Counting Lattice Points in Polytopes: The Ehrhart Theory

3 Counting Lattice Points in Polytopes: The Ehrhart Theory Ubi materia, ibi geometria. Johannes Kepler (1571 1630) Given the profusion of examples that gave rise to the polynomial behavior of the integer-point counting function L P (t) for special polytopes P, we now ask whether there is a general structure theorem. As the ieas unfol, the reaer is invite to look back at Chapters 1 an 2 as appetizers an inee as special cases of the theorems evelope below. 3.1 Triangulations an Pointe Cones Because most of the proofs that follow work like a charm for a simplex, we first issect a polytope into simplices. This issection is capture by the following efinition. A triangulation of a convex -polytope P is a finite collection T of - simplices with the following properties: P =. T For any 1, 2 T, 1 2 is a face common to both 1 an 2. We say that P can be triangulate using no new vertices if there exists a triangulation T such that the vertices of any T are vertices of P. Theorem 3.1 (Existence of triangulations). Every convex polytope can be triangulate using no new vertices. This theorem seems intuitively obvious but is not entirely trivial to prove. We carefully work out a proof in the appenix.

58 3 Counting Lattice Points in Polytopes: The Ehrhart Theory Fig. 3.1. Two (very ifferent) triangulations of the 3-cube. A pointe cone K R is a set of the form K = {v + λ 1 w 1 + λ 2 w 2 + + λ m w m : λ 1, λ 2,..., λ m 0}, where v, w 1, w 2,..., w m R are such that there exists a hyperplane H for which H K = {v}; that is, K \ {v} lies strictly on one sie of H. The vector v is calle the apex of K, an the w k s are the generators of K. The cone is rational if v, w 1, w 2,..., w m Q, in which case we may choose w 1, w 2,..., w m Z by clearing enominators. The imension of K is the imension of the affine space spanne by K; if K is of imension, we call it a -cone. The -cone K is simplicial if K has precisely linearly inepenent generators. Just as polytopes have a escription as an intersection of half-spaces, so o pointe cones: A rational pointe -cone is the -imensional intersection of finitely many half-spaces of the form { x R : a 1 x 1 + a 2 x 2 + + a x b } with integral parameters a 1, a 2,..., a, b Z such that the corresponing hyperplanes of the form { x R : a 1 x 1 + a 2 x 2 + + a x = b } meet in exactly one point. Cones are important for many reasons. The most practical for us is a process calle coning over a polytope. Given a convex polytope P R with vertices v 1, v 2,..., v n, we lift these vertices into R +1 by aing a 1 as their last coorinate. So, let w 1 = (v 1, 1), w 2 = (v 2, 1),..., w n = (v n, 1). Now we efine the cone over P as cone(p) = {λ 1 w 1 + λ 2 w 2 + + λ n w n : λ 1, λ 2,..., λ n 0} R +1. This pointe cone has the origin as apex, an we can recover our original polytope P (strictly speaking, the translate set {(x, 1) : x P}) by cutting cone(p) with the hyperplane x +1 = 1, as shown in Figure 3.2.

3.1 Triangulations an Pointe Cones 59 x 3 x 2 x 1 Fig. 3.2. Coning over a polytope. By analogy with the language of polytopes, we say that the hyperplane H = {x R : a x = b} is a supporting hyperplane of the pointe -cone K if K lies entirely on one sie of H, that is, K { x R : a x b } or K { x R : a x b }. A face of K is a set of the form K H, where H is a supporting hyperplane of K. The ( 1)-imensional faces are calle facets an the 1-imensional faces eges of K. The apex of K is its unique 0-imensional face. Just as polytopes can be triangulate into simplices, pointe cones can be triangulate into simplicial cones. So, a collection T of simplicial -cones is a triangulation of the -cone K if it satisfies: K = S. S T For any S 1, S 2 T, S 1 S 2 is a face common to both S 1 an S 2.

60 3 Counting Lattice Points in Polytopes: The Ehrhart Theory We say that K can be triangulate using no new generators if there exists a triangulation T such that the generators of any S T are generators of P. Theorem 3.2. Any pointe cone can be triangulate into simplicial cones using no new generators. Proof. This theorem is really a corollary to Theorem 3.1. Given a pointe - cone K, there exists a hyperplane H that intersects K only at the apex. Now translate H into the cone, so that H K consists of more than just one point. This intersection is a ( 1)-polytope P, whose vertices are etermine by the generators of K. Now triangulate P using no new vertices. The cone over each simplex of the triangulation is a simplicial cone. These simplicial cones, by construction, triangulate K. 3.2 Integer-Point Transforms for Rational Cones We want to encoe the information containe by the lattice points in a set S R. It turns out that the following multivariate generating function allows us to o this in an efficient way if S is a rational cone or polytope: σ S (z) = σ S (z 1, z 2,..., z ) := z m. m S Z The generating function σ S simply lists all integer points in S in a special form: not as a list of vectors, but as a formal sum of monomials. We call σ S the integer-point transform of S; the function σ S also goes by the name moment generating function or simply generating function of S. The integerpoint transform σ S opens the oor to both algebraic an analytic techniques. Example 3.3. As a warm-up example, consier the 1-imensional cone K = [0, ). Its integer-point transform is our ol frien σ K (z) = m [0, ) Z z m = m 0 Example 3.4. Now we consier the two-imensional cone K := {λ 1 (1, 1) + λ 2 ( 2, 3) : λ 1, λ 2 0} R 2 z m = 1 1 z. epicte in Figure 3.3. To obtain the integer-point transform σ K, we tile K by copies of the funamental parallelogram Π := {λ 1 (1, 1) + λ 2 ( 2, 3) : 0 λ 1, λ 2 < 1} R 2. More precisely, we translate Π by nonnegative integer linear combinations of the generators (1, 1) an ( 2, 3), an these translates will exactly cover

3.2 Integer-Point Transforms for Rational Cones 61 K. How can we list the integer points in K as monomials? Let s first list all vertices of the translates of Π. These are nonnegative integer combinations of the generators (1, 1) an ( 2, 3), so we can list them using geometric series: m=j(1,1)+k( 2,3) j,k 0 z m = j 0 z j(1,1)+k( 2,3) 1 = (1 z 1 z 2 ) ( 1 z1 2 ). z3 2 k 0 We now use the integer points (m, n) Π to generate a subset of Z 2 by aing to (m, n) nonnegative linear integer combinations of the generators (1, 1) an ( 2, 3). Namely, we let L (m,n) := {(m, n) + j(1, 1) + k( 2, 3) : j, k Z 0 }. It is immeiate that K Z 2 is the isjoint union of the subsets L (m,n) as (m, n) ranges over Π Z 2 = {(0, 0), (0, 1), (0, 2), ( 1, 2), ( 1, 3)}. Hence σ K (z) = ( 1 + z 2 + z2 2 + z1 1 z2 2 + z1 1 ) z3 2 z m m=j(1,1)+k( 2,3) j,k 0 = 1 + z 2 + z2 2 + z1 1 z2 2 + z1 1 z3 2 (1 z 1 z 2 ) ( 1 z 2 ). 1 z3 2 x 2 ( 2, 3) (1, 1) x 1 Fig. 3.3. The cone K an its funamental parallelogram.

62 3 Counting Lattice Points in Polytopes: The Ehrhart Theory Similar geometric series suffice to escribe integer-point transforms for simplicial -cones. The following result utilizes the geometric series in several irections simultaneously. Theorem 3.5. Suppose K := {λ 1 w 1 + λ 2 w 2 + + λ w : λ 1, λ 2,..., λ 0} is a simplicial -cone, where w 1, w 2,..., w Z. Then for v R, the integer-point transform σ v+k of the shifte cone v+k is the rational function σ v+k (z) = where Π is the half-open parallelepipe σ v+π (z) (1 z w1 ) (1 z w2 ) (1 z w ), Π := {λ 1 w 1 + λ 2 w 2 + + λ w : 0 λ 1, λ 2,..., λ < 1}. The half-open parallelepipe Π is calle the funamental parallelepipe of K. Proof. In σ v+k (z) = m (v+k) Z zm, we list each integer point m in v + K as the monomial z m. Such a lattice point can, by efinition, be written as m = v + λ 1 w 1 + λ 2 w 2 + + λ w for some numbers λ 1, λ 2,..., λ 0. Because the w k s form a basis of R, this representation is unique. Let s write each of the λ k s in terms of their integer an fractional part: λ k = λ k + {λ k }. So m = v+ ( {λ 1 } w 1 +{λ 2 } w 2 + +{λ } w ) + λ1 w 1 + λ 2 w 2 + + λ w, an we shoul note that, since 0 {λ k } < 1, the vector p := v + {λ 1 } w 1 + {λ 2 } w 2 + + {λ } w is in v +Π. In fact, p Z, since m an λ k w k are all integer vectors. Again the representation of p in terms of the w k s is unique. In summary, we have prove that any m v + K Z can be uniquely written as m = p + k 1 w 1 + k 2 w 2 + + k w (3.1) for some p (v + Π) Z an some integers k 1, k 2,..., k 0. On the other han, let us write the rational function on the right-han sie of the theorem as a prouct of series: σ v+π (z) (1 z w1 ) (1 z w = z k1w1 z k w. ) k1 0 k 0 p (v+π) Z z p If we multiply everything out, a typical exponent will look exactly like (3.1).

3.2 Integer-Point Transforms for Rational Cones 63 Our proof contains a crucial geometric iea. Namely, we tile the cone v+k with translates of v + Π by nonnegative integral combinations of the w k s. It is this tiling that gives rise to the nice integer-point transform in Theorem 3.5. Computationally, we therefore favor cones over polytopes ue to our ability to tile a simplicial cone with copies of the funamental omain, as above. Another reason for favoring cones over polytopes appears in Brion s theorem in Chapter 9. Theorem 3.5 shows that the real complexity of computing the integerpoint transform σ v+k is embee in the location of the lattice points in the parallelepipe v + Π. By milly strengthening the hypothesis of Theorem 3.5, we obtain a slightly easier generating function, a result we shall nee in Section 3.4 an Chapter 4. Corollary 3.6. Suppose K := {λ 1 w 1 + λ 2 w 2 + + λ w : λ 1, λ 2,..., λ 0} is a simplicial -cone, where w 1, w 2,..., w Z, an v R, such that the bounary of v + K contains no integer point. Then σ v+k (z) = where Π is the open parallelepipe σ v+π (z) (1 z w1 ) (1 z w2 ) (1 z w ), Π := {λ 1 w 1 + λ 2 w 2 + + λ w : 0 < λ 1, λ 2,..., λ < 1}. Proof. The proof of Theorem 3.5 goes through almost verbatim, except that v+π now has no bounary lattice points, so that there is no harm in choosing Π to be open. Since a general pointe cone can always be triangulate into simplicial cones, the integer-point transforms a up in an inclusion exclusion manner (note that the intersection of simplicial cones in a triangulation is again a simplicial cone, by Exercise 3.2). Hence we have the following corollary. Corollary 3.7. Given any pointe cone K = {v + λ 1 w 1 + λ 2 w 2 + + λ m w m : λ 1, λ 2,..., λ m 0} with v R, w 1, w 2,..., w m Z, the integer-point transform σ K (z) evaluates to a rational function in the coorinates of z. Philosophizing some more, one can show that the original infinite sum σ K (z) converges only for z in a subset of C, whereas the rational function that represents σ K gives us its meromorphic continuation. Later, in Chapters 4 an 9, we make use of this continuation.

64 3 Counting Lattice Points in Polytopes: The Ehrhart Theory 3.3 Expaning an Counting Using Ehrhart s Original Approach Here is the funamental theorem concerning the lattice-point count in an integral convex polytope. Theorem 3.8 (Ehrhart s theorem). If P is an integral convex -polytope, then L P (t) is a polynomial in t of egree. This result is ue to Eugène Ehrhart, in whose honor we call L P the Ehrhart polynomial of P. Naturally, there is an extension of Ehrhart s theorem to rational polytopes, which we will iscuss in Section 3.7. Our proof of Ehrhart s theorem uses generating functions of the form t 0 f(t) zt, similar in spirit to the ones iscusse in the beginning of Chapter 1. If f is a polynomial, this power series takes on a special form, which we invite the reaer to prove (Exercise 3.8): Lemma 3.9. If f(t) z t g(z) = (1 z) +1, t 0 then f is a polynomial of egree if an only if g is a polynomial of egree at most an g(1) 0. The reason we introuce generating functions of the form σ S (z) = m S Z zm in Section 3.2 is that they are extremely hany for lattice-point problems. The connection to lattice points is evient, since we are summing over them. If we re intereste in the lattice-point count, we simply evaluate σ S at z = (1, 1,..., 1): σ S (1, 1,..., 1) = 1 m ( = 1 = # S Z ). m S Z m S Z (Here we enote by 1 a vector all of whose components are 1.) Naturally, we shoul make this evaluation only if S is boune; Theorem 3.5 alreay tells us that it s no fun evaluating σ K (1) if K is a cone. But the magic of the generating function σ S oes not stop there. To literally take it to the next level, we cone over a convex polytope P. If P R has the vertices v 1, v 2,..., v n Z, recall that we lift these vertices into R +1, by aing a 1 as their last coorinate. So let Then w 1 = (v 1, 1), w 2 = (v 2, 1),..., w n = (v n, 1). cone(p) = {λ 1 w 1 + λ 2 w 2 + + λ n w n : λ 1, λ 2,..., λ n 0} R +1.

3.3 Expaning an Counting Using Ehrhart s Original Approach 65 Recall that we can recover our original polytope P by cutting cone(p) with the hyperplane x +1 = 1. We can recover more than just the original polytope in cone(p): By cutting the cone with the hyperplane x +1 = 2, we obtain a copy of P ilate by a factor of 2. (The reaer shoul meitate on why this cut is a 2-ilate of P.) More generally, we can cut the cone with the hyperplane x +1 = t an obtain tp, as suggeste by Figure 3.4. Fig. 3.4. Recovering ilates of P in cone(p). Now let s form the integer-point transform σ cone(p) of cone(p). By what we just sai, we shoul look at ifferent powers of z +1 : there is one term (namely, 1), with z+1 0, corresponing to the origin; the terms with z1 +1 correspon to lattice points in P (liste as monomials in z 1, z 2,..., z ), the terms with z+1 2 correspon to points in 2P, etc. In short,

66 3 Counting Lattice Points in Polytopes: The Ehrhart Theory σ cone(p) (z 1, z 2,..., z +1 ) = 1 + σ P (z 1,..., z ) z +1 + σ 2P (z 1,..., z ) z+1 2 + = 1 + σ tp (z 1,..., z ) z+1 t. t 1 Specializing further for enumeration purposes, we recall that σ P (1, 1,..., 1) = # ( P Z ), an so σ cone(p) (1, 1,..., 1, z +1 ) = 1 + t 1 σ tp (1, 1,..., 1) z t +1 = 1 + t 1 # ( tp Z ) z t +1. But by efinition, the enumerators on the right-han sie are just evaluations of Ehrhart s counting function, that is, σ cone(p) (1, 1,..., 1, z +1 ) is nothing but the Ehrhart series of P: Lemma 3.10. σ cone(p) (1, 1,..., 1, z) = 1 + t 1 L P (t) z t = Ehr P (z). With all this machinery at han, we can prove Ehrhart s theorem. Proof of Theorem 3.8. It suffices to prove the theorem for simplices, because we can triangulate any integral polytope into integral simplices, using no new vertices. Note that these simplices will intersect in lower-imensional integral simplices. By Lemma 3.9, it suffices to prove that for an integral -simplex, Ehr (z) = 1 + t 1 L (t) z t = g(z) (1 z) +1 for some polynomial g of egree at most with g(1) 0. In Lemma 3.10 we showe that the Ehrhart series of equals σ cone( ) (1, 1,..., 1, z), so let s stuy the integer-point transform attache to cone( ). The simplex has + 1 vertices v 1, v 2,..., v +1, an so cone( ) R +1 is simplicial, with apex the origin an generators w 1 = (v 1, 1), w 2 = (v 2, 1),..., w +1 = (v +1, 1) Z +1. Now we use Theorem 3.5: σ cone( ) (z 1, z 2,..., z +1 ) = σ Π (z 1, z 2,..., z +1 ) (1 z w1 ) (1 z w2 ) (1 z w +1 ), where Π = {λ 1 w 1 + λ 2 w 2 + + λ +1 w +1 : 0 λ 1, λ 2,..., λ +1 < 1}. This parallelepipe is boune, so the attache generating function σ Π is a Laurent polynomial in z 1, z 2,..., z +1.

3.4 The Ehrhart Series of an Integral Polytope 67 We claim that the z +1 -egree of σ Π is at most. In fact, since the x +1 - coorinate of each w k is 1, the x +1 -coorinate of a point in Π is λ 1 +λ 2 + + λ +1 for some 0 λ 1, λ 2,..., λ +1 < 1. But then λ 1 + λ 2 + + λ +1 < + 1, so if this sum is an integer it is at most, which implies that the z +1 -egree of σ Π (z 1, z 2,..., z +1 ) is at most. Consequently, σ Π (1, 1,..., 1, z +1 ) is a polynomial in z +1 of egree at most. The evaluation σ Π (1, 1, 1,..., 1) of this polynomial at z +1 = 1 is not zero, because σ Π (1, 1, 1,..., 1) = # ( Π Z +1) an the origin is a lattice point in Π. Finally, if we specialize z w k to z 1 = z 2 = = z = 1, we obtain z+1 1, so that σ cone( ) (1, 1,..., 1, z +1 ) = σ Π (1, 1,..., 1, z +1 ) (1 z +1 ) +1. The left han sie is Ehr (z +1 ) = 1 + t 1 L (t) z+1 t by Lemma 3.10. 3.4 The Ehrhart Series of an Integral Polytope We can actually take our proof of Ehrhart s theorem one step further by stuying the polynomial σ Π (1, 1,..., 1, z +1 ). As mentione above, the coefficient of z+1 k simply counts the integer points in the parallelepipe Π cut with the hyperplane x +1 = k. Let us recor this. Corollary 3.11. Suppose is an integral -simplex with vertices v 1, v 2,..., v +1, an let w j = (v j, 1). Then Ehr (z) = 1 + t 1 where h k equals the number of integer points in L (t) z t = h z + h 1 z 1 + + h 1 z + h 0 (1 z) +1, {λ 1 w 1 + λ 2 w 2 + + λ +1 w +1 : 0 λ 1, λ 2,..., λ +1 < 1} with last variable equal to k. This result can actually be use to compute Ehr, an therefore the Ehrhart polynomial, of an integral simplex in low imensions very quickly (a fact that the reaer may iscover in some of the exercises). We remark, however, that things are not as simple for arbitrary integral polytopes. Not only is triangulation a nontrivial task in general, but one woul also have to eal with overcounting where simplices of a triangulation meet. Corollary 3.11 implies that the numerator of the Ehrhart series of an integral simplex has nonnegative coefficients, since those coefficients count something. Although the latter cannot be sai of the coefficients of the Ehrhart series of a general integral polytope, the nonnegativity property magically survives.

68 3 Counting Lattice Points in Polytopes: The Ehrhart Theory Theorem 3.12 (Stanley s nonnegativity theorem). Suppose P is an integral convex -polytope with Ehrhart series Then h 0, h 1,..., h 0. Ehr P (z) = h z + h 1 z 1 + + h 0 (1 z) +1. Proof. Triangulate cone(p) R +1 into the simplicial cones K 1, K 2,..., K m. Now Exercise 3.14 ensures that there exists a vector v R +1 such that cone(p) Z = (v + cone(p)) Z (that is, we neither lose nor gain any lattice points when shifting cone(p) by v) an neither the facets of v + cone(p) nor the triangulation hyperplanes contain any lattice points. This implies that every lattice point in v + cone(p) belongs to exactly one simplicial cone v + K j : cone(p) Z = (v + cone(p)) Z = m ( (v + Kj ) Z ), (3.2) an this union is a isjoint union. If we translate the last ientity into generating-function language, it becomes σ cone(p) (z 1, z 2,..., z +1 ) = j=1 m σ v+kj (z 1, z 2,..., z +1 ). But now we recall that the Ehrhart series of P is just a special evaluation of σ cone(p) (Lemma 3.10): j=1 Ehr P (z) = σ cone(p) (1, 1,..., 1, z) = m σ v+kj (1, 1,..., 1, z). (3.3) It suffices to show that the rational generating functions σ v+kj (1, 1,..., 1, z) for the simplicial cones v + K j have nonnegative numerator. But this fact follows from evaluating the rational function in Corollary 3.6 at (1, 1,..., 1, z). This proof shows a little more: Since the origin is in precisely one simplicial cone on the right-han sie of (3.2), we get on the right-han sie of (3.3) precisely one term that contributes 1/(1 z) +1 to Ehr P ; all other terms contribute to higher powers of the numerator polynomial of Ehr P. That is, the constant term h 0 equals 1. The reaer might feel that we are chasing our tail at this point, since we assume from the very beginning that the constant term of the infinite series Ehr P is 1, an hence h 0 has to be 1, as a quick look at the expansion of the rational function representing Ehr P shows. The above argument shows merely that this convention is geometrically soun. Let us recor this: j=1

3.4 The Ehrhart Series of an Integral Polytope 69 Lemma 3.13. Suppose P is an integral convex -polytope with Ehrhart series Ehr P (z) = h z + h 1 z 1 + + h 0 (1 z) +1. Then h 0 = 1. For a general integral polytope P, the reaer has probably alreay iscovere how to extract the Ehrhart polynomial of P from its Ehrhart series: Lemma 3.14. Suppose P is an integral convex -polytope with Ehrhart series Ehr P (z) = 1 + t 1 Then ( ) ( t + t + 1 L P (t) = + h 1 Proof. Expan into a binomial series: L P (t) z t = h z + h 1 z 1 + + h 1 z + 1 (1 z) +1. Ehr P (z) = h z + h 1 z 1 + + h 1 z + 1 (1 z) +1 ) ( ) ( ) t + 1 t + + h 1 + h. = ( h z + h 1 z 1 + + h 1 z + 1 ) ( t + t 0 ( ) t + ( t + = h z t+ + h 1 t 0 t 0 ( ) t + + h 1 z t+1 + ( ) t + z t t 0 t 0 ( ) t ( ) t + 1 = h z t + h 1 z t + t t 1 ( ) t + 1 + h 1 z t + ( ) t + z t. t 1 t 0 ) z t ) z t+ 1 + In all infinite sums on the right-han sie, we can start the inex t with 0 without changing the sums, by the efinition (2.1) of the binomial coefficient. Hence Ehr P (z) = t 0 ( ( ) ( ) ( ) t t + 1 t + 1 h + h 1 + + h 1 + ( t + )) z t.

70 3 Counting Lattice Points in Polytopes: The Ehrhart Theory The representation of the polynomial L P (t) in terms of the coefficients of ( Ehr P can be interprete as the Ehrhart polynomial expresse in the basis t ( ), t+1 ) (,..., t+ ) (see Exercise 3.9). This representation is very useful, as the following results show. Corollary 3.15. If P is an integral convex -polytope, then the constant term of the Ehrhart polynomial L P is 1. Proof. Use the expansion of Lemma 3.14. The constant term is ( ) ( ) ( ) ( ) ( ) 1 1 0 L P (0) = + h 1 + + h 1 + h = = 1. This proof is exciting, because it marks the first instance where we exten the omain of an Ehrhart polynomial beyon the positive integers, for which the lattice-point enumerator was initially efine. More precisely, Ehrhart s theorem (Theorem 3.8) implies that the counting function L P (t) = # ( tp Z ), originally efine for positive integers t, can be extene to all real or even complex arguments t (as a polynomial). A natural question arises: are there nice interpretations of L P (t) for arguments t that are not positive integers? Corollary 3.15 gives such an interpretation for t = 0. In Chapter 4, we will give interpretations of L P (t) for negative integers t. Corollary 3.16. Suppose P is an integral convex -polytope with Ehrhart series Ehr P (z) = h z + h 1 z 1 + + h 1 z + 1 (1 z) +1. Then h 1 = L P (1) 1 = # ( P Z ) 1. Proof. Use the expansion of Lemma 3.14 with t = 1: ( ) ( ) ( ) ( ) + 1 2 1 L P (1) = + h 1 + + h 1 + h = + 1 + h 1. The proof of Corollary 3.16 suggests that there are also formulas for h 2, h 3,... in terms of the evaluations L P (1), L P (2),..., an we invite the reaer to fin them (Exercise 3.10). A final corollary to Theorem 3.12 an Lemma 3.14 states how large the enominators of the Ehrhart coefficients can be: Corollary 3.17. Suppose P is an integral polytope with Ehrhart polynomial L P (t) = c t +c 1 t 1 + +c 1 t+1. Then all coefficients satisfy! c k Z.

3.5 From the Discrete to the Continuous Volume of a Polytope 71 Proof. By Theorem 3.12 an Lemma 3.14, ( ) ( ) ( ) ( ) t + t + 1 t + 1 t L P (t) = + h 1 + + h 1 + h, where the h k s are integers. Hence multiplying out this expression yiels a polynomial in t whose coefficients can be written as rational numbers with enominator!. We finish this section with a general result that gives relations between negative integer roots of a polynomial an its generating function. This theorem will become hany in Chapter 4, in which we fin an interpretation for the evaluation of an Ehrhart polynomial at negative integers. Theorem 3.18. Suppose p is a egree- polynomial with the rational generating function t 0 p(t) z t = h z + h 1 z 1 + + h 1 z + h 0 (1 z) +1. Then h = h 1 = = h k+1 = 0 an h k 0 if an only if p( 1) = p( 2) = = p ( ( k)) = 0 an p ( ( k + 1)) 0. Proof. Suppose h = h 1 = = h k+1 = 0 an h k 0. Then the proof of Lemma 3.14 gives ( ) ( ) ( ) t + t + k + 1 t + k p(t) = h 0 + + h k 1 + h k. All the binomial coefficients are zero for t = 1, 2,..., + k, so those are roots of p. On the other han, all binomial coefficients but the last one are zero for t = + k 1, an since h k 0, + k 1 is not a root of p. Conversely, suppose p( 1) = p( 2) = = p ( ( k)) = 0 an p ( ( k + 1)) 0. The first root 1 of p gives 0 = p( 1) = h 0 ( 1 ) +h 1 ( 2 ) ( ) 0 + +h 1 ( ) ( ) 1 1 +h = h, so we must have h = 0. The next root 2 forces h 1 = 0, an so on, up to the root + k, which forces h k+1 = 0. It remains to show that h k 0. But if h k were zero then, by a similar line of reasoning as in the first part of the proof, p( + k 1) = 0, a contraiction. 3.5 From the Discrete to the Continuous Volume of a Polytope Given a geometric object S R, its volume, efine by the integral vol S := x, is one of the funamental ata of S. By the efinition of the integral, say S

72 3 Counting Lattice Points in Polytopes: The Ehrhart Theory in the Riemannian sense, we can think of computing vol S by approximating S with -imensional boxes that get smaller an smaller. To be precise, if we take the boxes with sie length 1/t then they each have volume 1/t. We might further think of the boxes as filling out the space between gri points in the lattice ( 1 t Z). This means that volume computation can be approximate by counting boxes, or equivalently, lattice points in ( 1 t Z) : 1 vol S = lim t t # ( S ( 1 t Z ) ) It is a short step to counting integer points in ilates of S, because ( ( ) ) 1 # S t Z = # ( ts Z ). Let us summarize: Lemma 3.19. Suppose S R is -imensional. Then 1 vol S = lim t t # ( ts Z ). We emphasize here that S is -imensional, because otherwise (since S coul be lower-imensional although living in -space), by our current efinition vol S = 0. (We will exten our volume efinition in Chapter 5 to give nonzero relative volume to objects that are not full-imensional.) Part of the magic of Ehrhart s theorem lies in the fact that for an integral -polytope P, we o not have to take a limit to compute vol P; we nee to compute only the + 1 coefficients of a polynomial. Corollary 3.20. Suppose P R is a integral convex -polytope with Ehrhart polynomial c t + c 1 t 1 + + c 1 t + 1. Then c = vol P. Proof. By Lemma 3.19, vol P = lim t c t + c 1 t 1 + + c 1 t + 1 t = c. On the one han, this shoul not come as a surprise, because counting integer points in some object shoul grow roughly like the volume of the object as we make it bigger an bigger. On the other han, the fact that we can compute the volume as one term of a polynomial shoul be very surprising: the polynomial is a counting function an as such is something iscrete, yet by computing it (an its leaing term), we erive some continuous ata. Even more, we can at least theoretically compute this continuous atum (the volume) of the object by calculating a few values of the polynomial an then interpolating; this can be escribe as a completely iscrete operation! We finish this section by showing how to retrieve the continuous volume of an integer polytope from its Ehrhart series..

3.6 Interpolation 73 Corollary 3.21. Suppose P R is an integral convex -polytope, an Ehr P (z) = h z + h 1 z 1 + + h 1 z + 1 (1 z) +1. Then vol P = 1! (h + h 1 + + h 1 + 1). Proof. Use the expansion of Lemma 3.14. The leaing coefficient is 1! (h + h 1 + + h 1 + 1). 3.6 Interpolation We now use the polynomial behavior of the iscrete volume L P of an integral polytope P to compute the continuous volume vol P an the iscrete volume L P from finite ata. Two points uniquely etermine a line. There exists a unique quaratic passing through any three given points. In general, a egree- polynomial p is etermine by + 1 points (x, p(x)) R 2. Namely, evaluating p(x) = c x + c 1 x 1 + + c 0 at istinct inputs x 1, x 2,..., x +1 gives where so that p (x 1 ) p (x 2 ). p (x +1 ) = V c c 1. c 0 x 1 x 1 1 x 1 1 x 2 x 1 2 x 2 1 V =........, x +1 x 1 +1 x +1 1 c c 1. c 0 = V 1 p (x 1 ) p (x 2 ). p (x +1 ), (3.4). (3.5) (Exercise 3.16 makes sure that V is invertible.) The ientity (3.5) gives the famous Lagrange interpolation formula. This gives us an efficient way to compute L P, at least when im P is not too large. The continuous volume of P will follow instantly, since it is the leaing coefficient c of L P. In the case of an Ehrhart polynomial L P, we know that L P (0) = 1, so that (3.4) simplifies to

74 3 Counting Lattice Points in Polytopes: The Ehrhart Theory L P (x 1 ) 1 x 1 x 1 1 x 1 c L P (x 2 ) 1. = x 2 x 1 2 x 2 c 1..... L P (x ) 1 x x 1 x c 1 Example 3.22 (Reeve s tetraheron). Let T h be the tetraheron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), an (1, 1, h), where h is a positive integer (see Figure 3.5). z 2h (2, 2, 2h) h (1, 1, h) 1 2 y 1 x 2 Fig. 3.5. Reeve s tetraheron T h (an 2T h ). To interpolate the Ehrhart polynomial L Th (t) from its values at various points, we use Figure 3.5 to euce the following: 4 = L Th (1) = vol (T h ) + c 2 + c 1 + 1, h + 9 = L Th (2) = vol (T h ) 2 3 + c 2 2 2 + c 1 2 + 1. Using the volume formula for a pyrami, we know that

3.7 Rational Polytopes an Ehrhart Quasipolynomials 75 vol (T h ) = 1 3 (base area)(height) = h 6. Thus h + 1 = h + 2c 2 1, which gives us c 2 = 1 an c 1 = 2 h 6. Therefore L Th (t) = h ( 6 t3 + t 2 + 2 h ) t + 1. 6 3.7 Rational Polytopes an Ehrhart Quasipolynomials We o not have to change much to stuy lattice-point enumeration for rational polytopes, an most of this section will consist of exercises for the reaer. The structural result paralleling Theorem 3.8 is as follows. Theorem 3.23 (Ehrhart s theorem for rational polytopes). If P is a rational convex -polytope, then L P (t) is a quasipolynomial in t of egree. Its perio ivies the least common multiple of the enominators of the coorinates of the vertices of P. We will call the least common multiple of the enominators of the coorinates of the vertices of P the enominator of P. Theorem 3.23, also ue to Ehrhart, extens Theorem 3.8, because the enominator of an integral polytope P is one. Exercises 3.21 an 3.22 show that the wor ivies in Theorem 3.23 is far from being replaceable by equals. We start the path towar a proof of Theorem 3.23 by stating the analogue of Lemma 3.9 for quasipolynomials (see Exercise 3.19): Lemma 3.24. If t 0 f(t) z t = g(z) h(z), then f is a quasipolynomial of egree with perio iviing p if an only if g an h are polynomials such that eg(g) < eg(h), all roots of h are p th roots of unity of multiplicity at most + 1, an there is a root of multiplicity equal to + 1 (all of this assuming that g/h has been reuce to lowest terms). Our goal is now evient: we will prove that if P is a rational convex - polytope with enominator p, then Ehr P (z) = 1 + t 1 L P (t) z t = g(z) (1 z p ) +1, for some polynomial g of egree less than p( + 1). As in Section 3.3, we will have to prove this only for the case of a rational simplex. So suppose the - simplex has vertices v 1, v 2,..., v +1 Q, an the enominator of is p. Again we will cone over : let

76 3 Counting Lattice Points in Polytopes: The Ehrhart Theory then w 1 = (v 1, 1), w 2 = (v 2, 1),..., w +1 = (v +1, 1) ; cone( ) = {λ 1 w 1 + λ 2 w 2 + + λ +1 w +1 : λ 1, λ 2,..., λ +1 0} R +1. To be able to use Theorem 3.5, we first have to ensure that we have a escription of cone( ) with integral generators. But since the enominator of is p, we can replace each generator w k by pw k Z +1, an we re reay to apply Theorem 3.5. From this point, the proof of Theorem 3.23 procees exactly like that of Theorem 3.8, an we invite the reaer to finish it up (Exercise 3.20). Although the proofs of Theorem 3.23 an Theorem 3.8 are almost ientical, the arithmetic structure of Ehrhart quasipolynomials is much more subtle an less well known than that of Ehrhart polynomials. 3.8 Reflections on the Coin-Exchange Problem an the Gallery of Chapter 2 At this point, we encourage the reaer to look back at the first two chapters in light of the basic Ehrhart-theory results. Popoviciu s theorem (Theorem 1.5) an its higher-imensional analogue give a special set of Ehrhart quasipolynomials. On the other han, in Chapter 2 we encountere many integral polytopes. Ehrhart s theorem (Theorem 3.8) explains why their lattice-point enumeration functions were all polynomials. Notes 1. Triangulations of polytopes an manifols are an active source of research with many interesting open problems; see, e.g., [68]. 2. Eugène Ehrhart lai the founation for the central theme of this book in the 1960s, starting with the proof of Theorem 3.8 in 1962 [78]. The proof we give here follows Ehrhart s original lines of thought. An interesting fact is that he i his most beautiful work as a teacher at a lycée in Strasbourg (France), receiving his octorate at age 60 on the urging of some colleagues. 3. Given any linearly inepenent vectors w 1, w 2,..., w R, the lattice generate by them is the set of all integer linear combinations of w 1, w 2,..., w. Alternatively, one can efine a lattice as a iscrete subgroup of R, an these two notions can be shown to be equivalent. One might woner whether replacing the lattice Z by an arbitrary lattice L throughout the statements of the theorems requiring now that the vertices of a polytope be in L gives us any ifferent results. The fact that the theorems of this chapter

Exercises 77 remain the same follows from the observation that any lattice can be mappe to Z by an invertible linear transformation. 4. Richar Stanley evelope much of the theory of Ehrhart (quasi-)polynomials, initially from a commutative-algebra point of view. Theorem 3.12 is ue to him [169]. The proof we give here appeare in [30]. An extension of Theorem 3.12 was foun by Takayuki Hibi; he prove that if h > 0 then h k h 1 for all 1 k 1 (using the notation of Theorem 3.12) [97]. 5. The tetraheron T h of Example 3.22 was use by John Reeve to show that Pick s theorem oes not hol in R 3 (see Exercise 3.18) [153]. Incientally, the formula for L Th also proves that the coefficients of an Ehrhart polynomial (of a close polytope) are not always positive. 6. There are several interesting questions (some of which are still open) regaring the perios of Ehrhart quasipolynomials. Some particularly nice examples about what can happen with perios were given by Tyrrell McAllister an Kevin Woos [125]. 7. Most of the results remain true if we replace convex polytope by polytopal complex, that is, a finite union of polytopes. One important exception is Corollary 3.15: the constant term of an Ehrhart polynomial of an integral polytopal complex C is the Euler characteristric of C. 8. The reaer might woner why we o not iscuss polytopes with irrational vertices. The answer is simple: noboy has yet foun a theory that woul parallel the results in this chapter, even in imension two. One notable exception is [11], in which irrational extensions of Brion s theorem are given; we will stuy the rational case of Brion s theorem in Chapter 9. On the other han, Ehrhart theory has been extene to functions other than strict lattice-point counting; one instance is escribe in Chapter 11. Exercises 3.1. To any permutation π S on elements, we associate the simplex π := conv { 0, e π(1), e π(1) + e π(2),..., e π(1) + e π(2) + + e π() }, where e 1, e 2,..., e enote the unit vectors in R. (a) Prove that { π : π S } is a triangulation of the unit -cube [0, 1]. (b) Prove that all π are congruent to each other, that is, each one can be obtaine from any other by reflections, translations, an rotations. (c) Show that for all π S, L π (t) = ( ) +t.

78 3 Counting Lattice Points in Polytopes: The Ehrhart Theory 3.2. Suppose T is a triangulation of a pointe cone. Prove that the intersection of two simplicial cones in T is again a simplicial cone. 3.3. Fin the generating function σ K (z) for the following cones: (a) K = {λ 1 (0, 1) + λ 2 (1, 0) : λ 1, λ 2 0} ; (b) K = {λ 1 (0, 1) + λ 2 (1, 1) : λ 1, λ 2 0} ; (c) K = {(3, 4) + λ 1 (0, 1) + λ 2 (2, 1) : λ 1, λ 2 0}. 3.4. Let S R m an T R n. Show that σ S T (z 1, z 2,..., z m+n ) = σ S (z 1, z 2,..., z m ) σ T (z m+1, z m+2,..., z m+n ). 3.5. Let K be a rational -cone, an let m Z. Show that σ m+k (z) = z m σ K (z). 3.6. For a set S R, let S := { x : x S}. Prove that ( 1 σ S (z 1, z 2,..., z ) = σ S, 1,..., 1 ). z 1 z 2 z 3.7. Given a pointe cone K R with apex at the origin, let S := K Z. Show that if x, y S then x + y S. (In algebraic terms, S is a semigroup, since 0 S an associativity of the aition in S follows trivially from associativity in R.) 3.8. Prove Lemma 3.9: If f(t) z t g(z) = (1 z) +1, t 0 then f is a polynomial of egree if an only if g is a polynomial of egree at most an g(1) 0. 3.9. Prove that ( ) ( x+n n, x+n 1 ) ( n,..., x ) n is a basis for the vector space Poln of polynomials (in the variable x) of egree less than or equal to n. 3.10. For a polynomial p(t) = c t + c 1 t t 1 + + c 0, let H p (z) be efine by p(t) z t = H p(z) (1 z) +1. t 0 Consier the map φ : Pol Pol, p H p. (a) Show that φ is a linear transformation. (b) Compute the matrix escribing φ for = 0, 1, 2,.... (c) Deuce formulas for h 2, h 3,..., similar to the one in Corollary 3.16. 3.11. Compute the Ehrhart polynomials an the Ehrhart series of the simplices with the following vertices:

(a) (0, 0, 0), (1, 0, 0), (0, 2, 0), an (0, 0, 3); (b) (0, 0, 0, 0), (1, 0, 0, 0), (0, 2, 0, 0), (0, 0, 3, 0), an (0, 0, 0, 4). 3.12. Define the hypersimplex (, k) as the convex hull of {e j1 + e j2 + + e jk : 1 j 1 < j 2 < < j k }, Exercises 79 where e 1, e 2,..., e are the stanar basis vectors in R. For example, (, 1) = (, 1) are regular ( 1)-simplices. Compute the Ehrhart polynomial an the Ehrhart series of (, k). 3.13. Suppose H is the hyperplane given by H = { x R : a 1 x 1 + a 2 x 2 + + a x = 0 } for some a 1, a 2,..., a Z, which we may assume to have no common factor. Prove that there exists v Z such that n Z ( (nv + H) Z ) = Z. (This implies, in particular, that the points in Z \ H are all at least some minimal istance away from H; this minimal istance is essentially given by the ot prouct of v with (a 1, a 2,..., a ).) 3.14. A hyperplane H is rational if it can be written in the form H = { x R : a 1 x 1 + a 2 x 2 + + a x = b } for some a 1, a 2,..., a, b Z. A hyperplane arrangement in R is a finite set of hyperplanes in R. Prove that a rational hyperplane arrangement H can be translate so that no hyperplane in H contains any integer points. 3.15. The conclusion of the previous exercise can be strengthene: Prove that a rational hyperplane arrangement H can be translate such that no hyperplane in H contains any rational points. 3.16. Show that, given istinct numbers x 1, x 2,..., x +1, the matrix x 1 x 1 1 x 1 1 x 2 x 1 2 x 2 1 V =.... x +1 x 1 +1 x +1 1 is not singular. (V is known as the Vanermone matrix.) 3.17. Let P be an integral -polytope. Show that ( vol P = 1 ( ) ) ( 1) + ( 1) k L P (k).! k k=1

80 3 Counting Lattice Points in Polytopes: The Ehrhart Theory 3.18. As in Example 3.22, let T n be the tetraheron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), an (1, 1, n), where n is a positive integer. Show that the volume of T n is unboune as n, yet for all n, T n has no interior an four bounary lattice points. This example proves that Pick s theorem oes not hol for a three-imensional integral polytope P, in the sense that there is no linear relationship among vol P, L P (1), an L P (1). 3.19. Prove Lemma 3.24: If t 0 f(t) z t = g(z) h(z), then f is a quasipolynomial of egree with perio iviing p if an only if g an h are polynomials such that eg(g) < eg(h), all roots of h are p th roots of unity of multiplicity at most + 1, an there is a root of multiplicity equal to + 1 (all of this assuming that g/h has been reuce to lowest terms). 3.20. Provie the etails for the proof of Theorem 3.23: If P is a rational convex -polytope, then L P (t) is a quasipolynomial in t of egree. Its perio ivies the least common multiple of the coorinates of the vertices of P. ( 3.21. Let T be the rational triangle with vertices (0, 0), 1, p 1 p ), an (p, 0), where p is a fixe integer 2. Show that L T (t) = p 1 2 t 2 + p+1 2 t + 1; in particular, L T is a polynomial. 3.22. Prove that for any 2 an any p 1, there exists a -polytope P whose Ehrhart quasipolynomial is a polynomial (i.e., it has perio 1), yet P has a vertex with enominator p. 3.23. Prove that the perio of the Ehrhart quasipolynomial of a 1-imensional polytope is always equal to the lcm of the enominators of its vertices. 3.24. Let T be the triangle with vertices ( 1 2, ( 2) 1, 1 2, ( 2) 1, an 0, 3 2). Show that L T (t) = t 2 + c(t) t + 1, where { 1 if t is even, c(t) = 0 if t is o. (This shows that the perios of the coefficients of an Ehrhart quasipolynomial o not necessarily increase with ecreasing power.) Fin the Ehrhart series of T. 3.25. Prove the following extension of Theorem 3.12: Suppose P is a rational -polytope with enominator p. Then Ehr P (z) = f(z) (1 z p ) +1, where f is a polynomial with nonnegative integral coefficients.

Exercises 81 3.26. Fin an prove a statement that extens Lemma 3.14 to Ehrhart quasipolynomials. 3.27. Prove the following extension of Corollary 3.15 to rational polytopes. Namely, the Ehrhart quasipolynomial L P of the rational convex polytope P R satisfies L P (0) = 1. 3.28. Prove the following analogue of Corollary 3.17 for rational polytopes: Suppose P is a rational polytope with Ehrhart quasipolynomial L P (t) = c (t) t + c 1 (t) t 1 + + c 1 (t) t + c 0 (t). Then for all t Z an 0 k, we have! c k (t) Z. 3.29. Prove that Corollary 3.20 also hols for rational polytopes: Suppose P R is a rational convex -polytope with Ehrhart quasipolynomial c (t) t + c 1 (t) t 1 + + c 0 (t). Then c (t) equals the volume of P; in particular, c (t) is constant. 3.30. Suppose f an g are quasipolynomials. Prove that the convolution F (t) := t f(s) g(t s) s=0 is also a quasipolynomial. What can you say about the egree an the perio of F, given the egrees an perios of f an g? 3.31. Given two positive, relatively prime integers a an b, let { { 1 if a t, 1 if b t, f(t) := an g(t) := 0 otherwise, 0 otherwise. Form the convolution of f an g. What function is it? 3.32. Suppose P R m an Q R n are rational polytopes. Prove that the convolution of L P an L Q equals the Ehrhart quasipolynomial of the polytope given by the convex hull of P {0 n } {0} an {0 m } Q {1}. Here 0 enotes the origin in R. 3.33. We efine the unimoular group SL (Z) as the set of all matrices with integer entries an eterminant ±1. (a) Show that SL (Z) acts on the integer lattice Z as a one-to-one, onto map. That is, fix any A SL (Z). Then A maps Z to itself in a bijective fashion. (b) For any open simplex R an any A SL (Z), consier the image of uner A, efine by A ( ) := {A x : x }. Show that # { Z } = # { A ( ) Z }.

82 3 Counting Lattice Points in Polytopes: The Ehrhart Theory (c) Let P be an integral polytope, an let Q := A (P), where A SL (Z), so that P an Q are unimoular images of each other. Show that L P (t) = L Q (t). (Hint: Write P as the isjoint union of open simplices.) 3.34. Search on the Internet for the program LattE: Lattice-Point Enumeration [65, 114]. You can ownloa it for free. Experiment. Open Problems 3.35. How many triangulations are there for a given polytope? 3.36. What is the minimal number of simplices neee to triangulate the unit -cube? (These numbers are known for 7.) 3.37. Classify the polynomials of a fixe egree that are Ehrhart polynomials. This is completely one for = 2 [159] an partially known for = 3 an 4 [24, Section 3]. 3.38. Stuy the roots of Ehrhart polynomials of integral polytopes in a fixe imension [24, 37, 41, 93]. Stuy the roots of the numerators of Ehrhart series. 3.39. Come up with an efficient algorithm that computes the perio of an Ehrhart quasipolynomial. (See [187], in which Woos escribes an efficient algorithm that checks whether a given integer is a perio of an Ehrhart quasipolynomial.) 3.40. Suppose P an Q are integer polytopes with the same Ehrhart polynomial, that is, L P (t) = L Q (t). What aitional conitions on P an Q o we nee to ensure that integer translates of P an Q are unimoular images of each other? That is, when is Q = A (P) + m for some A SL (Z) an m Z? 3.41. Fin an Ehrhart theory for irrational polyopes.

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