Ivrii’s conjecture for some cases in outer and symplectic billiards

Let $l$ be a tangent line with tangency point $z_{1}=z$ and assume we have two 3-periodic orbits $(x_{1},x_{2},x_{3})$ and $(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime})$ with tangency points $z_{1},z_{2},z_{3}$ and $z_{1},z_{2}^{\prime},z_{3}^{\prime}$ as in the figure 2. Since the domain is strongly convex, it lies inside both triangles $\triangle x_{1}x_{2}x_{3}$ and $\triangle x_{1}^{\prime}x_{2}^{\prime}x_{3}^{\prime}$ by their construction. In particular the midpoints (which are tangency points) of one triangle lie inside the other.

Without loss of generality assume that $x_{1}$ lies between $x_{1}^{\prime}$ and $z_{1}$, which implies that $x_{2}$ lies between $x_{2}^{\prime}$ and $z_{1}$. The midpoints $z_{2}^{\prime}$ and $z_{3}^{\prime}$ lie inside $\triangle x_{1}x_{2}x_{3}$, hence inside $\triangle z_{3}z_{2}x_{3}$ (otherwise $z_{2}^{\prime}$ lies in quadrilateral $x_{1}x_{2}z_{2}z_{3}$ and $z_{2}$ is outside the angle $\angle x_{1}^{\prime}x_{2}^{\prime}z_{2}^{\prime}=\angle x_{1}^{\prime}x_{2}^{\prime}x_{3}^{\prime}$ while it should be in $\triangle x_{1}^{\prime}x_{2}^{\prime}x_{3}^{\prime}$, the same argument for $z_{3}^{\prime}$).

So, the segment $z_{2}^{\prime}z_{3}^{\prime}$ is inside $\triangle z_{3}z_{2}x_{3}$ and parallel to its side $z_{2}z_{3}$ since they are both midsegments (for $\triangle x_{1}^{\prime}x_{2}^{\prime}x_{3}^{\prime}$ and $\triangle x_{1}x_{2}x_{3}$ respectively) and parallel to $l$. Then the length of $z_{2}^{\prime}z_{3}^{\prime}$ is at most length of $z_{2}z_{3}$, but, on the other hand, we have

and get a contradiction. ∎

Let $l$ be a tangent line with tangency point $z_{1}=z$ and assume we have two 4-periodic orbits $(x_{1},x_{2},x_{3},x_{4})$ and $(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime},x_{4}^{\prime})$ with tangency points $z_{1},z_{2},z_{3},z_{4}$ and $z_{1},z_{2}^{\prime},z_{3}^{\prime},z_{4}^{\prime}$ respectively. Since the domain is strongly convex, it lies inside both quadrilaterals $x_{1}x_{2}x_{3}x_{4}$ and $x_{1}^{\prime}x_{2}^{\prime}x_{3}^{\prime}x_{4}^{\prime}$. In particular the midpoints (which are tangency points) of one quadrilateral lie inside the other and form a parallelogram (e.g. $z_{1}z_{2}\parallel x_{1}x_{3}\parallel z_{3}z_{4}$ and $z_{2}z_{3}\parallel x_{2}x_{4}\parallel z_{4}z_{1}$).

Without loss of generality assume that $x_{1}$ lies between $x_{1}^{\prime}$ and $z_{1}$, which implies that $x_{2}$ lies between $x_{2}^{\prime}$ and $z_{1}$. Then midpoints $z_{2}^{\prime}$ and $z_{4}^{\prime}$ lie inside the angle $\angle z_{2}z_{1}z_{4}$ (from strong convexity of $X$), hence the angle $\angle z_{2}^{\prime}z_{1}z_{4}^{\prime}$ is less than angle $\angle z_{2}z_{1}z_{4}$.

Now we prove that a strongly convex body $X$ cannot contain two different inscribed parallelograms such that they have a common vertex $z$ and the angle at this vertex of one parallelogram lies inside the other one (strictly inside for both sides). Let $z$ be an origin. Denote vectors of sides from $z$ by $v_{1},v_{2}$ for a parallelogram with a bigger angle and $w_{1},w_{2}$ for a parallelogram with a smaller one. Take a basis $v_{1},v_{2}$. Consider vectors from point $z$ which are inside the angle between $v_{1}$ and $v_{2}$ and with ends on $\partial X$. They all, except $v_{1}+v_{2}$, have one of the coordinates bigger than 1 and the other bigger than 0 but smaller than 1 (from strong convexity of $X$).

If $w_{1}=(a_{1},b_{1})$ and $w_{2}=(a_{2},b_{2})$, then $w_{1}+w_{2}=(a_{1}+a_{2},b_{1}+b_{2})$ and we know it also lies on $\partial X$. Notice that $w_{1}+w_{2}\not=v_{1}+v_{2}$ otherwise the parallelogram with a smaller angle is inside the parallelogram with a bigger angle. So, without loss of generality $a_{1}+a_{2}>1$ and $b_{1}+b_{2}<1$. Hence $b_{1}<1$ and $b_{2}<1$. Then we write

For one of them both coefficients are positive (they cannot be zero as $w_{1}$, $w_{2}$ are linearly independent and not parallel to $v_{2}$). Without loss of generality it holds for $w_{1}$. Then their sum is

because $a_{2}b_{1}-a_{1}b_{2}<a_{2}b_{1}<a_{2}$. So, we get that $w_{1}$ lies in the convex hull of $w_{1}+w_{2}$ and $v_{2}$ and this contradicts that they all lie on $\partial X$. ∎

Assume there is an open set $U$ of $n$-periodic points. Take any $x_{1}\in U$ and let $l$ be a tangent line from $x_{1}$ to $X$ and $z$ be a tangency point. Then $U\cap l$ contains an nonempty open interval of $n$-periodic points (any open ball with center in $x_{1}$ intersects with $l$ by an open interval) and hence there is an infinite set of $n$-periodic trajectories through $z$ which contradicts for $n=3$ and $n=4$ to propositions.

Moreover, the set of 3 and 4-periodic points has measure zero. Indeed, let $P$ be the set of 3-periodic points, and assume by contradiction that it has Lebesgue measure $>0$. There exists a point $x_{1}\in P$ such that any neighborhood of $x_{1}$ intersects $P$ with positive measure. Since $x_{1}$ lies outside of the domain $X$, consider a tangent line to $\partial X$ going through $x_{1}$, more precisely let a point $z\in\partial X$ such that $T_{z}\partial X$ contains $x_{1}$.

Now consider a neighborhood $V$ of $x_{1}$. For any $z^{\prime}\in\partial X$, endow $T_{z^{\prime}}\partial X$ with the one-dimensional Lebesgue measure, and consider the intersection $T_{z^{\prime}}\partial X\cap V$ as a subset of the latter line. If $V$ is sufficiently small, there is a neighborhood $W\subset\partial X$ of $z$ such that the family of sets $(T_{z^{\prime}}\partial X\cap V)_{z^{\prime}\in W}$ is a partition of $V$ . By Fubini’s theorem, there is a point $z^{\prime}\in W$ such that $T_{z^{\prime}}\partial X\cap V\cap P$ has non-zero measure as a subset of the line $T_{z^{\prime}}\partial X$. In particular, $T_{z^{\prime}}\partial X$ intersects $P$ infinitely many times which contradicts Proposition 2.1. The proof for period 4 is similar.

∎

Consider all halflines $l$ through the origin and define a quasi-direction of a halfline as follows:

Each shear matrix of above form rotates a halfline clockwise by some angle and preserves its quasi-direction. Each rotation rotates halflines counterclockwise by angles $0<\alpha_{i}<\pi$ and does not decrease quasi-direction except for the cases when the line goes from 3 to 0 or 1. Consider a halfline $l_{0}$ spanned by the vector $(1,0)^{T}$. After the first shear and the first rotation $l_{0}$ has a quasi-direction 1. The second shear preserves the quasi-direction of $l_{0}$ equaled 1 and rotates $l_{0}$ clockwise by nonzero angle $\varphi_{2}$. Having $R(\alpha_{n})A_{n}...R(\alpha_{1})A_{1}=Id$ means that after all shears and rotations $l_{0}$ should return to quasi-direction 0. But to return back to 0, $l_{0}$ should be rotated at least for one whole cycle of quasi-directions 0,1,2,3,0. If $\sum_{i=1}^{n}\alpha_{i}\leq 2\pi$ then $\sum_{i=1}^{n}\alpha_{i}-\sum_{i=2}^{n}\varphi_{i}<2\pi$ and the angle of whole rotation of $l_{0}$ is less then $2\pi$. ∎

Since we have a one-to-one correspondence between 3-periodic outer billiard orbits and 3-periodic symplectic billiard orbits, the planar case follows from corollary 2.4.

In higher dimensions ($n>1$) if the set of 3-periodic points of symplectic billiard has nonempty interior then from the one-to-one correspondence described above we have an injective continuous map from some open set of 3-periodic points $U$ (which is $(4n-2)$-dimensional manifold as it is an open set of the phase space for symplectic billiard) to the phase space of outer billiard (exterior of the domain which is $2n$-dimensional manifold). Hence, for $n>1$ we have an injective continuous map from a manifold of higher dimension to a manifold of lower dimension, which is impossible. Thus, the set of 3-periodic points has an empty interior which implies empty interior for 3-periodic orbits.

∎

We give a general proof which works for any dimension (both planar and higher). Notice that for any 4-periodic trajectory $ABCD$ ($A$, $B$, $C$, $D$ are points on the boundary of the body) one has that the characteristic line through point $A$, the characteristic line through point $C$ and the segment $BD$ are parallel from the definition of the symplectic billiard map. Similarly, the characteristic line through point $B$, the characteristic line through point $D$ and the segment $AC$ are parallel. From strict convexity of the body for any point $A$ there is exactly one point $C$ with the opposite characteristic direction and there are unique points $B$ and $D$ with characteristic directions parallel to $AC$. Thus there is at most one 4-periodic trajectory through point $A$. ∎