Question

prove that kronecker delta is a mixed tensor of rank2

Answer 1

Explanation:

the Kronecker Delta is a second-order mixed tensor with one contravariant and one covariant index as indicated:

δij={10i=ji≠j

By definition, if it were such a tensor we would have:

δ′ij=δkl∂xi′∂xk∂xl∂xj′

using the summation convention. I have seen a "proof" of this equation somewhere else that goes on to say:

δ′ij=δkk∂xi′∂xk∂xk∂xj′=∂xi′∂xj′={10i′=j′i′≠j′

supposedly verifying it by using the fact that δkl=0 if k≠l.

However, since k is a dummy index that gets summed over, shouldn't we have

δ′ij={n0i′=j′i′≠j′

where n is the dimension of the space?

For example, if i' = 1, j' = 1, n = 2, wouldn't the transformation equation above give:

δ′11=δ11∂x1′∂x1∂x1∂x1′+δ22∂x1′∂x2∂x2∂x1′=2

where I've used δ21=δ12=0? Where am I going wrong here?