2

Statement: "Let L be a lattice in $R^n$ and $S\subset R^n$ be a convex, bounded set symmetric about the origin. If $Volume(S) > 2^ndet(L)$, then S contains a nonzero lattice vector.

Moreover, if S is closed, then it suffices to take $Volume(S) \geq 2^ndet(L)$."

I am wondering if the condition for boundedness can be relaxed. According to my text, it can, only for the first part with strict inequality, but not for the second part when we take into account the possibility for equality.

i.e. If the text is true, I will be able to find a subset S with volume = det(L) that does not contain any lattice points. I am trying to imagine how such a subset will look like.

First I considered a line passing through the origin with irrational gradient. Then it won't pass through any lattice point. However, the 2-volume of this subset is not very well defined...(I feel that there is something naive about the way volume is defined for unbounded sets)

To have a well defined volume, I imagine some shape looking like a hyperbola. Then it has a proper volume and does not pass through any point. However, it is not convex.

Martin Sleziak
  • 56,060
Jan Lynn
  • 1,243

1 Answers1

1

As explained in this MO answer, for unbounded sets which fulfill the assumptions of Minkowski's theorem $V(S)>0$ implies $V(S)=+\infty$. Using this fact we get that the version for bounded sequence implies the version for unbounded sets.

Let me give a different explanation why the set $S$ must contain some neighborhood of zero. (This fact is used in the linked answer.)

Let $S$ be an unbounded, convex, symmetric (and measurable) set with $V(S)>0$.

The $S$ cannot be subset of a line, since for a line $L$ we have $V(L)=0$.

Therefore there are points $(x_1,y_1)$, $(x_2,y_2)$ such that $\frac{y_2}{x_2}\ne\frac{y_1}{x_1}$. (I.e., these points do not belong to the same line going through the origin.) W.l.o.g. we may assume that $x_1<0<y_1$.

The intersection of the line between $(x_1,y_1)$ and $(x_2,y_2)$ with $x$-axis is some non-zero point. This implies that there exists some $\varepsilon_1>0$ such that points $(x,0)$ for $|x|<\varepsilon_1$ belong to $S$.

Similarly we get that $\{0\}\times(-\varepsilon_2,\varepsilon_2)$ is a subset of $S$ for some $\varepsilon_2$.

Since $S$ is unbounded, there exists a sequence of points $(a_n,b_n)\in S$ such that $a_n$ is unbounded or $b_n$ is unbounded.

Notice that $S$ contains the triangle determined by the points $(-\varepsilon_1,0)$, $(\varepsilon_1,0)$ and $(a_n,b_n)$ which has volume $\varepsilon_1a_n$. Similarly, it contains a triangle with volume $\varepsilon_2b_n$. At least one of these sequences is unbounded. Therefore $V(S)=+\infty$.

Community
  • 1
Martin Sleziak
  • 56,060
  • Based on what I understand from these two answers, Minkowski's theorem remains true for unbounded sets? It is shown that an unbounded convex set <=> infinite volume. We then take the intersection on the infinite volume with closed ball such that we can satisfy volume(S) > 2^n det(L). – Jan Lynn May 04 '15 at 01:07
  • That's how I understand it, too. Although I did not know the version with unbounded closed sets and strict inequality before finding it in that MO post. Still, I would assume something like this being mentioned in some book. – Martin Sleziak May 04 '15 at 05:50