components of mixed tensor with same indices If my tensor $a^{\mu\nu}=$ matrix of 4*4 size (let's say, in 1+3 dimensions with mostly negative convention for the metric), what is $a^{\mu}_{\mu}$ ? Is it the trace or the vector of diagonal elements or the matrix containing only diagonal elements as the non-zero elements ?
 A: The general convention is that all repeated ("dummy") indices (and they had better only be repeated in pairs, one up, one down) are summed over. All remaining indices -- the "free" indices -- vary over all possible values.


*

*With no free indices, you have a rank-0 tensor (a scalar).

*With one free index, you have a rank-1 tensor (a vector), which can be written as a column or a row.

*With two free indices, you have a rank-2 tensor, which can always be written as a matrix.


So $a^{\mu\nu}$ is representable as a matrix not simply because it has two indices, but because it has two free indices. On the other hand, $a^\mu_\mu$ has no free indices and so must be a scalar. In particular, it is the trace of the rank-2 tensor $a$.
By the way, only symmetric tensors can unambiguously use overstacked indices in general. That is, unless $a^{\mu\nu} = a^{\nu\mu}$, you don't know whether $a^\mu_\nu$ refers to ${a^\mu}_\nu \equiv g_{\nu\lambda} a^{\mu\lambda}$ or ${a_\nu}^\mu \equiv g_{\nu\lambda} a^{\lambda\mu}$. In the special case of taking the trace it turns out not to matter, since
$$ {a^\mu}_\mu \equiv g_{\mu\lambda} a^{\mu\lambda} \stackrel{\text{symmetry of $g$}}{\equiv} g_{\lambda\mu} a^{\mu\lambda} \equiv {a_\lambda}^\lambda \equiv {a_\mu}^\mu $$
always, but it's nonetheless rare to see $a^\mu_\mu$ written.
To summarize, $a^{\mu\nu}$, $a_{\mu\nu}$, ${a^\mu}_\nu$, and ${a_\mu}^\nu$ are all representable by (generally different) $4\times4$ matrices, while ${a^\mu}_\mu \equiv {a_\mu}^\mu \equiv a^\mu_\mu$ is simply a scalar, equal to the the trace of the matrices that represent ${a^\mu}_\nu$ or ${a_\mu}^\nu$ (these two matrices would be transposes of one another in any consistent way of assigning indices to rows/columns, and so have the same trace).
