I am stuck on a homework problem. The metric is an invariant tensor. $\Lambda^T \cdot g \cdot \Lambda=g$. Considering the infinitesimal lorentz transformation $\Lambda= 1+ \epsilon \Omega$, i have to show $\Omega_{\alpha \beta}=\Omega_{\beta \alpha} $.

I did the following expansion:

g$_{\alpha \beta} (1^\alpha_\mu+\epsilon\Omega^\alpha_\mu )(1^\beta_\nu+\epsilon \Omega^\beta_\nu)= (g_{\beta \mu}+\epsilon g_{\beta \mu})(1^\beta_\nu+\epsilon \Omega^\beta_\nu) $

g$_{\mu \nu}+ \epsilon g_{\mu \nu}+ \epsilon g_{\mu \nu}=g_{\mu \nu}$ neglecting $\epsilon^2 terms$

This step now goes nowhere. Could u please help me out?

  • 1
    $\begingroup$ 1. The metric does not change, 2. How is the transpose implemented in your notation? Can you relate the transpose to the inverse? $\endgroup$ Commented Feb 13, 2017 at 13:35

1 Answer 1


The infinitesimal Lorentz transformation takes the form $\Lambda^\mu{}_\nu = \delta^\mu{}_\nu + \epsilon\Omega^\mu{}_\nu$. So, neglecting terms of order $\epsilon^2$ we have $$ g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu = \left(g_{\mu\beta} + \epsilon\Omega_{\beta\mu}\right)\left(\delta^\beta{}_\nu + \epsilon\Omega^\beta{}_\nu\right) = g_{\mu\nu} + \epsilon\Omega_{\mu\nu} + \epsilon\Omega_{\nu\mu}, $$ where $\Omega_{\mu\nu} = g_{\mu\alpha}\Omega^\alpha{}_{\nu}$. Since, as you say, $g_{\alpha\beta}\Lambda^\alpha{}_{\mu}\Lambda^\beta{}_{\nu} = g_{\mu\nu}$ we must have $\Omega_{\mu\nu} = -\Omega_{\nu\mu}$.

You stated that you should show that $\Omega_{\mu\nu} = \Omega_{\nu\mu}$, but the above result is the correct one.


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