Metric Tensor notation

Question

I am studying the case of infinitesimal proper Lorentz transformations, and I stumble at this notation result : $\Delta\omega^\alpha_\gamma g_{\alpha\beta} g^\beta_\delta = \Delta \omega_{\delta\gamma}$?

I know that the following is true : $x^\mu = g^{\mu\sigma} x_\sigma$. So what happens here is that the metric tensor raises the index and then the raised index changes to the Greek letter which is not contracted, μ in this case. How does the same applies in the above equation which I cannot understand ?

Edit : Prahar was correct I had made a mistake. I solved it and this is how :

$g_{\alpha\beta} g^\beta_\delta = g_{\alpha\delta}$

$\Delta\omega^\alpha_\gamma\cdot g_{\alpha\delta}=\Delta\omega_{\delta\gamma}$

Answer 1

First, I think that $g^{\beta}_{\delta}$ notation should be changed to $\delta^{\beta}_{\delta}$, representing the Kronecker symbol, which by definition is equal to 1 if $\beta=\delta$ and equal to 0 if $\beta\neq\delta$. So, the equation reduces to the case when $\beta=\delta$ and thus we have $\Delta\omega^{\alpha}_{\gamma} g_{\alpha\delta}$ which is then equal to $\Delta\omega_{\delta\gamma}$ by the transformation you are given. I hope this helps.