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Rigidity for totally integrable convex billiards

ICMAT, Madrid, November 11-15, 2013

Michael Bialy

bialy@post.tau.ac.il

Tel Aviv University

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Introduction

1.Birkhoff billiards

• 2. Definitions:

Billiard ball map Caustics, Invariant curves, Phase portraits. Lazutkin, R.Douady– KAM type result. Integrable billiards. Elliptic versus circular billiards Total integrability. 3.Birkhoff conjecture: The only integrable billiards are circles and Ellipses.

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Approaches:

1) Bolotin: Polynomial integrals-requires smoothness of certain complex algebraic curve. 2) Delshams, Ramirez-Ros: Small perturbation of ellipse–splitting of separatrices. 3) Kaloshin and Sorrentino: If an integrable billiard is C2 conjugate to an ellipse (resp. a circle) in a neighborhood of the boundary, then it is an ellipse (resp. a circle). 4) Treschev: Computer experiments on local integrability. 5) Baryshnikov,Zharnitsky and later Glutsyuk: Birkhoff distributions My approach is based on the so called E.Hopf rigidity phenomenon: If geodesics on the torus have no conjugate points then the metric is flat. No conjugate points condition turns out to be almost ”equivalent” to total integrability. The result for case of K=0, I proved many years ago. Later on M. Wojtkowski suggested the use

• f Mirror formula for an alternative proof. I shall show that Mirror formula works well for constant

curvature surfaces and for magnetic billiards.

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Results

Theorem 1. Let γ be smooth simple closed curve on a constant curvature surface S having positive geodesic curvature k. If the billiard ball orbits have no conjugate points. Then γ is a circle. Theorem 2. Let γ be smooth simple closed curve as above on S. If the billiard ball map is totally integrable then γ is a circle. Theorem 3. Theorems 1,2 can be generalized to magnetic billiards on constant curvature surface S. Remark: Magnetic billiards in ellipses seem to be non-integrable: Robnik, M, Berry, M V, 1985, J.Phys.A 18, 1361-1378, ’Classical billiards in magnetic fields’. In contrast with the 2D case: Theorem 4. Billiard orbits for D ≥ 3 always have conjugate points.

• Example. Billiard in a ball. North and South poles are conjugate points.

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Mirror equation on surfaces

• 1. Mirror formula in the plane:

1 a + 1 b = 2k(x) sin φ .

2.Mirror formula on surface:

Y

′

Y (a) + Y

′

Y (b) = 2k(x) sin φ , Y (t) = sinh t (K = −1); Y (t) = t (K = 0); Y (t) = sin t (K = 1).

• 3. Mirror formula for magnetic field β.

The Jacobi fields change according to the effective curvature K + β2:

Yβ = 1

• K + β2 sin(
• K + β2 t), for K + β2 > 0,

Yβ = t, for K + β2 = 0, Yβ = 1

• −(K + β2)

sinh(

• −(K + β2)t), for K + β2 < 0.

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Magnetic mirror formula

Y

′

β

Yβ (a) + Y

′

β

Yβ (b) = 2(k(x) − β cos φ) sin φ

• Theorem. If the billiard has no conjugate points, then there exists a measurable function on the

phase cylinder a : Ω → R such that 0 < a(x, Φ) < L(x, Φ) which satisfies the mirror equation:

Y

′

Y (a(x, Φ)) + Y

′

Y (L(x−1, Φ−1) − a(x−1, Φ−1)) = 2k(x) sin φ .

Proof by the following picture:

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Proof of rigidity for Hyperbolic plane

Mirror equation reads:

coth (a(x, Φ)) + coth (L(x−1, Φ−1) − a(x−1, Φ−1)) = 2k(x) sin φ

For t > 0, coth t is a convex function on t with coth t > 1. Take φ = π/2 to get that

k(x) > 1

for any x. So the domain must be convex with respect to horocycles on the hyperbolic plane. Moreover, using convexity one has

coth a(x, Φ) + L(x−1, Φ−1) − a(x−1, Φ−1) 2 ≤ k(x) sin φ.

This can be written in the equivalent form

a(x, Φ) + L(x−1, Φ−1) − a(x−1, Φ−1) 2 ≥ arctanh sin φ k(x)

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Integrate the last inequality with respect to the invariant measure dµ to get

• L dµ ≥ 2

P dx π arctanh sin φ k(x)

• sin φ dφ =

= 4 P dx π/2 arctanh sin φ k(x)

• sin φ dφ.

Here P is the perimeter of the boundary curve γ. For every x compute the inner integral on the right hand side integrating by parts

π/2 arctanh sin φ k(x)

• sin φ dφ = k(x)

π/2 cos2 φ k2(x) − sin2 φ dφ = = π 2 (k(x) −

• k2(x) − 1).

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Using Santalo’ formula

• L dµ = 2πA (A is the area of the domain) we obtain the following

inequality

A ≥ P (k(x) −

• k2(x) − 1) dx.

Using Gauss-Bonnet we can write it in the form

A ≥ 2π + A − P

• k2(x) − 1 dx

and therefore

P

• k2(x) − 1 dx ≥ 2π.

However, it then follows from the next lemma stating the opposite inequality that the curve γ must be a circle.

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Lemma for K=-1

• Lemma. For any simple closed curve γ on the hyperbolic plane which is convex with respect to

horocycles the following inequality holds true

I = P

• k2(x) − 1 dx ≤ 2π,

where the equality is possible only for circles.

• Proof. The integral can be estimated from above by Cauchy-Schwartz

I ≤ P (k(x) − 1) dx 1

2 P

(k(x) + 1) dx 1

2

= ((A + 2π − P)(A + 2π + P))

1 2 ,

where Gauss-Bonnet formula is applied. The last expression gives

((A + 2π − P)(A + 2π + P))

1 2 = (A2 + 4πA − P 2 + 4π2) 1 2 ≤ 2π,

since by the isoperimetric inequality on the hyperbolic plane

A2 + 4πA − P 2 ≤ 0.

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Proof of Rigidity for the Hemisphere

Write the mirror equation for the hemisphere:

cot (a(x, Φ)) + cot (L(x−1, Φ−1) − a(x−1, Φ−1)) = 2k(x) sin φ

Claim

cot a + b 2

• ≤ cot a + cot b

2 ,

for all a, b in the range (0; π) satisfying a+b

2

≤ π/2.

• Proof. This becomes obvious in the case when both a and b belong to (0; π/2] simply by the

convexity of cot on this interval. In the remaining case when one of the numbers, say a lies in

(0; π/2) and b in (π/2; π) and the average is ≤ π/2 we can write a = π/2 − x − δ and b = π/2 + x, where x, δ are non-negative and x + δ < π/2. We need to prove tan δ 2 ≤ tan(x + δ) − tan x 2 .

Indeed we have by the trigonometric formula

tan(x + δ) − tan x 2 = (tan δ)(1 + tan x tan(x + δ)) 2 ≥ tan δ 2 ≥ tan δ 2.

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Apply the Claim to the following a = a(x, Φ), b = L(x−1, Φ−1) − a(x−1, Φ−1) :

cot a(x, Φ) + L(x−1, Φ−1) − a(x−1, Φ−1) 2 ≤ k(x) sin φ.

This can be written in the equivalent form:

a(x, Φ) + L(x−1, Φ−1) − a(x−1, Φ−1) 2 ≥ arctan sin φ k(x)

• .
• L dµ ≥ 2

P dx π arctan sin φ k(x)

• sin φ dφ =

= 4 P dx π/2 arctan sin φ k(x)

• sin φ dφ.

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Compute the inner integral on the right hand side integrating by parts

π/2 arctan sin φ k(x)

• sin φ dφ = k(x)

π/2 cos2 φ k2(x) + sin2 φ dφ = = π 2 (

• k2(x) + 1 − k(x)).

Using Santalo’ formula

• L dµ = 2πA again we obtain the following inequality

A ≥ P (

• k2(x) + 1 − k(x)) dx.

Using Gauss-Bonnet we write it in the form

A ≥ P

• k2(x) + 1 dx − (2π − A),

which leads to

P

• k2(x) + 1 dx ≤ 2π.

However, the following lemma then implies that the curve γ must be a circle, thus completing the proof of the main theorem for the hemisphere.

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Lemma for K=+1 case

• Lemma. For any simple closed curve on the hemisphere the following inequality holds

I = P

• k2(x) + 1 dx ≥ 2π,

where the equality happens only for circles. Proof

P (

• k2(x) + 1+1) dx ·

P (

• k2(x) + 1−1) dx ≥

P k(x) dx 2 = (2π −A)2.

Then this can be rewritten as

(I − P)(I + P) ≥ (2π − A)2,

and hence

I2 ≥ P 2 + A2 − 4πA + 4π2 ≥ 4π2.

In the above inequality I used the isoperimetric inequality on the sphere:

P 2 + A2 − 4πA ≥ 0.

Thus I ≥ 2π. The proof is completed.

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Remarks on the Lemmas

I1 = P

• k2(x) − 1 dx ≤ 2π,

for Horocyclicly convex γ ⊂ H2,

I2 = P

• k2(x) + 1 dx ≥ 2π,

for convex γ ⊂ S2. Notice that the second inequality follows from Fenchel inequality for integral curvature of curves in

R3.

In fact it was understood by S.Tabachnikov that the analog of Fenchel inequality (with oposit sign) exists also for spacelike curves in Lorentzian R1,2.

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Rigidity for magnetic billiards

Magnetic Santalo formula:

• L dµ = 2πA

Notice Yβ looks differently in the cases K + β2 is negative , zero or positive. So the mirror equation also depends. Let us stick to K = −1 and β ∈ (0, 1) so that K + β2 < 0 (other cases are treated analogously). Mirror equation gives

coth(

• 1 − β2a(x, φ)) + coth(
• 1 − β2(L(x−1, φ−1) − a(x−1, φ−1))

2 = k(x) − β cos φ

• 1 − β2 sin φ

.

One can see that for given x, the minimum of the right hand side equals

• k2(x) − β2/
• 1 − β2, which is attained for some angle φ. Comparing this value with the left

hand side, which is obviously strictly greater than 1, we get

• 1 − β2 <
• k2(x) − β2.

Thus, k(x) ≥ 1 for all x, so that the curve γ must be convex with respect to horocycles. Moreover, by the convexity of coth we have:

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• 1 − β2 coth
• 1 − β2[a(x, φ) + L(x−1, φ−1) − a(x−1, φ−1)]

2 ≤ k(x) − β cos φ sin φ .

Or equivalently

a(x, φ) + L(x−1, φ−1) − a(x−1, φ−1) ≥ 2

• 1 − β2 arctanh
• 1 − β2 sin φ

k(x) − β cos φ

Integrating:

• L dµ ≥

P dx π 2

• 1 − β2 arctanh
• 1 − β2 sin φ

k(x) − β cos φ sin φdφ.

Amazingly both LHS and RHS of the last do not depend on β. So we end with the same inequality as before

P

• k2(x) − 1 dx ≥ 2π,

which is possible only for circles by the lemma.

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Open questions

Are there totally integrable outer billiards different from ellipses? Does this type rigidity extends to variable magnetic fields? How about billiards on curved surfaces? Does totally integrable curved billiard table necessarily admits S1 symmetry group? Given a twist symplectic map of the cylinder. One can show that the condition of no conjugate points is in fact equivalent to existence of measurable field of non-vertical tangent lines which is invariant under the twist map. Does it follow that this field of lines is necessarily smooth? How about higher dimensional convex billiards? Is there a billiard characterization of spheres?

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