Intuitions regarding Einstein Summation Convention results

Question

This is my first question on the site, so I apologise in advance if it isn't very good. I've recently learnt about the Einstein summation convention and I'm somewhat familiar with the basics of the rules governing it - you sum over any repeated indices, and the rest are all free indices, and free indices should balance, lest you'd have inhomogeneity.

My issue is in being able to apply certain results in different contexts:

For example, the result $$ \delta_{ij}\delta_{jk} = \delta_{ik} $$ makes sense to me, but I can't quite justify to myself why you can use it to show that $$ \delta_{ij}\delta_{ji} = \delta_{ii} = 3 $$ The first result involves 2 free indices, but the second doesn't have any, and is a double sum over i and j - they're fundamentally different, but the former still applies to the latter.

Another example that confused me was using $$ \varepsilon_{ijk}\varepsilon_{ipq} = \delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp} $$ to show that $$ \varepsilon_{ijk}\varepsilon_{ij\ell} = \delta_{jj}\delta_{k\ell} - \delta_{j\ell}\delta_{jk} = 3\delta_{k\ell} - \delta_{k\ell} = 2\delta_{k\ell} $$ To me, this all just feels like a convenient coincidence, and I can't convince myself that applying these rules across different contexts is justified, though it is rather convenient for sure. As such, I'd really appreciate if anybody could help me find some better intuitions for the convention, or at least let me know for sure that doing the above is always justified, because it does seem to always work out.

Answer 1

The first case is easy to get an intuition through matrix algebra. In the matrix language, the identity $\delta_{ij}\delta_{jk}=\delta_{ik}$ means that $I_{3}I_{3}=I_{3}$, where $I_{3}$ denotes the identity matrix of size $3$. Thus, $\delta_{ij}\delta_{jk}=\delta_{ik}$ implies $\delta_{ij}\delta_{ji}=\delta_{ii}=3$ means that $I_{3}I_{3}=I_{3}$ implies $\text{Tr}\left(I_{3}I_{3}\right)=\text{Tr}\left(I_{3}\right)=3$, where $\text{Tr}$ denotes the trace of a square matrix.

Unfortunately, the second case is harder to translate into the matrix language.

Answer 2

I think I've understood what's confusing you because I got confused by the same thing! Essentially we have an identity, $\delta _{ij} \delta _{jk}= \delta_{ik} (*) $. You're probably thinking about it the same way I did - this identity is talking about the $ik^{th}$ component of something that's described by two components.

If we replace the $k$ on both sides by anything other than $i$ and $j$ the identity is means exactly the same thing so no issues there. If we happen to pick $i$, then why do we magically end up summing - i.e. how does it 'know' to sum. Perhaps you're thinking why isn't it just $\delta _{ij} \delta _{ji}= \delta_{ii}$ without a sum, i.e. the $ii^{th}$ component. Well, summation notation is defined as if we have a repeated suffix then we sum over it - that's just what we've decided it means when we write two of the same suffices in the same term. So if we run with this definition on the LHS and want to find $\delta _{ij} \delta _{ji}$ then what we mean by this is we want to sum over $i$ and $j$ . I like to think about it as doing two separate sums to see why the identity works. Starting with $(*) $, this is a sum over $j$ already by definition of the notation. Since we've deliberately decided to set $k=i$ because we're summing over $i$ now as well, all this means is that we're taking the sum of whatever we had on each side over $i$ too, and we've chosen to write it as a double suffix. So essentially by setting $k=i$ in the (*) we've gone from saying the LHS=RHS to the $\Sigma_i$ LHS= $\Sigma_i$ RHS which makes sense as we've done the same thing to both sides. Because the suffices of $(*)$ match up on both sides (as they always have to), by converting a free suffix to a dummy suffix on one side we've just taken the sum over it, and so converting the same free suffix to a dummy suffix on the other side is just summing over it there too, which is legitimate. It's the same as doing, for example:

$$ (i-1)^2=i^2-2i+1 \\ \Rightarrow \Sigma_{i=1}^3 (i-1)^2= \Sigma_{i=1}^3 i^2-2i+1$$

Since the $i$'s 'match up' on both sides, i.e. an i means the same thing on both sides, we can sum over different values of i. Notice how, what you said about the two thing being 'fundamentally different' still applies - we've gone from an identity about $i$ that says two things are the same for any value of $i$ to just a number on each side. But it still 'works out', so the summation convention is just a neat shorthand and it works because of this fundamental idea.