HAL Id: hal-03151422
https://hal.archives-ouvertes.fr/hal-03151422
Preprint submitted on 24 Feb 2021
HAL is a multi-disciplinary open access
archive for the deposit and dissemination of
sci-entific research documents, whether they are
pub-lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d’enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.
Examples of reflective projective billiards and outer
ghost billiards
Corentin Fierobe
To cite this version:
Corentin Fierobe. Examples of reflective projective billiards and outer ghost billiards. 2021.
�hal-03151422�
Examples of reflective projective billiards and outer ghost billiards
Corentin Fierobe
June 14, 2020
Abstract
In the class of projective billiards, which contains the usual billiards, we exhibit counter-examples to Ivrii’s conjecture, which states that in any planar billiard with smooth boundary the set of periodic orbits has zero measure. The counter-examples are polygons admitting a 2-parameters family of n-periodic orbits, with n being either 3 or any even integer grower than 4.
Contents
1 Introduction 1
2 3-reflectivity of the right-spherical billiard 4
3 Two 4-reflective projective billiards 5
3.1 The converging mirrors . . . 5 3.2 The centrally-projective quadrilateral . . . 5
4 centrally-projective regular polygons 7
5 Centrally-projective polygons and dual billiards 8
1
Introduction
In the theory of billiard, describing properties of the periodic orbits of the billiard inside a domain is the subject of many studies. A famous conjecture, due to Ivrii states the following: in any smooth euclidean
planar billiard, the set of periodic orbits has zero measure. The proof of the conjecture was made in the case
of a billiard with a regular analytic convex boundary, see [15]. The conjecture is also true for 3-periodic orbits, and was proved in [1, 9, 10, 16, 17]. For 4-periodic orbits, it was proven in [7, 8] and a complete classification of 4-reflective (defined later) complex analytic billiards was presented in [6]. Ivrii’s conjecture was also studied in manifolds of constant curvature: it was proven to be true for k = 3 in the hyperbolic plane H2
, see [3]; the case of the sphere S2 was apparently firstly studied in [2], as quoted in [3] but we were
not able to find the correponding paper. The sphere is in fact an example of space were Ivrii’s conjecture is not true, and we can find a classification of all counter-examples in [3].
b
γ γ
O
Figure 1: Left: a curve b endowed with a field of transverse line. Center: a convex closed curve γ with a field of transverse lines. Right: the same curve γ with a so-called centrally-projective field of transverse lines.
In this paper we study a generalization of usual billiards, as it is described in [11]. We consider the so called projective billiards, which are billiards having their boundaries endowed with a field of transverse lines, called projective field of lines (see Figure 1), and defining at each point a new reflection law which we call
projective reflection law: suppose that at a point p of a curve γ, the transverse line at p is L. A line ` hitting γ at p will be reflected into a line `0 if the lines (`, `0, L, Tpγ) are harmonic, meaning that their cross-ratio is
−1. We recall that the cross-ratio of four lines through the same point p is defined as the cross-ratio of their intersection with a fifth line not passing through p, and this doesn’t depend on the line taken. Note that this definition makes the projective plane RP2 a convenient space to study projective billiards.
We can therefore study the billiard dynamics on projective billiard chosing this new law to determine the way a particle bouces inside a table. Note that, given any closed curve γ, the usual billiard dynamic inside γ is also a projective one by choosing the transverse lines to be the normal lines to γ. In this paper we study examples of projective billiards inside polygons. They are defined as follows:
EXAMPLE 1. Consider a smooth convex planar curve γ and an other point O inside γ called the origin. Define a transverse field of lines on γ by taking the lines passing through the origin, see Figure 2. As in [11], this billiard is called centrally-projective and has many interesting properties, including the fact that the billiard transformation has an invariant area form ([11], Theorem D).
In the case when γ is a polygon P with n-vertices, P0, . . . , Pn−1, we call this billiard a centrally-projective
polygon, and denote it by CPol(O; P0, . . . , Pn−1). When the poygon P is regular and the origin is the
intersection of its great diagonals, we call it centrally-projective regular polygon.
γ O P0 P1 P2 P3 O L0 L1 L2 L3
Figure 2: Left: a convex closed curve γ with a so-called centrally-projective field of transverse lines through an origin O. Right: a centrally-projective quadrilateral P0P1P2P3 with origin O.
EXAMPLE 2. Consider a triangle P0P1P2 in the (projective) plane. On a side PiPi+1, define a field of
transverse lines by taking the lines passing through the remaining vertex of the triangle, Pi−1, see Figure 3.
We call this projective billiard right-spherical billiard based at (P0, P1, P2). and denote it by S(P0, P1, P2).
P2 P0 P1 L2 L0 L1
Its name comes from a famous construction on the two-dimensionnal sphere S2: consider a triangle
Q0Q1Q2 on S2 having right angles at all its vertices (its area is one eighth of the total surface of S2). This
triangle is a well-known example of 2-reflective billiard on the sphere, as explained in [2, 3]. Now consider the projection of S2from its center to a plane P tangent to the sphere at a point located inside Q0Q1Q2: Q0Q1Q2
is projected on a triangle P0P1P2of P, and the normal geodesics to the sides of Q0Q1Q2 are projected into
the projective field of lines making P0P1P2 a right-spherical billiard.
Here we consider a simplified version of the billiard dynamic, where the name ghost come from, and which is enough to study examples of curves having the reflectivity property (defined later). In the particular case of a polygon P = P0· · · Pn−1 endowed with a field of transverse lines on each of its sides, we extend the
definition of edges of P to be the whole lines supporting the usual edges. We force the successive bounces of a billiard trajectory to be on successive edges of P, by forgetting any possible obstacles on the ball trajectory between two consecutive edges. More precisely, fix two distinct points M0 ∈ P0P1 and M1 ∈ P1P2 on two
consecutive edges. We define the orbit of (M0, M1) as follows:
Definition 1.1. The orbit of (M0, M1) is the sequence (Mk)k∈Z where for each k, Mk is a point on PkPk+1
(where k is taken modulo n) and the lines Mk−1Mkand MkMk+1are symmetric with respect to the projective
reflection law at Mk.
Let m = kn be a positive mutliplier of n. The orbit is called m-periodic if MmMm+1 = M0M1. The
corresponding projective billiard is said to be m-reflective if there is a non-empty open subset U × V ⊂
P0P1× P1P2 of (M0, M1) whose orbits are m-periodic.
In the following, we use these examples to construct reflective projective billiards (having the additional particular property that U × V = P0P1× P1P2, which can be seen as a consequence of analyticity of the
billiard map). First, since the right-spherical billiard is obtained by projecting a certain 3-reflective billiard of S2on the plane, we expect that it is also 2-reflective (see Figure 4). Indeed, we have the
Proposition 1.2. The right-spherical billiard, S(P0, P1, P2), is 3-reflective.
P1 P0 P2 M0 M1 M2 L0 L1 L2
Figure 4: The spherical billiard S(P0, P1, P2) and a triangular orbit (M0, M1, M2) obtained by reflecting any
segment M0M1two times
In Section 3, we construct classes of 4-reflective projective billiards which cannot be deduced from one another. More precisely, we build the first one by gluing two right-spherical billiards. The second one is a centrally-projective polygon, as stated by
Proposition 1.3. Let P0, P1, P2, P3 be a non-degenerate quadrilateral and O the intersection point of its
diagonals. The centrally-projective polygon CPol(O; P0, P1, P2, P3) is 4-reflective.
In Section 4, we generalize the case n = 4 to any even integer n ≥ 4, with restrictions on the polygon however:
Proposition 1.4. Let n ≥ 4 be even. Let P0· · · Pn−1be a regular polygon and O be the intersection point of
This result is not true for n even. However, if one allows two times more bounces, it becomes true, and not only for regular polygons. More precisely, we state the following
Proposition 1.5. Let n ≥ 3 be odd. Let P0· · · Pn−1 be a polygon and O be any point not lying on the edges
of P . The centrally-projective billiard CPol(O; P0, . . . , Pn−1) is 2n-reflective.
The results are proved in the following order: we prove Proposition 1.2 in Section 2, Proposition 1.3 in Section 3 (and present another example of 4-reflective billiard derived from the right-spherical model), Proposition 1.4 in Section 4, and Proposition 1.5 in Section 5.
Knowing these examples, many questions arise: are there polygonal examples of n-reflective projective billiards with n odd? Can we find centrally-projective billiards which are n-reflective but not built on polygons? Another more difficult question would be: what are the n-reflective projective billiards (with the requirement that the boundary has a certain class of smoothness)? Answering this question would be a great step in solving Ivrii’s conjecture. In [4], we show that the only 3-reflective projective billiard table with analytic boundary is the right-spherical billiard, as described in example described in Section 2.
We would like to conclude this introduction by saying that you can find, on the webpage of the author (which was [5] at the time of writing), a simulation of projective billiards in polygons in which you can visualize their dynamics.
2
3-reflectivity of the right-spherical billiard
In this section, we prove Proposition 1.2: the right-spherical billiard is 3-reflective. We first introduce some notations:
Let P0, P1, P2 be three points not on the same line. For i = 0, 1, 2, let γi be the line PiPi+1. For any
M ∈ γi, let Li(M ) be the line M Pi+2 (i is seen modulo 3), that is Li(M ) is line joining M and the only
point Pj which do lie on γi. The projective billiard table defined by the γi and the Li is the right-spherical
billiard S(P0, P1, P2) (see Figure 3).
Proposition 2.1. Any (M0, M1) ∈ γ0× γ1 with M0 6= M1 determines a 3-periodic orbit inside the
right-spherical billiard S(P0, P1, P2).
Proof. This proof was found by Simon Allais in a talk we had about harmonicity in a projective space. Let M2 ∈ γ2 be such that M0M1, M1M2, γ1, L1(M1) are harmonic lines. Define M20 ∈ γ2 similarly: M0M1,
M0M20, γ0, L0(M0) are harmonic lines.
P1 P0 P2 M0 M1 M2 L0 L1 A M20
Figure 5: As in the proof of Proposition 2.1, both quadruples of points (M2, A, P0, P2) and (M20, A, P0, P2)
Let us first show that necessarily M2 = M20 (see Figure 5). Consider the line γ2 and let A be its point
of intersection with M0M1. Let us consider harmonic quadruples of points on γ2. By harmonicity of the
previous defined lines passing through M1, the quadruple of points (A, M2, P2, P0) is harmonic. Doing the
same with the lines passing through M0, the quadruple of points (A, M20, P2, P0) is harmonic. Hence M2= M20
since the projective transformation defining the cross-ratio is one to one.
Now let us prove that the lines M1M2, M0M2, γ2, L2(M2) are harmonic lines. Consider the line γ0:
M1M2 intersects it at a certain point denoted by B, M2M0 at M0, γ2 at P0 and L2(M2) at P1. But the
quadruple of points (B, M0, P0, P1) is harmonic since there is a reflection law at M1 whose lines intersect γ0
exactly in those points.
3
Two 4-reflective projective billiards
In this section we construct two classes of 4-reflective projective billiards: the first one is obtained by gluing together two right-spherical billiards, the second one is obtained from a centrally-projective quadrilateral whose origin is the intersection point of its diagonals.
3.1
The converging mirrors
γ0 γ P2 P1 M0 M2 M0 0 M20 M1 L0 L
Figure 6: billiard deduced from the right-spherical billiard S(P0, P1, P2), where the point P0 is at infinity,
hence the lines P0P1, P0P2 and all tranverse lines L1(M ) on P1P2are parallel, making the reflection law on
P1P2 as the usual one.
In R2, consider two distinct parallel lines γ and γ0. Choose a normal line intersecting γ at a point P1 and
γ0 at a point P2. On γ, define a field of transverse lines by taking the lines L(M ) going from a point M ∈ γ
to P2. Do the same on γ0 with P1 to form the field of lines L0. Consider the projective billiard dynamics
made by a particle bouncing alternatively γ and γ0 with their respective fields of transverse lines. Call this construction the converging mirrors. We claim that
Proposition 3.1. The converging mirrors are 4-reflective.
Proof. Complete R2 with a line at infinity ` to form the projective plane. Consider the point P
0∈ ` where
γ and γ0 intersect. Choose any orbit (M0, M1, M2) of the spherical billiard S(P0, P1, P2), where M0 ∈ γ,
M1 ∈ P1P2 and M2 ∈ γ0 (see Figure 6). Now consider the axial reflection with respect to the line P1P2.
It leaves P1P2, γ, γ0 and P0 invariant, hence the orbit (M0, M1, M2) is transformed into another orbit
(M00, M1, M20) of S(P0, P1, P2), where M00 ∈ γ and M20 ∈ γ0.
But since the transverse lines on the side P1P2of S(P0, P1, P2) are orthogonal to the line P1P2(they pass
through P0, as well as γ and γ0), the reflection law on P1P2 is the usual one: the lines M0M1 and M1M2
make the same angle with P1P2. Hence M0, M1, M20 are on the same line, and so are M2, M1, M00. Therefore,
M0, M20, M00, M2 is a 4-periodic orbit.
3.2
The centrally-projective quadrilateral
Let P0, P1, P2, P3 be four points, no three of them being on the same line. For i = 0 . . . 3 (seen modulo
4), let γi be the line PiPi+1. Write O to be the point of intersection of the lines P0P2 and P1P3(diagonals).
For any M ∈ γi, let Li(M ) be the line OM . The projective billiard table defined by the γi and the Li is
what we call the centrally-projective quadrilateral CPol(O; P0, P1, P2, P3) (see Figure 7).
P0 P1 P2 P3 O L0 L1 L2 L3
Figure 7: The centrally-projective quadrilateral CPol(O; P0, P1, P2, P3) with each one of its fields of transverse
lines
Proposition 3.2. Any (M0, M1) ∈ γ0× γ1 with M0 6= M1 determines a 4-periodic orbit of the
centrally-projective quadrilateral CPol(O; P0, P1, P2, P3).
Proof. Let M2∈ γ2 such that M0M1 is reflected into M1M2 by the reflection law at M1. Let M3∈ γ3 such
that M1M2is reflected into M2M3by the reflection law at M2. Let M30 ∈ γ3such that M0M1is reflected into
P0 P1 P2 P3 O M0 M1 M2 A B M3= M30 C0 D0 A0 B0 L1 L2 L0 L3
Figure 8: The centrally-projective quadrilateral CPol(O; P0, P1, P2, P3) with a periodic orbit obtained by
M0M30 by the reflection law at M0. Denote by d the line reflected from M2M3 by the projective reflection
law at M3. We have to show that d = M0M30.
First, let us introduce a few notations (see Figure 8). Consider the line M0M1; it intersects: the line P0P2
at a point B and the line P1P3 at a point A0. Now consider the line M1M2; it intersects: the line P0P2 at a
point B0 and the line P1P3at a point A. Finally let C0 be the intersection point of M2M3with P1P3and D0
the intersection point of M0M30 with P0P2.
Then, notice that by the projective law of reflection at M1, the quadruple of points (A, A0, P1, O) is
harmonic. Since the points P1, A0, O correpond to the lines γ0, M0M1, L0(M0), the previously defined
reflected line M0M30 needs to pass through A in order to form a harmonic quadruple of lines. The same
remark on the other diagonal leads to note that M2M3passes through B.
Now by the reflection law at M2, one observe that the quadruple of points (A, C0, O, P3) is harmonic.
But γ3 passes through P3, M3M2 through C0 and L3(M3) through O. Hence d needs to pass through A.
Then, by the reflection law at M0, one observe that the quadruple of points (B, D0, O, P0) is harmonic. But
γ3 passes through P0, M3M2 through B and L3(M3) through O. Hence d needs to pass through D0.
Therefore we conclude that d = AD0= M0M30.
Remark 3.3. Notice that the spherical billiard cannot be deduced from a usual billiard on the sphere by the
same construction as described for the right-spherical billiard.
4
centrally-projective regular polygons
In the following, we prove Proposition 1.4. We first introduce some notations:
Let n = 2k, k ≥ 2, an even integer. Let P be an n-sided regular polygon of radius 1, whose vertices are clockwise denoted by P0, P1, . . . , Pn−1. In the following, each time we refer to the index i, we will consider it
modulo n, that is we identify i and its rest when divided by n. Define O to be the intersection of the great diagonals of the polygon, which are the lines PiPi+k, i = 0, . . . , k−1. Construct a field of transverse lines Lion
γi:= PiPi+1 by setting Li(M ) = M O (the line joining the basepoint M and O). The projective billiard table
defined by the γi and the Li is what we call the centrally-projective regular polygon CPol(O; P0, . . . , Pn−1),
see Figure 9.
Let (M0, M1) ∈ P0P1× P1P2 with M06= M1. We want to show that the orbit (Mm)m∈Zof (M0, M1) is
n-periodic. We first prove the
P1 P2 P3 P0 P5 P4 O M0 M1 M2 M3 M4 M5 L1 L2 L3 L4 L5 L0
Figure 9: A centrally-projective polygon CPol(O; P0, . . . , P5) and a piece of trajectory after four projective
bounces.
Lemma 4.1. Fix an m ∈ N and consider the great diagonal Lm= PmPm+k. Then for any r ∈ N, the lines
Pm Pm−r Pm−r−1 Pm−r−2 Pm+r Pm+r+1 Pm+r+2 Mm−r−1 Mm−r Mm−r−2 Mm+r−1 Mm+r+1 Mm+r Lm
Figure 10: As in the proof of Proposition 4.2, since the lines Mm−r−1Mm−r and Mm+r−1Mm+rintersect Lm
at the same point, the lines Mm−r−2Mm−r−1and Mm+rMm+r+1also intersect Lmat a same point.
Proof. Let us prove Lemma 4.1 by induction on r.
Case when r = 0: Fix any m ∈ Z. Let A be the intersection point of Mm−2Mm−1 with Lm, A0 of
MmMm+1 with Lm and B of Mm−1Mm. Consider harmonic quadruples of points on Lm: (A, B, Pm, O)
is harmonic by the reflection law in Mm−1, and (A0, B, Pm, O) is harmonic by the reflection law in Mm+1.
Hence A = A0 which concludes the proof for r = 0.
Inductive step: suppose Lemma 4.1 is true for any r0 < r and let us prove it for r. See Figure 10 for
a detailled drawing of the situation. Fix an m ∈ Z. By assumption, we know that Mm−r−1Mm−r and
Mm+r−1Mm+r intersect Lm at a same point A. Furthermore by symmetry of the regular polygon through
Lm, the lines Pm−r−1Pm−r and Pm+rPm+r+1intersect Lmat the same point. Now, the three following lines
through Mm−r−1, Mm−r−1Mm−r, Pm−r−1Pm−r, Mm−r−1O intersect Lm through the same points than the
three following lines through Mm+r, Mm+r−1Mm+r, Pm+rPm+r+1, Mm+rO. Hence in order to sastisfy the
projection law at Mm−r−1 and Mm+r respectively, the lines Mm−r−1Mm−r−2 and Mm+rMm+r+1 should
intersect at the same point. Hence the inductive step is over and this conclude the proof.
Proposition 4.2. Any (M0, M1) ∈ P0P1 × P1P2 with M0 6= M1 determines an n-periodic orbit of the
centrally-projective regular polygon CPol(O; P0, . . . , Pn−1).
Proof. We have to show that M−1M0 = Mn−1Mn. We will use Lemma 4.1. First, by setting m = k and
r = k − 1, we conclude that the lines M0M−1and the lines Mn−1Mn intersect Lkat a same point denoted by
A. Then, by setting m = k + 1 and r = k − 2 we get that the lines M1M2and Mn−1Mn intersect Lk+1= L1
at a same point B. Now it is also true that M0M−1intersect L1at B, by Lemma 4.1 and setting m = 1 and
r = 0. Hence we have shown that Mn−1Mn= AB = M0M−1 which concludes the proof.
Remark 4.3. We note that the assumptions that P is regular and that O is on the intersection of the great
diagonals is fundamental to guarantee the reflectivity. On a simulation, see for example [5], the orbits are destroyed after small perturbations of P’s vertices or of the origin. However, the next example we describe is much more stable in this sense.
5
Centrally-projective polygons and dual billiards
In this section we prove Proposition 1.5. We first introduce some notations:
Let n = 2k + 1, k ≥ 1, be an odd integer. Let P be an n-sided polygon, whose vertices are clockwise denoted by P0, P1, . . . , Pn−1. We suppose that any three consececutive vertices are not on the same line.
Choose another point O, which do not lie on any line of the type γi := PiPi+1 (i is taken modulo n).
Construct a field of transverse lines Li on γi by setting Li(M ) = M O (the line joining the basepoint M and
O). The projective billiard table defined by the γi and the Li is what we call the centrally-projective polygon
We will prove the
Proposition 5.1. Let n = 2k +1 ≥ 3. Any (M0, M1) ∈ P0P1×P1P2with M06= M1determines a 2n-periodic
orbit of the centrally-projective polygon CPol(O; P0, . . . , Pn−1).
P0 P2 P1 M4 M5 M0 O M1 M2 M3
Figure 11: A 6-periodic orbit (Mk)k on a centrally-projective triangle P0P1P2 with origin O. The dotted
lines are representatives of the transverse fields of lines on the sides of the triangle.
Outer ghost billiard associated to a centrally pojective billiard
We can associate to any projective billiard a billiard called dual, or outer billiard, and this construction is well-explained in [11]. We will explain from the beginning the constructions which we will use, but for now, the already enlightened reader should keep in mind that the version of outer billiard we will use is very simplified, and the latter should be better called outer ghost billiard. Usual outer billiards are much more complicated, see [12, 14] for more details. Let us first recall some notions about polar duality.
Fix a point O ∈ R2
and complete R2
with a line at infinity λ to form RP2. Consider the polar duality
with respect to the point O. It sends any point p ∈ RP2 to a line p∗
and any line ` ⊂ RP2 to a point `∗
in a bijective way, such that O∗ = λ and λ∗ = O. It is involutive in the sense that if you apply the polar duality with respect to O two times, you get the identity: p∗∗= p and `∗∗= `. Finaly polar duality has the incidence property: p ∈ ` if and only if `∗∈ p∗.
Now as in Figure 12, suppose that we are given four lines of R2, `, `0, T and L such that
– L passes through O;
– The four lines `, `0, L, T pass through the same point p 6= O and are harmonic.
By the incidence property, the points `∗, `0∗, L∗ and T∗ belong to the line p∗, and are harmonic since the corresponding four lines are. But also L∗ ∈ O∗ = λ is at infinity. By harmonicity, both vectors −−→T∗`∗ and
−−−→
T∗`0∗ are opposite.
We can apply this to a centrally-projective polygon CPol(O; P0, . . . , Pn−1): denote by Q0, . . . , Qn−1 the
dual points of its sides, where Qk = PkPk+1∗ (k is tajen modulo n). Choose a projective orbit (Mk)k∈Z of
CPol(O; P0, . . . , Pn−1), an consider, for all k, the dual point Nk of the line Mk−1Mk; we call (Nk)k the dual
orbit of (Mk)k. Now we are ready to define the outer ghost billiard on Q0· · · Qn−1(see Figure 13):
Definition 5.2. To any point N0 distinct from Q0, . . . , Qn−1, we associate the sequence of points (Nk)k,
called ghost outer orbit of the polygon Q0· · · Qn−1, and uniquely determined by the relation
−−−→
NkQk =
−−−−−→
QkNk+1
O L T p ` `0 `∗ `0 ∗ T∗
Figure 12: On the left: four harmonic lines `, `0, L and T through p. On the right: their polar duals with respect to O, T∗is in the middle of the segment joining `∗ and `0∗, L∗ is at infinity.
Q0 Q1 Q2 N0 N1 N2 N3 N4
Figure 13: Five successive points, Nk with k = 0 . . . 4, of an outer ghost orbit of Q0Q1Q2.
By previous discussion, we have the
Proposition 5.3. If (Mk)k is an orbit of CPol(O; P0, . . . , Pn−1), its dual orbit (Nk)k is an outer ghost orbit
of Q0· · · Qn−1. Furthermore, (Mk)k is m-periodic if and only if (Nk)k is m-periodic.
Therefore, to prove Proposition 5.1, we only need to show the
Proposition 5.4. Let n = 2k + 1 ≥ 3. Any outer ghost orbit of Q0· · · Qn−1is 2n-periodic.
Proof. Let (Nk)k be an outer ghost orbit of Q0· · · Qn−1. We first show the
Lemma 5.5. For k ∈ Z, we have −−−−−→NkNk+2= 2
−−−−−→
QkQk+1.
Proof. Here we are in a Thales configuration and the proof follows. Indeed, by Chasles relation and the
equalities−−−→NkQk= −−−−−→ QkNk+1, −−−−−−−→ Nk+1Qk+1= −−−−−−−→
Qk+1Nk+2, we can write that
−−−−−→ NkNk+2= −−−→ NkQk+ −−−−−→ QkNk+1+ −−−−−−−→ Nk+1Qk+1+ −−−−−−−→ Qk+1Nk+2= 2 −−−−−→ QkNk+1+ 2 −−−−−−−→ Nk+1Qk+1= 2 −−−−−→ QkQk+1.
To conclude the proof, we need to show that N0= N2n. Indeed by Lemma 5.5 we have
−−−−→ N0N2n= n−1 X j=0 −−−−−−→ N2jN2j+2= 2 n−1 X j=0 −−−−−−→ Q2jQ2j+1
and we want to show that this is equal to 0. By splitting the latter sum at k, where n − 1 = 2k, we get n−1 X j=0 −−−−−−→ Q2jQ2j+1= k X j=0 −−−−−−→ Q2jQ2j+1+ n−1 X j=k+1 −−−−−−→ Q2jQ2j+1
and the latter sum can be rewritten with the change of index j0= j − k since−−−−−−→Q2jQ2j+1=
−−−−−−−−−−→ Q2j−nQ2j+1−n= −−−−−−−→ Q2j0−1Q2j0, to get n−1 X j=0 −−−−−−→ Q2jQ2j+1= k X j=0 −−−−−−→ Q2jQ2j+1+ k+1 X j0=1 −−−−−−−→ Q2j0−1Q2j0 = 0.
which concludes the proof.
Proof of Proposition 5.1. Since any outer ghost orbit of Q0· · · Qn−1 is 2n-periodic (Proposition 5.4), so is
any projective orbit of CPol(O; P0, . . . , Pn−1) by Proposition 5.3.
Acknowledgments
I’m very grateful to Simon Allais for his explanations about harmonic lines and how to prove Proposition 2.1 with this method, and Alexey Glutsyuk for its very helpful advices.
References
[1] Y.M. Baryshnikov, V. Zharnitsky, Billiards and nonholonomic distributions, J. Math. Sci. 128 (2005), 2706–2710.
[2] Y.M. Baryshnikov, Spherical billiards with periodic orbits, preprint.
[3] V. Blumen, K. Kim, J. Nance, V. Zharnitsky, Three-Period Orbits in Billiards on the Surfaces of Constant Curvature, International Mathematics Research Notices (2012), 10.1093/imrn/rnr228.
[4] C. Fierobe, On 3-reflective projective billiards with analytic boundary, in preparation. [5] Personal page of Corentin Fierobe, Section Projective billiards
http://perso.ens-lyon.fr/corentin.fierobe/projective_billiards.php
[6] A. Glutsyuk, On 4-reflective complex analytic billiards, Journal of Geometric Analysis 27 (2017), 183–238.
[7] A.A. Glutsyuk, Yu.G. Kudryashov, On quadrilateral orbits in planar billiards, Doklady Mathematics 83 (2011), No. 3, 371–373.
[8] A.A. Glutsyuk, Yu.G. Kudryashov, No planar billiard possesses an open set of quadrilateral trajectories,
J. Modern Dynamics 6 (2012), No. 3, 287–326.
[9] M. R. Rychlik, Periodic points of the billiard ball map in a convex domain, Journal of Differential
Geometry 30 (1989), 191–205.
[10] L. Stojanov, Note on the periodic points of the billiard, Journal of Differential Geometry 34 (1991), 835–837.
[11] S. Tabachnikov, Introducing projective billiards, Ergodic Theory and Dynamical Systems 17 (1997), 957–976.
[12] S. Tabachnikov, Dual billiards, Russian Math. Surveys 48 (1993), 81–109.
[14] S. Tabachnikov, On the dual billiard problem, Adv. in Math. 115 (1995), 221–249.
[15] D. Vasiliev, Two-term asymptotics of the spectrum of a boundary value problem in interior reflection of general form, Funct Anal Appl. 18 (1984), 267–277.
[16] Ya. B. Vorobets, On the measure of the set of periodic points of the billiard, Math. Notes 55 (1994), 455–460.
[17] M.P. Wojtkowski, Two applications of Jacobi fields to the billiard ball problem, Journal of Differential
