# Volume of convex lattice polytopes with one interior lattice point

Let $P$ be a convex polytope in $\mathbb{R}^3$ whose every vertex lies in the $\mathbb{Z}^3$ lattice.

Question: If $P$ contains exactly one lattice point in its interior, what is the maximum possible volume of $P$?

Notice that a convex lattice polytope with no interior lattice point can have arbitrarily large volume, but if the number of interior lattice points in it is a given finite number, then its volume cannot be arbitrarily large. In two dimensions the answer is known: A convex lattice polygon with exactly one interior point has volume at most $9/2$, which follows from a more general inequality by P.R. Scott (see On Convex Lattice Polygons, Bull. Austral. Math. Soc., vol 15; 1976, 395-399).

The best example I know in that respect is the tetrahedron with vertices $(0,0,0),\ (4,0,0),\ (0,4,0)$ and $(0,0,4)$, with volume ${32}\over3$. Is this perhaps the maximum volume? Of course, the same question can be asked in higher dimensions as well, with an analogous example to consider as a possible candidate for an answer.

In view of several examples presented in answers and comments below, it seems that the optimal polytope should be a simplex, in every dimension. Has this been conjectured already?

• If P is centrally symmetric, then the center of symmetry must be a lattice point, and then the Minkowski Theorem provides the answer in every dimension $n$, namely the volume cannot exceed $2^n$, and this bound is reached only by parallelepipeds. – Wlodek Kuperberg Oct 17 '17 at 3:14
• Could you elaborate on why the volume of a convex polytope with one interior point is bounded? – Mikhail Tikhomirov Oct 17 '17 at 7:16
• The Scott paper is available at faculty.math.illinois.edu/~reznick/Scott1976.pdf – Gerry Myerson Oct 17 '17 at 8:58
• The simplex with vertices $(0,0,0), \ (2,0,0), \ (0,3,0)$ and $(0,0,12)$ has volume $12 > 32 / 3$ and it has only one interior lattice point, namely $(1,1,1)$. – js21 Oct 17 '17 at 10:52
• Moreover the discussion under cor. $3.5$ of msp.org/pjm/1983/105-1/pjm-v105-n1-p10-s.pdf yields an upper bound $\mathrm{vol} \leq 42^3$. It is surely far from optimal. – js21 Oct 17 '17 at 10:53

This is addressed in the 2013 paper (appeared in Advances in 2015) by Averkov, Krumpelmann, Nill. The give a sharp bound for the volume of a lattice simplex with one interior lattice point (Theorem 2.2 in the paper), and an improved bound for a general lattice polyhedron with the same property (theorem 2.7) (the two bounds are not the same, indicating that Wlodek's conjecture is either still open or false). The results are stated in terms of the Sylvester sequence:

$$s_1 = 2; s_i = 1 + \prod_{j=1}^{i-1} s_j.$$

With that, the volume of tbe biggest simplex in $d$ dimension with one lattice point is bounded (with equality achieved) by

$$\frac2{d!}(s_d-1)^2,$$

while for arbitrary polytopes, the bound is $$(s_{d+1}-1)^d,$$ so a lot worse.

• I would like to add that the same authors just today posted an extension of the work referred to in Igor's answer, where there are some fixed number of interior lattice points rather than just one. – Rob Davis Oct 25 '17 at 17:19

Depictions of the two lattice polyhedra mentioned so far (by Wlodek Kuperberg & js21):

Left: $(0,0,0),(4,0,0),(0,4,0),(0,0,4)$. Right: $(0,0,0), (2,0,0), (0,3,0), (0,0,12)$.

• Nice pictures :). The simplex with vertices $(0,0,0), \ (2,0,0), \ (0,6,0)$ and $(0,0,6)$ also yields volume $12$ (and has $(1,1,1)$ as its only lattice point). – js21 Oct 17 '17 at 12:44
• Anything analogous to the right one in 4 dimensions...? – Wlodek Kuperberg Oct 17 '17 at 15:59
• I would guess the vertices $(2,0,0,0)$, $(0,3,0,0)$, $(0,0,7,0)$ and $(0,0,0,84)$; the paper cited by @js21 seems to suggest something similar, although I cannot tell exactly... (The nontricial face passes through $(1,1,1,2)$.) – Ilya Bogdanov Oct 17 '17 at 17:54

Just to extend the examples so far: Given a non-decreasing list of $n$ integers $a_1 \leq a_2 \leq \cdots \leq a_n,$ consider the simplex with vertices at the origin and the points with all coordinates zero except $a_i$ in the $i$th position. The volume is $\frac{\prod a_i}{n!}.$ The point $(1,1,\cdots,1)$ is in the interior provided $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_{n-1}}+ \frac{1}{a_n}\lt 1$ and there are no other lattice points in the interior provided $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_{n-1}}+ \frac{2}{a_n}\ge 1.$ Equality here puts the point $(1,1,\cdots,1,2)$ on the "sloped" face.

Subject to these constraints, I find that the optimal edge lengths with corresponding volume are the previously mentioned:

$3,3\rightarrow \frac{9}{2}$

$2,3,12 \rightarrow 12$ and $2,6,6\rightarrow 12$

$2, 3, 7, 84\rightarrow 147$

along with

$2, 3, 7, 43, 3612 \rightarrow 54360 \frac35$

$2,3,7,43,1807,6526884 \rightarrow 29583482464 \frac9{10}.$

These seem,with two small exceptions, to be given by, $a_i=1+\prod_{j=1}^{i-1}a_j$ except that $a_n=2 \prod_{j=1}^{n-1}a_j.$

This puts the point $(1,1,\cdots,1,2)$ on the boundary.

this is a great question and fairly good answers are known for arbitrary dimensions. (I think there are works for $d=3$ but I am not sure.) The relevant papers are

On lattice polytopes having interior lattice points, by J.M. Wills; J. Zaks; M.A. Perles . Elemente der Mathematik (1982) Volume: 37, page 44-45 (lower bounds)

and

Bounds for lattice polytopes containing a fixed number of interior points by JC Lagarias, GM Ziegler - Canad. J. Math, 1991‏ (upper bounds)

Here are the lower bounds, $n$ is the number of points in the interior. ($v$ is the function.)