Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 25, 2000, 91–100

INVOLUTIONS AND SIMPLE CLOSED GEODESICS ON RIEMANN SURFACES

Paul Schmutz Schaller

Université de Genève, Section de mathématiques Case postale 240, CH-1211 Genève 24, Switzerland; paul.schmutz@math.unige.ch

Abstract. A new geometric characterization of Riemann surfaces with an orientation preserving involution is given. It is proved that a closed Riemann surface M of genus g ≥ 2 has an involution with exactly k = 2g + 2 − 4j fixed points, 0 ≤ j ≤ \frac{1}{2} g, if and only if M has a set F of 2g simple closed geodesics which all intersect in the same two points A and B (such that among the elements of F there are no further intersection points). Moreover, A and B are fixed points of the involution and F partitions M into 2g hyperbolic quadrilaterals such that exactly k −2 of them are symmetric (opposite sides of a symmetric quadrilateral are contained in the same element of F ).

1. Introduction

Let M be a closed Riemann surface of genus g ≥ 2 equipped with a metric of constant curvature −1. Assume that M has an orientation preserving isometric involution ϕ ≠ id. By the Riemann–Hurwitz relation ϕ has k = 2g + 2 − 4j different fixed points for an integer j with 0 ≤ j ≤ \frac{1}{2}(g + 1); we will always exclude the case that ϕ has no fixed points. Let A and B be two fixed points of ϕ, A ≠ B. Let u be a simple geodesic segment from A to B. Then u ∪ ϕ(u) is a simple closed geodesic of M. In this manner we can construct a maximal set F of simple closed geodesics of M such that (every) two elements of F intersect only in A and B; we shall see that the order of F is always 2g. Let M(F) be the surface obtained by cutting M along all elements of F. It will be showed that M(F) has exactly 2g connected components which all are (hyperbolic) quadrilaterals. The following questions are then natural.

(1) Let M be a closed Riemann surface of genus g and assume that M has a set F of 2g simple closed geodesics such that all elements of F intersect in the same two points A and B and such that there are no further intersection points among the elements of F. Does this imply that M has an orientation preserving involution ϕ such that A and B are among the fixed points of ϕ?

(2) If the answer is yes, in which way the topological properties of F determine the number of fixed points of ϕ?

(3) If the answer to the first question is yes, is the number 2g best possible or is a smaller set F already sufficient in order to determine an involution?

1991 Mathematics Subject Classification: Primary 30F20; Secondary 30F10.

Theorem A. The answer to the first question is yes.

Moreover (this concerns the second question), if k > 0 is the number of fixed points of ϕ and s the number of connected components of M(F) which are symmetric quadrilaterals, then k − 2 = s. (In a symmetric quadrilateral the opposite sides are parts of the same element of F.)

Theorem B. Concerning the third question, the number 2g is best possible. Namely, for every integer g ≥ 2 there exists a closed surface M of genus g with a set F of 2g − 1 elements which does not induce an involution.

The proofs are given in Section 2.

One may ask whether the intersection points of the geodesics in Theorem A and Theorem B are Weierstrass points (for general references on Weierstrass points see [4], [2]). By Lewittes [6], the fixed points of an orientation preserving involution ϕ are ordinary (or 1-fold) Weierstrass points if ϕ has more than four fixed points. If ϕ has four fixed points, then, by Accola [1], the fixed points are at least 2fold Weierstrass points (but, in general, not ordinary Weierstrass points). Finally, if the involution ϕ has only two fixed points, then these fixed points may miss the dense set of q -fold Weierstrass points, q = 1, 2, 3, …, as has been showed by Guerrero [5]. The latter may well be true also for the intersection points of the “counter-examples” in Theorem B.

In the hyperelliptic case we also have the following related result.

Theorem C (Schmutz Schaller). Let M be a closed surface of genus g. Then M is hyperelliptic if and only if M has a set G of at least 2g − 2 simple closed geodesics which all intersect in the same point such that among the elements of G there are no further intersection points.

This result has first been proved in [7] (see also the survey paper [8] and [9]). Note that by sets of simple closed geodesics which all intersect in a unique point (as in Theorem C), other involutions than the hyperelliptic one cannot be characterized.

For some results related to those of this paper see Birman and Series [3].

Acknowledgment. I thank the referee for helpful comments.

2. Geometric characterization of involutions

Definition (i) A surface is a Riemann surface equipped with a metric of constant curvature −1.

(ii) A (g, n)-surface is a surface of genus g with n boundary components which are simple closed geodesics. A closed surface is a compact surface without boundary.

(iii) A simple geodesic is one without selfintersections.

(iv) An involution is an isometry ϕ ≠ id with ϕ^{2} = id (id is the identity).

(v) Let M be a closed surface. A geodesic 2-set F of order k > 0 on M is a set of k different simple closed geodesics of M which all intersect in the same two points, the intersection points of F, such that among the elements of F, there are no further intersection points. Define by M(F) the surface obtained by cutting M along all elements of F.

(vi) Let M be a closed surface and F a geodesic 2-set. Let A and B be the intersection points of F. Let u ∈ F. Then u is separated by A and B into two parts, the segments of u.

Remark and Definition. Let N be (the closure of) a connected component of M(F) (F and M are defined as above). Then the boundary of N consists of a number of simple closed curves which are called boundary components of N; they are considered as disjoint, taking different copies of A and B on each boundary component. If N has genus zero and only one boundary component, then I call N a polygon and treat the vertices of this polygon as different copies of A and B, respectively. The same convention is used in related cases.

Lemma 1. Let M be a closed surface of genus g with an orientation preserving involution ϕ with exactly k fixed points. If g is even, then k ≡ 2 mod(4). If g is odd, then k ≡ 0 mod(4).

Proof. This is a consequence of the Riemann–Hurwitz relation.

Remark. Let M be a closed surface of genus g which has an orientation preserving involution ϕ with fixed points. It then follows by Lemma 1 that ϕ has at least two fixed points. This fact will be used throughout without comment.

Lemma 2. Let M be a closed surface of genus g. Let A and B be two different points on M. Let F be a set of 4g simple geodesic segments starting in A and ending in B which are all mutually disjoint in M \ {A, B}. Then the elements of F cut M into exactly 2g connected components which all are hyperbolic quadrilaterals.

Proof. Let F′ ⊃ F such that all elements of F′ are simple geodesic segments starting in A and ending in B and such that all elements of F′ are mutually disjoint in M \ {A, B}. Assume further that F′ is maximal with respect to these conditions. Let M(F′) be the surface obtained by cutting M along all elements of F′. Define M(F) analogously.

Let N be a connected component of M(F′). Then each boundary component of N contains an even number of geodesic segments. Assume that N has two different boundary components b_{1} and b_{2} . Let A_{1} be a copy of A on b_{1} and let B_{2} be a copy of B on b_{2} . Since N is convex, N contains a simple geodesic segment v_{N} from A_{1} to B_{2} . Since v(N) is not in F′, this contradicts the maximality of F′.

Assume now that N has only one boundary component b and that N has genus g(N) > 0. Then N has a simple geodesic segment w_{N} ⊄ b from a copy of A on b to a copy of B on b. This again contradicts the maximality of F′.

Assume finally that N has only one boundary component b and that g(N) = 0. It follows that N is a polygon. Assume that b has at least six vertices (recall that the number of vertices must be even). Then b contains a copy A_{1} of A and a copy B_{1} of B such that N has a simple geodesic segment t(N) from A_{1} to B_{1} , t(N) ⊄ b. This contradicts the maximality of F′.

We therefore have proved that each connected component of M(F′) is a quadrilateral (note that a connected component of M(F′) cannot be a polygon with two sides).

Assume that M(F′) has q connected components Q_{1}, Q_{2}, …, Q_{q}. Let S be the sum of all (inner) angles of the quadrilaterals Q_{i}, i = 1, …, q. Then S = 4π since all vertices of Q_{i} are copies of A and B. We obtain

∑_{i=1}^{q} vol(Q_{i}) = 2qπ − 4π = vol(M) = 4(g − 1)π

where vol is the (hyperbolic) volume. This implies that q = 2g. Therefore, there are 8g segments as sides of the quadrilaterals Q_{i}, i = 1, …, 2g. It follows that F′ has order 4g. This proves F = F′ and hence the lemma.

Corollary 1. Let M be a closed surface of genus g. Let F be a geodesic 2-set of M of order 2g. Then M(F) has 2g connected components which all are hyperbolic quadrilaterals.

Proof. Clear by Lemma 2.

Corollary 2. Let M be a closed surface of genus g which has an orientation preserving involution ϕ with fixed points. Let A and B be fixed points of ϕ, A ≠ B. Then M has a geodesic 2-set of order 2g with fixed points A and B.

Proof. Let F be a maximal geodesic 2-set on M with intersection points A and B. It is clear that F is not empty since M has a simple geodesic segment u_{1} from A to B which implies that u = u_{1} ∪ ϕ(u_{1}) is a simple closed geodesic passing through A and B. Let N be a connected component of M(F). Assume that N has a simple geodesic segment v starting in a copy of A (on a boundary component of N) and ending in a copy of B (on a boundary component of N) such that v is not a segment of an element of F. Then v ∪ ϕ(v) is a simple closed geodesic, v ∉ F, and F ∪{v} is a geodesic 2-set. This contradicts the maximality of F. It therefore follows analogously as in the proof of Lemma 2 that N must be a quadrilateral, that the number of connected components of M(F) is 2g, and that the order of F is 2g.

Definition. Let F be a geodesic 2-set of order 2g in a closed surface M of genus g.

(i) A quadrilateral of F is a connected component of M(F).

(ii) Let Q be a quadrilateral of F. Then the sides s_{i} (in the natural order) of Q are segments of elements u_{i} of F, i = 1, 2, 3, 4. If u_{1} = u_{3} and u_{2} = u_{4} , then Q is called symmetric.

Lemma 3. Let F be a geodesic 2-set of order 2g in a closed surface M of genus g. Let Q be a quadrilateral of F. Let u_{1}, …, u_{4} be the elements of F which form the boundary of Q. Then there is a quadrilateral Q′ of F such that u_{i}, i = 1, …, 4, form the boundary of Q′ and Q ≠ Q′ if and only if Q is not symmetric. Moreover, Q′ has the same inner angles as Q.

Figure 1. The quadrilaterals Q and Q′.

Proof. (i) We may assume that the notation is such that the segments s_{i} ⊂ u_{i} which form the boundary of Q, appear in the natural order, counter-clockwise say (compare Figure 1). Let α_{i} be the directed angle from s_{i} to s_{i+1} , i = 1, …, 4 (taking the indices modulo 4), such that α_{i} is an inner angle of Q. Then all four angles α_{i}, i = 1, …, 4, are measured clockwise. Let A and B be the intersection points of F where the notation is such that A is a vertex in Q between s_{1} and s_{2} as well as between s_{3} and s_{4} .

(ii) For each i ∈ {1, 2, 3, 4} let t_{i} ⊂ u_{i} be the segment of u_{i} which is different from s_{i}. Denote by β_{i} the directed angle from t_{i} to t_{i+1} , measured clockwise, i = 1, …, 4. Then α_{i} = β_{i}, i = 1, …, 4. It follows that there is a quadrilateral Q′ of F, containing t_{1} and t_{2} as sides, such that β_{1} is an inner angle of Q′. Since the vertex between t_{1} and t_{2} in Q′ is a copy of A, t_{2} will end in a copy B_{1} of B. In order to obtain the other side of Q′ ending in B_{1} , we have to turn clockwise around B from u_{2} to the next element of F. But this must be u_{3} (by the existence of Q), more precisely, this third side of Q′ is t_{3} . The same argument proves that t_{4} is the fourth side of Q′. It also follows that β_{1}, …, β_{4} are the inner angles of Q′, therefore, Q′ and Q have the same inner angles.

(iii) By the existence of Q, u_{2} is the next element of F when we turn clockwise around A from u_{1} , and u_{4} is the next element of F when we turn clockwise around A from u_{3} . It follows that u_{1} = u_{3} if and only if u_{2} = u_{4} . Assume that Q is symmetric. It follows by (ii) that t_{i} and s_{i+2} is the same segment of u_{i}, i = 1, …, 4 (taking the indices modulo 4), which shows that Q = Q′.

On the other hand, if Q = Q′ then t_{i} must be a side of Q, i = 1, …, 4. Since u_{i} is simple, it follows that t_{i} equals s_{i+2} , i = 1, …, 4, and hence u_{1} = u_{3} and u_{2} = u_{4} .

Corollary 3. Let F be a geodesic 2-set of order 2g in a closed surface M of genus g. Let Q be a quadrilateral of F, let u_{1}, …, u_{4} be the elements of F which form the boundary of Q. Then either all elements u_{i}, i = 1, 2, 3, 4, are different or Q is symmetric. In the latter case, M has an embedded (1, 1)-surface S(Q) which contains Q.

Proof. It was already shown during the proof of Lemma 3 that either all four elements u_{i}, i = 1, 2, 3, 4, are different or Q is symmetric. Assume that Q is symmetric. Let the notation be such that u_{1} = u_{3} and u_{2} = u_{4} . Let s_{2} and s_{4} be the segments of u_{2} . Cut M along u_{1} yielding a (g − 1, 2)-surface M′; denote the boundary geodesics of M′ by v_{1} and w_{1} . It then follows that M′ contains a unique simple closed geodesic z and an embedded (0, 3)-surface Y with boundary geodesics z, v_{1}, w_{1} such that s_{2} ⊂ Y (in M, the subsurface Y is an embedded (1, 1)-surface). Since s_{4} is freely homotopic to s_{2} , it follows that s_{4} is contained in Y .

Corollary 4. Let M be a closed surface M of genus g which has an orientation preserving involution ϕ with k > 0 fixed points. Let A and B be fixed points of ϕ. Let F be a geodesic 2-set of order 2g with intersection points A and B. Then among the 2g quadrilaterals of F, there are exactly k − 2 which are symmetric.

Proof. Let s be the number of symmetric quadrilaterals of F.

If k > 2, then ϕ has a fixed point C ∉ {A, B}. C lies in the interior of a quadrilateral Q_{C} of F (C cannot lie on an element of F since A and B already lie on each element of F). It follows that ϕ(Q_{C}) = Q_{C} which implies that Q_{C} is symmetric. This proves k − 2 ≤ s.

On the other hand, let Q be a quadrilateral of F which is symmetric. By Corollary 3, M has an embedded (1, 1)-surface S(Q) which contains Q. To Q correspond two elements u_{1} and u_{2} of F which lie in S(Q). Every (1, 1)-surface S has a (hyperelliptic) involution ψ with three fixed points, and if v and w are two simple closed geodesics of S which intersect twice, then both intersection points are among the fixed points of ψ. It follows that the hyperelliptic involution ψ of S(Q) is the restriction of ϕ and therefore, ϕ has a third fixed point in S(Q) which lies in the interior of Q. This proves s ≤ k − 2.

Definition. Let F be a geodesic 2-set of order 2g in a closed surface M of genus g. Let u ∈ F. Then u is called symmetric if the two segments of u have equal length.

Corollary 5. Let F be a geodesic 2-set of order 2g in a closed surface M of genus g. Let Q be a quadrilateral of F which is symmetric. Let u ∈ F such that the segments of u are sides of Q. Then u is symmetric.

Proof. By Corollary 3, Q is contained in an embedded (1, 1)-surface S(Q) of M. As already noted in the proof of Corollary 4, S(Q) has a hyperelliptic involution ψ and the intersection points of F are among the fixed points of ψ. This proves the corollary.

Lemma 4. Let Q_{1} and Q_{2} be quadrilaterals with sides a_{i}, b_{i}, c_{i}, d_{i}, i = 1, 2 (in the natural order). Let Q_{1} and Q_{2} have the same inner angles (the angle between a_{1} and b_{1} equals the angle between a_{2} and b_{2} , and so on). Then

(i) L(a_{1}) = L(a_{2}) if and only if Q_{1} and Q_{2} are isometric, and

(ii) L(a_{1}) > L(a_{2}) ⟺ L(b_{1}) < L(b_{2})

(where L(x) is the length of x).

Proof. (i) is obvious by hyperbolic trigonometry so assume that L(a_{1}) > L(a_{2}). It then follows by (i) that L(b_{1}) ≠ L(b_{2}). Assume that L(b_{1}) > L(b_{2}). Let R_{i} be the vertex of Q_{i} between a_{i} and b_{i}, i = 1, 2. In the hyperbolic plane place Q_{2} on Q_{1} such that R_{1} = R_{2} and such that a_{2} ⊂ a_{1} and b_{2} ⊂ b_{1} . Then c_{1} and c_{2} cannot intersect (since Q_{1} and Q_{2} have the same angles). By the same argument also d_{1} and d_{2} cannot intersect. It follows that Q_{2} ⊂ Q_{1} . But since Q_{1} and Q_{2} have the same volume, this yields a contradiction.

Corollary 6. Let F be a geodesic 2-set of order 2g in a closed surface M of genus g. Let Q be a quadrilateral of F which is not symmetric. Let u_{1}, …, u_{4} be the four elements of F which form the boundary of Q. Then either all u_{i}, i = 1, …, 4, are symmetric or none.

Proof. Let Q′ be defined as in Lemma 3. Assume that one of the u_{i} is symmetric. Since Q and Q′ have different segments, it follows by Lemma 4(i) that Q and Q′ are isometric and therefore, all u_{i} are symmetric.

Theorem 1. Let M be a closed surface of genus g. Then M has a geodesic 2-set F of order 2g if and only if M has an orientation preserving involution with fixed points.

Proof. One direction has already been proved by Corollary 2. Assume now that M has a geodesic 2-set F of order 2g with intersection points A and B. Assume that F has a symmetric quadrilateral. It then follows by Corollary 5 and Corollary 6 (and by the fact that all elements of F intersect in A) that every element of F is symmetric.

Let ϕ be the π -rotation around A. It follows that, by ϕ, the quadrilaterals of F are mapped into quadrilaterals of F. More precisely, ϕ(Q) = Q if Q is symmetric (by the proof of Corollary 4) and ϕ(Q) = Q′ if Q is not symmetric where Q′ is defined as in Lemma 3. It follows that ϕ is an involution of M.

We therefore can assume that none of the quadrilaterals of F is symmetric. Denote the elements of F by u_{1}, …, u_{2g} such that the u_{i} lie in the natural order around A. Denote the segments of u_{i} by v_{i} and v_{i+2g}, i = 1, …, 2g, such that the v_{j} lie in the natural order around A (j = 1, …, 4g). If an element of F is symmetric, then all elements of F are symmetric by Corollary 6. Assume that the elements of F are not symmetric and that v_{1} > v_{2g+1} . It then follows by Lemma 4 that v_{2} < v_{2g+2} . The same argument shows that v_{3} > v_{2g+3} . By repeating this argument we obtain that v_{2g} < v_{4g} and hence v_{2g+1} > v_{1} , a contradiction. We have therefore proved that all elements of F are symmetric. This implies that the π -rotation around A is an involution of M.

Theorem 2. For every integer g ≥ 2 there exists a closed surface M of genus g which has a geodesic 2-set F of order 2g − 1 with intersection points A and B, but no involution such that A and B are among the fixed points of M.

Proof. Let ε > 0 be small. For 1 ≤ t < 2, let T (t) be a (hyperbolic) triangle with (inner) angles

α(t) = \frac{tπ}{4(2g − 1)} − ε,  β(t) = \frac{tπ}{4(2g − 1)} + ε,  γ = \frac{π}{2g − 1}.

Denote by A, B, C the vertices of T (t) and by a, b, c the sides of T (t) (with the usual convention of notation: a is opposite to A and to α(t), and so on). Take 4g − 2 copies of T (t) and glue them along a or along b such that the vertex C is the same for all 4g − 2 copies. We obtain a 4g − 2-gon P (t) where all sides have the length L(c) and where 2g − 1 angles are 2α(t) and 2g − 1 angles are 2β(t). Denote the sides of P (t) by c_{i}, i = 1, …, 4g − 2, in the natural order.

(i) Assume now that g is odd. Let t = 1. Let S be a triangle of (hyperbolic) area \frac{1}{2}π such that two sides x and y of S have the same length while the third side z of S has length L(c). It is clear that S exists and is unique up to isometry. Glue a copy S_{1} (with sides x_{1}, y_{1}, z_{1} ) of S along z_{1} and along c_{1} of P (t) such that the interior of P (t) is not intersected by S_{1} . Glue a copy S_{2} (with sides x_{2}, y_{2}, z_{2} ) of S along z_{2} and along c_{2g} of P (t) such that the interior of P (t) is not intersected by S_{2} and such that x_{2} is opposite to x_{1} and y_{2} is opposite to y_{1} (the orientation of S_{1} and S_{2} is the same). Thereby P (t) has been enlarged to a 4g -gon R(t). By construction, the area of R(t) is 4π(g − 1). R(t) is the fundamental domain of a closed surface M(t) of genus g and we obtain M(t) by the following identifications of the sides of R(t) (the identification is symbolized by a +).

x_{1} + x_{2} , y_{1} + y_{2} ,

c_{4m−2} + c_{4m} (m = 1, …, \frac{1}{2}(g − 1)),
c_{4m} + c_{4m+2} (m = \frac{1}{2}(g + 1), …, g − 1),
c_{4m−1} + c_{4m+1} (m = 1, …, g − 1).

Let T be one of the copies of T (t) in R(t). By construction, there is a copy T′ of T (t) in R(t) such that the side b_{1} of T′ is the prolongation of the side a_{1} of T. Let u_{1} = a_{1} ∪ b_{1} . It is then easy to verify that u_{1} is a simple closed geodesic in M(t). Since we have 4g − 2 copies of T (t) we obtain 2g − 1 simple closed geodesics u_{i}, i = 1, …, 2g − 1, in M(t) which all intersect in C and in V where V corresponds to the 4g vertices of R(t) (which all are identified in M(t)). Therefore, {u_{1}, …, u_{2g−1}} is a geodesic 2-set of order 2g − 1. Since L(a) < L(b), M(t) has not an involution with fixed points C and V .

(ii) Assume now that g is even. Let W be a triangle with three sides of equal length L(c). Denote by δ an (inner) angle of W. Glue a copy W_{1} of W along c_{1} and glue a copy W_{2} of W along c_{2g} (such that the interior of P (t) is not intersected by W_{i}, i = 1, 2). Thereby, P (t) has been enlarged to a 4g -gon R(t). Denote the new sides of R(t) by x_{1}, y_{1} (coming from W_{1} ) and by x_{2}, y_{2} (coming from W_{2} ) such that x_{1} is a neighbour of c_{4g−2} and x_{2} is a neighbour of c_{2g−1} .

Let us now assume that t is chosen such that the area of R(t) is

(1)  (4g − 2)π − 6δ − (4g − 2)(α(t) + β(t)) = 4π(g − 1).

R(t) is then the fundamental domain of a closed surface M(t) of genus g and we obtain M(t) by the following identifications of the sides of R(t).

x_{1} + c_{4g−3} , y_{1} + c_{3} , x_{2} + c_{2g−2} , y_{2} + c_{2g+2} , and, if g ≠ 2,

c_{4m−2} + c_{4m} (m = 1, …, \frac{1}{2}(g − 2)), c_{4m} + c_{4m+2} (m = \frac{1}{2}(g + 2), …, g − 1), c_{4m+1} + c_{4m+3} (m = 1, …, g − 2). It is now easy to see (as above in (i)) that M(t) has a geodesic 2-set of order

2g − 1 with intersection points C and V , but C cannot be a fixed point of an involution since L(a) < L(b).

It remains to show that (1) is possible. Note first that when (1) holds, then

(2)  δ = \frac{1}{6}π(2 − t).

Let t = 1. It then follows (by a calculation) that cosh(L(c)) > 21 (if ε is small enough) which yields cos δ < 21/22 and δ is too small (by (2) δ should equal π/6). Let now t ⟶ 2. Then L(c) is shorter than in the case t = 1 and therefore, δ becomes larger than in the case t = 1. But now δ is too large since, by (2), δ should tend to zero. This proves that (1) is possible.

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100 Paul Schmutz Schaller

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Received 7 April 1998