Define $$g_{\mu\nu}=g^{\mu\nu}=\left( \begin{array}{cc} 1 & 0 & 0 &0 \\
0 & -1 & 0 &0 \\ 0 & 0 & -1 &0 \\ 0 & 0 & 0 &-1 \end{array} \right)$$to be the standard metric tensor in flat spacetime. Assume the summation rule for four index. It was asked to evaluate $$g^{\mu\nu}g_{\nu\rho}g^{\rho\mu}.$$ Then what I got is $$\delta^{\mu}_{\rho}g^{\rho\mu}=g^{\mu\mu}.$$
Actually, since there is no free index, I would like to sum it, but it gets both index up, which cannot be summed conventionally. What exactly is $g^{\mu\mu}$, is it still a second rank tensor or is it a scalar? Thank you.
Evaluation of the multiplication of the metric tensors
tensor-productstensors
Best Answer
As Toffomat says, that looks like a typo. At least, it's not something that one would normally write. If the author doesn't say anything about it, it is probably a typo. However, if I were forced to interpret the expression $g^{\mu\mu}$ without any commentary, I'd say that the symbol would represent the four entries of the diagonal, one for each possible value of $\mu$, so in that sense it would be a set of four scalars.