Lattice Polytopes with Distinct Pair-Sums
نویسندگان
چکیده
Let P be a lattice polytope in Rn , and let P∩Z n = {v1, . . . , vN }. If the N+ ( N 2 ) points 2v1, . . . , 2vN ; v1 + v2, . . . , vN−1 + vN are distinct, we say that P is a “distinct pairsum” or “dps” polytope. We show that if P is a dps polytope in Rn , then N ≤ 2n , and, for every n, we construct dps polytopes in Rn which contain 2n lattice points. We also discuss the relation between dps polytopes and the study of sums of squares of real polynomials. Let P be a lattice polytope in Rn , the convex hull of a finite set in Zn , and let L(P) := P ∩ Z = {v1, . . . , vN }, where N = N (P) := |L(P)|. Suppose the N + (N2 ) points in L(P)+ L(P), 2v1, . . . , 2vN ; v1 + v2, v1 + v3, . . . , vN−1 + vN , are distinct. In this case we say thatP is a distinct pair-sum or dps polytope. Our interest in dps polytopes comes from the study of the representation of polynomials as a sum of squares of polynomials. The following lemma offers two other geometrical characterizations of dps polytopes. Lemma 1. Let P be a lattice polytope. Then the following are equivalent: (1) L(P) is a dps polytope. 66 M. D. Choi, T. Y. Lam, and B. Reznick (2) L(P) does not contain the vertices of a (nondegenerate) parallelogram, and does not contain three collinear points. (3) Suppose v 6= v′ and w 6= w′ are in L(P). Then v′ − v and w′ − w are parallel only if {v, v′} = {w,w′}. Proof. (1) ⇒ (2) Suppose v1, v2, v3, v4 ∈ L(P) are the vertices of a parallelogram. Then v1 − v2 = v3 − v4 implies v1 + v4 = v2 + v3, so that P is not dps. Now suppose v1, v2, v3 ∈ L(P), and v2 is interior to the line segment v1v3. If v2 is the midpoint of the segment, then v2 + v2 = v1 + v3, so P is not dps. Otherwise, we may assume that v2 is closer to v1 than to v3. Then v4 = v2 + (v2 − v1) will also be a lattice point on the line segment v1v3, and v2 is the midpoint of v1v4; again, P is not dps. (2) ⇒ (3) For u ∈ Zn , let g(u) = gcd(u1, . . . , un). Suppose g(u′ − u) = d > 1. Then u′ − u = du′′ for u′′ ∈ Zn , and the line segment uu′ contains the lattice points u, u + u′′, . . . , u + du′′ = u′. Thus, if (2) holds and u, u′ ∈ L(P), u 6= u′, we have g(u′ − u) = 1. Suppose w′ − w = α · (v′ − v). Then α = p/q for nonzero integers p, q, and q(w′ − w) = p(v′ − v). Hence |q| = g(q(w′ − w)) = g(p(v′ − v)) = |p|, so α = ±1. Now the parallelogram condition in (2) implies that {v, v′} = {w,w′}. (3)⇒ (1) If (3) holds forP , and vi , vj , vk, v` ∈ L(P)with i / ∈ {k, `}, then vi −vk 6= v` − vj , and so vi + vj 6= vk + v`. This proves (1). Our main results are these: if P in Rn is a dps polytope, then N (P) ≤ 2n , and, for every n, we construct dps polytopes in Rn for which N (P) = 2n . Example 1. Let P ⊂ R2 be the triangle with vertices {(0, 1), (1, 2), (2, 0)}. Then P is a dps polytope, because L(P) = {(0, 1), (1, 2), (2, 0), (1, 1)} and L(P)+L(P) = {(0, 2), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 0)}. We can viewP as the projection onto the first two coordinates of the triangle with vertices {(0, 1, 2), (1, 2, 0), (2, 0, 1)}, which lies in the hyperplane x1 + x2 + x3 = 3. (In this example, we could have just as well taken the triangle with vertices {(0, 0), (1, 2), (2, 1)}; again, L(P) will consist of the vertices of P and (1, 1).)
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ورودعنوان ژورنال:
- Discrete & Computational Geometry
دوره 27 شماره
صفحات -
تاریخ انتشار 2002