Simple Closed Geodesics on a Polyhedron

Geodesics on a surface are locally shortest curves. They play a role analogous to that of straight lines in the plane. However, they may possess properties different from those exhibited by lines in the plane. For example, geodesics can have many points of intersection, and they can be closed curves. Closed geodesics have important applications in dynamical systems and differential geometry. In 1905, Henri Poincaré conjectured that on every convex smooth surface in ${{\mathbb {R}}}^3$ there exist at least three simple (i.e., non-self-intersecting), closed geodesics, in other words, that there are at least three ways to pull a rubber band around a rock so that it will not slip off. The conjecture was proved in 1930 by Lazar Lyusternik and Lev Schnirelmann [19], and later [4, 10], it was shown that every convex smooth surface admits infinitely many simple closed geodesics. Why was the hypothesis formulated only for three geodesics? Perhaps this can be explained by noting the general belief that the ellipsoid with axes of three distinct lengths admits exactly three geodesics, namely the ellipsoid’s intersections with its three planes of symmetry. This mistake was made even by David Hilbert and Stefan Cohn-Vossen in their book Geometry and the Imagination [17]. Elementary examples of a fourth geodesic on an ellipsoid were constructed much later; see, for example, [18].

Geodesics on polyhedra have been studied systematically since the 1990s and have by now become the subject of an extensive literature. The geometric properties of geodesics are analyzed for regular polyhedra in [2, 3, 8, 9, 12, 14] and for general polyhedra in [15, 16, 20, 21]. The relation between geodesics and billiards was addressed in [11, 12], and geodesics in the spherical and Lobachevsky geometries are considered in [5, 6].

Most convex polyhedra do not admit any simple closed geodesics at all. This fact does not contradict the Lyusternik–Schnirelmann theorem, since the surface of a polyhedron is not smooth but only piecewise smooth. If such a geodesic exists, then it divides the surface into two domains, and for each of them, the sum of the face angles over the vertices in that domain must be a multiple of $2\pi$. This is a direct consequence of the Gauss–Bonnet theorem, although there are more elementary proofs [15, 16]. For example, a tetrahedron can have a simple closed geodesic only if the sum of the angles at two of its vertices is $2\pi$. That is why a right regular triangular pyramid distinct from a regular tetrahedron has no simple closed geodesics at all. On the other hand, all regular polyhedra possess simple closed geodesics [14]. The geodesics on the cube have only three distinct lengths, while those on the regular octahedron have two. The squares of these lengths (for unit edge length) are as follows:

• Cube: 16, 18, 20;

• Octahedron: 9, 12;

• Icosahedron: 25, 27, 28;

• Dodecahedron: $27 + 18\phi$ , $28 + 20\phi$, $29 + 18\phi$, $29+ 19\phi$, $25 + 25\phi$,

where $\phi = (\sqrt{5}+1)/{2}$ is the golden ratio. A regular tetrahedron admits arbitrarily long simple closed geodesics. For example, a regular tetrahedron with edge length one centimeter possesses a closed geodesic longer than a kilometer. Without self-intersections! The same is true for every disphenoid [14, 21].¹ A natural question arises: are there other polyhedra that admit arbitrarily long geodesics? For convex polyhedra, the answer turns out to be negative. We proved this in 2008 in [21], where we also conjectured that the disphenoid is unique in enjoying this property among not only polyhedra but all convex surfaces. The conjecture was proved in 2018 by Arseniy Akopyan and Anton Petrunin [1]. What can be said about nonconvex polyhedra? Can they have arbitrarily long geodesics?

Since geodesics on polyhedra are broken lines with vertices on edges, they can be studied using the tools of elementary geometry, which is what we are going to do in this paper.

First, we recall the compete classification of simple closed geodesics on the surface of a disphenoid, in particular, that the disphenoid admits arbitrarily long geodesics. Then we show that no other convex polyhedra possess this property. To this end, we present a new, elementary proof. Finally, we solve this problem for the nonconvex case.

We shall also formulate several open problems. One of them concerns geodesics in spherical and Lobachevsky spaces. Do they possess polyhedra (possibly nonconvex) with infinitely many simple closed geodesics? It is known that regular tetrahedra in those spaces possess this property [5, 6], while to the best of our knowledge, in the spherical space, the problem is still open.

Geodesics on the Surface of a Polyhedron

We begin with some definitions. Let ${\mathbb {T}}$ denote the unit circle. A closed geodesic on a surface S is a rectifiable curve $\ell : {\mathbb {T}}\rightarrow S$ that possesses the following property: for every $t \in {\mathbb {T}}$, there exists $\varepsilon >0$ such that for every $\tau _1, \tau _2 \in (t - \varepsilon , t + \varepsilon )$, the arc of $\ell$ is the shortest path on S between the points $\ell (\tau _1)$ and $\ell (\tau _2)$. We always assume that a geodesic is simple, that is, without self-intersections.

If P is a polyhedron, then every geodesic is a broken line with vertices on edges. Indeed, on each face, $\ell$ must be a straight line as the unique locally shortest curve in the plane. We call the vertices of $\ell$ nodes, not to confuse them with the vertices of P. The sides of $\ell$ are links, and several consecutive links form a chain.

Every node V is either a vertex of P or an interior point of an edge with the “reflection property”: two links passing through V form equal angles with that edge (all angles are oriented, with the links directed in succession). This means that after unfolding the two faces onto the plane, these links become collinear.

Can a geodesic $\ell$ pass through a vertex V of a polyhedron P? If the sum of the face angles at V is less than $2\pi$, then it cannot. Otherwise, if $\ell$ contains V, we cut the surface of the polyhedral angle V into two parts along the links KV, LV that meet at V. The sum of the angles in one of those parts is less than $\pi$. We unfold this part onto the plane and take two close points on different links (points M and N in Figure 1). Then the shortest path between those points is the segment MN, which does not pass through V.

Thus, a geodesic does not contain vertices at which the sum of face angles is less than $2\pi$. In particular, if a polyhedron is convex, then the geodesic cannot pass though any of the vertices.

Figure 1.

A geodesic on a polyhedron cannot pass through the vertices.

For a convex polyhedron, the aforementioned properties are sufficient to define a geodesic.

Proposition 1.

A simple closed curve on a convex polyhedron is a geodesic if and only if the following hold:

• 1. It is a broken line with nodes on edges.

• 2. It has the reflection property at every node.

• 3. It does not pass through any vertices.

This result enables us to study geodesics on polyhedra by elementary geometric methods. The convexity is significant: for nonconvex polyhedra, a simple closed geodesic can contain vertices and can pass along edges.

If one develops the faces of P onto the plane along the geodesic $\ell$, then $\ell$ becomes a straight line. This line will be denoted by the same symbol $\ell$. The unfolding begins at some face h and ends in its copy $h'$. The polygonal boundary of $h'$ is a translation of h by a vector parallel to $\ell$ of length equal to the perimeter of $\ell$; see Figure 2. We call two geodesics equivalent if they intersect the same edges in the same order.

Figure 2.

Equivalent geodesics $AA'$ and $CC'$ on the diagram.

Proposition 2.

If a convex polyhedron admits a simple closed geodesic, then it has infinitely many equivalent geodesics of the same length.

Proof.

We unfold the faces of the polyhedron onto the plane of the net along the geodesic $\ell$. We begin at some node $A \in \ell$ and move along the geodesic, unfolding the faces that $\ell$ meets. Having traversed the whole geodesic, we arrive at the same node A, which becomes a point $A'$ on the net. Thus the geodesic $\ell$ becomes a segment $AA'$ that does not pass through the images of vertices on the net. Move the point A slightly along the edge to obtain a point C (Figure 2). Then the segment $CC'$ connects two images of the same node and does not pass through the vertices, provided that C is close enough to A. Since the segments AC and $A'C'$ are equal and parallel, $ACC'A'$ is a parallelogram. Therefore, $CC'$ is equal and parallel to $AA'$. Thus, the segment $CC'$ also represents a geodesic, which is equivalent to $\ell$ and has the same length. $\square$

Isosceles Tetrahedra: Arbitrarily Long Geodesics

The disphenoid possesses an exceptional property: it has a full net that fills the whole plane. To see this, take a disphenoid ABCD and consider the triangular lattice generated by the face ABC. This is the full net, with knots corresponding to the vertices of ABCD. This means that one can label the vertices of the lattice with the letters A, B, C, D so that for any development, the labeling of the vertices of the tetrahedron will match the labeling of the vertices of the lattice; see Figure 3. Nothing like that can be done for the cube (or the octahedron or the icosahedron).

Introduce a system of coordinates in the plane with origin at A and with basis vectors 2AB, 2AC. Thus $B =\bigl (\frac{1}{2}, 0 \bigr )$ and $C = \bigl (0, \frac{1}{2}\bigr )$. Each point X on the surface of the tetrahedron has its full preimage in the plane: $\bigl \{ \pm X + {\varvec{u}},\ {\varvec{u}}\in {{\mathbb {Z}}}^2\bigr \}$, where ${{\mathbb {Z}}}^2$ is the set of integer vectors. In particular, the vertex A corresponds to all integer points in the plane,

$$B \mapsto \left(n + \frac{1}{2} , m \right ), C \mapsto \left (n , m + \frac{1}{2}\right ), D \mapsto \left (n + \frac{1}{2} , m + \frac{1}{2}\right )$$

for $m,n \in {{\mathbb {Z}}}$.

Figure 3.

Geodesics of types (1, 0), (3, 2), and (1, 1) on the disphenoid.

Now we apply the full net to analyze the geodesics on a disphenoid. Two geodesics on a polyhedron are said to be isomorphic if one of them is equivalent to the other under an automorphism (an isometric map to itself) of the polyhedron.

Theorem 3.

Consider the full net of the disphenoid. Every straight line on this net parallel to an integer vector and not containing the knots represents a simple closed geodesic. All geodesics parallel to one vector are isomorphic and have the same length.

Proof.

Let $\ell$ be parallel to an integer vector ${\varvec{u}}=(n,m)$. We may assume that $n \ge m \ge 0$ and n, m are coprime unless $m=0$. If the points $X, Y \in \ell$ represent the same point on the tetrahedron, then either they are symmetric about a knot or XY is an integer vector. The former is impossible, since otherwise, $\ell$ would contain a knot (the center of symmetry). In the latter case, the vector XY must be an integer multiple of ${\varvec{u}}$, since ${\varvec{u}}$ has coprime coordinates. After division, it may be assumed that $XY = {\varvec{u}}$. Hence the segment XY is folded onto a closed geodesic, which is simple, since all interior points of that segment represent distinct points of the surface of the tetrahedron.

It remains to prove that parallel lines on the net produce isomorphic geodesics. Among all straight lines on the net parallel to $\ell$ and passing through knots, choose the closest ones from each side; call them $\ell _1, \ell _2$. On each of those lines we take a segment that connects two knots and does not contain other knots. Since $\ell _2$ is obtained from $\ell _1$ by an integer translation leaving the net invariant, those two segments are equal. They are opposite sides of a parallelogram that does not contain other knots, since there are no knots in the entire strip between $\ell _1$ and $\ell _2$ (we denote this strip by $\ell _1\ell _2$). The vertices of this parallelogram represent four different vertices of the tetrahedron, for otherwise, the midpoint of the segment between the two corresponding knots would be a knot. So we may denote this parallelogram by $A'B'C'D'$, where $X'$ represents a point X on the net. Any other line a parallel to $\ell$ on the net can also be put into a strip $a_1a_2$ whose sides contain vertices of a parallelogram $A''B''C''D''$. A suitable integer translate or symmetry about a knot maps the strip $\ell _1\ell _2$ to $a_1a_2$, and the knot $A'$ to $A''$. Consequently, we can assume that a lies in the same strip $\ell _1\ell _2$. Since there are no knots inside this strip, it follows that the lines $\ell$ and a intersect the same edges in the same order; i.e., these geodesics are isomorphic. $\square$

Remark 4.

In the proof of Theorem 3, we actually show that every geodesic on a surface of a dispheniod is simple. Indeed, within each face, a geodesic creates a family of parallel lines.

The correspondence between simple closed geodesics on the isosceles tetrahedron and the straight lines with rational directions in the full triangular net is well known; see, for example, [7, 14]. The isomorphism between all trajectories parallel to a given rational vector was observed in [20]. This gives a combinatorial classification of simple closed geodesics on a disphenoid. Each of them is unfolded to a straight line parallel to an integer vector (n, m) with coprime components or with $n=1$, $m=0$. Conversely, every such vector produces a class of isomorphic geodesics of equal lengths $|\ell | = \sqrt{n^2 + m^2 + mn}$. It is easily shown that $\ell$ has m, n, and $n+m$ nodes on the edges AB, AC, and BC respectively, i.e., $4(n+m)$ nodes in total. Figure 3 shows the geodesics of types $(n,m) = (1,0)$, (3, 2), and (1, 1). The geodesic (3, 2) has $4(3+2) = 20$ links.

As we have mentioned, a right regular triangular pyramid has no simple closed geodesics, provided that the lateral edge is not equal to the side of the base. Hence, if we slightly move the vertex of the regular tetrahedron along the altitude, then all its (infinitely many!) geodesics will slip off.

Remark 5.

Our classification actually holds for geodesics on an arbitrary tetrahedron, not only on an isosceles one. Namely, if we associate vertices of an arbitrary tetrahedron $\Delta$ with the vertices of a disphenoid, then every simple closed geodesic on $\Delta$ is isomorphic to one on the disphenoid [20]. This means that if $\Delta$ admits a simple closed geodesic, then it intersects the edges of the tetrahedron in the same order as does one of the geodesics on the disphenoid. Therefore, every simple closed geodesic on the tetrahedron is of combinatorial type (n, m) as described above. Every nonisosceles tetrahedron has only finitely many (possibly zero) inequivalent geodesics; we prove this in the next section. All of them have the aforementioned combinatorial types.

Let us recall that a generic tetrahedron does not possess any simple closed geodesics [16]. On the other hand, it is still unknown (although it looks plausible) whether a generic tetrahedron possesses at least one closed geodesic, simple or not.

Other Convex Polyhedra

Do there exist convex polyhedra apart from disphenoids that admit infinitely many nonisomorphic simple closed geodesics, or equivalently, arbitrarily long geodesics? We notice at once that such examples can be obtained merely by slight perturbations of a disphenoid. For instance, by cutting off its vertices (replacing them by small triangular faces), one obtains a polyhedron that is very close to a disphenoid. It will clearly inherit many inequivalent geodesics from the disphenoid, provided that the cuts are small enough. However, it may not have infinitely many. It turns out that in fact, it never has infinitely many. The following theorem, established in [21], asserts that the existence of arbitrarily long simple closed geodesics is an exceptional property of disphenoids. Here we give an elementary proof.

Theorem 6.

If a convex polyhedron has arbitrarily long simple closed geodesics, then it is a disphenoid.

The roadmap of the proof is the following. Every simple closed geodesic splits the surface into two pieces $\Gamma$ and ${\tilde{\Gamma }}$. We show that if $\ell$ is sufficiently long, then on unfolding $\Gamma$ to the plane, we obtain a very long and narrow straight strip with almost parallel lateral sides. Then we prove that this strip ends by hooking a vertex of P. This implies that $\Gamma$ is folded from one strip and contains two vertices. The same is true for ${\tilde{\Gamma }}$. Hence P has four vertices, i.e., is a tetrahedron. Then we show that this tetrahedron is isosceles. The proof is divided into four steps.

1. The domain $\Gamma$ possesses a net, a part of which is a straight strip with many nodes on the sides.

Every edge of the polyhedron P can contain several nodes of the geodesic $\ell$. At most two of them are extreme, that is, closest to the ends of the edge. All the others are said to be intermediate. Since the total number of extreme nodes does not exceed twice the number of edges of P, there are pieces of the geodesic $\ell$ that contain only intermediate nodes $A_1, \ldots , A_N$. The number N can be as large as we want, provided that $\ell$ is long enough. Every node $A_i$ lies on an edge of P, where it has two neighbors. For one of them (call it $B_i$), the segment $A_iB_i$ lies in $\Gamma$. Thus we obtain a sequence of nodes $B_1, \ldots , B_N$ such that each segment $A_iB_i$ is a part of one edge of P. We call the link $A_iA_{i+1}$ incorrect if the nodes $B_i$ and $B_{i+1}$ are not connected by a link.

In this case, they are ends of two different links on the same face on one half-plane with respect to $A_iA_{i+1}$. If $\ell$ contains one incorrect link connecting two nodes on the same edges, then it lies on the other half of that face, and hence there are no other incorrect links. Therefore, each pair of edges of P produces at most two incorrect links. Thus, the total number of incorrect links of $\ell$ is bounded above by twice the number of pairs of edges of P.

Consequently, there are arbitrarily long sequences of the nodes $A_i$ without incorrect links. We may assume that all links in the chain $A_1 \ldots A_N$ are correct. Hence, $B_1 \ldots B_N$ is also a chain, and the piece of $\Gamma$ between those chains is a belt wound around P. Denote this belt by $S_N$. Unfolding it onto the plane yields a straight strip on the net bounded by the segments $A_1A_N$ and $B_1B_N$.

2. Those strips are long and narrow, and their lateral sides are almost parallel. Thus for arbitrarily large N, there is a belt $S_N$ bounded by two chains $A_1\ldots A_N$ and $B_1\ldots B_N$ of correct links. We denote by M the maximal number of edges of P starting at one vertex, and by d the smallest distance between disjoint edges of P. Let us show that the perimeter of every subchain $A_{i} \ldots A_{i+M}$ of M links is at least d. This will imply that the length of the chain $A_1\ldots A_N$ is large, provided that N is large enough.

Suppose the contrary: the perimeter of some subchain $A_{i} \ldots A_{i+M}$ is smaller than d. Hence every two nodes $A_j, A_k$ of this subchain belong to neighboring edges of P. Indeed, if they are from disjoint edges, then the length $A_jA_k$ is at least the distance between those edges, which is at least d. This is impossible, since the arc of $\ell$ between $A_j$ and $A_k$ is shorter than d. Therefore, all edges containing the nodes $A_{i}, \ldots , A_{i+M}$ are neighboring, i.e., they begin from the same vertex of P. Denote this vertex by V.

Let m be the total number of edges starting at V. Since $m \le M$, we see that two of those nodes must be on the same edge. Let them be $A_i$ and $A_{i+k}$, $k\le m$, and all nodes between them are on different edges. This means that the line $\ell$ traverses k edges, intersecting them at the nodes $A_{i}, \ldots , A_{i+k-1}$, and returns to the first edge (containing $A_i$) at the node $A_{i+k}$. This implies that $\ell$ passes through all edges starting at V, i.e., $k=m$.

If the segment $VA_{i+m}$ is shorter than $VA_{i}$, then the line $\ell$ must go beyond $A_{i+m}$ around the same edges and intersect them at nodes located closer to the vertex V, since otherwise, $\ell$ would have self-intersections. In this way, $\ell$ will make infinitely many loops and never end. If $VA_{i+m}$ is longer than $VA_{i}$, we reach the same conclusion by moving along $\ell$ in the opposite direction, from the node $A_i$ to $A_{i-1}$, etc.

Thus, taking N large enough, one can make the belt $S_N$ arbitrarily long. Hence the quadrangle $A_1A_N B_N B_1$ in the plane of the net has long sides $A_1A_{N}$ and $B_1B_{N}$. Its area, however, does not exceed the surface area of P. Hence those sides are almost parallel and close to each other. This means that the angle and the distance between them are both very small.

3. The domain ${\varvec{\Gamma }}$ contains exactly two vertices, each of whose face angles have sum close to ${\varvec{\pi }}.$

The domain $\Gamma$ contains a belt $S_N$ with arbitrarily long and “almost parallel” lateral sides. This belt ends when two links $A_{N}A_{N+1}$ and $B_{N}B_{N+1}$ arrive at the nodes $A_{N+1}$ and $B_{N+1}$ located on different edges of a single face F of P. The part of F between those links is in $\Gamma$, and hence it contains no other nodes. The segment $A_{N+1}B_{N+1}$ is very short, since otherwise, $S_N$ would have a large area, provided that it were long enough. Therefore, the arc $A_{N+1}B_{N+1}$ of the boundary of F contains a unique vertex V. We draw a ray a on F from this vertex parallel to $A_{N}A_{N+1}$. We then draw a ray b from V on another face of P such that a and b split the surface of the solid angle V into two parts with equal sums of the face angles of the faces. Each sum is less than $\pi$; denote this sum by $\pi - \varepsilon$.

Making a cut along b (after which b becomes two rays $b_1$ and $b_2$) and unfolding the faces around V onto the plane, we obtain $\angle a b_1 = \angle a b_2 = \pi - \varepsilon$. Since the links $A_{N}A_{N+1}$ and $B_{N}B_{N+1}$ are almost parallel to a (the first one is exactly parallel), it follows that the straight lines containing those links on the net intersect the rays $b_1$ and $b_2$ respectively (Figure 4).

Figure 4.

The geodesic $\ell$ hooks the vertex V.

This means that the arc of the geodesic $\ell$ connecting the points $A_{N+1}$ and $B_{N+1}$ intersects all the edges starting at V and intersects the ray b. Thus the line $\ell$ makes a loop around V, changing direction from the vector $A_{N}A_{N+1}$ to the vector $B_{N+1}B_{N}$. Those vectors are almost parallel, and consequently, the sum of face angles at the vertex V is close to $\pi$.

Thus the line $\ell$ “hooks up” the vertex V. Extending the belt $S_N$ in the opposite direction, we prove that it hooks up another vertex, and thus the domain $\Gamma$ closes. Therefore, $\Gamma$ contains exactly two vertices, and the sum of the face angles at each of them is close to $\pi$.

The polyhedron ${\varvec{P}}$ is a disphenoid. Since we can apply the same line of reasoning to both domains $\Gamma$ and ${\tilde{\Gamma }}$, we see that each of them contains two vertices. Hence P is a tetrahedron. The sum of the face angles at each vertex can be made arbitrarily close to $\pi$; hence it is equal to $\pi$. It remains to invoke the following well-known fact: a tetrahedron is isosceles if and only if the sum of face angles at each vertex is equal to $\pi$. This completes the proof.

Nonconvex Polyhedra

Can a nonconvex polyhedron P have infinitely many simple closed geodesics? The proof of Theorem 6 fails for only one reason: there may be vertices at which the sum of face angles exceeds $2\pi$. If P does not possess such vertices, then all face angles of its faces are less than $\pi$, and hence all faces are convex polygons. Thus all faces of P are convex, and the sum of face angles at each vertex is less than $2\pi$. In this case, the proof of Theorem 6 can be directly extended to P, and we have the following proposition.

Proposition 7.

If for a nonconvex polyhedron, the sum of the face angles at each vertex is less than $2\pi$, then it does not admit arbitrarily long simple closed geodesics.

If the polyhedron has a vertex V with the sum of face angles at least $2\pi$, then the geodesic can pass through V and, moreover, can go from V along an edge. Even if the geodesic forms a narrow strip, it now may not hook the vertex V. We are going to see that this is enough to make Theorem 6 false for nonconvex polyhedra.

Before we formulate the main result, we clarify one aspect. Let a surface S bound a compact domain $G \subset {{\mathbb {R}}}^3$. If a geodesic $\ell$ on S is locally the shortest not only among curves lying on S but also among all curves not intersecting the interior of G, then $\ell$ is said to be strong. In other words, every short arc of a strong geodesic is the shortest path between its ends among all paths not intersecting the interior of G. This stronger property possesses the physical interpretation of a geodesic as a rubber band on the surface. We call the geodesics possessing this property strong. Geodesics that are not strong may not have the form of an elastic band. For convex surfaces, all geodesics are strong.

Proposition 7 is true for all geodesics, not just for those that are strong. Now we present a nonconvex polyhedron with arbitrarily long strong geodesics. The construction consists of seven equal cubes.

Theorem 8.

There exists a nonconvex polyhedron that possesses arbitrarily long closed geodesics that are simple and strong.

The main idea of the construction is to assemble four equal cubes face-to-face as in Figure 5 (left). Choose an arbitrary point M on the edge EF and connect it with the vertex D by a geodesic line MKD. Then draw a segment AL parallel to MK with $L\in EF$. We make two key observations:

Figure 5.

Construction of the polyhedron.

• 1. The broken line LABCDKM is a geodesic connecting the points M and L. This is shown in a straightforward manner by unfolding.

• 2. The strip between KM and AL can be made arbitrarily narrow by moving the point M closer to E.

To carry out the construction, we put two figures from Figure 5 (left) together so that they share a common middle cube. We thereby obtain the polyhedron P, which consists of seven cubes; see Figure 5 (right). The line $\ell$ makes a loop around the upper cube along the path ABCD, then makes several turns around the middle cube, and then forms a symmetric loop $A'B'C'D'$. Since $\ell$ can form an arbitrarily narrow strip, it can make arbitrarily many turns around the middle cube without self-intersections.

Figure 6.

Arbitrarily long geodesics.

Now we present the formal construction. Consider the net obtained by k consecutive unfoldings of the middle cube onto the plane; see Figure 6. It consists of $4k+1$ square faces, the first and the last ones representing the same face $ADA'D'$. Two parallel segments $AD'$ and $DA'$ and two letters $\Pi$ on the upper and lower squares form the image of the closed geodesic $\ell$. Taking k large enough, we can make $\ell$ as long as necessary.

Discussion and Open Problems

The geodesics constructed in the proof of Theorem 8 contain entire edges of the polyhedron. Can this be avoided?

Problem 1.

Can a nonconvex polyhedron admit arbitrarily long simple closed geodesics not containing edges?

Similarly to the convex case, one can show that each of the two parts of the surface bounded by a geodesic possesses long and narrow straight strips with almost parallel sides. The only difference is that now the line $\ell$ may hook several vertices. If the sum of face angles at those vertices is $\pi k$, where $k \ge 3$ is an odd number, then the sides of the strip will be parallel. The question is whether those sides can be arbitrarily close to each other, which is necessary for long geodesics.

Problem 2.

What is the minimal number of vertices of a nonconvex polyhedron that admits arbitrarily long simple closed geodesics? For example, can it have five, six, or seven vertices?

It is known that regular tetrahedra in Lobachevsky spaces have infinitely many simple closed geodesics [5, 6]. A natural question arises: are there other polyhedra (possibly nonconvex) in Lobachevsky and spherical spaces with infinitely many geodesics?

Problem 3.

Do there exist nonconvex polyhedra in spherical and Lobachevsky spaces that admit infinitely many simple closed geodesics?