Lattice polygons. P : lattice polygon in R 2 (vertices Z 2, no self-intersections)

Transcription

1 Lattice polygons P : lattice polygon in R 2 (vertices Z 2, no self-intersections)

2 A, I, B A = area of P I = # interior points of P (= 4) B = #boundary points of P (= 10)

3 Pick s theorem Georg Alexander Pick ( ) A = 2I + B 2 2 = = 9

4 Higher dimensions? Pick s theorem (seemingly) fails in higher dimensions. Example. Let T 1 and T 2 be the tetrahedra with vertices v(t 1 ) = {(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)} v(t 2 ) = {(0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1)}.

5 Two tetrahedra

6 Two tetrahedra Then I(T 1 ) = I(T 2 ) = 0 B(T 1 ) = B(T 2 ) = 4 vol(t 1 ) = 1/6, vol(t 2 ) = 1/3.

7 Dilation Let P be a convex polytope (convex hull of a finite set of points) in R d. For n 1, let np = {nα : α P}.

8 Dilation Let P be a convex polytope (convex hull of a finite set of points) in R d. For n 1, let np = {nα : α P}. P 3P

9 i(p, n) Let i(p, n) = #(np Z d ) = #{α P : nα Z d }, the number of lattice points in np.

10 ī(p, n) Similarly let P = interior of P = P P ī(p, n) = #(np Z d ) = #{α P : nα Z d }, the number of lattice points in the interior of np.

11 An example P 3P i(p,n) = (n + 1) 2 ī(p,n) = (n 1) 2 = i(p, n)

12 Reeve s theorem lattice polytope: polytope with integer vertices Theorem (Reeve, 1957). Let P be a three-dimensional lattice polytope. Then the volume V (P) is a certain (explicit) function of i(p, 1), ī(p, 1), and i(p, 2).

13 The Ehrhart-Macdonald theorem Theorem (Ehrhart 1962, Macdonald 1963). Let P = lattice polytope in R N, dim P = d. Then i(p,n) is a polynomial (the Ehrhart polynomial of P) in n of degree d.

14 The Ehrhart-Macdonald theorem Theorem (Ehrhart 1962, Macdonald 1963). Let P = lattice polytope in R N, dim P = d. Then i(p,n) is a polynomial (the Ehrhart polynomial of P) in n of degree d. Note. Eugène Ehrhart ( ): taught at lycées in France, received Ph.D. in 1966.

15 Reciprocity, volume Moreover, i(p, 0) = 1 ī(p,n) = ( 1) d i(p, n), n > 0 (reciprocity).

16 Reciprocity, volume Moreover, If d = N then i(p, 0) = 1 ī(p,n) = ( 1) d i(p, n), n > 0 (reciprocity). i(p,n) = V (P)n d + lower order terms, where V (P) is the volume of P.

17 Reciprocity, volume Moreover, If d = N then i(p, 0) = 1 ī(p,n) = ( 1) d i(p, n), n > 0 (reciprocity). i(p,n) = V (P)n d + lower order terms, where V (P) is the volume of P. ( relative volume for d < N)

18 Generalized Pick s theorem Corollary (generalized Pick s theorem). Let P R d and dim P = d. Knowing any d of i(p,n) or ī(p,n) for n > 0 determines V (P).

19 Generalized Pick s theorem Corollary (generalized Pick s theorem). Let P R d and dim P = d. Knowing any d of i(p,n) or ī(p,n) for n > 0 determines V (P). Proof. Together with i(p, 0) = 1, this data determines d + 1 values of the polynomial i(p,n) of degree d. This uniquely determines i(p, n) and hence its leading coefficient V (P).

20 Reeve s theorem redux Example. When d = 3, V (P) is determined by i(p, 1) = #(P Z 3 ) i(p, 2) = #(2P Z 3 ) ī(p, 1) = #(P Z 3 ), which gives Reeve s theorem.

21 The Birkhoff polytope Example (magic squares). Let B M R M M be the Birkhoff polytope of all M M doubly-stochastic matrices A = (a ij ), i.e., i a ij 0 a ij = 1 (column sums 1) a ij = 1 (row sums 1). j

22 Integer points in B M Note. B = (b ij ) nb M Z M M if and only if b ij N = {0, 1, 2,... } b ij = n b ij = n. i j

23 Integer points in B M Note. B = (b ij ) nb M Z M M if and only if b ij N = {0, 1, 2,... } b ij = n b ij = n. i j (M = 4, n = 7)

24 H M (n) H M (n) := #{M M N-matrices, line sums n} = i(b M,n).

25 H M (n) H M (n) := #{M M N-matrices, line sums n} = i(b M,n). H 1 (n) = 1 H 2 (n) = n + 1

26 H M (n) H M (n) := #{M M N-matrices, line sums n} = i(b M,n). H 1 (n) = 1 H 2 (n) = n + 1 [ a n a n a a ], 0 a n.

27 More examples H 3 (n) = ( ) n ( ) n ( ) n M 0 (MacMahon) H M (0) = 1 H M (1) = M! (permutation matrices) H M (2) xm M! 2 = e x/2 1 x (Anand-Dumir-Gupta, 1966)

28 Anand-Dumir-Gupta conjecture Theorem (Birkhoff-von Neumann). The vertices of B M consist of the M! M M permutation matrices. Hence B M is a lattice polytope.

29 Anand-Dumir-Gupta conjecture Theorem (Birkhoff-von Neumann). The vertices of B M consist of the M! M M permutation matrices. Hence B M is a lattice polytope. Corollary (Anand-Dumir-Gupta conjecture). H M (n) is a polynomial in n (of degree (M 1) 2 ).

30 H 4 (n) Example. H 4 (n) = ( 11n n n n n n n n n ).

31 H 4 (n) Example. H 4 (n) = ( 11n n n n n n n n n ). Open: positive coefficients

32 Positive magic squares Reciprocity ±H M ( n) = #{M M matrices B of positive integers, line sum n}. But every such B can be obtained from an M M matrix A of nonnegative integers by adding 1 to each entry.

33 Reciprocity for magic squares Corollary. H M ( 1) = H M ( 2) = = H M ( M + 1) = 0 H M ( M n) = ( 1) M 1 H M (n) (greatly reduces computation)

34 Reciprocity for magic squares Corollary. H M ( 1) = H M ( 2) = = H M ( M + 1) = 0 H M ( M n) = ( 1) M 1 H M (n) (greatly reduces computation) Applications e.g. to statistics (contingency tables) by Diaconis, et al.

35 Zeros (roots) of H 9 (n) Zeros of H_9(n)

36 Explicit calculation For what polytopes P can i(p,n) be explicitly calculated or related to other interesting mathematics? Main topic of subsequent talks.

37 Zonotopes Let v 1,...,v k R d. The zonotope Z(v 1,...,v k ) generated by v 1,...,v k : Z(v 1,..., v k ) = {λ 1 v λ k v k : 0 λ i 1}

38 An example Example. v 1 = (4, 0), v 2 = (3, 1), v 3 = (1, 2) (1,2) (3,1) (0,0) (4,0)

39 i(z, 1) Theorem. Let where v i Z d. Then Z = Z(v 1,...,v k ) R d, i(z, 1) = X h(x), where X ranges over all linearly independent subsets of {v 1,...,v k }, and h(x) is the gcd of all j j minors (j = #X) of the matrix whose rows are the elements of X.

40 An example Example. v 1 = (4, 0), v 2 = (3, 1), v 3 = (1, 2) i(z, 1) = gcd(4, 0) + gcd(3, 1) +gcd(1, 2) + det( ) = = 24.

41 Decomposition of a zonotope i(z, 1) = gcd(4, 0) + gcd(3, 1) +gcd(1, 2) + det( ) = = 24.

42 Decomposition of a zonotope i(z, 1) = gcd(4, 0) + gcd(3, 1) +gcd(1, 2) + det( ) = = 24.

43 Ehrhart polynomial of a zonotope Theorem. Let Z = Z(v 1,...,v k ) R d where v i Z d. Then i(z,n) = X h(x)n #X, where X ranges over all linearly independent subsets of {v 1,...,v k }, and h(x) is the gcd of all j j minors (j = #X) of the matrix whose rows are the elements of X.

44 Ehrhart polynomial of a zonotope Theorem. Let Z = Z(v 1,...,v k ) R d where v i Z d. Then i(z,n) = X h(x)n #X, where X ranges over all linearly independent subsets of {v 1,...,v k }, and h(x) is the gcd of all j j minors (j = #X) of the matrix whose rows are the elements of X. Corollary. The coefficients of i(z, n) are nonnegative integers.

45 Ordered degree sequences Let G be a graph (with no loops or multiple edges) on the vertex set V (G) = {1, 2,...,n}. Let d i = degree (# incident edges) of vertex i. Define the ordered degree sequence d(g) of G by d(g) = (d 1,...,d n ).

46 An example Example. d(g) = (2, 4, 0, 3, 2, 1)

47 Number of distinct d(g) Let f(n) be the number of distinct d(g), where V (G) = {1, 2,...,n}.

48 Number of distinct d(g) Let f(n) be the number of distinct d(g), where V (G) = {1, 2,...,n}. Example. If n 3, all d(g) are distinct, so f(1) = 1, f(2) = 2 1 = 2, f(3) = 2 3 = 8.

49 f(4) For n 4 we can have G H but d(g) = d(h), e.g.,

50 f(4) For n 4 we can have G H but d(g) = d(h), e.g., In fact, f(4) = 54 < 2 6 = 64.

51 The polytope of degree sequences Let conv denote convex hull, and D n = conv{d(g) : V (G) = {1,...,n}} R n, the polytope of degree sequences (Perles, Koren).

52 The polytope of degree sequences Let conv denote convex hull, and D n = conv{d(g) : V (G) = {1,...,n}} R n, the polytope of degree sequences (Perles, Koren). Easy fact. Let e i be the ith unit coordinate vector in R n. E.g., if n = 5 then e 2 = (0, 1, 0, 0, 0). Then D n = Z(e i + e j : 1 i < j n).

53 The Erdős-Gallai theorem Theorem (Erdős-Gallai). Let α = (a 1,...,a n ) Z n. Then α = d(g) for some G if and only if α D n a 1 + a a n is even.

54 A generating function Fiddling around leads to: Theorem. Let F(x) = n 0 f(n) xn n! Then = 1 + 1x + 2 x2 2! + 8x3 3! + 54x4 4! +.

55 A nice formula ( F(x) = n n xn 2 n! n 1 ( 1 ) n 1 xn (n 1) n! n 1 ) 1/2 + 1 ] exp n 1 n n 2 xn n!

56 A bad Ehrhart polynomial Let P denote the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 13). Then i(p,n) = 13 6 n3 + n n + 1.

57 A bad Ehrhart polynomial Let P denote the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 13). Then i(p,n) = 13 6 n3 + n n + 1. Thus in general the coefficients of Ehrhart polynomials are not nice. Is there a better basis?

58 Diagram of the bad tetrahedron y x z

59 The h-vector of i(p, n) Let P be a lattice polytope of dimension d. Since i(p,n) is a polynomial of degree d taking Z Z, h i Z such that n 0 i(p,n)x n = h 0 + h 1 x + + h d x d (1 x) d+1.

60 The h-vector of i(p, n) Let P be a lattice polytope of dimension d. Since i(p,n) is a polynomial of degree d taking Z Z, h i Z such that n 0 Definition. Define i(p,n)x n = h 0 + h 1 x + + h d x d (1 x) d+1. h(p) = (h 0,h 1,...,h d ), the h-vector of P (as an integral polytope).

61 An example Example. Recall i(b 4,n) = (11n9 +198n n n n n n n n ).

62 An example Example. Recall i(b 4,n) = (11n9 Then +198n n n n n n n n ). h(b 4 ) = (1, 14, 87, 148, 87, 14, 1, 0, 0, 0).

63 Properties of h(p) h(b 4 ) = (1, 14, 87, 148, 87, 14, 1, 0, 0, 0). Always h 0 = 1. Trailing 0 s i(b 4, 1) = i(b 4, 2) = i(b 4, 3) = 0. Palindromic property i(b 4, n 4) = ±i(b 4,n).

64 Nonnegativity and monotonicity Theorem A (nonnegativity), McMullen, RS: h i 0

65 Nonnegativity and monotonicity Theorem A (nonnegativity), McMullen, RS: h i 0 Theorem B (monotonicity), RS: If P and Q are lattice polytopes and Q P, then for all i: h i (Q) h i (P). B A: take Q =.

66 Proofs Both theorems can be proved geometrically. There are also elegant algebraic proofs based on commutative algebra. Related to toric varieties.

67 Fractional lattice polytopes Example. Let S M (n) denote the number of symmetric M M matrices of nonnegative integers, every row and column sum n. Then S 3 (n) = { 1 8 (2n3 + 9n n + 8), n even 1 8 (2n3 + 9n n + 7), n odd = 1 16 (4n3 + 18n n ( 1) n ).

68 Fractional lattice polytopes Example. Let S M (n) denote the number of symmetric M M matrices of nonnegative integers, every row and column sum n. Then S 3 (n) = { 1 8 (2n3 + 9n n + 8), n even 1 8 (2n3 + 9n n + 7), n odd = 1 16 (4n3 + 18n n ( 1) n ). Why a different polynomial depending on n modulo 2?

69 The symmetric Birkhoff polytope T M : the polytope of all M M symmetric doubly-stochastic matrices.

70 The symmetric Birkhoff polytope T M : the polytope of all M M symmetric doubly-stochastic matrices. Easy fact: S M (n) = # ( nt M Z M M) = i(t M,n).

71 The symmetric Birkhoff polytope T M : the polytope of all M M symmetric doubly-stochastic matrices. Easy fact: S M (n) = # ( nt M Z M M) = i(t M,n). Fact: vertices of T M have the form 1 2 (P + P t ), where P is a permutation matrix.

72 The symmetric Birkhoff polytope T M : the polytope of all M M symmetric doubly-stochastic matrices. Easy fact: S M (n) = # ( nt M Z M M) = i(t M,n). Fact: vertices of T M have the form 1 2 (P + P t ), where P is a permutation matrix. Thus if v is a vertex of T M then 2v Z M M.

73 S M (n) in general Theorem. There exist polynomials P M (n) and Q M (n) for which S M (n) = P M (n) + ( 1) n Q M (n), n 0. Moreover, deg P M (n) = ( M 2 ).

74 S M (n) in general Theorem. There exist polynomials P M (n) and Q M (n) for which S M (n) = P M (n) + ( 1) n Q M (n), n 0. Moreover, deg P M (n) = ( M 2 ). Difficult result (W. Dahmen and C. A. Micchelli, 1988): deg Q M (n) = { ( M 1 ) 2 1, M odd ) 1, M even. ( M 2 2

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