Dec 13, 2017 Differential Geometry and Differential Equations

On Mahler’s conjecture

– a symplectic aspect –

Hiroshi Iriyeh (Ibaraki University)

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)

1 Convex body and its polar

Definition.

• K ⊂ Rn : convex bodydef⇐⇒ K : compact convex set in Rn with

nonempty interior

• K ⊂ Rn : centrally symmetricdef⇐⇒ K = −K

• polar body of K

K◦ := {Q ∈ Rn | Q · P ≤ 1 for ∀P ∈ K}

• K◦ = {Q ∈ Rn | Q · P ≤ 1 for ∀P ∈ K}y

x

K

O

(1,0)

(0,1)

P

Q

y

x

K◦

O

(1,0)

(0,1)

Q

P

Rem. K ⊂ Rn is a cent. sym. convex body.

⇒ K◦ is also c. s. convex body, (K◦)◦ = K.

volume product K ⊂ Rn : c. s. convex body

P(K) := |K||K◦| (|K| : volume of K)

Rem. P(AK) = P(K) for ∀A ∈ GLn(R).

volume product K ⊂ Rn : c. s. convex body

P(K) := |K||K◦| (|K| : volume of K)

Rem. P(AK) = P(K) for ∀A ∈ GLn(R).

Upper bound (Blaschke-Santalo ineq., 1949)

K ⊂ Rn : centrally symmetric convex body

⇒P(K) ≤ P(B) = |B|2, B : unit ball

“ = ” ⇐⇒ K is an ellipsoid.

How about the lower bound ?

2 Mahler’s conjecture

Theorem (Mahler’s inequality, 1939)

K ⊂ Rn : centrally symmetric convex body

⇒4n

(n!)2≤ |K||K◦| ≤ 4n.

• He proved this inequality to apply Minkowski’s

“Geometry of numbers”.

• This inequality is not sharp, but it has many

applications.

• Mahler conjectured the following the sharp

lower bound estimate.

Mahler’s conjecture (1939).

K ⊂ Rn : centrally symmetric convex body

⇒ P(K) = |K||K◦| ≥ 4n

n!.

• Mahler conjectured the following the sharp

lower bound estimate.

Mahler’s conjecture (1939).

K ⊂ Rn : centrally symmetric convex body

⇒ P(K) = |K||K◦| ≥ 4n

n!.

• n = 1 · · · trivial

• n = 2 · · · Mahler (1938), many proofs

• n ≥ 3 · · · open

equality

K = [−1, 1]n : n-cube ⇒ K◦: l1-unit ball

P(K) = 2n · 2n

n!=

4n

n!.

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.

• Kn≤8 : c. s. convex polytope with at most

2n+ 2 vertices (M.A. Lopez and Reisner, 1998)

=⇒ P(K) ≥ 4n/n!.

Asymptotic estimate (Bourgain-Milman, 1987)

K ⊂ Rn : centrally symmetric convex body

⇒ There exists a constant c > 0, indep. of n

s.t. P(K) ≥ cn|Bn(1)|2.

Mahler’s conjecture has a very long history.

Ref. T.Tao, Open question: The Mahler

conjecture on convex bodies, 2007,

https://terrytao.wordpress.com

Main result

Theorem 1 (Shibata-I., arXiv:1706.01749v2)

K ⊂ R3 : centrally symmetric convex body

⇒ P(K) = |K||K◦| ≥ 43

3!.

• “ = ” ⇐⇒ K or K◦ is a parallelepiped.

Main result

Theorem 1 (Shibata-I., arXiv:1706.01749v2)

K ⊂ R3 : 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).

3 A simple proof for 2-dim. case

Theorem 2 (Mahler, 1938)

K ⊂ R2 : centrally symmetric convex body

⇒ P(K) = |K||K◦| ≥ 8.

Proof. (Shibata-I.)

By Schneider’s approximation theorem (1984),

∃{Kn} : seq. of c. s. strongly convex bodies

(def⇐⇒ ∂Kn: C

∞ and curvature> 0)

s.t. Kn → K w.r.t. Hausdorff distance.

• P(K) is continuous w.r.t. Hausdorff dist.

We may assume K is strongly convex.y

x

K

O

A = (a, 0)

−A

(0, b) = B

−B

K1K2

K3

K4

y

x

K◦

O

A◦

=

(

1

a, c

)

−A◦

(

d,1

b

)

= B◦

−B◦

K◦

1

K◦

2

K◦

3 K◦

4

Put A,B,A◦, B◦ and divide K,K◦ as above.

• a, b > 0, c, d ∈ R. • A ·A◦ = B ·B◦ = 1.

By symmetry, |K1| = |K3|, |K2| = |K4|.

Since P is invariant under rotations around O,

we may assume |K1| = |K2|.

As a result,

|K1| = |K2| = |K3| = |K4| =|K|4

.

Key inequality ∀Q = (x, y) ∈ K◦,

O

A◦ = (1/a, c)

B◦ = (d, 1/b)

K◦

1

Q

O

A◦ = (1/a, c)

B◦ = (d, 1/b)

K◦

1

Q

The signed area of the polygon OA◦QB◦

12

∣∣∣∣1/a c

x y

∣∣∣∣+ 12

∣∣∣∣x y

d 1/b

∣∣∣∣ ≤ |K◦1 |.

12

∣∣∣∣1/a c

x y

∣∣∣∣+ 12

∣∣∣∣x y

d 1/b

∣∣∣∣ ≤ |K◦1 |.

⇔ 1

2|K◦1 |

(1

b− c,

1

a− d

)·Q ≤ 1 (∀Q ∈ K◦).

⇔ 1

2|K◦1 |

(1

b− c,

1

a− d

)∈ (K◦)◦ = K.

Similarly,1

2|K◦2 |

(−1

b− c,

1

a+ d

)∈ K.

Repeat the same argument for K:1

2|K1|(b, a) ,

1

2|K2|(−b, a) ∈ K◦.

By definition, P ·Q ≤ 1 (∀P ∈ K, ∀Q ∈ K◦).

2− bc− ad ≤ 4|K1||K◦1 |,

2 + bc+ ad ≤ 4|K2||K◦2 |.

• |K1| = |K2| = |K3| = |K4| = |K|/4.

P(K) = |K||K◦| = |K| 2(|K◦1 |+ |K◦

2 |)= 2|K||K◦

1 |+ 2|K||K◦2 |

= 8|K1||K◦1 |+ 8|K2||K◦

2 |≥ 2 ((2− bc− ad) + (2 + bc+ ad)) = 8.

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

R2n = (R2)n with ωstd =n∑

i=1

dxi ∧ dyi.

symplectic capacity on (R2n, ωstd)

c : R2n ⊃ U : domain 7→ c(U) ∈ [0,∞] with

(A1) U ⊂ V ⇒ c(U) ≤ c(V ),

(A2) c(φ(U)) = |α|c(U) if φ∗ωstd = αωstd,

(A3) c(B2nr ) = c(B2

r × R2(n−1)) = πr2.

• B2nr : ball of radius r

• “s→” denotes a symplectic embedding.

Ex. (Gromov width)

U ⊂ (R2n, ωstd) : domain

• cGr(U) := sup{πr2 | φ : (B2nr , ωstd)

s→ U}

− (A1), (A2) are easily verified.

− (A3) is nontrivial.

(non-squeezing theorem)

− cGr ≤ c ≤ ccyl (ccyl: cylindrical capacity)

for any capacity c.

Viterbo’s conjecture (2000).

∀c : symplectic capacity,

∀Σ ⊂ R2n: convex body,

we havec(Σ)

c(B2n1 )

≤(

|Σ||B2n

1 |

)1/n

.

• This conjecture is widely open.

Viterbo’s conjecture (2000).

∀c : symplectic capacity,

∀Σ ⊂ R2n: convex body,

we havec(Σ)

c(B2n1 )

≤(

|Σ||B2n

1 |

)1/n

.

• This conjecture is widely open.

Theorem (Artstein-Karasev-Ostrover)

Viterbo’s conjecture implies Mahler’s

conjecture.

Theorem (Artstein-Karasev-Ostrover)

Viterbo’s conjecture implies Mahler’s

conjecture.

Theorem (A-K-O, 2014)

K ⊂ Rn : centrally symmetric convex body

⇒ cHZ(K ×K◦) = 4,

where cHZ: Hofer-Zehnder capacity

• Σ := K ×K◦ ⊂ (R2n, ωstd): symp. domain

• cHZ(Σ) = the minimal action of closed

characteristics on ∂Σ.

Assume: Viterbo’s conjecture is true.

4n

πn=

cHZ(K ×K◦)n

πn≤ |K ×K◦|

|B2n(1)|=

|K||K◦|πn

n!

∴ 4n

n!≤ |K||K◦|.

• Viterbo’s conjecture is much stronger than

Mahler’s conjecture.

• The converse ?

Assume: Mahler’s conjecture is true.

cHZ(K ×K◦)n

πn

AKO=

4n

πn≤ |K ×K◦|

πn

n!

.

Hence, we find Viterbo’s conjecture is true in

the case where c = cHZ and Σ = K ×K◦.

Assume: Mahler’s conjecture is true.

cHZ(K ×K◦)n

πn

AKO=

4n

πn≤ |K ×K◦|

πn

n!

.

Hence, we find Viterbo’s conjecture is true in

the case where c = cHZ and Σ = K ×K◦.

Corollary 3 (Shibata-I.)

K ⊂ R3 : centrally symmetric convex body

⇒ Viterbo’s conjecture is true for

Σ = K ×K◦ ⊂ (R6, ωstd) w.r.t. c = cHZ.

5 A symplectic aspect of Mahler’s conj.

Open Problem (∗)Whether all symplectic capacities coincide for

convex bodies Σ ⊂ R2n ?

Lemma 4

If Problem (∗) is solved affirmatively, then

Viterbo conjecture is true.

Proof. Because Viterbo’s conj. is trivial for

c = cGr as follows.

Let Σ ⊂ R2n: convex body with cGr(Σ) = πR2.

Note that R2 =cGr(Σ)

cGr(B2n1 )

.

For 0 < ∀ε < R, ∃φ : (B2nR−ε, ωstd)

s→ Σ.

Because φ is volume preserving,

|Σ| ≥ |φ(B2nR−ε)| = |B2n

R−ε| = (R− ε)2n|B2n1 |.

Hence we have

(R− ε)2 ≤(

|Σ||B2n

1 |

)1/n

, 0 < ∀ε < R.

c(Σ)

c(B2n1 )

(∗)=

cGr(Σ)

cGr(B2n1 )

= R2 ≤(

|Σ||B2n

1 |

)1/n

• If Problem (∗) is solved for the class

B := {K×K◦ ⊂ (R2n, ωstd) | K ⊂ Rn : 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.

• Is there any approach to Problem (∗) ?Recall (A-K-O). cHZ(K ×K◦) = 4 for all

centrally symmetric convex body K ⊂ Rn.

• Is there any approach to Problem (∗) ?Recall (A-K-O). cHZ(K ×K◦) = 4 for all

centrally symmetric convex body K ⊂ Rn.

In fact, they proved the following stronger result:

cHZ(K ×K◦) = ccyl(K ×K◦) = 4.

• cGr ≤ ∀c ≤ ccyl on {convex body of R2n}.• cGr ≤ ∀c ≤ cHZ = ccyl = 4 on B.Lemma 5. If cGr(K ×K◦) = 4 for any centrally

symmetric convex body K ⊂ Rn, then Mahler’s

conjecture is true.

Conjecture.

For any centrally symmetric convex body

K ⊂ Rn, cGr(K ×K◦) = 4.

Ex 1. (Latschev-McDuff-Schlenk (2014)+

Gluskin-Ostrover(’16)) cGr([−1, 1]n× l1ball)= 4.

Ex 2. (Choi-Gardiner-Frenkel-Hutchings-Ramos

(2014) +Ramos (2017)) cGr(B21 ×B2

1) = 4.

Rem. K,K◦: Lagrangian in (R2n, ωstd).

For D21 ×D2

1 ⊂ (R2, ω)× (R2, ω) ⊂ (R4, ωstd),

cGr(D21 ×D2

1) = π. (easy !)

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.

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 !