How do I compute the trace and the inverse of this tensor?

Question

$$[X^{\mu\nu}] = [X] = \begin{pmatrix} 2 & 3 & 1 & 0 \\ 1 & 1 & 2 & 0 \\ 0 & 1 & 1 & 3 \\ 1 & 2 & 3 & 0 \end{pmatrix}$$

How to compute the trace of $X^{\mu \nu}$? I mean, by eye I can see it is $4$, but I don't know how to do it by writing with the metric tensor.

I calculated for example

$$X^{\mu}_{\phantom{a}\nu}= X^{\mu\sigma}\eta_{\sigma\nu} $$

And the result is

$$X^{\mu}_{\phantom{a}\nu}= \left( \begin{array}{cccc} 2 & -3 & -1 & 0 \\ 1 & -1 & -2 & 0 \\ 0 & -1 & -1 & -3 \\ 1 & -2 & -3 & 0 \\ \end{array} \right)$$

Its trace is zero. How can I obtain instead the trace of $X^{\mu\nu}$, which is $4$?

Also I understood that $X_{\mu\nu}$ would be, in terms of matrices, the inverse of $X^{\mu\nu}$. But when I compute it I have

$$X_{\mu\nu} = X^{\mu\nu} g_{\mu\alpha} g_{\nu\beta}$$

But $g_{\mu\alpha} g_{\nu\beta}$ is the identity matrix.

Yet the inverse of $X^{\mu\nu}$ is not itself.

Answer 1

I think you are confusing contravariant, covariant, and mixed tensors. The trace of $X^{\mu \nu}$ as is written in matrix form above is $\sum_{\mu=1}^4 X^{\mu \mu} = 4$. On the other hand the trace of $X^\mu_{\;\nu}$ is when you have $\mu = \nu$ which is $X^\mu_{\;\mu} = 0$.

I think the confusion stems from interpreting tensor indices as like the matrix indices and then also interpreting contraction in the same sense. The trace of a tensor $X$ is defined as $X^\mu_{\;\mu}$, not the sum of the diagonal elements of $X^{\mu \nu}$.

As for the inverse, it is the same confusion as I have stated. It just so happened that the Minkowski metric $\eta_{\mu \nu}$ is inverse to $\eta^{\mu \nu}$. Nobody said that if you contract a tensor with the metric it will give you the inverse. Again, contraction is a tensor operation NOT necessarily some matrix operation that will give you the inverse. In fact, lowering the index of a contravariant tensor $X^{\mu \nu}$ gives you the covariant tensor $X_{\mu \nu}$. I'm not sure where you got the idea that the covariant tensor is the inverse of the contravariant tensor. These are in a sense different representations of the tensor $X$. If you want the inverse, better calculate it in the standard approach.