Posted by: matheuscmss | April 16, 2020

## Maximal entropy measures and Birkhoff normal forms of certain Sinai billiards

Sinai billiards are a fascinating class of dynamical systems: in fact, despite their simple definition in terms of a dynamical billiard on a table given by a square or a two torus with a certain number of dispersing obstacles, they present some “nasty” features (related to the existence of “grazing” collisions) placing them slightly beyond the standard theory of smooth uniformly hyperbolic systems.

The seminal works by several authors (including Sinai, Bunimovich, Chernov, …) paved the way to establish many properties of Sinai billiards including the so-called fast decay of correlations for the Liouville measure ${\mu_{\text{SRB}}}$ (a feature which is illustrated in this numerical simulation [due to Dyatlov] here).

Nevertheless, some ergodic-theoretical aspects of (certain) Sinai billiards were elucidated only very recently: for instance, the existence of an unique probability measure of maximal entropy ${\mu_{\text{max}}}$ was proved by Baladi–Demers in this article here.

An interesting remark made by Baladi–Demers is the fact that the maximal entropy measure ${\mu_{\text{max}}}$ should be typically different from the Liouville measure ${\mu_{\text{SRB}}}$: in fact, ${\mu_{\text{max}}=\mu_{\text{SRB}}}$ forces a rigidity property for periodic orbits, namely, the Lyapunov exponents of all periodic orbits coincide (note that this is a huge number of conditions because the number of periodic orbits of period ${\leq n}$ grows exponentially with ${n}$) and, in particular, there are no known examples of Sinai billiards where ${\mu_{\text{max}}=\mu_{\text{SRB}}}$.

In a recent paper, De Simoi–Leguil–Vinhage–Yang showed that ${\mu_{\text{max}}=\mu_{\text{SRB}}}$ forces another rigidity property for periodic orbits, namely, the Birkhoff normal forms (see also Section 6 of Chapter 6 of Hasselblatt–Katok book and Appendix A of Moreira–Yoccoz article) at periodic orbits whose invariant manifolds produce homoclinic orbits are all linear. In particular, they proposed in Remark 5.6 of their preprint to prove that ${\mu_{\text{max}}\neq\mu_{\text{SRB}}}$ for certain Sinai billiards by computing the Taylor expansion of the derivative of the billiard map at a periodic orbit of period two and checking that its quadratic part is non-degenerate.

In this short post, we explain that the strategy from the previous paragraph can be implemented to verify the non-linearity of the Birkhoff normal form at a periodic orbit of period two of the Sinai billiards in triangular lattices considered by Baladi–Demers.

Remark 1 In our subsequent discussion, we shall assume some familiarity with the basic aspects of billiards maps described in the classical book of Chernov–Markarian.

1. Sinai billiards in triangular lattices

We consider the Sinai billiard table on the hexagonal torus obtained from the following picture (extracted from Baladi–Demers article):

Here, the obstacles are disks of radii ${1}$ centered at the points of a triangular lattice and the distance between adjacent scatterers is ${d}$.

As it is discussed in Baladi–Demers paper, the billiard map ${F}$ has a probability measure of maximal entropy whenever ${2 (where the limit case ${d=2}$ corresponds to touching obstacles and the limit case ${d=4/\sqrt{3}}$ produces “infinite horizon”).

2. Billiard map near a periodic orbit of period two

We want to study the billiard map ${F}$ near the periodic orbit of period two given by the horizontal segment between ${p_0=(1,0)\in\mathbb{R}^2}$ and ${p_1=(d-1,0)\in\mathbb{R}^2}$.

A trajectory leaving the point ${e^{i\theta}}$ in the direction ${e^{i\psi}}$ travels along the line ${e^{i\theta}+se^{i\psi}}$ until it hits the boundary of the disk of radius ${1}$ and center ${(d-1,0)}$ for the first time ${t}$. By definition, the time ${t}$ is the smallest solution of ${|e^{i\theta}+se^{i\psi}-d|=1}$, i.e.,

$\displaystyle s^2+2(\cos(\psi-\theta)-d\cos\psi)s+(d^2-2d\cos\theta)=0.$

Hence, ${t=B(\theta,\psi)-\sqrt{B(\theta,\psi)^2-A(\theta,\psi)}}$ where ${B(\theta,\psi):=d\cos\psi-\cos(\psi-\theta)}$ and ${A(\theta,\psi):=d^2-2d\cos\theta}$, the billiard trajectory starting at ${e^{i\theta}}$ in the direction ${e^{i\psi}}$ hits the obstacle at

$\displaystyle e^{i\theta}+(B(\theta,\psi)-\sqrt{B(\theta,\psi)^2-A(\theta,\psi)})e^{i\psi}=:d+e^{i\Theta}$

where ${\Theta=\Theta(\theta,\psi)\approx\pi}$ and, after a specular reflection, it takes the direction ${e^{i(2\Theta-\psi-\pi)}:= e^{i\Psi}}$.

As it is explained in page 35 of Chernov–Markarian’s book, this geometric description of the billiard map near the periodic orbit of period two allows to compute ${DF^2}$ along the following lines. If we adopt the convention in Chernov–Markarian’s book of using arc-length parametrization ${x}$ on the obstacle and describing the angle ${y\in[-\pi/2,\pi/2]}$ with the normal direction pointing towards the interior of the table using a sign determined by the orientation of the obstacles so that this normal vector stays at the left of the tangent vectors to the obstacles, then the billiard map becomes

$\displaystyle F(x,y):=(X(x,y),Y(x,y))=(\pi-\arcsin\left(\sin(x) + t(x, y) \sin(x-y)\right), -X(x,y)+x-y+\pi)$

where

$\displaystyle t(x,y)=B(x,y)-\sqrt{B(x,y)^2-A(x,y)},$

$\displaystyle B(x,y)=d\cos(x-y)-\cos(-y), \quad A(x,y)=d^2-2d\cos(x),$

(and ${\arcsin(z)\in[-\frac{\pi}{2},\frac{\pi}{2}]}$ for ${z\in[-1,1]}$) and, moreover,

$\displaystyle D_{(x,y)}F=\left(\begin{array}{cc} M_{11}(x,y) & M_{12}(x,y) \\ M_{21}(x,y) & M_{22}(x,y)\end{array}\right)$

and

$\displaystyle D_{(X,Y)}F = \left(\begin{array}{cc} M_{11}(\pi-X(x,y),Y(x,y)) & M_{12}(\pi-X(x,y),Y(x,y)) \\ M_{21}(\pi-X(x,y),Y(x,y)) & M_{22}(\pi-X(x,y),Y(x,y))\end{array}\right),$

where

$\displaystyle \begin{array}{rcl} M_{11}(x,y)&=&-\frac{1}{\cos(Y(x,y))}(t(x,y)+\cos(y)) \\ &=& (1 - d) + \frac{1}{2} (2 d^2 - d^3) y^2+ (-d - d^2 + d^3) xy + \frac{1}{2} (d - d^3) x^2 + O(x^3+x^2y^2+y^3), \end{array}$

$\displaystyle \begin{array}{rcl} M_{12}(x,y)&=&-\frac{1}{\cos(Y(x,y))}t(x,y) \\ &=& (2 - d) + \frac{1}{2} (-2 d + 3 d^2 - d^3) y^2 + (-2 d^2 + d^3) xy + \frac{1}{2} (d + d^2 - d^3) x^2 + O(x^3+x^2y^2+y^3), \end{array}$

$\displaystyle \begin{array}{rcl} M_{21}(x,y)&=&-\frac{1}{\cos(Y(x,y))}(t(x, y) + \cos(Y(x, y)) + \cos(y)) \\ &=& -d + \frac{1}{2} (2 d^2 - d^3) y^2+ (-d - d^2+d^3) xy+\frac{1}{2} (d - d^3)x^2 + O(x^3+x^2y^2+y^3), \end{array}$

$\displaystyle \begin{array}{rcl} M_{22}(x,y)&=&-\frac{1}{\cos(Y(x,y))} (t(x, y) + \cos(Y(x, y))) \\ &=& (1 - d) + \frac{1}{2} (-2 d + 3 d^2 - d^3) y^2 + (-2 d^2 + d^3) xy + \frac{1}{2} (d + d^2 - d^3) x^2 + O(x^3+x^2y^2+y^3). \end{array}$

Since we also have

$\displaystyle \pi-X(x,y)=(2 - d) y+(-1 + d)x+O((y^2+x^2)(x+y))$

and

$\displaystyle -Y(x,y)=(-1 + d) y-dx+O((y^2+x^2)(x+y)),$

it follows that

$\displaystyle D_{(x,y)}F^2 = \left(\begin{array}{cc}2d^2-4d+1 & 2d^2-6d+4 \\ 2d^2-2d & 2d^2-4d+1\end{array}\right)+ Q(x,y)+ O(x^4+x^3y+y^4+x^2y^2+xy^3),$

where

$\displaystyle Q(x,y) = \left(\begin{array}{cc}Q_{11}(x,y) & Q_{12}(x,y) \\ Q_{21}(x,y) & Q_{22}(x,y)\end{array}\right)$

and

$\displaystyle \begin{array}{rcl} Q_{11}(x,y)&=&d (4 - 4 d - 17 d^2 + 33 d^3 - 20 d^4 + 4 d^5) y^2 + d (2 - 4 d + d^2 + 9 d^3 - 12 d^4 + 4 d^5) x^2 \\ &+& 2 d (-3 + 5 d + 4 d^2 - 19 d^3 + 16 d^4 - 4 d^5) x y, \end{array}$

$\displaystyle \begin{array}{rcl} Q_{12}(x,y)&=&d (6 + 5 d - 42 d^2 + 51 d^3 - 24 d^4 + 4 d^5) y^2 + d (3 - 5 d - 4 d^2 + 19 d^3 - 16 d^4 + 4 d^5) x^2 \\ &+& 2 d (-4 + 4 d + 17 d^2 - 33 d^3 + 20 d^4 - 4 d^5) x y, \end{array}$

$\displaystyle \begin{array}{rcl} Q_{21}(x,y)&=&d (4 - 6 d - 16 d^2 + 33 d^3 - 20 d^4 + 4 d^5) y^2 + d(1 - 2 d)^2 (1 - 2 d^2 + d^3) x^2 \\ &+& 2 d (-2 + 6 d + 3 d^2 - 19 d^3 + 16 d^4 - 4 d^5) x y , \end{array}$

$\displaystyle \begin{array}{rcl} Q_{22}(x,y)&=&d (8 + 2 d - 41 d^2 + 51 d^3 - 24 d^4 + 4 d^5) y^2 + d (2 - 6 d - 3 d^2 + 19 d^3 - 16 d^4 + 4 d^5) x^2 \\ &+& 2 d (-4 + 6 d + 16 d^2 - 33 d^3 + 20 d^4 - 4 d^5) x y. \end{array}$

In particular, the determinant of the quadratic part of the Taylor expansion of ${DF^2}$ near the periodic orbit of period two ${\{p_0, p_1\}}$ is

$\displaystyle \begin{array}{rcl} \frac{1}{d^2}\det Q(x,y) &=& (1 - 3 d - 2 d^2 + 10 d^3 - 8 d^4 + 2 d^5) x^4 - 2 (-2 + d)^2 (1 - d - 4 d^2 + 4 d^3) x^3 y \\ &+& (22 - 27 d - 68 d^2 + 126 d^3 - 68 d^4 + 12 d^5) x^2 y^2 - 2 (-2 + d)^2 (3 + d - 8 d^2 + 4 d^3) x y^3 \\ &+& 2 (-2 + d)^2 (1 - 2 d^2 + d^3) y^4 \end{array}$

Thus, for ${y=-x}$, one has

$\displaystyle \frac{1}{d^2}\det Q(x,-x) = x^4 (63 - 70 d - 172 d^2 + 320 d^3 - 176 d^4 + 32 d^5),$

so that the quadratic part ${Q}$ of the Taylor expansion of ${DF^2}$ is non-degenerate because ${63 - 70 d - 172 d^2 + 320 d^3 - 176 d^4 + 32 d^5\geq 3}$ for ${2\leq d\leq 4/\sqrt{3}}$.

Remark 2 ${\det Q(x,y)= 12 x^4}$ in the limit case ${d=2}$.

This site uses Akismet to reduce spam. Learn how your comment data is processed.