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This article is cited in 1 scientific paper (total in 1 paper) Billiard with slipping by an arbitrary rational angle V. N. Zav'yalovab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
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     § 1. IntroductionDuring the last two years a considerable progress has been achieved in the theory of integrable billiards. For instance, Glutsyuk [1], Mironov and Byalyi [2] and Kaloshin and Sorrentino [3] established several analogues and special cases of the famous Birkhoff conjecture. Leaving aside some details, the condition that the smooth boundary arcs of the billiard table belong to a family of confocal ellipses and hyperbolae or to a degenerate analogue of such a family (for instance, to concentric circles and radii of such circles) turns out to be not just sufficient for integrability but also necessary. In other words, for planar tables the class of integrable billiards (without potential) turns out no be quite narrow and closely related to families of quadrics. Vedyushkina [4], [5] constructed an integrable generalization of planar confocal and circular billiards, namely, piecewise planar table complexes $X$ with commuting permutations. Such tables are called billiard books: 2-cells (sheets of the book) are isometric to tables of planar billiards, and they are attached together by gluing isometrically smooth arcs on their boundaries (which produces spines of the book). Each spine is endowed with a cyclic permutation $\sigma$ (acting on the sheets incident to this spine) which describes the passage from sheet $i$ to sheet $\sigma(i)$ after hitting the gluing arc or part of the boundary in question. The permutations at each vertex commute, which ensures that the billiard flow is continuous. An important class of such systems consists of topological billiards (see [6] and [7]): their tables are piecewise planar 2-manifolds and all cyclic permutations have length 1 or 2. If the projection of the table thus glued onto the plane takes all spines to arcs of quadrics with the same foci, then the billiard system on this table complex (or 2-manifold) is integrable similarly to the planar billiards in domains equal to the isometric projections of 2-cells of this table (and all these billiards have the same quadratic additional first integrals). The same holds for degenerate cases of a confocal family, for instance, for concentric circles and radii of such circles (in which case the additional integral is linear). The simplest example of an integrable billiard table with linear additional first integral is a table homeomorphic to a sphere and glued of two equal discs along their common boundary circle. One system considered in our paper is a modification of another topological billiard, whose table is glued of two equal (round) annuli along their common inner circle. The phase space of an integrable billiard has the structure of a foliation by common level sets of the Hamiltonian $H = |\vec{v}|^2$ and the additional integral, although, generally speaking, it is just a topological manifold glued of pieces endowed with smooth and symplectic structures. For a plane domain $\Omega \subset \mathbb{R}^2(x,y)$ with piecewise smooth boundary such that the angles at all corners are equal to $\pi/2$, the phase space $M^4$ is defined by taking account of the law of elastic reflection, when the incidence and reflection angles are equal:
$$
\begin{equation*}
M^4 = \{(P, \vec{v})\mid P \in \Omega,\, \vec{v}\in T_P\mathbb {R}^2,\, |\vec{v}|>0\}/\sim ,
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
(P_1, v_1) \sim (P_2, v_2) \quad\Longleftrightarrow\quad P_1=P_2 \in \partial \Omega, \quad |v_1|=|v_2|, \quad v_1+v_2\in T_{P_1} \partial \Omega.
\end{equation*}
\notag
$$
If $\Omega$ is a disc of radius $R$ with centre $O$, then, as a first integral, we can take the oriented angle $\varphi = \varphi(P, \vec{v}) \in [0, \pi]$ between the tangent ray to $\partial \Omega$ (oriented clockwise) and the ray $P + \alpha \vec{v}$, $\alpha > 0$, at the point of reflection. For $\varphi \notin \{0, \pi/2, \pi\}$ the billiard ball moves in the annulus between the circle $\partial \Omega$ (the boundary of the table) of radius $R$ and the caustic of radius $r=R \cos \varphi$, while for $\varphi = 0$ and $\varphi = \pi$ it moves along the boundary of the table. For $0 \leqslant \varphi < \pi/2$ the ball goes clockwise about the centre of the disc, while for $\pi/2 < \varphi \leqslant \pi$ it goes anticlockwise. For $\varphi = \pi/2$ all trajectories consists of two links, each of which passes through $O$. Fix $H = |\vec{v}|^2 = 1$; then in the nonsingular isoenergy manifold $Q^3\colon H = 1$ each level surface of the first integral $\varphi$, where $0 < \varphi < \pi$, is connected and homeomorphic to a 2-torus, while the singular values $\varphi=0$ and $\varphi=\pi$ correspond to levels homeomorphic to a circle. Recall that for Liouville integrable Hamiltonian systems on a smooth symplectic manifold $M^4$ Fomenko and his school developed a theory of topological classification; see [8]. Two systems are said to be Liouville equivalent in nonsingular energy ranges if there exists a diffeomorphism between nonsingular energy levels of the two systems $Q_1^3\colon H_1 = h_1$ and $Q_2^3\colon H_2 = h_2$ that preserves the Liouville foliation and the orientation of certain critical circles. This approach is based on the theory of bifurcations of Liouville tori due to Fomenko (see [9] and [10]), which enables one to classify nondegenerate (Bott) singularities of corank 1 occurring in integrable systems. Such singularities are called 3-atoms. In the systems under consideration here only simplest atoms of types $A$ and $B$ occur. They look like the Cartesian products of a fibred two-dimensional base (called a 2-atom $A$ or $B$, respectively, and a circle. The fibration on the base contains a unique critical point, which is Morse nondegenerate: it can be elliptic (minimal or maximal) or saddle. In Figure 1 we show 2-atoms and 3-atoms $A$ and $B$, and also represent an atom $A$ as a maximum or a minimum of the function $z$ on a surface. The Fomenko-Zieschang invariant, which classifies systems up to Liouville equivalence (see [11]), is a finite framed graph, whose edges correspond to one-parameter families of regular Liouville tori and whose vertices correspond to bifurcations of these families. The type of the bifurcation atom is assigned to each vertex, a pair of numerical marks $r$ and $\varepsilon$ to each edge, and an integer mark $n$ is assigned to some connected subgraph not containing elliptic atoms $A$ (a family). The Fomenko-Zieschang invariant is also called the marked molecule, while without the numerical marks $r$, $\varepsilon$ and $n$ it is called the molecule or rough molecule. Many integrable cases in dynamics and mathematical physics and the systems of geodesic flows on two-dimensional surfaces can be analysed using these invariants, and some nontrivial equivalences between such systems have been discovered (see [8], Vol. 2). We can calculate the numerical marks by choosing some admissible bases $\lambda_i$, $\mu_i$ on the boundary tori of 3-atoms and finding the gluing matrices, which express the cycles in an admissible basis on a boundary torus of one 3-atom in terms of similar cycles on another 3-atom. The gluing matrix $C = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$ is formed by the integers $a$, $b$, $c$, $d$ such that $\det C = -1$. Then the marks $r$ and $\varepsilon$ have the following expressions:
$$
\begin{equation*}
b \ne 0 \quad \Longrightarrow \quad r = \frac ab, \quad \varepsilon =\operatorname{sgn}b; \qquad b = 0 \quad \Longrightarrow \quad r = \infty, \quad \varepsilon =\operatorname{sgn}a.
\end{equation*}
\notag
$$
The rules for choosing cycles in the general case were presented in [8], Vol. 1, Ch. 4. We consider the case of an atom $A$ and a boundary torus if it. The cycle $\lambda$ can be contracted to a point in the solid torus. The cycle $\mu$ completes $\lambda$ to a basis and is homologous to the critical circle of the $3$-atom oriented in the direction of the Hamiltonian field of $H$. The orientation of $\lambda$ is chosen from the condition that $Q^3$ is oriented by the pair of tangent vectors and the normal vector to the 3-atom. Note that there are several ways to choose $\mu$: we can also take $\mu + k \lambda$ for $k \in \mathbb{Z}$ as a cycle $\mu'$. The case $r = \infty$ means that for two 3-atoms joined by an edge the (uniquely defined) $\lambda$-cycles are homologous in $Q^3$. The simplest example of a molecule is $A$–$A$. Then the isoenergy manifold $Q^3$ is glued of two solid tori. Proposition. An isoenergy surface $Q^3$ of an integrable system which carries a Liouville foliation with molecule $A$–$A$ is homeomorphic to $\bullet$ the 3-sphere $S^3$ if $r=0$; $\bullet$ the three-dimensional projective space $\mathbb {R}P^3$ if $r=1/2$; $\bullet$ the Cartesian product $S^1\times S^2$ of a circle and a sphere if $r=\infty$; $\bullet$ the lens space $L(q, p)$ if $r=p/q$ for $0<p<q$. (Note that $L(2,1)$ is homeomorphic to $\mathbb {R}P^3$.) For the billiard in a disc the molecule has the form $A$–$A$ ($Q^3$ is glued of two solid tori), and the marks are $r= 0$ and $\varepsilon = 1$. In fact, let us cut $Q^3$ along the preimage $\pi^{-1}(S^1\colon \{r = R_0\})$ (with respect to the projection $\pi\colon Q^3 \to \mathbb{R}^2$) of the circle with centre $O$ and radius $ R_0$, $0 < R_0 < R$. The preimages of the disc $r < R_0$ and the annulus $R_0 < r \leqslant R$ under this projection are two solid tori. It is easy to see that in the torus $\pi^{-1}(S^1 : \{r = R_0\})$ the cycle $\lambda$ in an admissible basis on one solid torus is glued to the cycle $\mu$ in an admissible basis on the other solid torus and vice versa. Thus, $Q^3$ if homeomorphic to $S^3$ (the well-known Hopf fibration), so that ${r = 0}$. Although confocal and circular billiards are not smooth systems in general, one can calculate analogues of Fomenko-Zieschang invariant for their Liouville foliations. For some confocal planar billiards these were calculated by Dragović and Radnović [12], while for arbitrary confocal planar billiards (see [13]) and arbitrary topological confocal billiards (see [6] and [14]) they were calculated by Vedyushkina. Fomenko and Vedyushkina constructed billiards realizing the Fomenko-Zieschang invariants of various systems in mechanics and physics (see [15] and [16]), that is, they modelled mechanical systems (including ones with first integrals of degrees 3 and 4 in components of the momentum) by means of billiard with ‘canonical’ quadratic first integrals (see [17]). Furthermore, topological billiards and billiard books were used to model all geodesic flows on orientable two-dimensional surfaces which have an additional first integral that is linear or quadratic in components of the momentum (by means of circular or confocal billiards, respectively; see [18]). Considerable advancements have also been made in the proof of Fomenko’s general conjecture on billiards which was stated in [19]: Vedyushkina and Kharcheva showed that using billiard books one can realize arbitrary Bott 3-atoms [4], [5] and the base of a Liouville foliation with these singularities [20], so that any molecule without numerical marks can be realized. Fomenko, Vedyushkina and Kibkalo investigated the ‘local’ version of the general conjecture (see [21]): they showed that arbitrary values of the marks $r$ and $\varepsilon$ on an edge of a molecule (see [22]) and the integer mark $n$ on a family subgraph (see [23]) can be realized, as well as certain combinations of marks (see [24]). Also, an isoenergy surface $Q^3$ of a billiard book was shown to be homeomorphic to a 3-manifold [25], and billiards were constructed for which $Q^3$ is homeomorphic to a 3-torus, an arbitrary lens space $L(q,p)$ for coprime $p$ and $q$, $0 < p < q$ (see [26]), or a connected sum of arbitrary lens spaces $L(q_i, p_i)$ and products $S^1 \times S^2$ (see [27]). As a result, for a billiard system $Q^3$ is not necessarily a Seifert manifold: one example is the connected sum of three lens spaces $L(2,1)$ (which are homeomorphic to the three-dimensional projective space). Another new generalization of classical planar billiards, billiards with slipping, was recently proposed by Fomenko [28]. Using such billiards one can develop further the results obtained in [18], where billiards on table complexes were used to model integrable geodesic flows on the two-dimensional sphere $S^2$ and torus $T^2$, and can extend them to nonorientable two-dimensional configuration manifolds. Recall that in the analytic category, according to Kozlov’s famous result [29], integrable geodesic flows exist on a two-dimensional manifold $M^2$ only when $M^2$ is a sphere $S^2$, a torus $T^2$, a projective plane $\mathbb{R}P^2$, or a Klein bottle $\mathrm{Kl}^2$. Circular billiards with slipping by an angle of $\pi$ are integrable and enable one to realise an arbitrary geodesic flow on the projective plane $\mathbb{R}P^2$ or the Klein bottle $\mathrm{Kl}^2$ that is integrable by a linear first integral (see [30]). In a similar way slipping by an angle of $\pi$ along elliptic boundaries of confocal billiards was defined in [28]. In the same paper such billiards were used to model examples of geodesic flows with quadratic first integrals on $\mathbb{R}P^2$ and $\mathrm{Kl}^2$. Such results extend the results in [18] to nonorientable manifolds in a natural way. Let $\alpha \in[0,2\pi]$ be commensurable with $\pi$. Assume that the boundary of the table $\Omega$ consists of circles with common centre $O$. We introduce slipping on one circle: after hitting the boundary at a point $P \in \partial \Omega$ with incidence angle $\phi$ the particle continues its motion with the same reflection angle and the same modulus of velocity, but not from $P$: it continues the motion from the image of $P$ under the rotation of the circle through $\alpha$ in the direction of the orientation induced by the oriented tangents (here this means clockwise). Such billiards are integrable. Consider a tangent to the caustic. By symmetry each rotation about the common centre takes the caustic to itself and takes a tangent to the caustic to another tangent to it. All links of the trajectory remain tangent to the caustic (the same for all links) after the rotation. Thus the resulting billiard is integrable (Figure 2). Remark 1. The condition that the slipping angle must be commensurable with $\pi$ is a reasonable restriction. This is because for $\alpha$ incommensurable with $\pi$, on the singular level corresponding to a minimum or a maximum of the first integral ($\varphi = 0$ or $\varphi = \pi$, respectively) all points $\varphi_0 + k \alpha$, $k \in \mathbb{Z}$, which differ by one or several rotations by $\alpha$ are identified. The set of identified points is dense on the singular circle but some points on it do not belong to this set. For example, fix a point on the circle. Let us identify it with all images of rotations of this point through $\alpha$ around the centre of the disc. Next we rotate this point through $\pi$. This image does not belong to the set of points identified with the original one. For let it be otherwise. Then $\pi$ can be represented in the form $p\alpha-2\pi q$, so that $\alpha$ can be expressed as a rational fraction of $\pi$. This is a contradiction. Any invariant neighbourhood of the set of points identified with the original one contains all points on the circle. Hence the complex obtained as the quotient by the action of the group generated by the rotation through $\alpha$ is a non-Hausdorff topological space. It follows, in particular, that neither this level nor the corresponding isoenergy set is homeomorphic to a smooth manifold. § 2. Billiard in a disc with slipping by an arbitrary rational angleWe state and prove the central result of this work. Theorem 1. Let $m$ and $k$ be coprime integers such that $m< 2k$. Let $l$ be an integer such the for the solution $l$, $i$ of the Diophantine equation $2ki-lm=(m,2k)$, where $(m,2k)$ is the GCD of $m$ and $2k$, the vector $\begin{pmatrix} l \\ i \end{pmatrix}$ has the minimum length among all solutions of this Diophantine equation. Consider a billiard in a disc with slipping by an angle of ${m\pi}/{k}$ along the boundary circle. Then its Fomenko-Zieschang invariant has the form $A$–$A$ with marks and the mark $r$ is equal to $(1+kl)/{k^2}$ if $m$ is even and to $(1+2kl)/{(4k^2)}$ if $m$ is odd.It follows from the form of the molecule that the isoenergy manifold $Q^3$ is homeomorphic to the lens space $L(4k^2,(m,2k)[(m,2k)+2kl])$. Consider the change of variables
$$
\begin{equation*}
\alpha=\frac{2k}{(m,2k)}, \qquad \beta=\frac{m}{(m,2k)}.
\end{equation*}
\notag
$$
Then out Diophantine equation becomes $\alpha i-\beta l=1$, and the $r$-mark takes the form $r=(1+\alpha l)/{\alpha^2}$. Note that $\alpha$ and $l$ have no common divisors since the Diophantine equation is solvable. Set $q=4k^2$ and $p=(m,2k)[(m,2k)+2kl]$. Corollary. Let $q\ne s^2$ for $s\in\mathbb{N}$, and let $0<p<q$, where $p,q \in \mathbb{N}$ are coprime. Then $L(q, p)$ cannot be an isoenergy manifold for a billiard in a disc with slipping by an angle commensurable with $\pi$. Now consider a billiard with slipping by an angle $\psi={m\pi}/{k}$ commensurable with $\pi$ along the boundary of a simply connected plane domain $\Phi(D^2) \subset \mathbb{R}^2$ obtained from the disc $D^2$ by a diffeomorphism $\Phi$. We parametrize the boundary circle: $\gamma(t) = \partial D^2$, $t \in [0, 2\pi)$. Inside the domain $\Phi(D^2)$ the particle moves along geodesics of the plane metric. Assume that reflections are elastic, so that the modulus $H = \vec{v}^2$ of the velocity vector is preserved in reflections. Thus the isoenergy surface $Q^3$ is invariant. After hitting the boundary $\partial \Phi(D^2)$ at a point $P = \Phi(\gamma(t_0))$ the particle continuous its motion from the point $P'=\Phi(\gamma(t_0+\psi))$ so that the incidence angle at $P$ is equal to the reflection angle at $P'$. Theorem 2. The isoenergy manifold $Q^3$ of a billiard with slipping by an angle $\psi = {m\pi}/{k}$ commensurable with $\pi$ along the boundary of a simply connected domain which is a diffeomorphic image of a circle is the lens space $L(q,p)$, where $q$ and $p$ are calculated from $k$ and $m$ by Theorem 1. Remark 2. The diffeomorphism $\Phi$ does not change the topological type of $Q^3$, so that we can prove the more general result of Theorem 2 (in which the billiard system is not assumed to be integrable) based on the theory of topological invariants of Liouville foliations of integrable systems, as applied to the proof of Theorem 1. Proof of Theorem 1. Fix a value of the additional integral. From each point in the annulus between the caustic and the boundary of the billiard table we can draw precisely two tangents to the caustic. At each fixed moment of time the velocity vector of the point mass is directed along one of them. There are two vectors parallel to a tangent line: one pointing towards the caustic and the other pointing away from it. On the other hand two vector correspond to clockwise motion and the other two correspond to anticlockwise motion. Clockwise motion corresponds to values $<{\pi}/{2}$ of the additional integral, while values $>{\pi}/{2}$ correspond to anticlockwise motion. Any pair point-vector (a point on the table and the velocity vector of a particle at this point) corresponds to a point in the phase space.
We can draw a unique tangent line to the caustic from a point on the caustic. Thus, the two velocity vectors corresponding to anticlockwise motion glue together at this point, and the same holds for the two vectors corresponding to clockwise motion. On the boundary a pair point-vector, where the velocity vector points away from the caustic, is glued with the following pair: the point is obtained by the rotation of the radius vector through ${m\pi}/{k}$, and the velocity vector at this point is directed towards the caustic. We endow points in the annulus between the two circle (namely, the caustic and the boundary of the billiard table) by velocity vectors corresponding to clockwise motion. Then we obtain two annuli on the phase manifold, one corresponding to pairs point-vector with vectors pointing toward the caustic, and the other corresponding to pairs with velocity vectors pointing away from it. Since on the boundary of the table and the caustic some pairs are glued together, we see that a regular level surface of the integrals $H$ and $\varphi$ is a torus if the additional integral takes a value $0<\varphi<{\pi}/{2}$. The same holds for ${\pi}/{2}<\varphi<\pi$. Because two annuli are glued into a torus in this way, we have a natural projection of the torus onto the billiard table. We call the annulus formed by the pairs point-vector with vector pointing away from the caustic the ‘bottom’ half of the torus, and we call the second annulus the ‘top’ half. Curves on the torus can also be shown on these annuli (Figure 3). Consider the value $\varphi={\pi}/{2}$ of the additional integral. In the limit the caustic contracts into a point (the centre of the circle) and all trajectories pass through it. Consider a small value $\varepsilon$ of the radius; to each point on the circle obtained we can assign two vectors, pointing towards the centre or away from it. Then we obtain two circles on the phase manifold, one is formed by the points of the circle of radius $\varepsilon$ with velocity vector pointing towards the centre and the other consists of points with velocity vectors pointing away from the centre. Then for $\varepsilon=0$ these circle are glued together as follows: a point with velocity vector pointing towards the centre is glued to the diametrically opposite point (relative to the centre) with velocity vector pointing away from the centre (Figure 4). Hence for $\varphi={\pi}/{2}$ the regular level is homeomorphic to a torus. For $\varphi=0$ or $\varphi=\pi$ there can only be one velocity vector at each point, so that a level surface is a circle. Thus, the molecule has the form $A$–$A$, that is, the isoenergy surface is glued of two solid tori. To describe fully the topology of the foliation we must understand how atoms are glued to one another. On the boundary tori of an atom $A$ one introduces admissible systems of coordinates with basis cycles denoted by $\lambda$ and $\mu$. The cycle $\lambda$ is uniquely chosen and can be contracted to a point inside the solid torus. As $\mu$ we take an arbitrary cycle complementing $\lambda$ to a basis. In approaching a singular level $\mu$ tends to the critical circle. This allows one to define its orientation uniquely (as coinciding with the orientation of the critical circle). We show these cycles in the projection onto the billiard table. The meridian of the torus (in the projection onto the billiard table) is by definition the preimage of the caustic (or a point, for $\varphi={\pi}/{2}$) with the orientation coinciding with that of the boundary, where the boundary is oriented clockwise. The system has $S^1$-symmetry with respect to rotations about the centre of the circles. All tori are surfaces of revolution with respect to this symmetry, and by a parallel on a Liouville torus of the system we mean the cycle obtained by the action of the symmetry group on some point. In the case of a billiard without slipping a parallel on a torus projects onto the billiard table as a radial segment connecting the caustic and the boundary circle of the table and endowed with velocity vectors pointing towards and away from the caustic alike (both vectors correspond to clockwise motion for $0< \varphi < \pi/2$ and anticlockwise motion for $\pi/2 < \varphi < \pi$). As already noticed, in a billiard with slipping the ‘bottom’ half of the torus (corresponding to points with velocity vectors pointing away from the caustic) makes a rotation in projection onto the billiard table, so the projection of a parallel onto this table consists of a radial segment from the boundary to the caustic with velocity vectors pointing towards the caustic and a radial segment obtained by the reverse rotation of the former segment, with velocity vectors pointing away from the caustic (Figure 5). These two segments make up indeed a cycle on the torus. Recall that the torus is glued of two annuli consisting of point-vector pairs. The first annulus is formed by pairs where the vectors point towards the caustic, and the interior of this annulus remains fixed. The second annulus is formed by pairs with velocity vectors pointing away from the caustic, and this annulus is rotated before gluing the torus. The radial segment with velocity vectors pointing towards the caustic remains fixed on the torus, while the segment with velocity vectors pointing away from the caustic has been obtained by the reverse rotation, so on the torus it goes to a segment similar to the one with velocity vectors pointing towards the caustic, and it is glued to it at the points whose projections onto the billiard table lie on the boundary and caustic, respectively. First let an atom $A$ correspond to clockwise motion. Consider the vertical line segment from the boundary to the caustic that begins at $(0,R)$. Now we rotate this segment through $m\pi/k$ around the centre of the disc. Since points on the boundary are also identified after the rotation of radius vectors by $m\pi/k$, the endpoints of this segment and the result of its rotation are identified on the boundary. Hence their union is homeomorphic to a line segment. We repeat this operation of rotation of the vertical segment through angles that are multiples of $m\pi/k$, combining the results with the segments obtained previously. Since $2m\pi$ is also a multiple of $m\pi/k$, some rotation of the vertical segment takes it to itself. We endow points of the resulting segment by velocity vectors in the direction of clockwise motion. Every such segment corresponds to two curves on the phase manifold (on one of which the velocity vectors point away from the caustic, while on the other they point towards the caustic). The boundary point with velocity vector pointing away from the caustic is identified on the phase manifold with the boundary point obtained from it by the rotation through $m\pi/k$ around the centre of the disc, with velocity vector pointing towards the caustic. That is, while we identify boundary points obtained by the rotation through $m\pi/k$, on the phase manifold we identify two point-vector pairs. We mentioned already in Remark 1 that velocity vectors pointing away from the caustic and towards the caustic are glued together at points on the caustic. Thus, the preimages of radial segments on the phase manifold are glued at their endpoints with the endpoints of the preimages of other segments (which are different for different endpoints). Thus we obtain a circle on the phase manifold, which lies on the torus equal to a common level surface of the energy integral and the first integral $\varphi$. This is the required cycle $\lambda$, because, as the integral $\varphi$ decreases, the radius of the caustic increases to the radius of the boundary circle of the billiard table, and the segments reduce to points. Thus, the cycle obtained contracts to a point inside the solid torus (Figure 6). Assume that $(m,k)=1$. Then $(m,2k)$ is equal to 1 or 2. As a point traverses one segment in the projection of the cycle onto the table, the cycle shifts by an angle of $m\pi/k$ along the meridian of the torus and makes one revolution along the parallels. To return to the initial position the cycle must make a full revolution along the meridian. For $(m,2k)=1$ we see that the number of segments is $2k$, and for $(m,2k)=2$ it is $k$. That is, the number of segments is ${2k}/{(m,2k)}$. The number of revolutions made along the meridian is
$$
\begin{equation*}
\frac{m\pi}{k}\cdot\frac{2k}{(m,2k)}\colon 2\pi=\frac{m}{(m,2k)}.
\end{equation*}
\notag
$$
Hence the cycle $\lambda$ has the following decomposition with respect to a parallel and the meridian of the torus:
$$
\begin{equation*}
\begin{pmatrix} \dfrac{2k}{(m,2k)} \\ \dfrac{m}{(m,2k)} \end{pmatrix}.
\end{equation*}
\notag
$$
Next consider the cycle $\mu$. It intersect $\lambda$ in one point. The segments of $\lambda$ partition the interior of the annulus into sectors. We construct $\mu$ as follows: taking two boundary points in adjacent sectors we connect them by a curve intersecting $\lambda$ in one point. Now we connect these two points by segments similar to those forming $\lambda$, by using slipping. The number of sectors is equal to the number of segments forming $\lambda$, namely, to ${2k}/{(m,2k)}$. We rotate the cycle $\lambda$ through an angle of clockwise around the centre of the disc; this takes the cycle to itself. Each sector goes to an adjacent sector. Note that the rotation through ${(m,2k)\pi}/{k}$ is a rotation through an angle which is a multiple of $m\pi/k$ modulo $2\pi$. This is because the integer $(m,2k)$ is a linear combination of $m$ and $2k$, that is, there exist integers $u$ and $v$ such that $(m,2k)=um-2kv$. Substituting this linear combination for $(m,2k)$ into the expression for the angle of rotation we obtain
$$
\begin{equation*}
\frac{(m,2k)\pi}{k}=\frac{(um-2kv)\pi}{k}=u\frac{m\pi}{k}-2\pi v.
\end{equation*}
\notag
$$
A rotation through $2\pi$ does not change the position of points, so a multiple of $2\pi$ can be dropped, which yields $um\pi/k$. Now we construct $\mu$. Consider two points on the boundary of the billiard table which differ by a rotation by ${(m,2k)\pi}/{k}$ around the centre of the disc (so that they lie in adjacent sectors) but do not belong to radial segments of $\lambda$. We connect these two distinguished boundary points by a curve passing through the point of intersection of the caustic and the segment of $\lambda$ lying between these two points and disjoint from $\lambda$ otherwise. Consider the distinguished point that is taken to the marked point by a clockwise rotation through ${(m,2k)\pi}/{k}$ around the centre of the disc; we call it the initial point of the cycle to be constructed. We rotate the second point clockwise by $m\pi/k$ around the centre and draw a radial segment connecting the boundary with the caustic and going out of the resulting point. We endow this segment with velocity vectors (like radial segments in $\lambda$). Next we rotate this segment clockwise through angles which are multiples of $m\pi/k$, until we obtain a radial segment containing the initial point. Then our cycle consists of the union of all these segment and the curve constructed previously and connecting points in two adjacent sectors. We endow points on these segments by velocity vectors in the direction of clockwise motion which point towards and away from the caustic. We endow points on the above curve with velocity vectors pointing at the caustic before the point of intersection with the caustic and pointing away from the caustic after this point. We claim that by choosing the orientation of this cycle to be compatible with the directions of these vectors we obtain the required cycle $\mu$. First of all, it intersects $\lambda$ in a unique point because the curve intersects it only at the point lying on the caustic and radial segments in $\mu$ do not intersect $\lambda$ since they lie inside sectors formed by $\lambda$. Second, the preimage of the above curve in the phase manifold is a homeomorphic to a line segment. The endpoints of this curve and the segments endowed with vectors corresponding to clockwise motion glue pairwise into a circle on the phase manifold. Hence $\mu$ is a cycle on a torus in the isoenergy manifold. As the radius of the caustic tends to that of the boundary circle, the segments forming $\mu$ contract to a point and the curve converges to an arc of the boundary circle with length ${(m,2k)R}/(2k)$. Since points on the boundary circle distinct by rotations through $m\pi/k$ are identified, which (as shown above) mean that points different by a rotation through ${(m,2k)\pi}/{k}$ are glued together, any arc of length ${(m,2k)R}/(2k)$ contains all points of the boundary circle. Hence $\mu$ becomes a critical circle in the limit, and the orientation of $\mu$ coincides with that of this circle. Thus the cycle $\mu$ satisfies all conditions defining the complementary cycle in a homology basis. Assume that $\mu$ contains $l-1$ segments. In traversing one segment the cycle shifts by an angle of $m\pi/k$ along the meridian and makes one revolution along the parallels. The curve it its projection onto the table makes one revolution along the parallels and a shift by ${(m+(m,2k))\pi}/{k}$ along the meridian since this curve can be represented as a segment and a part of the caustic, the shift due to the segment is $m{\pi}/{k}$ in the clockwise direction and the part of the caustic corresponds to a shift by ${(m, 2k)\pi}/{k}$. Thus, the total number of revolutions made along the parallels is $l$, and the total shift along the meridian is Since the cycle makes several revolutions along the meridian, this number is a multiple of $2\pi$. Hence Multiplying this by $k$ and dividing by $\pi$ we obtain $lm+(m,2k)=2ki$. Hence $2ki-lm=(m,2k)$ is a Diophantine equation with respect to $l$ and $i$, where $l$ and $i$ must be the solution of this equation with the minimum modulus for otherwise we traverse the cycle $\mu$ several times, which corresponds to several intersections with $\lambda$. Hence the expansion of $\mu$ with respect to the meridian and a parallel has the form $\begin{pmatrix} l \\ i \end{pmatrix}$.Now consider two vertical radial segments from the boundary circle of the table to the caustic, one going out of the point $(0,R)$ and the other going out of $(0,-R)$. We endow the ‘overlying’ segment (the one from $(0,R)$) by velocity vectors pointing towards the caustic in the direction of clockwise motion and endow the ‘underlying’ segment with vectors pointing away from the caustic in the direction of clockwise motion. Consider the endpoints of these segments on the caustic. As already noted, vectors pointing towards the caustic are glued with ones pointing away from it at points on the caustic. Now we connect the two diametrically opposite points under conideration by the half of the caustic going clockwise from the upper point to the lower one and endow this half with velocity vectors directed clockwise. Then we obtain a complex homeomorphic to a line segment (Figure 7). Next we rotate this complex through angles that are multiples of $m\pi/k$, and take its union with the results of these rotations until the complex returns to the initial position after a rotation. On the torus the complex (endowed with vectors as described) is homeomorphic to a line segment, and taking its union with its images under rotations means that the endpoints of the preimages of these complexes are glued pairwise, thus producing a circle. This is an auxiliary cycle $\eta$. Remark 3. The cycle $\eta$ is a curve with self-intersections on the torus. To avoid this we can perturb slightly points in segments forming successive rotations of the complex so as to increase the angles between their radius vectors and the horizontal coordinate axis. Next consider a small neighbourhood of the caustic and modify the half of the caustic included in the complex so that it intersects the caustic only at the endpoints of segments forming the complex. The resulting curve is no longer a part of the caustic. In traversing each image of the complex the cycle $\eta$ makes one revolution along the parallels and it rotates through an angle of along the meridian. If $m$ is even, then we cannot rotate a segment in $\lambda$ through $\pi$ by combining rotations through $m\pi/k$. Hence $\eta$ contains twice as many radial segments as $\lambda$, so that $\eta$ makes twice as many revolutions along the parallels as $\lambda$. If $m$ is odd, then the situation is different and $\lambda$ and $\eta$ make the same number of revolutions along the parallels. So this number is equal to $2k$. On the other hand the number of revolutions along the meridian is Therefore, the expansion of the cycle $\eta$ has the form $\begin{pmatrix} 2k \\ k+m \end{pmatrix}$.Remark 4. Let us look separately at the case when $m$ and $k$ are odd. Then in considering the expansion of $\eta$ we traverse this cycle twice, so the coefficients of the decomposition can be reduced twofold, which cannot be done in other cases. This doubling does not affect the result because the cycles $\eta$ corresponding to the atoms $A$ describing clockwise and anticlockwise motion pass one into the other, and their expansions with respect to $\lambda$ and $\mu$, which we need to calculate marks, also double. As both sides of the equality increase twofold, the final result remains the same. Also consider the cycle $\nu$, the caustic with velocity vectors directed clockwise (Figure 8). It clearly has the expansion $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$, because precisely one revolution along the meridian is made. Now we calculate the expansions of $\eta$ and $\nu$ with respect to $\lambda$ and $\mu$. First we consider $\eta$:
$$
\begin{equation*}
\begin{pmatrix} \dfrac{2k}{(m,2k)} & l \\ \dfrac{m}{(m,2k)} & i \end{pmatrix}\begin{pmatrix} x_\eta \\ y_\eta \end{pmatrix}= \begin{pmatrix} 2k \\ k+m \end{pmatrix}.
\end{equation*}
\notag
$$
The matrix on the left has determinant $(2ki-lm)/(m,2k)=1$. Therefore,
$$
\begin{equation*}
\begin{pmatrix} x_\eta \\ y_\eta \end{pmatrix} = \begin{pmatrix} i & -l \\ -\dfrac{m}{(m,2k)} & \dfrac{2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} 2k \\ k+m \end{pmatrix}=\begin{pmatrix} (m,2k)-lk \\ \dfrac{2k^2}{(m,2k)} \end{pmatrix}.
\end{equation*}
\notag
$$
In a similar way the expansion of $\nu$ is
$$
\begin{equation*}
\begin{pmatrix} x_\nu \\ y_\nu \end{pmatrix}=\begin{pmatrix} -l \\ \dfrac{2k}{(m,2k)} \end{pmatrix}.
\end{equation*}
\notag
$$
These expansions combine into a matrix:
$$
\begin{equation*}
\begin{pmatrix} \eta^+ \\ \nu^+ \end{pmatrix} = \begin{pmatrix} (m,2k)-lk & \dfrac{2k^2}{(m,2k)} \\ -l & \dfrac{2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^+ \\ \mu^+ \end{pmatrix},
\end{equation*}
\notag
$$
where $\lambda^+$ and $\mu^+$ are the cycles $\lambda$ and $\mu$ in the atom $A$ corresponding to clockwise motion. Here we denote the cycles $\eta$ and $\nu$ by $\eta^+$ and $\nu^+$. Now consider the atom $A$ corresponding to anticlockwise motion. In $\lambda^+$ we replace the velocity vectors by ones corresponding to anticlockwise motion and denote this cycle by $\lambda^-$. Hence $\lambda^-$ has the same decomposition with respect to the meridian and parallels as $\lambda^+$. We make the same operation with the cycle $\mu^+$, which tends to the critical circle in $A$. The cycle and the critical circle are oriented anticlockwise. Reversing this orientation we obtain a cycle $\mu^-$. It has the same expansion as $\mu^+$, but with the opposite signs (Figure 9). In the definition of $\eta^+$ we used a complex of two segments and half the caustic. To construct $\eta^-$ we consider a similar complex including the same segments, but the other half of the caustic. (Note that the other half also connects the endpoints of the segments that occur on the caustic.) We endow this complex with vectors similarly to $\eta^+$. Namely, we endow the ‘overlying’ segment with velocity vectors pointing toward the caustic and corresponding to anticlockwise motion and the ‘underlying’ segment with vectors pointing away from the caustic and corresponding to clockwise motion. The half of the caustic is endowed with vectors corresponding to anticlockwise motion. Then we rotate the complex by angles that are multiples of $m\pi/k$ around the centre of the disc and take the union of all images. This produces an auxiliary cycle $\eta^-$. The cycle $\eta^-$ involves the same number of revolutions along the parallels as $\eta^+$, namely, $2k$. In making one revolution along the parallels it makes a shift by an angle of along the meridian. Accordingly, the number of revolutions made along the meridian is $m-k$.On the other hand the cycle $\nu^-$ has the following expansion taking its orientation into account: $\begin{pmatrix} 0 \\ -1 \end{pmatrix}$. Now we can calculate the expansions of $\eta^-$ and $\nu^-$ with respect to $\lambda^-$ and $\mu^-$:
$$
\begin{equation*}
\begin{pmatrix} \dfrac{2k}{(m,2k)} & -l \\ \dfrac{m}{(m,2k)} & -i \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} 2k \\ m-k \end{pmatrix}.
\end{equation*}
\notag
$$
The left-hand matrix has determinant $-1$ because it is different from the corresponding matrix for clockwise motion by the signs in one row. Therefore,
$$
\begin{equation*}
\begin{pmatrix} x \\ y \end{pmatrix} = -\begin{pmatrix} -i & l \\ -\dfrac{m}{(m,2k)} & \dfrac{2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} 2k \\ m-k \end{pmatrix}=\begin{pmatrix} (m,2k)+lk \\ \dfrac{2k^2}{(m,2k)} \end{pmatrix}.
\end{equation*}
\notag
$$
Correspondingly,
$$
\begin{equation*}
\begin{pmatrix} \eta^- \\ \nu^- \end{pmatrix} = \begin{pmatrix} (m,2k)+lk & \dfrac{2k^2}{(m,2k)} \\ l & \dfrac{2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^- \\ \mu^- \end{pmatrix}.
\end{equation*}
\notag
$$
Consider now the ‘images’ of the auxiliary cycles for $\varphi= {\pi}/{2}$. The complexes forming $\eta^+$ and $\eta^-$ become segments joining two diametrically opposite points. The velocity vectors at points on these segments point towards the centre and away from it alike. Hence $\eta^+$ is transformed into $\eta^-$. The cycles $\nu^+$ and $\nu^-$ go to the centre, where there is a whole circle of velocity vectors. The orientations of the two circles are different because $\nu^+$ and $\nu^-$ have opposite orientations. Hence the cycle $\nu^+$ is transformed into $-\nu^-$. Having found the relations between $\eta^+$, $\nu^+$ and $\eta^-$, $\nu^-$ we can describe the ones between $\lambda^+$, $\mu^+$ and $\lambda^-$, $\mu^-$ by using the expressions for $\eta^+$, $\nu^+$ and $\eta^-$, $\nu^-$ in terms of these cycles:
$$
\begin{equation*}
\begin{pmatrix} (m,2k)-lk & \dfrac{2k^2}{(m,2k)} \\ -l & \dfrac{2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^+ \\ \mu^+ \end{pmatrix} = \begin{pmatrix} (m,2k)+lk & \dfrac{2k^2}{(m,2k)} \\ -l & -\dfrac{2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^- \\ \mu^- \end{pmatrix}.
\end{equation*}
\notag
$$
The left-hand matrix has determinant Therefore,
$$
\begin{equation*}
\begin{aligned} \, \begin{pmatrix} \lambda^+ \\ \mu^+ \end{pmatrix} &=\dfrac{1}{2k}\begin{pmatrix} \dfrac{2k}{(m,2k)} & \dfrac{-2k^2}{(m,2k)} \\ l & (m,2k)-lk \end{pmatrix}\begin{pmatrix} (m,2k)+lk & \dfrac{2k^2}{(m,2k)} \\ -l & -\dfrac{2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^- \\ \mu^- \end{pmatrix} \\ &= \frac{1}{2k}\begin{pmatrix} \dfrac{2k(m,2k)+4k^2l}{(m,2k)} & \dfrac{8k^3}{(m,2k)^2} \\ 2kl^2 & \dfrac{4k^2l-2k(m,2k)}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^- \\ \mu^- \end{pmatrix} \\ &= \begin{pmatrix} \dfrac{(m,2k)+2kl}{(m,2k)} & \dfrac{4k^2}{(m,2k)^2} \\ l^2 & \dfrac{2kl-(m,2k)}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^- \\ \mu^- \end{pmatrix} . \end{aligned}
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
r=\frac{(m,2k)[(m,2k)+2kl]}{4k^2},\qquad \varepsilon = 1.
\end{equation*}
\notag
$$
Theorem 1 is proved. We can introduce slipping by an angle commensurable with $\pi$ on ‘free’ circular boundaries of circular topological billiards and billiard books. Then saddle atoms can arise in the Liouville foliations, so we can wonder if we can realize family subgraphs and marks $n$ on them. We examine an example of such a billiard in what follows. § 3. Billiard with slipping by an arbitrary rational angle in a pair of glued annuliTheorem 3. Consider an annulus obtained by removing from a round disc the interior of another disc bounded by a circle in the same concentric family. Glue two copies of such an annulus along the inner circles and consider slipping by angles ${m_1\pi}/{k_1}$ and ${m_2\pi}/{k_2}$ on the respective outer circles. Then the Fomenko-Zieschang invariant of the billiard obtained is as shown in Figure 10. Proof. In this proof we also denote cycles corresponding to clockwise motion by ‘$+$’ and ones corresponding to anticlockwise motion by ‘$-$’. The cycles $\lambda$ and $\mu$ for an atom $A$ were described in Theorem 2. The cycles for an atom $B$ are as shown in Figure 11.
Also, the cycle $\lambda$ coincides with $\nu$ in Theorem 2. Therefore,
$$
\begin{equation*}
\nu^+= \begin{pmatrix} -l & \dfrac{2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^+ \\ \mu^+ \end{pmatrix}\quad\text{and} \quad \nu^-=\begin{pmatrix} l & \dfrac{2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^- \\ \mu^- \end{pmatrix}.
\end{equation*}
\notag
$$
The expansions of $\mu^+$ and $\mu^-$ with respect to the meridian and parallels on the torus corresponding to a fixed value of the first integral are of the form $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. We can express them in terms of the cycles for $A$:
$$
\begin{equation*}
\mu^+=\begin{pmatrix} i-l & \dfrac{2k-m}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^+ \\ \mu^+ \end{pmatrix}\quad\text{and} \quad \mu^-=\begin{pmatrix} i-l & \dfrac{m-2k}{(m,2k)} \end{pmatrix}\begin{pmatrix} \lambda^- \\ \mu^- \end{pmatrix}.
\end{equation*}
\notag
$$
Then we obtain the gluing matrix
$$
\begin{equation*}
\begin{pmatrix} -l & \dfrac{2k}{(m,2k)} \\ i-l & \dfrac{2k-m}{(m,2k)} \end{pmatrix}
\end{equation*}
\notag
$$
for clockwise motion and the matrix
$$
\begin{equation*}
\begin{pmatrix} l & \dfrac{2k}{(m,2k)} \\ i-l & \dfrac{m-2k}{(m,2k)} \end{pmatrix}
\end{equation*}
\notag
$$
for anticlockwise motion. Hence the marks $r$ for clockwise and anticlockwise motion are ${-l(m,2k)}/(2k)$ and ${l(m,2k)}/(2k)$, respectively, and $\varepsilon=1$. To describe the last edge note that in going over the value $\varphi=\pi/2$ the cycles $\lambda_A$ transform one into the other, while the cycle $\lambda_B$ transform one into the other with opposite orientation. Note that
$$
\begin{equation*}
\lambda^+_A=\frac{m-2k}{(m,2k)}\lambda^+_B+\frac{2k}{(m,2k)}\mu^+_B\quad\text{and} \quad \lambda^-_A=\frac{-m+2k}{(m,2k)}\lambda^-_B+\frac{2k}{(m,2k)}\mu^-_B.
\end{equation*}
\notag
$$
Then we obtain
$$
\begin{equation*}
\frac{m-2k}{(m,2k)}\lambda^+_B+\frac{2k}{(m,2k)}\mu^+_B = \frac{-m+2k}{(m,2k)}\lambda^-_B+\frac{2k}{(m,2k)}\mu^-_B.
\end{equation*}
\notag
$$
Since $\lambda^+_B=-\lambda^-_B$, we have $\mu^+_B=\mu^-_B$. Hence the gluing matrix is $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$. The contribution to the mark $n$ from the edges corresponding to clockwise motion is $\theta^+=[-1+{m}/(2k)]$, and for anticlockwise motion it is $\theta^+=[1-{m}/(2k)]$, while the central edge makes no contribution. It follows from the statement of the problem that $m<2k$, because we consider angles modulo $2\pi$. Hence the contribution from the edges corresponding to clockwise motion is $-1$, and for anticlockwise motion it is 0. Thus, $n=-2$. Theorem 3 is proved. We make a substitution similar to the one in Theorem 1, namely, we set $\alpha={2k}/{(m,2k)}$. Then the mark $r$ on the edge between atoms $A$ and $B$ has the expression $r={l}/{\alpha}$. As already noted, $\alpha$ and $l$ cannot have common divisors. Therefore, setting ${l}/{\alpha}$ to be an arbitrary rational number between 0 and 1 we can solve the Diophantine equation $\alpha i -\beta l=1$ (this time with respect to $\beta$ and $i$). The numbers $\alpha$ and $\beta$ determine the slipping angle, which can be calculated from the value of the rational mark $r$. Thus, there is a correspondence between the marks $r$ and the angles by which slipping in the system occurs. This means that any value of $r$ can be obtained. AcknowledgementThe author is grateful to A. T. Fomenko, V. V. Vedyushkina, V. A. Kibkalo and I. G. Rochev for valuable discussions. 
Citation in format AMSBIB
\Bibitem{Zav23} |
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