Illumination problems on translation surfaces with planar infinities
Abstract
In the current article we discuss an illumination problem proposed by Urrutia and Zaks. The focus is on configurations of finitely many twosided mirrors in the plane together with a source of light placed at an arbitrary point. In this setting, we study the regions unilluminated by the source. In the case of rational angles between the mirrors, a planar configuration gives rise to a surface with a translation structure and a number of planar infinities. We show that on a surface of this type with at least two infinities, one can find plenty of unilluminated regions isometric to unbounded planar sectors. In addition, we establish that the nonbijectivity of a certain circle map implies the existence of unbounded dark sectors for rational planar mirror configurations illuminated by a lightsource.
1 Introduction
Consider a planar domain with a light reflecting boundary. Place a source of light at a point inside the domain. Assume that the source emits rays in all directions. Each ray follows a straight line and whenever it reaches the boundary it is reflected according to the rule that the angle of incidence equals the angle of reflection. A point from the domain is considered illuminated by the source whenever there is a ray that reaches the point either directly or after a series of reflections. In this setting, one can ask the following questions, also known as illumination problems.
Question 1.
If we place the source of light at any point in the domain, will all of the domain be illuminated? If not, what could be said about the nonilluminated regions?
Question 2.
Is there a point from which the light source can illuminate the entire domain?
These problems are often attributed to E. Straus who posed them sometime in the early fifties and first published by V. Klee in 1969 [5]. Some famous examples and interesting results are Penrose’s room [1], Tokarsky’s example [5] as well as the article [3] by Hubert, Schmoll and Troubetzkoy on illumination on Veech surfaces.
In 1991, J. Urrita and J. Zaks proposed the following problem [6]. Assume we are given a finite number of disjoint compact line segments in the plane each representing a mirror that reflects light on both sides (a twosided mirror).
Let be any point on the plane not incident to any of the segments. Then, the complement of the set of mirrors is an unbounded domain with lightreflecting boundary and if we place a source of light at we can pose questions 1 and 2. Figure 1a depicts an example of a twosided mirror configuration with a light emitting source . The convex hull of the mirrors is a polygon. If is in the convex hull, one can construct a triangle unilluminated by , like the shaded one on figure 1b. To do that, it is sufficient for a mirror segment to be an edge of the convex hull.
In this paper we are interested in finite twosided mirror configurations with the following property: any pair of lines determined by the mirror segments are either parallel or intersect at an angle which is a rational multiple of We will call such a configuration a rational mirror configuration and the domain obtained as a complement of the mirrors will be called rational mirror domain. For those, we will find conditions that will guarantee the existence of unbounded unilluminated sectors in the plane (see definition 2).
A rational mirror domain can be ”unfolded” into a surface that carries a flat structure with conical singularities and trivial holonomy group [2], [4]. This means that the surface has a special atlas with the property that away from the cone points, the transition maps between two charts from the atlas are Euclidean translations. In the literature, such an object is called a translation surface. As a result, the piecewise linear trajectory of a light ray in the original domain becomes a smooth geodesic on the flat surface. Thus, one can think of a light source placed at a nonsingular point on the surface, emitting geodesic rays in all directions. Any other point on the surface is considered illuminated if there is a smooth geodesic connecting the source to the point. In this way, one can ask questions 1 and 2 for the translation surface. Notice that there are regions on it isometric to complements of compact sets in the plane. We will call a surface with such geometry a translation surface with planar infinities.
A translation surface with planar infinities gives rise to a pair where is a closed surface with a complex structure and is a meromorphic differential on with only double poles and zero residues. The zeroes of are the cone points of the flat structure [2],[4], and around each pole the surface looks like the complement of a compact set in the plane. The converse is also true. A pair of a closed Riemann surface and a meromorphic differential with only double poles and zero residues induces a translation structure on with planar infinities.
Definition 1.
The pair is called a translation surface with planar infinities whenever the following conditions hold:
(1) is a closed surface with a complex structure;
(2) is a meromorphic differential on ;
(3) Every pole of is of order exactly and the residue at that pole is zero. We will refer to the poles of as planar infinities.
In this study we will be interested in a special type of domains both on a translation surface with planar infinities and in the plane.
Definition 2.
a) Let and be two halflines in the plane both starting form a point and going to infinity. Let be the angle between and at the vertex , measured counterclockwise from to . Then, the open region bounded by and , whose internal angle at is , is called an infinite sector of angle (see figure 2a).
b) An open subdomain of a translation surface with planar infinities is called an infinite sector of angle whenever there exists a chart from the translation atlas of that maps isometrically to a planar infinite sector of angle like the one defined in point a.
On any translation surface one can always find an orientable foliation with singularities, whose leaves are geodesics. Indeed, let us foliate the Euclidean plane into horizontal straight lines, oriented as usual from left to right. Since each transition map between two charts is a Euclidean translation, it sends horizontal lines to horizontal lines (line orientation preserved). Thus, pulling back onto the surface the planar horizontal foliation from all translation charts defines globally the desired foliation . Moreover, the singularities of are the cone points of the surface , i.e. the zeroes of the differential We call the horizontal foliation of the surface and its leaves  the horizontal geodesics of the surface. At each nonsingular point of the oriented horizontal geodesic from defines a positive horizontal direction at . The counterclockwise angle between and an arbitrary oriented geodesic through is called the direction of at (see figure 2b). From now on, denotes the geodesic ray on starting from and going in the direction of angle . It is important to emphasize that, since we are working with a translation surface, the intersection of the geodesic with any other horizontal geodesic will always form the same angle as shown locally on figure 2b. In other words, just like in the plane, a geodesic on does not changes its angle with respect to the horizontal direction. Since a direction at any nonsingular point is defined as an angle , we can identify the set of all directions at with the unit circle . The point gives the horizontal direction .
2 Results
It is natural to ask questions about the behavior of the geodesics on a surface. The first question we will address is the following. On a translation surface with planar infinities, where do most geodesics emanating from a nonsingular point go? As it turns out, almost all of them fall onto the poles of the surface. Same is true for any rational mirror configuration in the plane.
Theorem 1.
The following two statements are true:
(1) Let be a translation surface with planar infinities and let be nonsingular. Then the set of all directions for which the geodesic passing through in direction of goes to one of the poles of is open and dense in the circle ;
(2) Assume we are given a rational mirror configuration in the plane and let be a point not lying on any of the mirrors. Then the set of all directions for which the piecewise linear reflected trajectory starting from in the direction of goes to infinity is open and dense in the circle .
The next result establishes the existence of infinite unilluminated sectors and large unbounded regions on translation surfaces with more than one planar infinity.
Theorem 2.
Let be a translation surface with at least two planar infinities. Then, for any point on there exists an infinite sector on unilluminated by , i.e. for any point there is no smooth geodesic on that connects to . Moreover, there exists a region on consisting of unilluminated, nonoverlapping infinite sectors of total angle , where is the number of poles of .
The main ideas used in the proof of theorem 2 can be adjusted to the study of illumination problems for rational mirror configurations in the plane. For instance, an interesting question put in an every day language, is the following. How big of an object can be hidden from a stationary observer in a rational mirror domain? Can we hide a car? A whole parking lot of cars? Precisely speaking, we would like to find a basic condition that will ensure the existence of an infinite unilluminated sector for a light source placed at a point inside a rational mirror domain.
Let be a rational mirror domain and let . Draw a large enough circle , so that its interior contains the mirrors from the configuration and the light source at the point . Denote by the open dense set of all directions which go to infinity, provided by theorem 1. For an angle follow the straight line starting form in direction of . Whenever the line reaches a mirror it is reflected, changing its direction. In this way, a piecewise linear trajectory is formed,
which at some point leaves the disc bounded by never to come back to it. Denote by the angle between the horizontal direction of and the portion of the trajectory that is outside the circle . As a result, we obtain a map . For a picture of the construction of see figure 3.
The map is defined almost everywhere on the unit circle. In fact, its domain is open and dense in . Moreover, is a rotation when restricted to any connected component of . Our hope is that finding ways to study the combinatorial properties of may facilitate the search for unbounded unilluminated sectors in rational mirror domains.
Theorem 3.
Assume we are given a rational mirror configuration. For an arbitrary point , not located on any of the mirrors, construct the circle map as explain in the previous two paragraphs (see also figure 3). If is not injective, then there exists an infinite sector in the plane unilluminated by .
3 Translation surfaces.
In the current section we discuss translation surfaces and show how to construct one from a rational mirror configuration. To illustrate the idea better, we apply the procedure to an example.
Various descriptions.
A translation surface is a closed surface with a finite set of points , called singularities, and a cover of by open charts
having the property that whenever the transition map between the two charts and is a Euclidean translation, i.e. . In our study, partitions into two subsets and . Each point from has a cone angle of , where is a positive integer. Each point form has an open neighborhood with a map such that is a translation chart from the atlas. Also, the set is compact. Thus, the collection contains all planar infinities on the surface.
Since all translations are holomorphic maps, the translation atlas induces a complex structure on (for details see [2] and [4]). Moreover, the differential in each can be pulled back as a holomorphic differential in the corresponding . But if
then . Hence, in any intersection which gives rise to a global holomorphic differential on . Moreover, extends to the singular set so that becomes the set of zeroes of and becomes the set of all poles of . The latter are all double and with residue . So we see that a translation surface with planar infinities induces a pair of a compact Riemann surface without boundary together with an appropriate meromorphic differential.
To recover the translation atlas from a pair , one can cover with topological discs . On each of them define the chart , where is fixed and varies in . As is either holomorphic or meromorphic with a double pole and residue inside the topological disc , the path of integration in is arbitrary. If then
for . Thus, we have obtained the desired translation atlas. As we can see, the description of a translation surface with planar infinities which we gave in the beginning of the current section is equivalent to definition 1.
The horizontal foliation on , mentioned in the introduction, is defined as follows. Let be the foliation of horizontal lines in oriented from left to right (see figure 2b). Define the pulledback local foliation in each . Observe that is invariant with respect to any translation, i.e. the translations map any horizontal line to a horizontal line. Hence, on each . Thus, all local foliations fit together in a global foliation on with geodesic leaves and singularities . The oriented leaves of determine globally a horizontal direction on . Since translations are Euclidean isometries, the Euclidean metric on induces a Euclidean metric on . In this metric geodesics that do not go through singularities are isometric to straight lines in . The notion of a direction at a nonsingular point is as defined in the introduction. It is the counterclockwise angle between the horizontal leaf and an oriented geodesic both passing through . Finally, an oriented geodesic always forms the same angle with any horizontal leaf it intersects, so it never selfintersects, except possibly to close up.
Construction.
Assume we have a configuration of disjoint compact line segments in the plane , which we regard as twosided mirrors. The angle between any two of them is a rationalmultiple of . Observe that if one of the mirrors forms a rational angle with the rest of the mirrors, then immediately follows that any pair of mirrors forms a rational angle. This is a consequence of the fact that in an Euclidean triangle the angles at the vertices sum up to
To understand better the construction that follows, one could have a simple toyexample in mind. Let us have two perpendicular mirrors and in the plane like the ones depicted on figure 4.
Begin by slicing along the segments to obtain a closed slitted domain in which every mirror segment is doubled in order to obtain two parallel copies and that form the boundary component of the surface around the slit . For an intuitive geometric picture of in the case of the toyexample, look at figure 4. Then is homeomorphic to a oncepunctured sphere with disjoint open discs removed, as shown on figure 5 for the case of two orthogonal mirrors. In particular, .
For each segment , fix the line through parallel to . Denote by the reflection of in . The group generated by all is a finite group. If is a generic direction in , then is an orbit of maximal length . In our example and a generic orbit has elements. Pick copies of each with a choice of a direction in it. If you prefer more formally, let . On figure 5, in the case of the toyexample, we can see a topological model of these four slitted planes with a choice of direction on each of them. We glue to if and only if there is a segment whose corresponding reflection satisfies . The gluing is done in the following way. Take and . Glue the edge to the edge and the edge to the edge of . On figure 4 of the toyexample, we have chosen and . The upper edge of the cut is glued to the lower edge of the cut . Analogously, the lower edge from is glued to upper edge from .
Both and are naturally translation surfaces with piecewise geodesic boundaries, global coordinates and , and differentials and respectively. Segments and are equal and parallel, hence the gluing map is a translation (see the gluing of the shaded pieces on figure 4). Therefore the resulting surface made out of and has a translation structure. Moreover, along the gluing locus, so there is a welldefined holomorphic differential on the new surface which extends meromorphically to both of its infinity points.
Now, follow the described gluing procedure for all cuts on the pieces , where . The final result is a closed Riemann surface and a meromorphic differential with only double poles and zero residues, as well as simple zeroes with cone angle . For the example of the two orthogonal mirrors, figure 5 illustrates how the four pieces fit together to form a compact torus with a complex structure and a meromorphic differential on . There are eight simple zeroes of and four double poles. The zeroes are obtained from identifying pairs of black vertices on the segments form figure 4. The cone angle at each zero is and the residue at each pole is as desired.
4 Proofs
Proof of theorem 1.
From now on is an arbitrary translation surface with planar infinities and any fixed point. The idea is to cut out a rectangle around each pole and replace it by a onehandle.
Indeed, choose a small topological disc around and map it to by where varies in and is fixed. Notice, is well defined as the residue at is , so the path of integration is irrelevant. The image is the complement of a compact set (the total shaded region on figure 6 stretching to infinity). Draw a rectangle as shown on figure 6 and remove its exterior (the darker region). On the surface, we remove the darker rectangular domain containing . Then glue together the lower horizontal edge of to the upper and the left to the right, like gluing a torus. The gluing maps are clearly a vertical and a horizontal translation respectively. Therefore we obtain a handle with a translation structure compatible with the structure on the rest of the surface (see figure 6). By doing this for each , we obtain a compact translation surface of , where is now holomorphic (has no poles). A lot is known about the behavior of the geodesics on such surfaces [2], [4], [7], so we use this knowledge in our advantage. Let be the set of all directions for which the geodesic on is closed or hits a zero of . Also, let be the set of all directions for which the geodesic flow of in direction of is minimal [4] (e.g. an ergodic flow is minimal [2],[4]). Then is countable but dense in (see [7]) and is dense and of full measure in (see [4], [2]). As a result, the set consists of all for which the geodesic ray is dense in . Moreover, is dense and of full measure in . Therefore, for any the corresponding geodesic ray on the original surface hits a pole of .
Let be the set of all directions with the property that the geodesic ray on in the direction of reaches a pole of . Since the geodesic flow on depends continuously on the initial point and direction, the condition that a geodesic ray reaches a planar infinity is open. Therefore, for each there exists an open circular interval that contains and for any the ray also reaches the same infinity. Hence, is open in . Moreover, the dense set of full measure is contained in . Therefore, is open and dense set of full measure in .
The second part of theorem 1 follows from the first one. If we are given a rational mirror configuration, unfold it into a translation surface with planar infinities as described earlier. Then, the infinity of the mirror domain lifts to the set of poles of on and we apply the first part of the theorem.
Proof of theorem 2.
As an open dense subset of , the constructed is a countable disjoint union of open circular intervals , i.e. . By construction, the geodesic rays emitted from in all directions go to the same pole of . Fix some and take a subinterval (it may even be convenient to choose ). Choose so that its measure is less than . Notice, that for every , each ray on goes to the same . In particular, on figure 7. As , take another and call it just like on our picture below. Choose a ”small” topological disc around with the property . Define the translation chart , where varies in and is fixed. The zero residue at guaranties independence of the integral on the path between and in . On figure 7 we have also provided an analogous chart around . From now on, we use the same notations in as the ones in . Thus, we identify with . In the domain looks like the complement of a compact set (the shaded region on figure 7). Let be a Euclidean circle in centered at and containing in its interior.
Abusing notation, let and be the two points on the circle such that the counterclockwise angles between the positive horizontal line through in and the radii and are respectively and . Let points and on be such that counterclockwise . Draw the lines and tangent to circle at and respectively. Then they bound an infinite sector , depicted on figure 7 as a darker shaded region.
We claim that that is not illuminated by . Assume that for some point there exists such that the geodesic staring from in the direction of passes through . Then, clearly goes to . As already commented in the introduction, any smooth geodesic on a translation surface forms the same angle with the horizontal direction at every point it passes through. In particular, the angle between and the horizontal direction in the chart as well as near the point is always . By looking at the picture of the chart on figure 7, we see that in . Hence at the point as well. By the choice of the circular interval , the geodesic ray should go to . But a geodesic ray can only reach one pole of , so we get to a contradiction. Therefore, the infinite sector on is not illuminated by .
To conclude the proof, notice that for each circular interval the unilluminated sector near can be also constructed around any other pole of , i.e. there are unilluminated copies of . Partition into disjoint subintervals for which we can apply the construction of unilluminated infinite sectors from the preceding two paragraphs. Thus, the the total sum of the angles of all unilluminated sectors constructed on is times the total measure of which is . Hence, the total angle is .
Proof of theorem 3.
Let be a rational mirror domain and (see figure 1 or 3). Recall the finite group generated by all reflections in the lines through parallel to the mirrors. It acts on by rotations. Let be the map described at the end of subsection ”Main results” (see also figure 3) and assume it is not injective. Then, there are from such that . Take the finite orbit . Then if and only if so . Hence, the restriction is not bijective and there is such that . Since is a restriction of a rotation on each connected component of , there is such that . Remember the circle from figure 3 that encompasses the mirrors and . Using the circular interval , we can carry out absolutely the same construction as the one in the chart described in the proof of theorem 2. For a picture of this construction look at the rightmost large shaded area on figure 7. Observe that the notations of the current proof match the picture’s notations so that we can use it directly, thinking that the set of mirrors is in the little white elliptic region containing the center . We claim that the infinite sector (the darker shaded area) is not illuminated by the source . Indeed, assume there is a light ray emitted by that reaches some . Then, from the picture, the direction of this ray is . But the light ray started from in some direction , so which is a contradiction.
References
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