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