# Kronecker Delta Expressions

I am trying to understand the Kronecker Delta and want to clarify. Considering the definition of the Kronecker Delta and assuming $$i=j=k$$ for the following situations:

I know that $$\delta _j^i \delta _i^j$$ is equal to $$N$$ where $$N$$ is the dimension of space. Would this mean that $$\delta _j^i \delta _k^j \delta _i^k$$ would also be equal to $$N$$?

Similarly, since $$\delta _i^i \delta _j^j$$ is equal to $$N^2$$, would $$\delta _i^i \delta _j^j \delta _k^k$$ be equal to $$N^3$$ ?

Last, is $$\delta _i^j \delta _j^k$$ just equal to $$\delta _i^k = 1$$ ?

• @AlvinJin This should be an answer not a comment – Eddy Sep 25 '18 at 20:51
• If $i=k$, wouldn't the $\delta_i^k$ be equal to 1, by definition though? – Cave Johnson Sep 25 '18 at 20:53

When you say $$\delta^i_j \delta^j_i = N$$, for example, this implicitly means $$\displaystyle \sum_{i=1}^N \displaystyle \sum_{j=1}^N \delta^i_j \delta^j_i = N$$ which is indeed true since the sum is non-vanishing whenever $$i=j$$ and this happens $$N$$ times.
Similarly, $$\delta^i_j \delta^j_k \delta^k_i = \displaystyle \sum_{i=1}^N \displaystyle \sum_{j=1}^N \displaystyle \sum_{k=1}^N \delta^i_j \delta^j_k \delta^k_i = N$$, as you surmise.
Also $$\delta^i_i \delta^j_j \delta^k_k = \displaystyle \sum_{i=1}^N \displaystyle \sum_{j=1}^N \displaystyle \sum_{k=1}^N \delta^i_i \delta^j_j \delta^k_k = N^3$$, as you suspected.
Finally, $$\delta^j_i \delta^k_j = \delta^k_i$$ but this is not a scalar quantity ($$\neq 1$$). Instead it has a separate value for each index $$i$$ and $$k$$. You can think of it as being represented by a matrix whose $$(i,k)$$th entry is $$1$$ if $$i=k$$ and $$0$$ otherwise (identity matrix in $$N$$ dimensions).