Say you have a complete homogeneous symmetric function $$h_4 = \sum_{1\leq i \leq j \leq k \leq l}q^{-i}q^{-j}q^{-k}q^{-l},$$ where $i = 1, 2, 3, \ldots$. There are 7 cases to consider, given by

$$h_4 = \sum_{1\leq i = j = k = l\\ 1\leq i = j < k < l\\ 1\leq i < j = k < l\\ 1\leq i < j < k = l\\ 1\leq i = j = k < l\\ 1\leq i < j = k = l\\ 1\leq i < j < k < l}q^{-i}q^{-j}q^{-k}q^{-l}.$$ Now, if we consider the case where $$1\leq i = j < k < l$$ and insert $i = j$ we obtain $$\sum_{1\leq i = j < k < l}q^{-2i}q^{-k}q^{-l}.$$ If we now do the sum the sum over $l$, we observe that since $k<l$ this is a geometric series with the first term being $q^{-(k+1)}$. I now say that because the sum $$\sum_{l=1}^{\infty}q^{-l} = \frac{1}{q-1},$$ and the terms in the sum over $l$ are written as $q^{-(k+1)} + q^{-(k+2)} + \ldots$, we get a contribution $$\sum_{1\leq i < k}q^{-2i}q^{-k}\frac{q^{-(k+1)}}{q-1}\\ = \frac{1}{q(q-1)}\sum_{1\leq i < k}q^{-2i}q^{-2k}$$ from the sum over $l$. Doing this again for $k$ since $k<i$, and so the sum over $k$ yields a contribution $$\frac{1}{q(q-1)}\sum_{1\leq i}q^{-2i}\frac{q^{-2(i+1)}}{q^2 - 1}\\ = \frac{1}{q(q-1)}\frac{1}{q^2(q^2 - 1)}\sum_{i = 1}^{\infty}q^{-4i}\\ = \frac{1}{q^3(q-1)(q^2 - 1)(q^4 - 1)}.$$

My question is about the contribution of the sum which has terms of the form $q^{-(i+1)} + q^{-(i+2)} + \ldots$, I have my doubts that it is correct that it yields a contribution $$q^{-(i+1)}/(q-1).$$ I am trying to find Schur functions so that I can calculate topological vertex amplitudes.

Any help is much appreciated!

  • $\begingroup$ Those are not really polynomials; they are rational functions. $\endgroup$ – Ben McKay Nov 11 '19 at 11:49
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    $\begingroup$ It is $\frac{q^6}{(q-1)(q^2-1)(q^3-1)(q^4-1)}$; do you need some kind of decomposition of this into sum of partial fractions? $\endgroup$ – მამუკა ჯიბლაძე Nov 11 '19 at 12:55
  • $\begingroup$ @მამუკაჯიბლაძე How did you get that answer? Did you use some other method? The expression you found is exactly what I'm looking for :) $\endgroup$ – QuantumMechanic Nov 11 '19 at 13:36
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    $\begingroup$ You redenote $i=1+a$, $j=1+a+b$, $k=1+a+b+c$, $l=1+a+b+c+d$ with $a,b,c,d\geqslant0$. Then you get sum of $q^{-(4+4a+3b+2c+d)}=q^{-4}\sum q^{-4a}\sum q^{-3b}\sum q^{-2c}\sum q^{-d}$ which is $q^{-4}(1-q^{-4})^{-1}(1-q^{-3})^{-1}(1-q^{-2})^{-1}(1-q^{-1})^{-1}$ $\endgroup$ – მამუკა ჯიბლაძე Nov 11 '19 at 13:48
  • $\begingroup$ @მამუკაჯიბლაძე Thanks! That was really helpful :) $\endgroup$ – QuantumMechanic Nov 11 '19 at 15:08