In the process of evaluating a "supersymmetric index", Bourget and Troost establish a rather elementary identity:
$$ \frac{N}{m} \sum_{d| N} \sum_{l=1}^{\mathrm{gcd}(d,m)} \mathrm{gcd}\left[ \mathrm{gcd}(d,m), n + \frac{ld}{\mathrm{gcd}(d,m)} \right] $$
but later they compute this same supersymmetric index to be another formula:
$$ N \sum_{d|N} \mathrm{gcd}\left[ d, m, \frac{N}{d}, \frac{N}{m}, n \right] $$
and the finally they count it come other way and get yet another formula:
$$ \sum_{d|N} \sum_{t = 0}^{d-1} \mathrm{gcd}\left[ N \frac{d}{m}, N \frac{m}{d}, N\left( \frac{t}{m} + \frac{n}{d} \right) \right] $$
These formulae should be equivalent for any $N, m, n$ with $m$ dividing $n$... (and possibly other hypotheses missing) Is there a conceptual proof this result?
As a special case they show:
$$ N \sum_{d|N} 1 = \sum_{d| N} \sum_{l=1}^d \mathrm{gcd}(d,l) $$
The supersymmetrc index counts just about everthing in hep-th
- what could it be counting here?
I can venture a guess these have something to do with the Lie groups they mention:
$$ (SU(N)/\mathbb{Z}_m)_n $$
where the meaning of the $n$ is unclear ( the paper says "dionic tilt"). In another section the Smith normal form is mentioned:
$$ \frac{\mathbb{Z}}{L \mathbb{Z}} \simeq \bigoplus_{i=1}^n \frac{\mathbb{Z}}{e_i \mathbb{Z}} $$
This looks quite like the chinese remainder theorem
This post imported from StackExchange MathOverflow at 2016-06-23 20:58 (UTC), posted by SE-user john mangual