Words in summation convention

In Einstein summation convention, repeated indices are summed over. For example, \delta_{ii} = \delta_{11} + \delta_{22} + \delta_{33} = 3. If a Kronecker delta has two different suffices, we can ‘contract’ them as follows: \delta_{ij} \delta_{jk} \delta_{ki} = \delta_{ij} \delta_{ji} = \delta_{ii} = 3, whereas \delta_{ii} \delta_{jj} \delta_{kk} = 27.

Vishal Patil considered contracting English words in this manner. For example, ‘intestines’ gives us the expression \delta_{in} \delta_{te} \delta_{st} \delta_{in} \delta_{es} = 9. In general, suppose we have a word W of 2n letters (n pairs of distinct letters in some permutation), and wish to evaluate it in summation convention. Then the product of the deltas evaluates to 3^c, where c is the number of cycles in the n-vertex (multi-)graph produced by joining two vertices with an edge if the corresponding indices share a delta. The graph corresponding to ‘intestines’ is shown below:


English words consisting entirely of repeated letters

I wanted to find other examples (other than ‘intestines’) of words in the English language which feature each letter precisely twice. To do so, I downloaded the official Scrabble dictionary SOWPODS (verifying that it was comprehensive by checking that it contains the word ‘boustrophedonic’) and wrote the following simple Mathematica program:


As you can see, there are a couple of 12-letter words in there, namely ‘trisectrices’ and ‘happenchance’. Conveniently, 12 is a multiple of 3, so we can evaluate the product of Levi-Civita symbols with these subscripts:

\epsilon_{hap} \epsilon_{pen} \epsilon_{cha} \epsilon_{nce} = \epsilon_{pha} \epsilon_{pen} \epsilon_{cha} \epsilon_{cen} = (\delta_{he} \delta_{an} - \delta_{hn} \delta_{ae})(\delta_{he} \delta_{an} - \delta_{hn} \delta_{ae}) = 9 + 9 - 3 - 3 = 12

Here I’ve used the fact that the Levi-Civita symbol is invariant under cyclic permutations, and then simplified the expression using epsilon-delta contractions. We can also evaluate ‘trisectrices’:

\epsilon_{tri} \epsilon_{sec} \epsilon_{tri} \epsilon_{ces} = -36

You can have a go yourself at evaluating expressions corresponding to other English words.

Sum of all words

Suppose we evaluate each of the (2n)! 2^{-n} permutations of aabbcc…dd (with 2n letters) using Kronecker deltas (as we did with intestines) and wish to sum them together. For n = 1, we just have one word with a value of 3. For n = 2, we have six words, two of which have a value of 9 and four of which have a value of 3, so the sum is 30. We can continue this sequence.

If we choose a random word on n letters, the probability generating function for the number of cycles is given below:


Setting s = 3 gives us the expected value of a random word. Multiplying this by the number of words will thus give us the total:


This sequence is already in the OEIS as A007019, but there is no mention of it being equal to the ‘sum of all words’ in Einstein summation convention.

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