# I Kronecker Delta summation (easy)

1. Aug 27, 2016

### joshmccraney

Hi PF!

As outlined in my book $\delta_{ij} \delta_{jk} = \delta_{ik}$ but don't we sum over repeated indices (and the $j$ is repeated)? Can someone explain why we do not sum in this situation?

Thanks!

2. Aug 27, 2016

### blue_leaf77

Yes it is summed over $j$. Written explicitly, the sum looks like
$$\delta_{i1} \delta_{1k} + \delta_{i2} \delta_{2k} + \ldots + \delta_{iN} \delta_{Nk}$$
You see that if $i\neq k$, all terms above will vanish. If $i=k$ there will be only one term surviving, therefore that sum can be written compactly as $\delta_{ik}$.

3. Aug 27, 2016

### joshmccraney

Thanks for responding blue_leaf77! Let's assume $i,j,k$ vary from $1,2,3$. Then we have $$\delta_{11} \delta_{11} + \delta_{22} \delta_{22} + \delta_{33} \delta_{33} = 3 \neq \delta_{ik}$$

Did I miss something?

4. Aug 27, 2016

### PeroK

Why would $i$ and $k$ be summed?

And, if you sum them on the right hand side you would get equality.

Last edited: Aug 27, 2016
5. Aug 27, 2016

### joshmccraney

That's what blue_leaf wrote.
Since $i$ and $k$ can range from 1,2,3 we would have$$\delta_{11} \delta_{11} + \delta_{21} \delta_{11} + \delta_{31} \delta_{11}+ \delta_{12} \delta_{21}+ \delta_{22} \delta_{21}+ \delta_{32} \delta_{21}+\cdots + \delta_{33} \delta_{33}\\=\delta_{11} \delta_{11}+\delta_{22} \delta_{22}+\delta_{33} \delta_{33}$$ which is why I was summing.

6. Aug 27, 2016

### PeroK

blue leaf summed only $j$, which is a repeated index.

7. Aug 27, 2016

### joshmccraney

Ohhhhh I see now! Shoot, yea I was doing it all wrong! Thank you both!!!