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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 $ ?

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  • $\begingroup$ @AlvinJin This should be an answer not a comment $\endgroup$ – Eddy Sep 25 '18 at 20:51
  • $\begingroup$ If $i=k$, wouldn't the $\delta_i^k$ be equal to 1, by definition though? $\endgroup$ – Cave Johnson Sep 25 '18 at 20:53
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It looks to me like you're inherently using the Einstein summation convention, familiar in general relativity. For the purposes of this question, this means that any indices seen twice are summed over.

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).

This is a bit of a simplification of the whole picture but the essence of the machinery is here.

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