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I need to express the following $$\varepsilon_{ijk}\varepsilon_{ipq}\delta_{jp}\delta_{kq}$$

This is how I solved it:

$$\varepsilon_{ijk}\varepsilon_{ipq}\delta_{jp}\delta_{kq} = \varepsilon_{ipk}\varepsilon_{ipq}\delta_{kq} = \varepsilon_{ipk}\varepsilon_{ipk}$$

Now this can be rewritten as $$\varepsilon_{ipk}\varepsilon_{ipk} = \sum_{i=1}^3\sum_{p=1}^3\sum_{k=1}^3\varepsilon_{ipk}\varepsilon_{ipk}=\sum_{i=1}^3\sum_{p=1}^3\sum_{k=1}^3(\varepsilon_{ipk})^2$$

Now we know that $\varepsilon_{ijk}$ has $27$ terms, depending on the index, of which $21$ are zero and then three terms are $1$ and three terms are $-1$. Hence this gives $$\varepsilon_{ipk}\varepsilon_{ipk}=1+1+1+1+1+1=6$$

Can I make those steps, in particular the one inside the three summations? And furthermore, it seems silly to have to go back to summations in order to reduce this expression.. Is there a cleverer way of doing it?

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    The way you calculated is correct. The summations are implicit if you use summation convention. You already used it in the second line. And by the way, the sum indices are wrong: You have to sum over $i,p,k$ and not over $i,i,k$. – 2017-02-02
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    @JakobElias I'll correct the typo now – 2017-02-02
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    @JakobElias and do you know a more formal or quicker way to show the same? Because it seems pedantic to have to go through the summations again in order to solve it – 2017-02-02
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    I wouldn't see how. The expression $\epsilon_{ipk}\epsilon_{ipk}$ already implies the three sums, one for each repeated index. Not writing them is a question of simplification of expressions. What you are doing is just to write everything explicitly which allows better insight. There is no simpler way to calculate expressions like this imo. – 2017-02-02

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