On Totally integrable magnetic billiards on constant curvature surface

We consider billiard ball motion in a convex domain of a constant curvature surface influenced by the constant magnetic field. We prove that if the billiard map is totally integrable then the boundary curve is necessarily a circle. This result is a manifestation of the so-called Hopf rigidity phenomenon which was recently obtained for classical billiards on constant curvature surfaces.


Introduction and the result
Let S be a surface of constant curvature K = 0, ±1. Let γ be a simple closed convex curve on S of class C 2 . We shall denote by k the geodesic curvature of γ and assume it is strictly positive everywhere. We consider the so-called magnetic billiard inside γ where the magnitude of the magnetic field is assumed to be constant β ≥ 0. This means that the billiard ball between elastic reflections from the boundary moves with a unite speed along curves of constant geodesic curvature β. The model of magnetic billiard was extensively studied (see incomplete list [2][1] [14] [12][10] [13] [8][7] [9]). Let me summarize the basic facts on the magnetic billiards which will be omitted. First of all the main geometric assumption which assures that the dynamics is well defined is the following saying that the field is not too large relative to the geodesic curvature of the boundary. In this case magnetic billiard ball map defines a smooth map T of the phase cylinder cylinder Ω = γ × (0, π), where we shall denote by x ∈ [0, P ) the arc-length coordinate along γ and ϕ ∈ (0, π) is the inward angle. Moreover T is a symplectic twist map, the form dx ∧ d(cos ϕ) is preserved. Remarkably, this is the same form which appears for classical billiards. We shall denote dµ = sin ϕdxdϕ the invariant measure.
An important question starting from [2] is when magnetic billiard map is integrable. The only known example of integrable magnetic billiard is the circular billiard, in contrast to the classical case where for any constant curvature surface ellipses are integrable also (see [15]).
We shall adopt the following definition of suggested by Andreas Knauf for geodesic flow on the torus ( [11]): Definition 1.1. The billiard ball map T is called totally integrable if through every point of the phase cylinder Ω = γ × (0, π) passes a closed non-contractible curve which is invariant under the map T .
Our main result is the following theorem: Theorem 1.2. If the magnetic billiard map is totally integrable then γ must be a circle. Remark 1.3. A more general result can be proved using the notion of conjugate points of twist maps ( [4], [3]). The more general statement is the following: any magnetic billiard on a constant curvature surface S which has no conjugate points is circular billiard. Remark 1.4. In view of the previous remark one can consider this result as a magnetic billiard analog of Hopf 's theorem on tori without conjugate points. It was proved in [5] that Hopf type rigidity holds true also for magnetic geodesic flows on tori, provided the metric is conformally flat. Notice that the magnetic field in [5] is not assumed to be constant. In higher dimensions it is not known if the conformal flatness assumption can be relaxed.

Magnetic versions of Santalo and mirror formula
One of the key observations for the result of theorem 1.2 is the fact that the classical Santalo formula for geodesics (for the proof see [6]) remains the same for magnetic geodesics as follows (it was known already to Santalo for horocycles;we need a very particular case, and refer to [10] and [12]) for the proof of the most general case): Next recall the Mirror formula for usual billiard on a surface S of constant where Y denotes the orthogonal Jacobi field along geodesics on the surface S satisfying initial conditions Y (0) = 0, Y ′ (0) = 1. Here x is a point on the mirror a is a distance from a point A inside the domain to the point x along the shortest ray and b is a distance along the reflected ray to the point B where the focusing of the reflected beam occurs, ϕ is the angle of reflection.
It is well known that the presence of the magnetic field results in adding to the curvature K the term β 2 , so that the Jacobi field Y should be changed in the Mirror formula to Y β where: There is also a change on the right hand side of the Mirror formula so that for any K = 0, ±1 the formula reads as follows: The main step in the proof of theorem 1.2 is the reduction to the case of nonmagnetic billiard on surface which was treated in [4]. This is done in the following way. First exactly as it was for non-magnetic billiards we have Theorem 2.2. If the billiard is totally integrable (or more generally 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: where l(x, ϕ) denotes the length of the magnetic geodesic segment which starts at the point x of γ with the inward angle ϕ).
The proof of this theorem is analogouse to the non-magnetic case and it is omitted.
In the sequel we shall distinguish between the cases of the Plane, K = 0; of the Sphere K = 1 and the Hyperbolic plane K = −1. In the last case three subcases naturally appear: β > 1; β = 1 and β ∈ (0, 1).

Planar and Spherical magnetic billiards
For the Plane and the Sphere the mirror equation reads: Notice that the geometric assumption β < min x∈γ k(x) implies that the right hand side is always positive and hence K + β 2 (a(x, ϕ) + l(x −1 , ϕ −1 ) − a(x −1 , ϕ −1 )) < π so that the lemma of [4] can be applied to get the inequality This can be written in equivalent way: Integrate (4) with respect to the invariant measure dµ = sin ϕ dxdϕ. We get For a given x denote by I(x) the inner integral in the right hand side of (5). Then we have: Proof. By "magnetic" Santalo formula, the integral on the left hand side equals 2πA independently of the magnetic field β. I claim that this fact implies without any additional calculations that the inner integral on the right hand side is independent on β also. Indeed, if the boundary curve is a circle of constant geodesic curvature k on S then one can easily see that there is equality in (5). Moreover due to the rotational symmetry (5) leads to the following equality for the circle of curvature k: So that the inner integral in (5) equals 2π/A(x) independently of β. This proves the claim. (Of course one could compute for any β the integral, but this is challenging even for MATHEMATICA.) Using the independence on β we can compute the right hand side of (5) putting β = 0. But then the inequality becomes identical to one obtained in a nonmagnetic case [4]. Namely consider first the case of the Sphere, K = 1. We This inequality implies [4] that γ must be a circle. This done by the following argument: Use Gauss-Bonnet to write it in the form

Denote this integral by I. On the other hand by Cauchy Schwartz one has
This can be rewritten as and since I ≤ 2π then Thus we end with the inequality which is opposite to the isoperimetric on the sphere. For the Plane, K = 0, the inequality (5) looks even simpler when one passes to the limit β → 0: which is possible only for circles as was observed in [16]. Because by Cauchy-Schwartz one has contradicting the isoperimetric inequality in the plane.

Magnetic billiards on the Hyperbolic plane
On the Hyperbolic plane, K = −1, we shall proceed in a similar way as before, dividing between the following cases where the mirror formula (1) looks differently depending on the magnitude of the magnetic field: Case1. Assume here that β > 1. In this case the mirror equation (3) looks exactly as in the Spherical and planar case (3) but with K = −1. In this case using again the lemma of [4] we get Or equivalently Integrate (8) with respect to the invariant measure dµ = sin ϕ dxdϕ. We get (9) l dµ ≥ Denote I 1 (x) the inner integral of (9) Case2. In this case β = 1. In this case the effective curvature K + β 2 vanishes and theorem 2.2 implies 1 a(x, ϕ) Using the convexity of the function 1 Integrating with respect to the invariant measure we end up with the inequality: (11) l dµ ≥ P 0 dx π 0 2 sin ϕ k(x) − cos ϕ sin ϕdϕ.
Case3. In this last case β ∈ (0, 1). By theorem 2.2 we have One can see that for the given x the minimum of the right hand side of (12) equals k 2 − β 2 which is attained for some angle ϕ ∈ (0, π). Comparing with the left hand side, which is obviously strictly greater than 1 − β 2 , we get or equivalently (13) k(x) ≥ 1 So that the curve γ must be convex with respect to horocycles. Moreover, by the convexity of coth Or equivalently Integrate (15) with respect to the invariant measure dµ = sin ϕ dxdϕ. We get (16) l dµ ≥ P 0 dx π 0 2 1 − β 2 arctanh 1 − β 2 sin ϕ k(x) − β cos ϕ sin ϕdϕ.
Denote I 3 (x) the inner integral of (16) Remarkably the following lemma holds true: Lemma 4.1. All three integrals I 1 , I 2 , I 3 are independent on β and Proof. This goes exactly like in the Spherical case. Indeed take a circle on the Hyperbolic Plane of curvature k = k(x), notice that such a circle exists since in the first two Cases k(x) > 1 just by the geometric assumption and in Case3 this is obtained in (13). For any circle both inequalities (9,16) becomes equalities. So using rotational symmetry both integrals I 1 , I 3 can be easily computed to be equal 2πA(x) P which shows independence of β. Moreover, it is clear that for β → 1 both integrals I 1 , I 3 tend to I 2 , which can be easily computed. This completes the proof of the lemma.
It is easy to finish the section. All three inequalities of the Cases1,2,3 (9),(11),(16) lead by the "magnetic" Santalo formula and by the lemma to the same inequality which does not contain β anymore. We proceed like in [4].
Use Gauss-Bonnet to write it in the form On the other hand the last integral can be estimated from above by the Cauchy-Schwartz where we applied Gauss Bonnet again. Thus we have the inequality ((A + 2π − P )(A + 2π + P )) 1 2 ≥ 2π which is equivalent to A 2 + 4πA − P 2 ≥ 0. But this is opposite to the isoperimetric inequality on the Hyperbolic plane. Thus γ must be a circle. This completes the proof for the Hyperbolic plane.