Impossible configurations for geodesics on negatively-curved surfaces

TL;DR

This paper generalizes a known example of a graph on a punctured sphere that cannot be realized by geodesics in negatively curved metrics to more complex surfaces with higher genus and punctures.

Contribution

It extends the concept of impossible geodesic configurations from a specific case to all surfaces of genus n with k punctures, broadening understanding of geometric constraints.

Findings

Identifies new impossible configurations on complex surfaces.

Generalizes previous specific examples to all genus and puncture cases.

Provides theoretical framework for understanding geodesic realization limitations.

Abstract

Hass and Scott's example of a 4-valent graph on the 3-punctured sphere that cannot be realized by geodesics in any metric of negative curvature is generalized to impossible configurations filling surfaces of genus $n$ with $k$ punctures for any $n$ and $k$.

Equations

$$\sum _{i=1}^p\sum _{j=1}^{n_i}{\alpha }_{i,j}<(N-2p)\pi .(\ast )$$

$$(2N-2q)\pi >\sum _{i=1}^q\sum _{j=1}^{m_i}{\beta }_{i,j}=2\sum _{i=1}^p\sum _{j=1}^{n_i}(\pi -{\alpha }_{i,j})=2N\pi -2\sum _{i=1}^p\sum _{j=1}^{n_i}{\alpha }_{i,j}$$

$$g_{\mathcal{P}}=\frac{1}{2}(2-\chi )=\frac{1}{2}(2-r+p)(\ast \ast ).$$

Full text

00footnotetext: Key words and phrases: surface of negative curvature, geodesic, configuration, genus, punctures.00footnotetext: 2000 Mathematics Subject Classification: Primary 53C22, Secondary 57M50,30F99.

1 Introduction

By a configuration we mean a surface $S$ together with a 4-valent, connected graph embedded in $S$. Going straight (neither right nor left) at each intersection decomposes such a graph canonically into a collection of closed curves (the tracks) intersecting themselves and each other transversally.

A basic question is whether or not there is a negative-curvature metric on $S$ such that the graph is isotopic to a collection of closed geodesics intersecting transversally. We will say in this case that the configuration can be realized by geodesics in a metric of negative curvature. It is an old but remarkable fact that the simple configuration shown in Figure 1 cannot be so realized. Joel Hass and Peter Scott discovered this phenomenon in 1999 [1]. As they remark, their proof of non-realizability can be by replaced by an argument, due to Ian Agol, using the Gauss-Bonnet Theorem.

In this paper Agol’s argument is generalized to produce an infinite family of non-realizable configurations, the polygonal impossible configurations, including a surface of genus $n$ with $k$ punctures for every $n\geq 2$ and $k\geq 0$. In some sense polygonal impossible configurations are all of the non-realizable examples that can be constructed using our general form of the argument.

1.1 Preliminaries

The Hass-Scott example, where $S$ is the 3-punctured sphere, is striking because at first inspection there seems to be no reason for it not to be realizable. Locally, it looks exactly like other, realizable configurations. In particular

• C1.

The curve segments do not enclose embedded or immersed contractible 1-gons or 2-gons (“monogons” or “bigons”): these are in fact outlawed in geodesic configurations in negative curvature.

• C2.

No curve represents a power ($>1$) in the free homotopy group $\pi_{1}S$, nor do two distinct curves represent powers ($\geq 1$) of the same element of $\pi_{1}S$. Since in negative curvature each free homotopy class contains a unique geodesic, a power curve collapses to multiple tracings of a single geodesic, and two homotopic curves collapse to the same geodesic; in either case the initial configuration is destroyed.

Furthermore, the Hass-Scott example fills its surface in the sense that

• C3.

The complement of the graph is a disjoint union of discs or singly punctured discs.

The configurations constructed here will satisfy the conditions C1, C2 and C3.

2 Polygonal impossible configurations

Definition 2.1**.**

A polygonal impossible configuration $\mathcal{P}$ is an orientable, connected 2-dimensional surface constructed as follows:

1. (1)

Choose a number $N\geq 3$, which will be the number of vertices in the configuration. 2. (2)

Choose a number $p$, with $N/4\leq p\leq N/3$, and $p$ polygons $A_{1},\dots,A_{p}$ which together have $N$ corners (This is possible since $p\leq N/3$). At least one $A_{i}$ must be a triangle: see the remark below. 3. (3)

Choose $q=N-2p$ even-sided polygons $B_{1},\dots,B_{q}$ which together have $2N$ corners. (This is possible since $p\geq N/4$ implies $4q=4N-8p\leq 2N$). Label the edges of the $B$-polygons alternately active and inactive. 4. (4)

Identify an edge of one of the $A_{i}$ with every active edge of each $B_{j}$, preserving orientations. Avoid forming a ring of squares: such a ring would lead to two parallel tracks.

Remark 2.2**.**

At least one of the $A_{i}$ must be a triangle. In fact, suppose first all the $A_{i}$ are squares; then $N=4p,q=N-2p=2p$, and $2N/q=4$, so all the $B_{i}$ must also be squares; and then the configuration $\mathcal{P}$ constructed by the algorithm will be made up of one or more sets of parallel curves, contradicting C2 above. On the other hand if all the $A_{i}$ have $\geq 4$ sides, and at least one has strictly more, then $N>4p$ contradicting $p\geq N/4$.

Theorem 2.3**.**

A polygonal impossible configuration cannot be given a metric of negative curvature so that the set of curves defined by its 1-skeleton is a set of geodesics.

Proof.

Suppose such a metric exists. Set $n_{i}$ to be the number of vertices of $A_{i}$, and $\alpha_{i,j}$, $j=1,\dots,n_{i}$ to be the interior angle at the $j$th vertex of $A_{i}$.

Likewise set $m_{i}$ to be the number of vertices of $B_{i}$, and $\beta_{i,j}$, $j=1,\dots,m_{i}$ to be the interior angle at the $j$th vertex of $B_{i}$.

Assume all the edges are geodesic arcs extending smoothly from polygon to polygon, so that each of the $\alpha_{i,j}$ is complementary to exactly two of the $\beta_{i,j}$.

The Gauss-Bonnet theorem [7] gives

\begin{array}[]{ccc}\alpha_{1,1}+\alpha_{1,2}+\cdots\alpha_{1,n_{1}}&<&(n_{1}-2)\pi\\ \dots&&\\ \alpha_{p,1}+\alpha_{p,2}+\cdots\alpha_{p,n_{p}}&<&(n_{p}-2)\pi.\end{array}

Adding these equations,

[TABLE]

Similarly, the sum of all the $\beta$s is strictly less than $(2N-2q)\pi$. On the other hand each $\beta$ is $\pi-\alpha$ for some $\alpha$, with each $\alpha$ occurring exactly twice.

So

[TABLE]

i.e. $\sum_{i=1}^{p}\sum_{j=1}^{n_{i}}\alpha_{i,j}>q\pi$. Since by the construction $q=N-2p$, this inequality contradicts $(*)$. ∎

3 The genus of a polygonal impossible configuration; minimal and unicursal configurations

A polygonal configuration $\mathcal{P}$ created from $N$, $A_{1},\dots,A_{p}$, $B_{1},\dots,B_{q}$ as above comes with a surface in which it is naturally embedded: the $N$ inactive edges of the $B$s are grouped by the identifications in step 4 into a collection $\gamma_{1},\dots,\gamma_{r}$ of closed curves; adding a disc $D_{i}$ along each $\gamma_{i}$ creates a closed orientable surface $S_{\mathcal{P}}$.

The surface $S_{\mathcal{P}}$ has Euler characteristic $\chi=N-2N+(p+q+r)=-p+r~{}$, and genus

[TABLE]

We can take $g_{\mathcal{P}}$ as the genus of $\mathcal{P}$; this matches the usual definition of the genus of a graph as the genus of the simplest surface on which it can be embedded so that its complement is topologically a set of discs.

The surface $S_{\mathcal{P}}$ does not necessarily admit a metric of negative curvature.

Preliminary punctures. To start, some of the discs $D_{1},\dots,D_{r}$ may be bounded by a single or exactly two curve segments. To satisfy condition C1 these discs must be punctured.

Further necessary punctures. The next steps depend on $g_{\mathcal{P}}$.

• •

$g_{\mathcal{P}}=0$. Since $p\geq 1$ the equation $(**)$ implies $r\geq 3$. Puncturing three of $D_{1},\dots,D_{r}$ if still necessary will give the 3-punctured sphere, a surface admitting a metric of negative curvature. Note that a sphere with one or two punctures admits such a metric, but in the first case there are no geodesics (so every configuration is impossible), and in the second case the only possible configuration is a circle.

• •

$g_{\mathcal{P}}=1$. Here $(**)$ implies $r\geq 1$; puncturing one of $D_{1},\dots,D_{r}$ if still necessary gives the punctured torus, a surface admitting a metric of negative curvature.

• •

$g_{\mathcal{P}}\geq 2$. In this case the surface admits a metric of negative curvature.

Definition 3.1**.**

A polygonal configuration on a surface a surface of genus $g$ with $k$ punctures is minimal if it fills the surface, and has the smallest possible number of vertices for such a configuration.

This minimal number depends on $g$ and $k$.

• •

If $k=0$, since $r\geq 1$ equation $(**)$ implies that $p$, the number of $A$-polygons, must satisfy $p\geq 2g-1$. Since each $A$-polygon is at least a triangle, the number $N$ of vertices for a polygonal impossible configuration filling the unpunctured surface of genus $g$ must satisfy $N\geq 6g-3$.

• •

Otherwise, since $r\geq k$ we have $p\geq 2g+k-2$ and $N\geq 6g+3k-6$.

Definition 3.2**.**

A configuration is unicursal if it has exactly one track, i.e., as described in the introduction, if it can be traversed by a single curve.

4 Examples

The smallest possible $N$ is $N=3$. Here $p$ and $q$ must equal 1, with $A_{1}$ a triangle and $B_{1}$ a hexagon. There are two ways to make the identification, with different results, as shown in Figure 2. Additional simple examples are shown in Figure 3.

5 Unicursal, filling configurations on surfaces of higher genus

5.1 Surfaces of

genus $\geq 2$

Proposition 5.1**.**

There exists a minimal unicursal impossible polygonal configuration filling the surface of genus $n$.

Proof.

We begin with genus $2$.

This construction can be extended to give an impossible configuration $\mathcal{P}_{n}$ filling the surface of genus $n$, for each $n\geq 2$.

Note that the configuration $\mathcal{P}_{n}$ is unicursal for all $n\geq 2$. Also, since $\mathcal{P}_{n}$ has $2n-1$ triangles and hence $6n-3$ vertices, it is minimal. ∎

5.2 Surfaces with punctures

The Hass-Scott example is minimal and unicursal. Generalizing it to surfaces with more punctures, using polygonal impossible configurations, can be done preserving both of these properties if the number of punctures is odd, but only one or the other if it is even.

Proposition 5.2**.**

The configuration $\mathcal{P}_{n}$ can be extended to become an impossible configuration filling the surface of genus $n$ with $k$ punctures, for any $k\geq 0$. If $k$ is odd, the new configuration can be minimal and unicursal. If $k$ is even, it can be minimal or unicursal but not both.

Proof.

Initially $\mathcal{P}_{n}$ has one complementary region, a disc. This disc can be punctured, yielding $k=1$. Splicing in $m$ copies of partial configuration c (Fig. 6) gives a minimal, unicursal polygonal configuration with the same genus and $2m+1$ punctures. On the other hand as remarked in Section 3 a minimal configuration of genus $n$ with $k$ punctures has $N=6n+3k-6$ vertices; if $k$ is even, so is $N$; and by Theorem 6.1 in the Appendix, such a configuration cannot be unicursal.

A unicursal configuration with genus $n$, with $2m$ punctures and one extra vertex can be obtained from $\mathcal{P}_{n}$ by splicing in $m-1$ copies of c and one of b. A minimal configuration with genus $n$, with $2m$ punctures and two tracks can be obtained by splicing in $m-1$ copies of c and one of a.

∎

6 Appendix: Unicursal configurations require an odd number of vertices

The examples in Figs. 2, 3 and 4 suggest the following statement.

Theorem 6.1**.**

The number of tracks of a polygonal impossible configuration is congruent mod 2 to the number of vertices. In particular, such a configuration can only be unicursal if the number of vertices is odd.

Preliminaries for the proof.

1. (1)

Taking the planar polygons $A_{i}$ and $B_{i}$ as in the construction of the configuration, give each one the standard (counterclockwise) orientation. Then give the segments of the configuration their inherited orientation, except segments shared by an $A$ and a $B$ keep their $A$-orientation. With this convention, the track-segments at each intersection are coherently oriented, and give a well-defined orientation on each track (Figure 7).

2. (2)

Project the configuration into the plane, and consider it as a collection of oriented immersed curves. The configuration depends only on the nature of the polygons $A_{i}$ and $B_{i}$ and the way they are connected. In particular, its projection can be displayed so that the $A_{i}$ and $B_{i}$ appear as in Figure 8.

Proof of Theorem 6.1

1. (1)

We show that the sum of the rotation numbers of the projected complex of curves is even. For each curve, the rotation number can be defined as the degree of the Gauss map, which takes a parameter value to the corresponding unit tangent vector, considered as a point on the unit circle. The degree of a smooth map is equal modulo 2 to the number of inverse images of a regular value [4]. For a regular value we choose $(-1,0)$, the horizontal unit vector pointing left.

First, inspection of Figure 8 shows that each $B$ polygon contributes exactly one to the count of inverse images of $(-1,0)$. With notation from the definition of polygonal impossible configuration, the contribution of the $B$-polygons is $q$.

Next we will show that $(***)$ the $n_{i}$-gon $A_{i}$ contributes $n_{i}-2$ to this count. It will follow that the total contribution of the $A$-polygons is $\sum_{i=1}^{p}(n_{i}-2)=N-2p$. Since $N-2p=q$, adding in the contribution of the $B$s gives as total the even number $2q$.

To prove $(***)$ note (Figure 9) that an oriented convex $n$-gon in general position (no horizontal sides) must have one of these three configurations:

1. (1)

${\bf t}_{L}+{\bf t}_{R}+{\bf b}_{L}+{\bf b}_{R}+(n-4)\{{\bf s}_{L},{\bf s}_{R}\}$ 2. (2)

${\bf e}_{L}+{\bf t}_{R}+{\bf b}_{R}+(n-3){\bf s}_{R}$ 3. (3)

${\bf e}_{R}+{\bf t}_{L}+{\bf b}_{L}+(n-3){\bf s}_{L}$.

In each of these cases, the number of inverse images of $-x$ is $n-2$. 2. (2)

The number of self-intersection points of an immersed oriented curve in the plane, counted mod 2, is one less than its rotation number. (Because the self-intersection number mod 2 is a regular homotopy invariant [6], and because any curve with rotation number $n$ is regularly homotopic to $n$ turns of a spiral, with the endpoints joined [5]: a curve with $|n|-1$ intersection points). So a curve with even rotation number must have an odd number of self-intersection points. 3. (3)

Let $\gamma_{1},\dots,\gamma_{k}$ be the $k$ tracks of the path through our configuration. We know that the sum of their winding numbers is even, so an even number $\ell$ of them have odd winding number; these $\ell$ tracks each have even self-intersection number. The other $k-\ell$ tracks have even winding number and therefore odd self-intersection number. The sum of their self-intersection numbers is therefore congruent to $k-\ell$ and therefore to $k$, since $\ell$ is even; it follows that the sum of all the self-intersection numbers of the $\gamma_{i}$ is congruent mod 2 to $k$. 4. (4)

Finally, the self-intersection points of the path through the configuration, drawn as in Figure 8, are of three types: those coming from the self-intersection numbers of $\gamma_{i}$ for $i=1,\dots,k$, those coming from intersections between $\gamma_{i}$ and $\gamma_{j}$ for $i\neq j$, and those coming from the intersections of the descending arms of the $B$-polygons. The second and third types come in pairs. So the total number of self-intersections is congruent mod 2 to $k$, the number of tracks; and this must also hold for the number of those of the first two types, which is the number of vertices of the configuration.