Sum of a product of four Kronecker Deltas

Question

The Kronecker delta has the following property:

$$\sum_{k} \delta_{ik}\delta_{kj} = \delta_{ij}. $$

Does anyone know whether the following formula is correct?

$$\sum_{i=1}^N \delta_{ij}\delta_{ik}\delta_{il}\delta_{im} = \delta_{jk}\delta_{lm}~?$$

Answer 1

It's not.

Take $j=k=1$ and $l=m=2$. Then $\delta_{jk}\delta_{lm}=1$, but

$$\sum_{i=1}^N \delta_{ij}\delta_{ik}\delta_{il}\delta_{im}=\delta_{11}\delta_{11}\delta_{12}\delta_{12}+\delta_{21}\delta_{21}\delta_{22}\delta_{22}=0+0=0.$$

What is true is that

$$\sum_{i=1}^N \delta_{ij}\delta_{ik}\delta_{il}\delta_{im}=\delta_{jklm},$$

i.e. the function that is $1$ exactly when $j=k=l=m$, and is otherwise $0$.