MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following question arises in passing in a joint paper that I am working on. Let $K$ be a centrally symmetric convex body in an $n$-dimensional real vector space $V$ which contains a lattice $L$. $L$ yields a natural volume scale fon $V$, and Minkowski established a well-known lower bound on the number of points in $L$ contained in $K$, namely $N \ge (\text{Vol}\ K)/2^n$. You could view this as a primitive estimate, but actually it's often roughly correct, after you take an $n$th root for a fair comparison.

Is there a similar upper bound that (a) is invariant under the $\text{GL}(n,\mathbb{Z})$ stabilizer of $L$, (b) behaves reasonably if you dilate $K$, and (c) can be regarded as useful or simple? Without any further restrictions on $K$, you obviously can't just use its volume, because it could be a thin rod that contains many lattice points of $L$. An extra restriction that holds in our case is that $K$ is a lattice polytope, i.e., the convex hull of finitely many lattice points. What we currently do is ask for basis of $L$, or a lattice that contains $L$, relative to which $K$ is contained in the $\ell^\infty$ ball of radius $c$. Then of course you get $N \le (2c+1)^n$. Or more generally, you could use a lattice parallelepiped. This works in the cases that we need it, but I don't know when it's a good estimate, again with the allowance of an $n$th root. It's also only $\text{GL}(n,\mathbb{Z})$-invariant by fiat.

share|cite|improve this question
Here is a primitive bound for lattice polytopes (not necessarily centrally symmetric) that is already interesting. It looks like a lattice polytope can be triangulated by lattice simplices with no internal vertices. This yields $N \le (n+1)!(\text{Vol} K)$. This bound is apparently refined by a bound on Ehrhart coefficients due to Betke and McMullen. This estimate and the Betke-McMullen bound might already be what I'm looking for. I will leave the question open for now in case there are further ideas. – Greg Kuperberg Jan 21 '13 at 17:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.