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Given the next tensor:

$X^{\mu \nu}= \left(\begin{array}{cccc} 2 & 0 & 1 & -1 \\ -1 & 0 & 3 & 2 \\ -1 & 1 & 0 & 0 \\ -2 & 1 & 1 & -2 \\ \end{array}\right)$ and the metric tensor $\eta_{\mu\nu} =\left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array}\right)$

I need to calculate the following tensors: $X^\mu$$_\nu$, $X_\mu$$^\nu$, $X^\alpha$$_\alpha$

So I get that for the first two i have to raise an lower the $\mu$ and $\nu$ indices using the metric tensor, and i get how, for example, $A^{\alpha\beta}=g^{\alpha\gamma}g^{\beta\delta}A_{\gamma\delta}$ and $A_{\alpha\beta}=g_{\alpha\gamma}g_{\beta\delta}A^{\gamma\delta}$, but what if I only want to lower(raise) one index ?

And for the third case ($X^\alpha$$_\alpha$) I imagine I'll need a Kronecker $\delta$ in order to have the same index, and then, due that it is a dummy index I can change the name.. right?

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    $\begingroup$ You use one metric tensor for every index you want to lower or raise. So for example $x^{\mu}_{\ \ \nu}= x^{\mu \sigma}g_{\sigma \nu}$. To calculate $x^{\mu}_{\ \ \mu}$ first lower one index and then contract with Kornecker delta. You will notice that it can be then simplified to yield simpler formula. $\endgroup$ – Blazej Sep 16 '16 at 21:48
  • $\begingroup$ So $X_\mu$$^\nu$ would be $X_\mu$$^\nu$=$g_{\mu\sigma}X^{\sigma\nu}$ ?? $\endgroup$ – C. Alexander Sep 17 '16 at 0:50
  • $\begingroup$ Yes, that's it. $\endgroup$ – udrv Sep 17 '16 at 6:30

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