If I have a tensor $X^{\mu}{}_{\nu} = \begin{bmatrix} a & b & c \\ d & e& f\\ g&h&i\\ \end{bmatrix}$ then what is $X^{\mu}_{\;\;\mu}$?

From what I understand it would be $(a,b,c)\cdot(a,b,c) +(d,e,f)\cdot(d,e,f)+(g,h,i)\cdot(g,h,i). $

Is this correct?

  • $\begingroup$ The left most index indicates the row number, the right most index indicates the column number. So would it be $(a,b,c)\cdot(a,d,g) +(d,e,f)\cdot(b,e,h)+(c,f,i)\cdot(g,h,i). $ $\endgroup$
    – Tim
    Mar 21, 2017 at 22:03
  • 2
    $\begingroup$ Related: math.stackexchange.com/q/1254041/72459 By the way, why do you think that you can count the indices up independently? When $\mu=1$, you have $X^1_1$, there's no way to get e.g. $X^1_2$ in $X^\mu_\mu$. $\endgroup$ Mar 21, 2017 at 22:12
  • $\begingroup$ The above mentioned is not a tensor: it is the components of a tensor in some particular basis. $\endgroup$
    – gented
    Mar 21, 2017 at 22:12
  • 4
    $\begingroup$ So then it would just be $a+e+i$? $\endgroup$
    – Tim
    Mar 21, 2017 at 22:40

1 Answer 1


$$ {\mathrm tr}\left(X\right) = X^{\mu}{}_{\mu} = X^{0}{}_{0} + X^{1}{}_{1} + X^{2}{}_{2} = a + e + i $$


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