# Levi-Cevita symbols: Why is $\epsilon_{ijk}\epsilon_{pjk}$ equal to $2\delta_{ip}$, but not $0$?

I'm learning vector calculus on my own and sometimes strange things happen that I don't know how I should explain them. We have this famous equality:

$$\epsilon_{ijk}\epsilon_{pqk}=\delta_{ip}\delta_{jq}-\delta_{iq}\delta_{jp}$$

Now, if we set $j=q$ we get $$\epsilon_{ijk}\epsilon_{pjk}=\delta_{ip}\delta_{jj}-\delta_{ij}\delta_{jp}=\delta_{ip}-\delta_{ip}=0$$

But apparently the correct equality is $$\epsilon_{ijk}\epsilon_{pjk}=2\delta_{ip}$$

Why is it so? Where's my mistake? :|

In the second equation, you sum over $j$, so $\delta_{ip}\delta_{jj}$ should be evaluated as $3\delta_{ip}$.
• So, whenever we have $\delta_{jj}$ it's equal to $3$ instead of $1$? because according to my book, whenever that we have repeated indices we sum over it. So, by keeping that in mind, $\delta_{jj}=3$. Right? – math.n00b Jul 20 '14 at 8:57
• It depends on whether $j$ is fixed or you sum over it. In your computations, you first substituted the fixed $q$ by a fixed $j$ and then summed over $j$. Clearly, $\delta_{11}+\delta_{22}+\delta_{33}=3$. (If you are not sure what to do, it might help to always write the $\sum$ symbols) – Peter Franek Jul 20 '14 at 9:06
• Yes, but according to my book, which is about elasticity theory, it says that whenever that we had something like $A_{jj}$ we'd sum over $j$. So, by this convention, we'll always get $\delta_{jj}=3$. Right? – math.n00b Jul 20 '14 at 9:09