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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 !