On Mahler s conjecture a symplectic aspect

Transcription

1 Dec 13, 2017 Differential Geometry and Differential Equations On Mahler s conjecture a symplectic aspect Hiroshi Iriyeh (Ibaraki University)

2 Contents 1. Convex body and its polar 2. Mahler s conjecture and main results 3. A simple proof for 2-dimensional case 4. Symplectic capacities 5. A symplectic aspect of Mahler s conj. joint work with Masataka Shibata(TIT)

3 1 Convex body and its polar Definition. K R n : convex body def K : compact convex set in R n with nonempty interior K R n : centrally symmetric def K = K polar body of K K := {Q R n Q P 1 for P K}

4 K = {Q R n Q P 1 for P K} y y (0,1) Q K (0,1) P K O P (1,0) x O Q (1,0) x Rem. K R n is a cent. sym. convex body. K is also c. s. convex body, (K ) = K.

5 volume product K R n : c. s. convex body P(K) := K K ( K : volume of K) Rem. P(AK) = P(K) for A GL n (R).

6 volume product K R n : c. s. convex body P(K) := K K ( K : volume of K) Rem. P(AK) = P(K) for A GL n (R). Upper bound (Blaschke-Santaló ineq., 1949) K R n : centrally symmetric convex body P(K) P(B) = B 2, B : unit ball = K is an ellipsoid. How about the lower bound?

7 2 Mahler s conjecture Theorem (Mahler s inequality, 1939) K R n : centrally symmetric convex body 4 n (n!) 2 K K 4 n. He proved this inequality to apply Minkowski s Geometry of numbers. This inequality is not sharp, but it has many applications.

8 Mahler conjectured the following the sharp lower bound estimate. Mahler s conjecture (1939). K R n : centrally symmetric convex body P(K) = K K 4n n!.

9 Mahler conjectured the following the sharp lower bound estimate. Mahler s conjecture (1939). K R n : centrally symmetric convex body P(K) = K K 4n n!. n = 1 trivial n = 2 Mahler (1938), many proofs n 3 open

10 equality K = [ 1, 1] n : n-cube K : l 1 -unit ball P(K) = 2 n 2n n! = 4n n!.

11 Known partial results K : symmetric w.r.t. all coordinate hyperplanes (J. Saint-Raymond, 1980) K : zonoid (S. Reisner, 1985, 86) def K is a convex body approximated by finite Minkowski-sum of line segments. K n 8 : c. s. convex polytope with at most 2n + 2 vertices (M. A. Lopez and Reisner, 1998) = P(K) 4 n /n!.

12 Asymptotic estimate (Bourgain-Milman, 1987) K R n : centrally symmetric convex body There exists a constant c > 0, indep. of n s.t. P(K) c n B n (1) 2. Mahler s conjecture has a very long history. Ref. T. Tao, Open question: The Mahler conjecture on convex bodies, 2007,

13 Main result Theorem 1 (Shibata-I., arxiv: v2) K R 3 : centrally symmetric convex body P(K) = K K 43 3!. = K or K is a parallelepiped.

14 Main result Theorem 1 (Shibata-I., arxiv: v2) K R 3 : centrally symmetric convex body P(K) = K K 43 3!. = K or K is a parallelepiped. Mahler s conjecture has been solved for n = 3. Rem. In this paper, we introduced a new very very simple proof for n = 2 (Mahler s theorem).

15 3 A simple proof for 2-dim. case Theorem 2 (Mahler, 1938) K R 2 : centrally symmetric convex body P(K) = K K 8. Proof. (Shibata-I.) By Schneider s approximation theorem (1984), {K n } : seq. of c. s. strongly convex bodies ( def K n : C and curvature > 0) s.t. K n K w.r.t. Hausdorff distance. P(K) is continuous w.r.t. Hausdorff dist.

16 We may assume K is strongly convex. y ( d, 1 ) b = B y A (0,b) = B K K 2 O K 1 K K 4 3 B x A = (a,0) A K K 2 K 3 O K 1 K 4 A = x ( 1 a,c ) B Put A, B, A, B and divide K, K as above. a, b > 0, c, d R. A A = B B = 1.

17 By symmetry, K 1 = K 3, K 2 = K 4. Since P is invariant under rotations around O, we may assume K 1 = K 2. As a result, K 1 = K 2 = K 3 = K 4 = K 4.

18 Key inequality Q = (x, y) K, B = (d,1/b) B = (d,1/b) Q K 1 Q K 1 O O A = (1/a,c) A = (1/a,c) The signed area of the polygon OA QB 1 1/a c 2 x y + 1 x y 2 d 1/b K 1.

19 1 1/a c 2 x y + 1 x y 2 d 1/b K 1. 1 ( 1 2 K1 b c, 1 ) a d Q 1 ( Q K ). 1 ( 1 2 K1 b c, 1 ) a d (K ) = K. ( 1 Similarly, 1b 2 K2 c, 1a ) + d K. Repeat the same argument for K: 1 2 K 1 (b, a), 1 2 K 2 ( b, a) K.

20 By definition, P Q 1 ( P K, Q K ). 2 bc ad 4 K 1 K1, 2 + bc + ad 4 K 2 K2. K 1 = K 2 = K 3 = K 4 = K /4. P(K) = K K = K 2( K1 + K2 ) = 2 K K1 + 2 K K2 = 8 K 1 K1 + 8 K 2 K2 2 ((2 bc ad) + (2 + bc + ad)) = 8.

21 4 Symplectic capacities Artstein-Karasev-Ostrover, From symplectic measurements to the Mahler conjecture (2014) Mahler s conjecture is closely related to another open problem in Symplectic Geometry : Viterbo s conjecture isoperimetric-type conjecture for c

22 R 2n = (R 2 ) n with ω std = n i=1 symplectic capacity on (R 2n, ω std ) dx i dy i. c : R 2n U: domain c(u) [0, ] with (A1) U V c(u) c(v ), (A2) c(φ(u)) = α c(u) if φ ω std = αω std, (A3) c(b 2n r ) = c(b 2 r R 2(n 1) ) = πr 2. Br 2n : ball of radius r

23 s denotes a symplectic embedding. Ex. (Gromov width) U (R 2n, ω std ) : domain c Gr (U) := sup{πr 2 φ : (B 2n r, ω std ) s U} (A1), (A2) are easily verified. (A3) is nontrivial. (non-squeezing theorem) c Gr c c cyl (c cyl : cylindrical capacity) for any capacity c.

24 Viterbo s conjecture (2000). c : symplectic capacity, Σ R 2n : convex body, ( c(σ) Σ we have c(b1 2n) B1 2n ) 1/n. This conjecture is widely open.

25 Viterbo s conjecture (2000). c : symplectic capacity, Σ R 2n : convex body, ( c(σ) Σ we have c(b1 2n) B1 2n ) 1/n. This conjecture is widely open. Theorem (Artstein-Karasev-Ostrover) Viterbo s conjecture implies Mahler s conjecture.

26 Theorem (Artstein-Karasev-Ostrover) Viterbo s conjecture implies Mahler s conjecture. Theorem (A-K-O, 2014) K R n : centrally symmetric convex body c HZ (K K ) = 4, where c HZ : Hofer-Zehnder capacity Σ := K K (R 2n, ω std ): symp. domain c HZ (Σ) = the minimal action of closed characteristics on Σ.

27 Assume: Viterbo s conjecture is true. 4 n π n = c HZ(K K ) n π n K K B 2n (1) = K K π n n! 4n n! K K. Viterbo s conjecture is much stronger than Mahler s conjecture. The converse?

28 Assume: Mahler s conjecture is true. c HZ (K K ) n π n AKO = 4n π n K K π n. n! Hence, we find Viterbo s conjecture is true in the case where c = c HZ and Σ = K K.

29 Assume: Mahler s conjecture is true. c HZ (K K ) n π n AKO = 4n π n K K π n. n! Hence, we find Viterbo s conjecture is true in the case where c = c HZ and Σ = K K. Corollary 3 (Shibata-I.) K R 3 : centrally symmetric convex body Viterbo s conjecture is true for Σ = K K (R 6, ω std ) w.r.t. c = c HZ.

30 5 A symplectic aspect of Mahler s conj. Open Problem ( ) Whether all symplectic capacities coincide for convex bodies Σ R 2n? Lemma 4 If Problem ( ) is solved affirmatively, then Viterbo conjecture is true. Proof. Because Viterbo s conj. is trivial for c = c Gr as follows.

31 Let Σ R 2n : convex body with c Gr (Σ) = πr 2. Note that R 2 = c Gr(Σ) c Gr (B1 2n). For 0 < ε < R, φ : (B 2n R ε, ω std) s Σ. Because φ is volume preserving, Σ φ(b 2n R ε) = B 2n R ε = (R ε) 2n B 2n 1. Hence we have (R ε) 2 ( Σ B 2n 1 ) 1/n, 0 < ε < R.

32 c(σ) ( ) c(b1 2n) = c Gr(Σ) c Gr (B1 2n) = R2 ( Σ B 2n 1 ) 1/n If Problem ( ) is solved for the class B := {K K (R 2n, ω std ) K R n : c.s.c.b.}, then Viterbo s conjecture is true for this class. This means that Mahler s conjecture is true for all dimension n.

33 Is there any approach to Problem ( )? Recall (A-K-O). c HZ (K K ) = 4 for all centrally symmetric convex body K R n.

34 Is there any approach to Problem ( )? Recall (A-K-O). c HZ (K K ) = 4 for all centrally symmetric convex body K R n. In fact, they proved the following stronger result: c HZ (K K ) = c cyl (K K ) = 4. c Gr c c cyl on {convex body of R 2n }. c Gr c c HZ = c cyl = 4 on B. Lemma 5. If c Gr (K K ) = 4 for any centrally symmetric convex body K R n, then Mahler s conjecture is true.

35 Conjecture. For any centrally symmetric convex body K R n, c Gr (K K ) = 4. Ex 1. (Latschev-McDuff-Schlenk (2014)+ Gluskin-Ostrover( 16)) c Gr ([ 1, 1] n l 1 ball)= 4. Ex 2. (Choi-Gardiner-Frenkel-Hutchings-Ramos (2014) +Ramos (2017)) c Gr (B1 2 B1) 2 = 4. Rem. K, K : Lagrangian in (R 2n, ω std ). For D1 2 D1 2 (R 2, ω) (R 2, ω) (R 4, ω std ), c Gr (D1 2 D1) 2 = π. (easy!)

36 Summary A longstanding open problem in Convex Geometry: Mahler s conjecture has been solved for n = 3. Mahler s conjecture for higher dimensional case can be reduced a new symplectic embedding problem.

37 Summary A longstanding open problem in Convex Geometry: Mahler s conjecture has been solved for n = 3. Mahler s conjecture for higher dimensional case can be reduced a new symplectic embedding problem. Thank you very much!