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Motivation - I was thinking about calculating the integrals from An interesting integral expression for $\pi^n$? using old plain Riemann sums. There, one needs integrating over that part of $[0,1]^n$ where the coordinates are more than $\varepsilon$ apart from each other, so I thought maybe I can choose a grid for the multidimensional Riemann integral with these $\varepsilon$s somehow built in.

Then a natural thing to try is this: $$ \left\{(k_1,...,k_n)\in\mathbb Z^n\mid k_i\ne k_j\mod n\text{ for $i\ne j$}\right\} $$ (and then scale down to $\varepsilon$)

The above set is clearly a(n affine) lattice. Have you seen it before? What symmetries does it have? Is it listed in any nomenclature and how to find it?

Specifically I would like to have a more manageable enumeration of this lattice, - say, a nice basis. All bases I can think of are very unnatural, break symmetry terribly and result in messy formulas when I am trying to calculate anything with them.

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    $\begingroup$ Two questions. First, you are using $n$ for both the dimension of your space and for the modulus. I'll assume that's a mistake and let $m$ denote the modulus. Second, what do you mean by an "affine lattice"? Your set is clearly not a subgroup of $\mathbb Z^n$, so not a lattice. For example, for $n=2$, the points $(1,0)$ and $(0,1)$ are in your set, but their sum is not. $\endgroup$
    – Joe Silverman
    May 4, 2017 at 20:39
  • $\begingroup$ @JoeSilverman I would be interested to know about the case when the modulus differs from the dimension but the case when they are equal suffices for me. It is sort of minimal. $\endgroup$
    – მამუკა ჯიბლაძე
    May 5, 2017 at 3:32
  • $\begingroup$ Affine lattices occur frequently, but I am not sure which terminology for them is established. Simplest definition probably is that it is a shift by some vector of a lattice in the ordinary sense. $\endgroup$
    – მამუკა ჯიბლაძე
    May 5, 2017 at 3:36
  • $\begingroup$ In terms of closure under sums, it is equivalent to define affine lattices as subsets closed under $(x,y,z)\mapsto x-y+z$ and finitely generated (in the appropriate sense) $\endgroup$
    – მამუკა ჯიბლაძე
    May 5, 2017 at 3:37

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