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Referring to the answer in the following question:

https://physics.stackexchange.com/a/349030/288587

I just cant figure out how to go from:

$$ \eta_{\mu\nu} = \Lambda^\alpha_{\;\mu}\Lambda^\beta_{\;\nu}\eta_{\alpha\beta} $$

to:

$$ η=Λ^TηΛ $$

I tried to do the following:

$$ \begin{align} \eta_{\mu\nu} &= \Lambda^\alpha_{\;\mu}\Lambda^\beta_{\;\nu}\eta_{\alpha\beta} \\ &= \Lambda^\alpha_{\;\mu}\Lambda_{\alpha\nu} \\ &= \Lambda^\alpha_{\;\mu}\eta_{\alpha\sigma}\Lambda^\sigma_{\;\nu} \\ &= \Lambda^\nu_{\;\mu}\eta_{\nu\mu}\Lambda^\mu_{\;\nu} \end{align} $$

Implying that the transpose of $\Lambda^\mu_{\;\nu}$ is $\Lambda^\nu_{\;\mu}$ but, as i know, this is not true, as the transpose of $\Lambda^\mu_{\;\nu}$ should be $\Lambda_{\nu\;}^{\;\mu}$.

Where have i gone wrong?

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  • $\begingroup$ By $\eta$ do you mean simply the trace of it? $\endgroup$ – DJA Feb 10 at 3:06
  • $\begingroup$ And by $\eta$ i mean the metric tensor $\endgroup$ – Jbfm Feb 10 at 11:47
  • $\begingroup$ Im sorry, but i still dont see how would $(\Lambda^T)_{\nu}{}^{\mu}$ := $\Lambda^\mu{}_\nu$ would be equal $\Lambda^\nu_{\;\mu}$. So i still cant really understand the conversion from matrix to tensor index notation. $\endgroup$ – Jbfm Feb 10 at 12:59
  • $\begingroup$ I think i got it!! The fact that both tensors on $\Lambda^\alpha_{\;\mu}\eta_{\alpha\sigma}$ have the summation index on the same slot (e.g. as $\alpha$ is summed over, the "columns" of both $\Lambda^\alpha_{\;\mu}$ and $\eta_{\alpha\sigma}$ are multiplied instead of "row"x"column") .This indicates that $\Lambda^\alpha_{\;\mu}$ is the transpose right? If we had $\eta_{\sigma\alpha}$, then we would see $\Lambda^\sigma_{\;\nu}$ as the transpose, correct? $\endgroup$ – Jbfm Feb 11 at 11:24

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