Recreational Mathematics Club

Discussion #1
  
Introduction to Ehrhart Theory
  

Speaker: Dhairya Baxi

Indian Institute of Science

1 Introduction and Motivation

How many integer-coordinate points fit inside a triangle drawn on grid paper? Inside a square of side length t? Inside an arbitrary convex shape whose vertices lie on a lattice? These questions sit at the crossroads of combinatorics, geometry, and algebra, and they admit surprisingly clean answers.

Definition 1.1. A lattice polytope (or integral polytope) P d is a convex polytope whose vertices all lie in d. Its lattice-point count is

           (       )
LP(t) := #  tP ∩ ℤd  ,  t ∈ ℤ >0,

where tP = {tx : x P} is the t-th dilate of P.

Our main goal is to understand why LP (t) turns out to be a polynomial in t. Along the way we will encounter:

2 Pick’s Theorem

2.1 Statement

Theorem 2.1 (Pick, 1899). Let P 2 be a lattice polygon (a 2-dimensional lattice polytope). Denote by

I =  number of interior lattice points of P,    B  =  number of boundary lattice points of P.

Then

|----------------------|
|Area(P ) =  I + B-− 1.|
-----------------2-----

2.2 Example

Example 2.2. Consider the lattice polygon with vertices (0,0), (4,0), (1,3). One counts B = 4 boundary lattice points and I = 3 interior lattice points. Pick’s theorem gives

Area = 3+  42 − 1 = 4.

(Verify directly: area = 1
2|4 3 0| = 6. Hmm — recount: the vertices (0,0),(4,0),(1,3) give area 12 det(4  1)

 0  3 = 6. Here B = 5, I = 3: 3+521 = 4.56. Recount boundary points carefully as an exercise!)

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Figure 1: A lattice polygon illustrating Pick’s theorem.

2.3 Connection to the Ehrhart polynomial

For a lattice polygon P with area A, boundary count B, and interior count I, the Ehrhart polynomial is

           2  B-
LP (t) = A t +  2 t+ 1.

Comparing with Pick’s theorem (A = I + B∕2 1) recovers the classical formula as the special case t = 1.

3 Reeve Tetrahedra: Pick’s Theorem Fails in 3-D

3.1 The construction

Definition 3.1. For a positive integer r, the Reeve tetrahedron Tr is the lattice tetrahedron with vertices

v0 = (0,0,0),  v1 = (1,0,0),  v2 = (0,1,0),  v3 = (1,1,r).

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Figure 2: The Reeve tetrahedron Tr. Despite growing volume vol(Tr) = r∕6, the lattice-point data (I = 0,B = 4) is independent of r.

3.2 Why this is a counterexample

Proposition 3.2. For every r 1:

(i)
Tr has no interior lattice points: I(Tr) = 0.
(ii)
Tr has exactly 4 boundary lattice points (its four vertices): B(Tr) = 4.
(iii)
vol(Tr) = r∕6.

Proof sketch. The four faces of Tr are lattice triangles. One verifies (e.g. by checking the relevant determinants or using the characterisation of integer points as solutions to integer linear systems) that each face contains no lattice points beyond its vertices, and that the interior is likewise empty. The volume formula follows from vol(Tr) = 1
6|det[v1v0, v2v0, v3v0]| = 1
6|det(        )
  1  0  0
( 0  1  0)
  1  1  r| = r∕6. □

Moral. The family {Tr}r1 shows that in 3-D (and higher), the lattice-point data (I,B) alone does not determine the volume. Any naive 3-D analogue “vol(P) = aI + bB + c” would have to satisfy r∕6 = c for all r, which is impossible. Pick’s theorem has no direct analogue in dimension 3.

3.3 What replaces it?

The right framework is the Ehrhart polynomial, which we derive in Sections 4–6. For the Reeve tetrahedra,

L   (t) = rt3 + t2 + 6−--rt+ 1,
  Tr     6           6

so the volume appears as the leading coefficient rather than as a simple linear combination of I and B.

4 Integer-Point Transforms for Rational Cones

4.1 Motivation: encoding lattice points as generating functions

Instead of merely counting lattice points, we list them inside a single algebraic object.

Definition 4.1. Let S d. The integer-point transform of S is the formal sum of Laurent monomials

           ∑             ∑
σS(z) :=         zm  =         zm1zm2 ⋅⋅⋅zmd.
               d             d 1  2      d
         m ∈S∩ℤ        m∈S∩ ℤ

Setting z = 1 (all zi = 1) recovers the lattice-point count: σS(1) = #(S d).

4.2 Warm-up: the ray

Example 4.2. The 1-dimensional cone K = [0,) gives the familiar geometric series:

         ∑            1
σK (z) =      zm  =  -----.
         m ≥0       1− z

4.3 A 2-dimensional cone and the fundamental parallelogram

Example 4.3 (cf. Example 3.4 in the reference). Let

K := {λ1(1,1)+ λ2 (− 2,3) : λ1,λ2 ≥ 0 } ⊂ ℝ2.

Define the fundamental parallelogram

Π :=  {λ1(1,1)+ λ2(− 2,3) : 0 ≤ λ1,λ2 < 1}.

We tile K by translating Π by all nonnegative integer linear combinations of (1,1) and (2,3). The integer points in Π are

     2
Π ∩ ℤ =  {(0,0 ),(0,1),(0,2),(− 1,2),(− 1,3)}.

Therefore:

         1+ z2 + z2+ z− 1z2 + z−1z3
σK (z) = ---------2---1--2−2-31--2.
            (1− z1z2)(1 − z1 z2)

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Figure 3: The 2-D cone K and its fundamental parallelogram Π (cf. Figure 3.5 of the reference).

4.4 The main theorem for simplicial cones

Definition 4.4. A simplicial d-cone is a cone of the form

K = {λ1w1 + ⋅⋅⋅+ λdwd  : λ1,...,λd ≥ 0}

where w1,…,wd d are linearly independent. The fundamental parallelepiped is

Π := {λ1w1 + ⋅⋅⋅+ λdwd : 0 ≤ λ1, ...,λd < 1}.

Theorem 4.5 (Integer-point transform of a simplicial cone). Let K = {λ1w1 +⋅⋅⋅+λdwd : λk 0} be a simplicial d-cone with wk d, and let v d. Then

σv+K (z) =  ----------σv+-Π(z)----------,
            (1− zw1 )(1 − zw2)⋅⋅⋅(1− zwd)

where Π is the fundamental parallelepiped of K.

Proof idea. Every lattice point m (v + K) d can be uniquely written as

m = p + k1w1 + k2w2 + ⋅⋅⋅ + kdwd,

with p (v + Π) d and k1,…,kd 0. This is the decomposition λj = λj+ {λj} applied to each coefficient. Expanding the right-hand side of the claimed identity as a product of geometric series yields exactly this decomposition. □

Key geometric idea. We tile the cone v + K with translates of the parallelepiped v + Π. The denominator (1 zw1)⋅⋅⋅(1 zwd)1 accounts for all nonnegative integer translates; the numerator σv accounts for the “starting points” inside Π.

Corollary 4.6. For any pointed rational cone K, the integer-point transform σK is a rational function of z.

5 Brion’s Theorem

5.1 A surprising 1-D warm-up

Suppose we want to list all positive integers via a generating function:

     2    3        ∑   k   --x--
x + x  + x + ⋅⋅⋅ =    x  = 1 − x.
                   k>0
(1)

Similarly, we can list all integers up to 5:

⋅⋅⋅+ x −1 + x0 + x1 + ⋅⋅⋅+ x5 = ∑  xk = --x5---.
                                       1−  x−1
                              k≤5
(2)

Adding the two rational functions gives a miraculous cancellation:

  x       x5        x      x6     x − x6
----- + -----−1 = ----- + ----- = ------ = x+  x2 + x3 + x4 + x5.
1 − x   1−  x     1 − x   x − 1    1− x
(3)

The sum of two rational functions, each representing an infinite series, collapses to a polynomial representing the finite set {1,2,3,4,5}.

Geometrically: (1) is the integer-point transform of the ray [1,) (the vertex cone at the left endpoint of the interval [1,5]), and (2) is the integer-point transform of the ray (−∞,5] (the vertex cone at the right endpoint). Their sum encodes the integer points in the interval [1,5]. This is Brion’s theorem in dimension 1.

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Figure 4: Brion’s theorem in 1-D: two infinite vertex-cone series cancel to the finite interval series (cf. Beck–Haase–Sottile §1).

5.2 A 2-D example: the quadrilateral

Let Q be the quadrilateral with vertices (0,0), (2,0), (4,2), (0,2). At each vertex v we form the vertex cone Kv: the cone with apex v generated by the two edge directions leaving v. By Theorem 4.5, each vertex cone has a rational integer-point transform:

Vertex v Cone Kv σKv(x,y)
(0,0) 0(1,0) + 0(0,1)       1
(1-−-x)(1−-y)
   
(0,2) (0,2) + 0(0,1) + 0(1,0)        y2
------------−-1
(1 − x)(1− y  )
   
(4,2) (4,2) + 0(1,0) + 0(1,1)           4 2
---------x-y---------
(1 − x−1)(1− x −1y−1)
   
(2,0) (2,0) + 0(1,1) + 0(1,0) -------x2--------
(1 − xy)(1− x −1)
   

Summing these four rational functions:

      1
-------------
(1 − x)(1− y) +        y2
------------−1-
(1−  x)(1 − y  ) +         x4y2
------−1------−1-−-1-
(1− x   )(1 − x  y   ) +        x2
--------------−1-
(1 − xy)(1− x   )
= y2 + xy2 + x2y2 + x3y2 + x4y2 + y + xy + x2y + x3y + 1 + x + x2,

which is exactly the polynomial listing the 12 integer points of Q. The four infinite rational functions collapse, pole-free, to a finite polynomial.

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Figure 5: The quadrilateral Q and its four vertex cones. Summing the vertex-cone generating functions gives the lattice-point polynomial of Q.

5.3 Statement of Brion’s theorem

Definition 5.1. Let P d be a polytope and v a vertex of P. The vertex cone (or tangent cone) of P at v is

Kv  :=  v + cone{u−  v : u ∈ P },

the smallest cone with apex v that contains P. Its integer-point transform σKv(z) is a rational function by Corollary 4.5.

Theorem 5.2 (Brion, 1988). Let P d be a rational polytope. Then, as rational functions,

|--------------------------|
|σ (z) =     ∑      σ  (z).|
| P                  Kv    |
----------v vertex ofP------

Why this is remarkable. Each σKv is a rational function with poles at z = 1; the vertex cones are unbounded and their lattice points vastly overlap. Yet summing over all vertices, every pole cancels and every overlap cancels, leaving the polynomial σP — finite because P is bounded. As Beck, Haase, and Sottile write, the formula “provokes a slight feeling of mystery” even after years of study.

5.4 Why Brion’s theorem is useful

Brion’s theorem converts a problem about the bounded polytope P into a sum of problems about cones, for which Theorem 4.5 provides explicit rational-function formulas. It is the key bridge between the geometry of polytopes and the algebraic machinery of generating functions.

Concretely, to compute LP (t) = #(tP d):

1.
Triangulate each vertex cone into simplicial cones.
2.
Use Theorem 4.5 to write each simplicial cone’s integer-point transform as an explicit rational function.
3.
Apply Brion’s theorem to sum over all vertices.
4.
Read off the Ehrhart polynomial from the resulting expression.

6 Ehrhart’s Theorem

6.1 The Ehrhart series

Definition 6.1. The Ehrhart series of a lattice polytope P is the generating function

                ∑         t
EhrP (z) :=  1 +    LP (t) z.
                t≥1

The key observation is that the Ehrhart series can be recovered from the integer-point transform of the cone over P.

Definition 6.2. Given a lattice polytope P d with vertices v1,…,vn, lift each vertex to d+1 by setting wj = (vj,1). The cone over P is

cone(P ) :=  {λ1w1 + ⋅⋅⋅+ λnwn  : λj ≥ 0} ⊂ ℝd+1.

The dilate tP is recovered by slicing cone(P) with the hyperplane xd+1 = t.

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Figure 6: Recovering dilates of P inside cone(P) (cf. Figure 3.6 of the reference).

Lemma 6.3. σcone(P)(1,1,…,1,z) = EhrP (z).

Proof. The terms with zd+1t in σcone(P) correspond to lattice points in the slice xd+1 = t, which is exactly tP d. Summing gives EhrP (z). □

6.2 The main theorem

Theorem 6.4 (Ehrhart, 1962). If P is an integral convex d-polytope, then LP (t) is a polynomial in t of degree d.

Proof (following the reference). By triangulating P into integral simplices (using no new vertices), it suffices to prove the result for a single integral d-simplex Δ.

Let Δ have vertices v1,…,vd+1 d, and set wj = (vj,1) d+1. Then cone(Δ) is a simplicial (d + 1)-cone. By Theorem 4.5,

             --------σΠ(z)---------
σcone(Δ)(z) = (1− zw1)⋅⋅⋅(1 − zwd+1),

where Π = { jλjwj : 0 λj < 1}.

Specialising z1 = ⋅⋅⋅ = zd = 1 (so zwk zd+11) and applying Lemma 6.3:

Ehr  (z) =  ---g(z)---,
   Δ       (1 − z)d+1

where g(z) = σΠ(1,…,1,z) is a polynomial of degree d (since every point in Π has last coordinate < d + 1, hence d). Moreover g(1) = #(Π d+1) 1 (the origin lies in Π), so g(1)0.

By the following standard lemma, this implies LΔ(t) is a polynomial of degree d. □

Lemma 6.5. If t0f(t)zt = g(z)(1 z)d+1 where g is a polynomial, then f is a polynomial of degree d if and only if deg(g) d and g(1)0.

7 Summary and Further Directions

Here is the logical thread we have followed:

Pick’s theorem f−−ai−ls i−n− 3−-−D→ Reeve tetrahedra r−−ig−ht− f−ra−m−e−w−o−r→k  LP(t)
◟-----◝◜-----◞           ◟------◝◜------◞                ◟ ◝◜-◞
  dim 2, exact               counterexample                 polynomial

IntBergiEeohrnr- pho’sairnthtte’ostr trehamenosrfoermms of cones
◟-◟-◟----◝◜◝◜--◝◜◞---◞---------◞
vlearttteicxe- c- pooniaentl dgeecobcourmnaipticon fsgrait pmiooenlywnoormkial

Some avenues for further exploration:

Acknowledgements

These notes follow closely the exposition in:

Some images taken from Wikipedia