^\beta \eta_{\beta\nu} = - \eta^{\mu\alpha} T^\beta{}_\alpha \eta_{\beta\nu} = - T_\nu{}^\mu$ which is the same as the condition on $\omega$. – Prahar Mar 29 '17 at 19:16
• @Prahar In this derivation, why is it acceptable to use $(T^{T})_{\alpha}^{\,\,\beta}=T^{\beta}_{\,\,\,\alpha}$ in the third equality, but not acceptable to use $\omega^{\mu}_{\,\,\nu}=(\omega^{T})_{\nu}^{\,\,\mu}$ to argue that $\omega^{T}=-\omega$? Clearly I'm misunderstanding something quite fundamental, so it would be good to understand this. – klgklm Mar 29 '17 at 20:09
• Can somebody tell me how do we derive $\eta=\Lambda^{T}\eta\Lambda$? – Naman Agarwal Dec 9 '17 at 4:36

I actually got confused by the third comment... The first equation in component form reads $$\Lambda^{\mu}{}_{\alpha}\, \eta_{\mu\nu}\, \Lambda^{\nu}{}_{\beta} = \eta_{\alpha\beta}$$ so that with the expansion near the identity: $$(\delta^{\mu}{}_{\alpha} + \omega^{\mu}{}_{\alpha})\, \eta_{\mu\nu}\, (\delta^{\nu}{}_{\beta} + \omega^{\nu}{}_{\beta}) + o(\omega) = \eta_{\alpha\beta} + \omega^{\mu}{}_{\alpha}\, \eta_{\mu\beta} + \eta_{\alpha\nu}\, \omega^{\nu}{}_{\beta} + o(\omega) = \eta_{\alpha\beta}$$

$$\Longrightarrow\quad \omega^{\mu}{}_{\alpha}\, \eta_{\mu\beta} = - \eta_{\alpha\nu}\, \omega^{\nu}{}_{\beta} \quad \Longleftrightarrow\quad \omega_{\beta\alpha} = - \omega_{\alpha\beta}$$ or also $$\omega^{\gamma}{}_{\alpha} = - \eta_{\alpha\nu}\, \omega^{\nu}{}_{\beta}\, \eta^{\beta\gamma} =: -\omega_{\alpha}{}^{\gamma} \qquad (Eq)$$

Let $M$ be a matrix with component $M^i{}_j$ at line $i$ and column $j$. $$\textbf{Inconsistent notation}\qquad (M^T)^i{}_j := M^j{}_i$$ One has to understand the transposition as a "dual" map: $M: E\rightarrow F$ then $M^T: F^* \rightarrow E^*,\ \lambda \mapsto \lambda \circ M$.

Now let $(\mathbf{e}_1,\cdots, \mathbf{e}_n)$ be a basis of $E$ and $(\boldsymbol{\epsilon}^1,\cdots, \boldsymbol{\epsilon}^n)$ its dual in $E^*$. A vector $\mathbf{x}\in E$, resp. $\boldsymbol{\varphi}\in E^*$ decomposes as $$\mathbf{x}= x^{\mu}\, \mathbf{e}_{\mu}\quad \text{and}\quad \boldsymbol{\varphi} = \varphi_{\nu}\, \boldsymbol{\epsilon}^{\nu}$$

This motivates $$(M^T)_i{}^j := M^j{}_i$$ this is the coefficient of the line $i$ column $j$ of $M^T$ matrix of $M^T: F^* \rightarrow E^*$ w.r.t. a basis of the domain and the target.

It thus seems that what you wrote in comment is true as an equality of maps from $F^*$ to $E^*$ (I distinguish the domain and target on purpose because if one says $\mathbf{R}^4$ then... Map defined by (Eq))

Other thing to understand: $\eta_{\mu\nu}$ as the matrix component of a map $\eta: E \rightarrow E^*$ w.r.t. a base in the domain and its dual in the target space.