How should $\delta_k^j$ be regarded? Is it a scalar that takes on variable values? A 3×3 identity matrix (in 3 dimensions)?
The wikipedia article on raising and lowering indices with the metric tensor includes the phrase:
"where $\delta_{ik}$ is the Kronecker delta or identity matrix''
however I think this is wrong.
Consider this:
$$
\delta_k^j a_j = \delta_k^1 a_1 + \delta_k^2 a_2 + \delta_k^3 a_3 = a_k
$$
In the pattern above $\delta$ is not equivalent to the identity matrix. Rather, it is a selection vector, i.e.,
$$
\delta_k^j a_j = [0, 1, 0] \left[ \begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array}\right]
$$
where the 1 is appearing in the $k$th position (here k=2).
However, as a $(\,_1^1)$ tensor it seems like it should be equivalent to a matrix
Best Answer
When you write $\delta_k^j a_j = \delta_k^1 a_1 + \delta_k^2 a_2 + \delta_k^3 a_3 = a_k$ it is essentially acting as the identity matrix, it's just renaming the index from j to k. It just seems weird because of the summation convention.