Evaluating the product of three Kronecker delta

Question

I am trying to evaluate the following product of these 3 Kronecker delta:

$\delta_{ij}\delta_{jk}\delta_{ki}$

I am not sure how to proceed. I understand that the Kronecker delta acts as a substitution operator; does this mean I have to work from the left since operators work on the thing to its right?

And if that is the case, how would the first Kronecker delta work on the next one? Thanks for any help.

Answer 1

The product will be $0$ if any of the $i,j,k$ are not equal, since then one of $$ \delta_{ij}\\ \delta_{jk}\\ \delta_{ki} $$ will be $0$.

It will be $1$ if $i=j=k$, since each term in the product will be $1$.

Answer 2

If you say "I understand that the Kronecker delta acts as a substitution operator", you probably mean that you use Einstein summation convention of the tensor calculus which allows to simplify $A_i \delta_{ij} = A_j$.

This is possible only in certain contexts that are briefly described in these lectures. Then you can substitute indices so that $\delta_{ij} \delta_{jk} \delta_{ki}= (\delta_{ij} \delta_{jk})\delta_{ki}=\delta_{ik}\delta_{ki} = \delta_{ii}$. But that is only allowed if $j,k$ indices are "dummy", i.e. you can sum over these indices. If also $i$ is dummy, then $\delta_{ii}$ is the trace of the unit matrix, i.e. it is equal to the dimension of the system, so in 3D Cartesian it is equal to $3$.

In general case of linear algebra you cannot use Kronecker delta as a substitution operator and the delta means a function that returns $1$ if its arguments are the same and $0$ otherwise. I.e. $\delta_{ij} \delta_{jk} \delta_{ki}=1$ if $i=j=k$ and $\delta_{ij} \delta_{jk} \delta_{ki}=0$ otherwise.