what is squared of a Kronecker ij?

Question

Is it right to write $\delta_{ij}\delta_{ij}=(\delta_{ij})^2=\delta_{ij}$?

Answer 1

Since $\delta_{ij}\in\{0,\,1\}$, it solves $x^2=x$. But you have to make explicit to the reader that $\delta_{ij}\delta_{ij}$ doesn't intend summation over repeated indices. (Even with $\delta_{ij}^2$, readers may think you intend such summation, especially if they're physicists.) If you sum over $i$, and each index has the same $n$ possible values, the result is $\sum_i\delta_{ij}^2=\delta_{jj}=1$; and if we also sum over $j$, we get $n$.

Answer 2

Kronecker symbol is just a convenient notation to say "1 if $i = j$ and 0 otherwise". In you expression, if $i = j$ you get $1 * 1= 1^2 = 1$ and if $i \neq j$ you get $0 * 0 = 0^2 = 0$, and both are correct, so the equlity hods