VOLUMES OF CONVEX POLYTOPES

Transcription

1 VOLUMES OF CONVEX POLYTOPES Richad P. Stanley Depatment of Mathematics M.I.T Cambidge, MA Tanspaencies available at: 1

2 P = convex polytope in R n intege polytope: vetices 2 Z n V(P) = volume of P If P is an intege polytope, let ev(p) = n! V (P) 2 Z; the nomalized volume of P. 2

3 Why compute V (P)? Let P = intege polytope = the nomal fan of P X = toic vaiety coesponding to ) deg(x ) = e V (P) (Gelfand, Kapanov, Zelevinsky) The numbe of linealy independent solutions to a geneic hypegeometic system with Newton polytope P is ev (P). 3

4 Two Renements of Volume Let P be an intege polytope and let 1. Dene P = fv : v 2 Pg i(p; ) = # (P \ Z n ) ; the Ehhat polynomial of P. i(p; ) is a polynomial in i(p; 0) = 1 If > 0, then i(p; ) = ( 1) dim P #(int(p)\z n ) i(p; ) = V (P) n + O( n 1 ). 4

5 Let dim P = n and X i(p; )x = h 0 + h 1 x + + h n x n (1 x) n+1 : 0 Then h j 2 Z, h j 0, and X h j = V e (P): j 5

6 Example. P = unit squae: 0 x 1; 0 y ev (P) = 2! 1 = 2 i(p; ) = ( + 1) 2 X 0 i(p; ) = ( 1) 2 i(p; )x = 1 + x (1 x) 3 6

7 Let P 1 ; : : : ; P k be convex polytopes (o any convex bodies) in R n. Let t 1 ; : : : ; t k 2 R 0 : Dene the Minkowski sum t 1 P 1 + +t k P k = ft 1 v 1 + +t k v k : v i 2 P i g: Theoem (Minkowski) Thee exist V (P a 1 ) 0 (mixed volumes) k 1 ; : : : ; Pa k such that V (t 1 P t k P k ) = X a 1 ++a k =n n a 1 ; : : : ; a k V (P a 1 1 ; : : : ; Pa k k )ta 1 1 ta k k : 7

8 Wite V (sp+tq) = so nx j=0 n j V j (P; Q)s n j t j ; V 0 (P; Q) = V (P); V n (P; Q) = V (Q): Minkowski poved V n 1 V n 1 V 0 n and conjectued V i 2 V i 1 V i+1 (Alexandov- Fenchel inequalities). Let B n be the unit ball in R n. Then V 1 (P; B n ) = 1 (suface aea of P): n 8

9 V 1 (P; B ) n V 0 (P; B ) n 1 V n (P; B ) ) aea(p) n V (P) n 1 V (B n ) n! isopeimetic inequality: 9

10 Theoem (Benstein). Let f 1 ; : : : ; f n be complex polynomials in the vaiables x 1 ; : : : ; x n. Let New(f j ) be the Newton polytope (convex hull of exponent vectos) of f j. If f 1 ; : : : ; f n ae othewise geneic, then the numbe Z(f 1 ; : : : ; f n ) of solutions to with no x i = 0 is f 1 = = f n = 0 Z(f 1 ; : : : ; f n ) = n! V (New(f 1 ); : : : ; New(f n )): Example. f 1 (x; y) = 1 + x + y 2, f 2 (x; y) = x + y 4. 10

11 2s+4t 4 4t 2 2s 1 t s+t 2 V (s New(f 1 )+t New(f 2 )) = s 2 + 2st 1 ) V (New(f 1 ); New(f 2 )) = 2! 2 = 4 11

12 A common genealization of Ehhat polynomials and mixed volumes (McMullen). Fo any convex body P, let N(P) = # (P \ Z n ) : Theoem. Let P 1 ; : : : ; P k be intege polytopes, and let Then t 1 ; : : : ; t k 2 N = f0; 1; : : :g: N(t 1 P t k P k ) 2 Q [t 1 ; : : : ; t k ] (mixed lattice point enumeato of P 1 ; : : : ; P k ). 12

13 The degee n pat of N(t 1 P t k P k ) is given by N(t 1 P t k P k )j n = V (t 1 P 1 + +t k P k ): 13

14 Catalan Numbes C n = 1 2n n + 1 n tiangulations of a convex (n+2)-gon into n tiangles by n 1 diagonals that do not intesect in thei inteios binay tees with 14

15 lattice paths fom (0; 0) to (n; n) with steps (0; 1) o (1; 0), neve ising above the line y = x sequences of n 1's and n 1's such that evey patial sum is nonnegative (with 1 denoted simply as below)

16 Fo 62 additional combinatoial intepetations of C n, see Execise 6.19 of R. Stanley, Enumeative Combinatoics, volume 2, Cambidge Univesity Pess (just published). 16

17 Flow Polytopes and Kostant's Patition Function (with A. Postnikov) Let B m denote the Bikho polytope of all m m doubly stochastic matices a ij, i.e., X Xi j a ij 0 a ij = 1 a ij = 1: Open: V (B m ) =??? (as a polytope of dimension (m 1) 2 ). 17

18 Chan and Robbins (1998) dened the Chan-Robbins polytope CR m by CR m = f a ij 2 Bm : a ij = 0 if j > i+1g (a face of B m ) dim CR m = m 2 Chan and Robbins conjectued that ev (CR m ) = C 1 C 2 C m 2 : 18

19 Flow polytopes. Let G a diected gaph with vetices 1; : : : ; m + 1 and edge set E such that if i! j is an edge, then i < j. Call G a ow gaph. Dene the ow polytope F G to be the set of all f 2 R E X 0 satisfying (1;j)2E X (j;m+1)2e f(1; j) = 1 f(j; m + 1) = 1 X f(i; j) = X f(j; k); i : (i;j)2e k : (j;k)2e fo 2 j m. 19

20 total ow out of 1 and into m + 1 is 1 ow into an intenal vetex = ow out 20

21 Fact: If E = f(i; j) : 1 i < j m + 1g; then F G is \unimodulaly equivalent" to CR m (so same volume and Ehhat polynomial). Call G the complete ow gaph on m + 1 vetices. 21

22 Kostant's patition function fo A n 1. Let e i = ith unit coodinate vecto in R n : Wite e ij = e i e j. Let A + n 1 = fe ij : 1 i < j ng: 22

23 v 2 N A + n 1 = 8 < : X 1i<jn a ij e ij : a ij 2 N 9 = ; : # K(v) = n aij 1i<jn : v = X a ij e ij o : If v = (v 1 ; : : : ; v n ) and x v = x v 1 1 xv n n, then K(v) = coef. of x v in Q 1i<jn 1 : 1 xi =x j 23

24 Example. K(2; 1; 0; 1) = 4, since (2; 1; 0; 1) = 2e 12 + e 23 + e 34 = 2e 12 + e 24 = e 12 + e 13 + e 34 = e 12 + e 14 : Gelfand: Evey subject has one \tanscendental aspect." Fo the epesentation theoy of semisimple Lie algebas, it is Kostant's patition function. 24

25 Moe geneally, if S A +, then den 1 ne the esticted Kostant's patition function by K S (v) = # n aij e ij 2S : v = X e ij 2S a ij e ij 9 > = >; : 25

26 Let G be a ow gaph on the vetex set 1; : : : ; m + 1 with edge set E. Let G i = estiction of G to vetices i; : : : ; m + 1: Let t i 0 and F G (t 1 ; : : : ; t m 1 ) = t 1 F G1 + +t m 1 F Gm 1 : Note. F G (t 1 ; : : : ; t m 1 ) is the polytope of all f 2 R E 0 satisfying X k : (j;k)2e f(j; k) X i : (i;j)2e f(i; j) = t j ; fo 1 j m 1. In othe wods, thee is an excess ow of t j out of vetex j. 26

27 Let S = S G = fe ij : (i; j) is an edge of Gg. Theoem. If each t j 2 N then N(F G (t 1 ; : : : ; t m 1 )) = K SG (t 1 + +t m 1 ; 0; t m 1 ; : : : ; t 1 ): Note: K SG (t 1 + +t m ; t m ; : : : ; t 1 ) = K S (t 1 + +t m 1 ; 0; t m 1 ; : : : ; t 1 ) if e 12 2 S G. (Othewise K SG (t t m ; t m ; : : : ; t 1 ) = 0 unless t m = 0.) Coollay. Fo t i 2 N we have K S (t t m 1 ; 0; t m 1 ; : : : ; t 1 ) 2 Q [t 1 ; : : : ; t m 1 ]: 27

28 Example: G = complete ow gaph on f1; : : : ; m + 1g, so F G = CRm and Then F Gi = CRm i+1 : N(F G (t 1 ; : : : ; t m 1 )) = K(t 1 + +t m 1 ; 0; t m 1 ; : : : ; t 1 ): E.g., K(a+b; 0; b; a) = 1 6 (a+1)(a+2)(a+3b+3) K(a + b + c; 0; c; b; a) = 1 (a + 1)(a + 2)(a + 3)(a + b + 3c + 3) 360 (a 2 + 5ab + 10b 2 + 9a + 30b + 20): 28

29 Theoem. K(t t m 1 ; 0; t m 1 ; : : : ; t 1 ) is divisible by (t 1 + 1) (t 1 + m 1). Conjectue. K(t t m 1 ; 0; t m 1 ; : : : ; t 1 ) is divisible by t 1 + t t m 2 + 3t m 1 + 3: Moe stongly, 3K(t 1 + +t m 1 ; 0; t m 1 ; : : : ; t 1 ) = (t t m 2 + 3t m 1 + 3) K no e23 (t 1 + +t m 2 ; 0; 0; t m 2 ; : : : ; t 1 ): 29

30 Poblem (not caefully looked at). Ae the coecients of the polynomial K S (t 1 + +t m 1 ; 0; t m 1 ; : : : ; t 1 ) nonnegative? 30

31 Recall: Let q = #E(G), so Then F G (t 1 ; : : : ; t m 1 ) R q : K SG (t t m 1 ; 0; t m 1 ; : : : ; t 1 ) q = V (t 1 F G1 + + t m 1 F Gm 1 ): Theoem. Let a 1 + +a m 1 = q, a i 0. Let Then i = outdeg(i) 1: ev (F a 1 ; : : : ; F a m 1 ) G 1 G m 1 = K SG (a 1 1 ; : : : ; a m 1 m 1 ): 31

32 Example ( 1 ; 2 ; 3 ) = (3; 1; 1) S = fe 12 ; e 13 ; e 14 ; e 23 ; e 34 g 1 s3 t 2 V (sf G1 + tf G2 + uf G3 ) = 3 tu 3! 2! +1 s 3! 1! 1! +2 s 4! 1! +1s 4! 1! +2s 5! : 4 t 4 u 5 32

33 K(0; 1; 1) = 1 K(0; 0; 0) = 1 K(1; 0; 1) = 2 K(1; 1; 0) = 1 K(2; 1; 1) = 2 33

34 Coollay. ev (F G ) = K SG (q m+1 1 ; 2 ; : : : ; m 2 ) Fo the complete ow gaph G with m + 1 vetices, m 1 K( 2 E.g., ev (F G ) = e V (CRm ) = ; (m 2); (m 3); : : : ; 1): K(6; 3; 2; 1) = C 1 C 2 C 3 = = 10: 34

35 Coollay. Let CT denote the constant tem of a Lauent seies. Then CT ny i=1 ev (CR n+2 ) = Y (1 x i ) 2 (x j x i ) 1 : Theoem (Mois). CT = 1 n! ny i=1(1 x i ) a n Y i=1 n 1 Y j=0 1i<jn x b i Y 1i<jn (x j x i ) 2c (a + b + (n 1 + j)c) (c) (a + jc) (c + jc) (b + jc + 1) : Coollay (Zeilbege) e V (CRn ) = C 1 C n 2. 35

36 Note: Let 0 a b. poduct fomula fo K b a 2 9 simple ; b; b + 1; : : : ; a : 36

37 One Futhe Flow Gaph E(G) = f(i; i + 1) : 1 i mg [ f(1; i) : 2 i mg [ f(i; m + 1) : 2 i mg: Theoem. V e (FG ) = C m 2. 37

38 The Catalanotope C n = conv(a + n [ f0g) R n+1 : e -e 2 3 e -e e -e

39 Let T be a tee with vetex set f1; : : : ; n + 1g and edge set E. Dene the simplex T = conv fe ij : ij 2 E; i < jg [ f0g : T = convfe 16 ; e 25 ; e 36 ; e 46 ; e 56 ; e 57 ; 0g 39

40 T is altenating if eithe evey neighbo of vetex i is less than i o is geate than i. T is noncossing if thee ae not edges ik and jl whee i < j < k < l

41 Theoem (A. Postnikov). The simplices T, whee T anges ove all noncossing altenating tees with vetex set f1; : : : ; n+1g, ae the maximal faces of a tiangulation of C n. e -e 2 3 e -e e -e 2 1 Easy: e V (T ) = 1. 41

42 Lemma. The numbe of noncossing altenating tees with vetex set f1; : : : ; n + 1g is the Catalan numbe C n Coollay. V e (Cn ) = C n. 42

43 Theoem. X i(c n ; )x = 0 P n 1 j=0 n 1 n j n j+1 x j (1 x) n+1 Hee 1 n n j n j+1 is a Naayana numbe. 43

44 Example (an ode polytope). Let O mn be the set of all points satisfying aij 1im 1jn R mn 0 a ij 1 a ij a i 1;j (1) a ij a i;j 1 : (2) Vetices ae the (0; 1)-matices in O mn. 44

45 Note: i(o mn ; ) is the numbe of matices a ij 1im 1jn and a ij satisfying (1), (2), 2 f0; 1; : : : ; g (plane patition with m ows, n columns, and lagest pat )

46 O mn has a tiangulation whose facets (maximal faces) T ae indexed by standad Young tableaux T of shape (n; : : : ; n) (m n's). Example. T = T : 0 a 11 a 12 a 21 a 13 a 22 a 23 Each T is pimitive, i.e., ev ( T ) = 1: 46

47 Coollay. We have ev (O mn ) = numbe of SYT of shape (n; : : : ; n) n! = Q Q m n i=1 j=1 (i + j 1) Theoem (MacMahon). We have i(o mn ; ) = my ny i=1 j=1 + i + j 1 i + j 1 : 47

48 Special case: m = 2. Then ev (O 2n ) = C n = 1 n + 1 the nth Catalan numbe. 2n n ; X 0 i(o 2n ; )x = P n 1 j=0 n 1 n j n j+1 x j (1 x) 2n+1 ; whee 1 n n j n j+1 is a Naayana numbe as above. 48