# Summation notation for Kronecker delta

I'm having some problems on notation for indices:

1. I've found in Goldstein, 3rd edition, that the Kronecker delta satisfies the following property: $$\delta_{ij}\delta_{ik}=\delta_{jk}$$

But imagine that $i \neq j$ and $j=k$. In this case, $$\delta_{ij}\delta_{ik}=0$$ but, $$\delta_{jk}=1.$$ So how does this work?

2. I've seen the following affirmation: $$\delta_{ii}=3$$ By the previous property isn't this possible?:

$$\delta_{ii}\delta_{jj}=9$$

But, when we make $\delta_{ij}\delta_{ik}=\delta_{jk}$, we can only have $\delta_{jk}=3$ for $j=k$ as maximum.

• All of your problems are solved by realizing that you're working in the Einstein summation convention. The first problem, e.g. Is resolved because you sum over the double i index, so you will always hit the i=j(=k) condition eventually. As for the second question, it's the same thing; sum over i and you get 1+1+1=3. Jun 5, 2015 at 21:40

You're getting tripped up by summation notation. Whenever you have a repeated index, this means that that index is to be summed from 1 to 3: $$\delta_{ij} \delta_{ik} \equiv \sum_{i=1}^3 \delta_{ij} \delta_{ik}.$$ You're right that there are two terms in this sum where $i \neq j$, and so the contribution to the sum from these terms is zero. But the remaining term has $i = j$, and so gives 1, and so the entire thing sums to 1 when $j = k$.
I think you're tripping on repeated indices/Einstein notation here. If an index is repeated, you're supposed to sum over it. So $\delta_{ij} \delta_{ik}$ seems like it would be zero, except that one term in the sum will have $i = j$ and another will have $i = k$. If they're the same, you get a $1 \cdot 1 = 1$ term, if they're different you get $1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0$, which is exactly the behavior of $\delta_{jk}$.