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Geometric subgroups of mapping class groups

This paper is a study of the subgroups of the mapping class groups of Riemann surfaces, called"geometric"subgroups, corresponding to the inclusion of subsurfaces. Our analysis includes surfaces with boundary and with punctures. The centres of all the mapping class groups are calculated. We determine the kernel of inclusion-induced maps of the mapping class group of a subsurface, and give necessary and sufficient conditions for injectivity. In the injective case, we show that the commensurability class of a geometric subgroup completely determines up to isotopy the defining subsurface, and we characterize centralizers, normalizers, and commensurators of geometric subgroups.


Introduction
Throughout the paper, M will denote a compact, connected, oriented surface. The boundary ∂M , if nonempty, is a finite collection of simple closed curves. Consider a finite subset P = {p 1 , . . . , p m }, of m distinct points (often called "punctures" or "marked points") in the interior of M . Define H(M, P ) to be the group of orientation-preserving homeomorphisms h : M → M such that h is the identity on each boundary component of M and h(P ) = P . Our main object of study is the mapping class group M(M, P ) = π 0 (H(M, P )), the set of isotopy classes of these mappings, with composition as the group operation. We emphasize that throughout an isotopy, the boundary, and also the points P remain fixed. It is clear that, up to isomorphism, these groups do not depend on the choice of P , but depend only on the cardinality m = |P |, so we may write (M, m) or M(M, m) in place of (M, P ) or M(M, P ), and simply M(M ) for M(M, ∅). M(M, P ) may equivalently be considered as the group of orientation-preserving diffeomorphisms of (M, P ), up to smooth isotopy. We refer the reader to the survey articles [2], [14], [15], [30] and [31] and their bibliographies for more information.
Let N ⊂ M be a subsurface, by which we mean a closed subset which is also a surface and for which we always assume the further properties: (1)  Our study of these subgroups depends on a careful analysis of curves in M and Dehn twists, which are the subject of Section 3. The mapping i * is often, but not always, injective. We determine its kernel in Section 4. Section 5 is devoted to the centres of mapping class groups. These are certainly well-known to specialists and many of them can be found in the literature (see [15] and [17]). However, we need the general result for the remainder of the paper and the proofs are straightforward applications of Section 3. Our main result is that, assuming injectivity of the i * , up to a finite number of exceptions, two geometric subgroups are commensurable if and only if they are equal if and only if their respective defining subsurfaces are isotopic (Theorem 6.5). Note that the assumption that ∂N lies entirely in the interior of M is necessary for the conclusion of Theorem 6.5; indeed, without this assumption, it is very easy to construct non-isotopic subsurfaces N and N ′ which define the same geometric subgroup, N satisfying ∂N ∩ ∂M = ∅ and N ′ satisfying ∂N ′ ∩ ∂M = ∅. From the main result, still assuming injectivity of the i * , we characterize the commensurator, the normalizer and the centralizer of a geometric subgroup in M(M, P ).
The above examples are essentially the only surface mapping class groups which are trivial.
The mapping class group of the annulus S 1 × I is generated by a Dehn twist, described in Section 3.
Under the isomorphism, σ j , 0 < j < m corresponds with the (class of the) diffeomorphism consisting of a " half-twist" interchanging the integers j and j + 1, and supported on a small neighborhood of the interval [j, j + 1] ⊂ C. See [1] and [7] for details, but beware some differences in choice of conventions. It is classical, but nontrivial, that for n < m the homomorphism B n → B m , taking σ j ∈ B n to σ j ∈ B m is injective, allowing us to write B n ⊂ B m . We chose notation so that under the isomorphisms, B n → B m corresponds to the inclusion-induced mapping i * : M(D n , P n ) → M(D m , P m ). We conclude that in this case i * is injective. Note that the closure of the complementary subsurface is an annulus with m − n punctures. The commensurator, the normalizer and the centralizer of M(D n , P n ) in M(D m , P m ) are characterized in [8] and [35]. (6) The following is an example of an inclusion map which is not injective on mapping class groups. Take M = S 2 a 2-sphere with P = 2 points in S 2 , and let D be a disk in S 2 which encloses the points P . We have the map i * : M(D, P ) → M(S 2 , P ). As already discussed, M(D, P ) is the braid group B 2 , which is infinite cyclic, generated by σ 1 . However, σ 2 1 is isotopic with a Dehn twist along ∂D. In the larger surface M = S 2 this twist is isotopic with the identity (rel P ). So the kernel of i * in this case is the infinite cyclic subgroup of index 2 in M(D, P ). M(S 2 , 2) is cyclic of order 2.
(7) For the torus T 2 = S 1 × S 1 with either zero or one puncture, the mapping class group is the modular group of invertible 2 × 2 matrices with integer entries: M(T 2 ) ∼ = M(T 2 , 1) ∼ = SL(2, Z). Dehn twists along the curves S 1 × * and * × S 1 generate the mapping class group. Note that if A is an annulus neighborhood of one of these curves, and happens to enclose the puncture of M(T 2 , 1), the map i * : M(A, 1) → M(T 2 , 1) fails to be injective.

Subgroups of mapping class groups
In this section we review some of the literature regarding subgroups of mapping class groups. Although there is little published on the geometric subgroups, which are the main concern of the present paper, certain other subgroups are quite well understood. First, we recall some general properties of the mapping class groups themselves. A more complete survey can be found in [30].
According to Grossman [10] and Ivanov [15], M(M g , m) is residually finitefor every nontrivial element, there is a homomorphism of the mapping class group onto a finite group which does not kill that element. This implies that M(M g , m) is Hopfian ( [24], Chapter IV) -every epimorphism M(M g , m) → M(M g , m) is an isomorphism. Conversely, in a recent paper [17] Ivanov and McCarthy proved that Although, it contains torsion elements, M(M g ) has a finite index subgroup which is torsion free (see [30] for a sketch of a proof.) It has recently been shown by Mosher [32] that the mapping class groups are automatic. This implies that the word problem is solvable (in quadratic time) and many other consequences [6].
The outer automorphism group of M(M g , m) has been determined by Ivanov [16] and McCarthy [28]. It is equal to Z/2Z under the assumptions g = 0, m ≥ 3 if g = 1, and m ≥ 1 if g = 2.
The abelianization of M(M g ) is cyclic of order 12, when g = 1, cyclic of order 10 when g = 2, and trivial (that is, M(M g ) is perfect) for g > 2 [34].
Finite subgroups of M(M g ) have been extensively studied. The so-called Nielsen realization problem [38] was solved by Kerckhoff [22]. It asserts that for any finite subgroup F of M(M g ), there is a complex structure on M g such that F is realized as a group of holomorphic automorphisms of M g . According to a classical result of Hurwitz [12], the orders of finite subgroups are bounded: McCarthy [27] showed that subgroups of M(M g ) satisfy the Tits alternative: every subgroup either contains a free group on two generators, or a solvable subgroup of finite index. Birman, Lubotzky and McCarthy [3] proved that solvable subgroups of M(M g ) are virtually abelian, and gave upper bounds for the rank of free abelian subgroups. Ivanov [15] proved these two results for M(M g , m) and showed that: if G is a subgroup of M(M g , m) which is not virtually abelian, then G contains an uncountably infinite number of maximal subgroups of infinite index. The Frattini subgroup φ(G) of a group G is the intersection of all its maximal subgroups. Ivanov also proved that the Frattini subgroup of a finitely generated subgroup of M(M g , m) is nilpotent. Note that this property also holds for finitely generated linear groups [33].
The centres of the M(M g , m) are well-known [15], [17]: cyclic of order two if g = 1 and m ≤ 2, and if g = 2 and m = 0, and trivial otherwise. We determine the centres in the more general situation, with boundary, in the present paper.
The Torelli subgroup of the mapping class group consists of classes of mappings which induce the identity on the homology of the surface. These subgroups have been studied extensively in the series of papers [18], [19], [20], [21]. In particular, they describe a finite set of generators for the Torelli groups. The cohomological properties of mapping class groups and the Torelli groups have been studied intensively and are also well-described in [30] and [31]. We shall not use any of these deeper properties -indeed our methods are quite elementary and self-contained, requiring as background only some basic properties of curves on surfaces due to Epstein [5].

Curves and Dehn twists
Working within the context of a surface M with punctures P as described above, we shall consider a simple closed curve in M \ P as an embedding c : S 1 → M \ P which does not intersect the boundary of M . Note that c has an orientation; the curve with the opposite orientation, but same image will be denoted c −1 . By abuse of notation, we also use the symbol c to denote the image of c. We will say that c is essential if it does not bound a disk in M disjoint from P , and that c is generic if it does not bound a disk in M containing 0 or 1 point of P .
Two simple closed curves a, b are isotopic if there exists a continuous family h t ∈ H(M, P ), t ∈ [0, 1] of homeomorphisms such that h 0 is the identity and h 1 • a = b. Isotopy of curves is an equivalence relation which we denote by a ≃ b. Following [7] the index of intersection of two simple closed curves a and b is: We note that: 1) I(a, b) = inf{|a ′ ∩ b|; a ′ ≃ a}; 2) If a is not generic, then I(a, b) = 0 for every simple closed curve b; 3) If a ≃ b then I(a, b) = 0.
A bigon cobounded by two simple closed curves a and b in M \ P is a disk D ⊂ M \ P whose boundary is the union of an arc of a and an arc of b.
Proposition 3.1 (Epstein [5]) Let a, b : S 1 → M \ P be two essential simple closed curves, and suppose a is isotopic to b.
i) If a∩b = ∅, then there exists an annulus in M \P whose boundary components are a and b.
ii) If a ∩ b = ∅, and they intersect transversely, then a and b cobound a bigon.
Proposition 3.2 Let a, b : S 1 → M \ P be two essential simple closed curves, which intersect transversely. Then if and only if a and b do not cobound a bigon.
Proof: It is clear that if a and b cobound a bigon, one can isotop one of the curves across the bigon and reduce the cardinality of the intersection by two. Now suppose they do not cobound a bigon, and choose a simple closed curve a ′ isotopic to a and transverse to a such that |a ′ ∩ b| = I(a, b).
We will argue by induction on |a ′ ∩ a| that |a ∩ b| = |a ′ ∩ b| = I(a, b).
If |a ′ ∩ a| = 0, then by Proposition 3.1, there is an annulus in M \ P with boundary components a and a ′ . Each arc of intersection of b with the annulus must run from one boundary component to the other (see Figure 1), since neither a, b nor a ′ , b cobound bigons. Therefore |a ∩ b| = |a ′ ∩ b| = I(a, b). Now suppose |a ′ ∩ a| > 0. By Proposition 3.1, a and a ′ cobound a bigon. We may assume a ′ ∩ a ∩ b is empty, so any arc of intersection of b with the bigon must have one endpoint in a and the other in a ′ (see Figure 1), again because neither a, b nor a ′ , b cobound bigons. Therefore we may push a ′ across the bigon to obtain a new curve a ′′ isotopic to a and satisfying: |a ′′ ∩ a| = |a ′ ∩ a| − 2 and |a ′′ ∩ b| = |a ′ ∩ b| = I(a, b).   Figure 1: Curves cobounding an annulus and a bigon.
Definition: If we parametrize S 1 as the unit circle in the complex plane, and the interval I = [0, 1], then the prototype Dehn twist τ : S 1 × I → S 1 × I is given by Note that τ is the identity on the boundary circles. More generally, let a : S 1 → M \ P be a simple closed curve, and let N ⊂ M \ P be an annulus regular neighborhood of the image of a, parametrized byã : S 1 × I → N . Define the Dehn twist along a to be (the isotopy class of) the homeomorphism A(x) =ãτã −1 (x) for x ∈ N, A(x) = x for x outside N .
We will use the convention throughout that a curve is denoted by a lower case letter and a Dehn twist along the curve is denoted by the corresponding capital letter. Note that, depending on parametrization chosen, there are two choices of Dehn twist along a, inverse to each other. Usually, the choice is immaterial, provided one is consistent throughout, but we will adopt the convention in illustrations that a curve crossing a will make a "right turn" at each encounter with a, after being acted on by A (see Figure 2). We also observe: 1) The Dehn twist along a −1 coincides with the Dehn twist along a.
2) The curve a is fixed by the Dehn twist A.   Proof: This is a special case of a formula in [7]. We outline a proof, leaving details to the reader. Assume |a ∩ b| = I(a, b). The cases n = 0 or I(a, b) = 0 being trivial, suppose they are nonzero. Construct the curve A n (b), which can be seen to cross b exactly |n|I(a, b) times at each point of intersection of a with b (see Figure 3). The proof is completed by noting that this is the minimal intersection of A n (b) with b, up to isotopy. For otherwise, by Proposition 3.2 there would be a bigon cobounded by them. One can see this would imply that a and b also cobound a bigon, which is impossible, again by Proposition 3.2.  Case 1, j = k: Then c 1 and c 2 are different components of the boundary of the connected surface N j . There is an arcb in N j with one endpoint in c 1 , the other endpoint being the point of c 2 having same projection in a i , and the interior ofb in the interior of N j and avoiding the puncture set P (see Figure 4). Then take b = ρ(b). Clearly a j ∩ b = ∅ if i = j; moreover a i and b are transverse and they cannot cobound a bigon, so |a i ∩ b| = 1 = I(a i , b). Case 2, j = k: Then c 1 is a boundary component of N j , and by hypothesis, if N j is a disk, then N j ∩ P contains at least two points, and if N j is an annulus, N j ∩ P contains at least one point. So in any case, there exists an arc b 1 in N j with both endpoints in c 1 , interior in the interior of N j and disjoint from P , and such that b 1 and c 1 do not cobound a bigon in N j \ P ∩ N j (see Figure 5). In the same way we choose an arc b 2 in N k , whose endpoints in c 2 project to ρ(b 1 ∩ c 1 ) in a i , whose interior is interior to N k \ P ∩ N k , and such that b 2 and c 2 do not cobound a bigon in and |a i ∩ b| = I(a i , b) = 2, because we have arranged that a i and b do not cobound a bigon in M \ P . Now consider a subsurface N ⊂ M ; recall this includes the assumption that P ∩ ∂N = ∅ and ∂N is interior to M . We will say that N is essential if each component of M \ N which is a disk has nonempty intersection with the puncture set P.
A component N ′ of M \ N will be called an exterior cylinder if N ′ is a cylinder (= annulus) disjoint from P , with both components of ∂N ′ also being components of ∂N (see Figure 6).  Proof: The only nontrivial part is to show that if a and b are isotopic in M \ P , then they are isotopic in N \ N ∩ P . We assume a and b intersect transversely and argue by induction on |a ∩ b|.
If |a ∩ b| = 0, then Proposition 3.1 implies there exists an annulus in M \ P with (unoriented) boundary a ∪ b. Since a is not isotopic to a boundary component of an exterior cylinder and N is essential, the annulus is disjoint from ∂N , and therefore it lies in N \ N ∩ P . It follows that a and b are isotopic in N \ N ∩ P .
If |a ∩ b| > 0, then a and b cobound a bigon in M \ P , by Proposition 3.1. Because N is essential, the bigon is disjoint from ∂N , and therefore it lies in N . Pushing across this bigon defines an isotopy in Proposition 3.6 Consider two generic simple closed curves a, b : S 1 → M \ P , and let A and B, respectively, denote Dehn twists along these curves. If j and k are integers, Remark: This proposition and the next one can be found in [17] for ∂M = ∅.
The general case needs an extra argument to consider Dehn twists along boundary curves.
Proof: We will argue that if a is not isotopic with b ±1 then A j = B k . It may be assumed that a and b meet transversely and that |a ∩ b| = I(a, b). First assume I(a, b) > 0. Then, by Proposition 3.3, Now suppose that I(a, b) = 0. If M has nonempty boundary, consider the larger closed surfaceM obtained by gluing a torus minus a disk to each boundary component of M (if ∂M = ∅, letM = M ). By Proposition 3.5 a is not isotopic with b ±1 inM \ P . By Proposition 3.4 there is a simple closed curve c inM \ P such that b ∩ c = ∅ and |a ∩ c| = I(a, c) > 0. Then Proposition 3.7 Consider two generic simple closed curves a, b : S 1 → M \ P , and let A and B, respectively, denote Dehn twists along these curves. If j and k are integers, j = 0 and k = 0, such that A j and B k commute in M(M, P ), then I(a, b) = 0.
Proof: Assuming A j and B k commute, we have where C is the Dehn twist along the curve c = B k (a). By Proposition 3.6 it follows that c is isotopic with a ±1 . Proposition 3.3 implies Remark: An alternative proof of Proposition 3.7 can be deduced from [13], where it is shown that A and B commute if I(a, b) = 0, A and B satisfy the braid relation ABA = BAB if I(a, b) = 1, and A and B generate a free group if I(a, b) ≥ 2.
Proposition 3.8 Suppose a 1 , . . . , a p : S 1 → M \ P are generic simple closed curves which are pairwise disjoint, and no curve a i is isotopic to a j or a −1 defined by h(n 1 , . . . , n p ) = A n 1 1 · · · A np p , where A i is the Dehn twist about a i . Then h is an injective homomorphism.
Proof: Because the curves are disjoint, the Dehn twists commute, and h is a homomorphism. To see it is injective, suppose A n 1 1 · · · A np p is the identity of M(M, P ) for some (n 1 , . . . , n p ). We again employ the trick of considering the closed surfacê M , which is M plus a copy of a torus minus a disk glued to each boundary component. Clearly, each a i : S 1 →M is generic, and by Proposition 3.5 a i is not isotopic to a ±1 j inM \ P , when i = j. Now fix an index i ∈ {1, . . . , p}. Proposition 3.4 supplies a simple closed curve b inM \ P disjoint from a j , i = j, with We calculate, using commutativity of the twists, b∩a j = ∅, i = j, and Proposition 3.3: Proposition 3.10 Let a 1 , . . . , a p , b 1 , . . . , b p : S 1 → M \ P be essential simple closed curves satisfying: Proof: We will use a double induction. First, induction on p. The proposition is obvious if p = 1, so we assume it is true for p − 1 pairs of curves. This means that, replacing each a i by h 1 • a i , we may assume that a i = b i for i = 1, . . . , p − 1. Then we have a p disjoint from a j = b j , j < p and also b p disjoint from a j = b j , j < p, and a p isotopic in M \ P to b p . We will be done if we show that there is a further isotopy taking a p to b p which does not move the curves a j = b j , j < p. Taking a p and b p to be transverse, we will argue by induction on |a p ∩ b p |.
If |a p ∩b p | = 0, then a p and b p cobound an annulus in M \P , by Proposition 3.1. Any simple closed curve in this annulus must be either inessential or parallel to a boundary component, so our hypotheses guarantee that the annulus is disjoint from all the curves a j = b j , j < p. Then there is an isotopy across the annulus taking a p to b p ; the isotopy may be taken to be the identity outside a small neighborhood of the annulus, so the other curves do not move. Suppose |a p ∩ b p | > 0, then by Proposition 3.1, the curves cobound a bigon in M \ P , and we may argue as above that the bigon is disjoint from the other curves. An isotopy taking a p across the bigon, fixed outside a neighborhood of the bigon, reduces the number |a p ∩ b p | and does not move the other curves. The inductive hypothesis now gives a final isotopy taking a p to b p .

Subsurfaces and injectivity
Define the pure mapping class group of M relative to P to be the subgroup PM(M, P ) of M(M, P ) consisting of all (classes of) diffeomorphisms which fix P pointwise. Letting Σ P denote the group of permutations of the set P , we have the exact sequence Definitions: A pair equivalent to (D 2 , 2) is called a pantalon of type I (see Figure  7). We have already observed that M(D 2 , 2) is infinite cyclic, generated by a "half-twist" σ which interchanges the two punctures. PM(D 2 , 2) is the subgroup generated by σ 2 , which is (up to isotopy) the same as a Dehn twist along the boundary.
A pair (S 1 × I, 1) is a pantalon of type II (see Figure 7). The mapping class groups M(S 1 × I, 1) and PM(S 1 × I, 1) coincide, and are isomorphic with Z 2 , generated by the Dehn twists along the two boundary components.
A pantalon of type III is a connected planar surface M with three boundary components, the puncture set P is taken to be empty (see Figure 7). Its mapping class group is M(M, 0) ∼ = Z 3 , generated by Dehn twists along the three boundary curves.
Type III Type I Type II Figure 7: The three types of pantalons.
We now define one of our basic tools, a pantalon decomposition of a pointed surface (M, P ) (see Figure 8). This consists of: 1) a collection (M i , P i ), i = 1, · · · , r, each pair being a pantalon of one of the three types, together with maps φ i : 2) a collection a 1 , · · · , a p of simple closed curves in M , disjoint from each other, from P and from ∂M satisfying the following.
Informally, we say that a 1 , · · · , a p determine a pantalon decomposition of (M, P ), cutting M open into pantalons (M i , P i ). It is straightforward to check that every connected compact orientable surface M with punctures P admits a pantalon decomposition, with the following exceptions: i) If N is a disk and |N ∩ P | ≤ 1 then M(N, N ∩ P ) is trivial, and therefore i * is injective.
ii) Suppose N is an annulus and N ∩ P = ∅. If N has an exterior pointed disk, then the kernel of i * is M(N, N ∩ P ); otherwise i * is injective.
iii) Assuming that (N, N ∩ P ) is not as in (i) or (ii), let a 1 , . . . , a r denote the boundary components of N which bound pointed disks exterior to N , and let b j , b ′ j , j = 1, . . . , s be the pairs of boundary components of N which cobound exterior cylinders (disjoint from P ) (see Figure 9). Denote by A i , B j , B ′ j the Dehn twists corresponding to the curves a i , b j , b ′ j , respectively. Then the kernel of i * is generated by Figure 9: Subsurface N and pantalon decomposition.
We have the following commutative diagram, with exact rows.
Let c 1 , . . . , c t denote the components of ∂N different from the a i and b j , b ′ j . In addition, let d 1 , . . . , d u : S 1 → N \ P be simple closed curves which determine a pantalon decomposition of (N, N ∩ P ) (see Figure 9). Note that all the curves we are considering are pairwise disjoint and non-isotopic. Since h is isotopic to the By Proposition 3.10 we may suppose that h • d i = d i for all i = 1, . . . , u, and that h is the identity on the boundary of each pantalon. Using the structure of the pure mapping class groups of pantalons we conclude that Conversely, it is clear that any [h] of this form is in the kernel of i * . Finally, Proposition 3.8 implies that Ker(i * ) is isomorphic to Z r+s .

Centres
We begin this section by considering a special mapping of the surface M of genus one, and with one boundary component; that is, M is a torus minus a disk, with empty puncture set P . We model M as a certain identification space of a planar surface, as follows (see Figure 10). Let is a pantalon (of type III) with boundary curves a 1 , a 2 , c : S 1 → ∂M which we parametrize as follows, 0 ≤ θ ≤ 2π: We consider M = (D \ (D 1 ∪ D 2 ))/ ∼ where we identify the points on the curves a 1 and a 2 by a 1 (e iθ ) ∼ a 2 (e iθ ).
Denote the natural projection by The "meridian" curve a : S 1 → M is defined by Now we define a homeomorphism R : D → D by the equation: Its class ρ = [R] ∈ M(M ) will be called a half-twist of M along c relative to a. The Dehn twists A and C about the respective curves a and c relate to ρ as follows: The latter equation follows since ρAρ −1 is the Dehn twist about ρ • a = a −1 , and the Dehn twist about a −1 is the same as the Dehn twist about a for any simple closed curve a. If h • a = a, by the structure of the mapping class group of the pantalon, for some integers m, n: It is clear that A and ρ are in G, so they generate G and we have already noted that they commute. If A n ρ m = 1, then The case of the torus is somewhat more interesting. Since we see algebraically that the centre is the cyclic group of order two, consisting of the two diagonal matrices, ±I, where I is the identity matrix. However, to warm up for the more complicated cases, whe will establish this fact geometrically. Consider the torus T 2 embedded in xyz-space as the set of points of distance 1 from the circle x 2 + y 2 = 4, z = 0 (see Figure 11). Let s : T 2 → T 2 be the (orientationpreserving) involution s(x, y, z) = (x, −y, −z). In the case of M(T 2 , 1), we suppose that P = {p} is a point on the x-axis, so that s(p) = p. Let σ = [s] ∈ M(T 2 ) and {p}). Since a is not isotopic to a −1 we conclude that σ = 1, and so σ has order 2. Letting Next we consider the torus with two marked points. Using the above model for T 2 , let P = {(0, 3, 0), (0, −3, 0)}, that it, a pair of points which are interchanged by the involution s : T 2 → T 2 described above. Let σ 2 ∈ M(T 2 , P ) = M(T 2 , 2) be the class represented by s. The following is proved in the same manner as the above.

Proposition 5.3
The centre of M(T 2 , 2) is the cyclic group of order 2 generated by σ 2 .
We now consider the closed surface M of genus 2, which we realize in R 3 as the boundary of a uniform regular neighborhood of the union of the circles (x + 1) 2 + y 2 = 1, z = 0 and (x − 1) 2 + y 2 = 1, z = 0.  Proof: Let a 1 , a 2 , b, c 1 , c 2 , c 3 be the simple closed curves as in Figure 12. In particular, b is the curve of intersection of M with the plane x = 0, c 1 and c 3 lie in the plane z = 0 and a 1 , a 2 , c 2 are in the plane y = 0. We have Since a 1 is not isotopic with a −1 1 , σ = 1 has order 2 in M(M ). Let A 1 ,A 2 ,B, C 1 , C 2 , C 3 be the Dehn twists about the respective curves, which generate M(M ), according to [23]. From the above equations we conclude that Let M 1 and M 2 be the closures of the two components of M \ b, with a 1 in M 1 , for some integers n 1 , n 2 , m 1 , m 2 . Consider Since [h] 2 is central it commutes with C 1 , as do A 2 and B. Therefore A 2m 1 1 also commutes with C 1 . But I(a 1 , c 1 ) = 1, so by Proposition 3.7, 2m 1 = 0. Similarly 2m 2 = 0 and n 1 + n 2 = 0. We note that either n 1 and n 2 are both odd or both even. If they are both even, then h • a 1 = a 1 and h • a 2 = a 2 , if both odd, then h • a 1 = a −1 1 and h • a 2 = a −1 2 . Case 1, n 1 and n 2 are even: write n 1 = 2k 1 and n 2 = 2k 2 . Then Because ξ and ρ commute with B, A p commutes with B. But I(a, b) = 1, so Proposition 3.7 implies that p = 0 and therefore ξ = ρ q .
Remark: Let M be a genus one oriented surface with one boundary component as above. According to [36] M(M ) is isomorphic to the Artin braid group A, B|ABA = BAB , by [4] its center is the infinite cyclic group generated by (ABA) 2 , and one can check directly that ρ = (ABA) 2 .
Following is the main result regarding the centre of M(M, P ), for a general Riemann surface. We have already noted that there are certain exceptional cases, so we list the hypotheses here for the generic result: 1) If M is a sphere, then assume |P | ≥ 4, 2) If M is a disk, then assume |P | ≥ 3, 3) If M is an annulus, then assume |P | ≥ 1, 4) If M is a torus, then assume |P | ≥ 3, 5) If M is a surface of genus one and one boundary component, then assume |P | ≥ 1, 6) If M is a surface of genus 2, then assume |P | ≥ 1.
Theorem 5.6 Let M be a connected compact orientable surface with marked points P ⊂ M and assume the hypotheses (1) -(6) above. Let c 1 , · · · , c q : S 1 → ∂M be the boundary curves of M and C 1 , · · · , C q the Dehn twists about these curves. Then the centre ZM(M, P ) of M(M, P ) is the subgroup generated by {C 1 · · · , C q } and is isomorphic with Z q . In particular, if ∂M is empty, the centre of M(M, P ) is trivial.
Proof: It is clear that C i is central in M(M, P ); the fact that the subgroup generated by C 1 , · · · , C q is isomorphic with Z q follows directly from Consider curves a 1 , · · · , a p which detemine a pantalon decomposition of (M, P ). (All the cases which do not admit a pantalon decomposition have been excluded.) Since ξ is central, the Dehn twist A i about the curves a i satisfy and since ξA i ξ −1 is a Dehn twist about h • a i we conclude from Proposition 3.6 that h • a i is isotopic with a i or a −1 i . By Proposition 3.10 we may suppose that and consequently, h permutes the pantalons. The pantalons (M j , P j ) are mapped to M by maps φ j . We call a curve a i separating if it is in the image of φ k and φ l , for some k = l (a separating curve in our sense need not separate the surface itself). First assume that a i is separating . We consider all the possibilities: If (M k , P k ) and (M l , P l ) are pantalons of type I, then M is a 2-sphere and |P | = 4 (see Figure 13.i). Consider the exact sequence: Since the centre of Σ 4 is trivial, ξ must act trivially on P , whereas h interchanges the points φ k (P k ) and φ l (P l ), so this case cannot occur.
If (M k , P k ) and (M l , P l ) are pantalons of type II, then φ k (M k ) ∩ φ l (M l ) = a i and there are other curves a µ and a ν which form the remaining boundary curves of these pantalons (see Figure 13.ii). Then, since h interchanges the pantalons, h(a µ ) = a ±1 ν . On the other hand, h(a µ ) = a ±1 µ , so we must have a µ = a ±1 ν and conclude that M is a torus and |P | = 2. This case has been excluded. Figure 13: First cases, a i separating.
If (M k , P k ) and (M l , P l ) are pantalons of type III with identifications, so that φ k (M k ) and φ l (M l ) are genus 1 surfaces with one boundary component a i , then M is a closed surface of genus 2 and P is empty (see Figure 14.i). This case also has been excluded.
Finally suppose (M k , P k ) and (M l , P l ) are pantalons of type III, mapped homeomorphically by φ k and φ l (see Figure 14.ii). Then we argue as in the type II case that their other boundary components must be identified, and that M is closed, has genus 2, and P is empty, an excluded case. Figure 14: a i separating type III pantalons.
Thus we have shown that if a i is separating, then h • a i = a i . In addition we have seen that h(φ k (M k )) = φ k (M k ) for all k.
Recall the notation m = |P | and q is the number of boundary components of M . We complete the proof by considering three cases: Case 1: m + q ≥ 2. Then we may assume all the a i are separating (see Figure  15). If M is not a sphere or disk, then each pantalon can be taken to be of type II or III. If M is either a sphere or disk, the exact sequence 1 → PM(M, P ) → M(M, P ) → Σ P → 1, and the fact that Σ P has trivial centre under the assumption |P | ≥ 3, shows that ξ ∈ PM(M, P ), or in other words, h fixes P pointwise. So we have, by the structure of the (pure) mapping class groups of the pantalons of type I, II and III: for some integers r 1 , . . . , r p , s 1 , . . . , s q . Now fix i ∈ {1, . . . , p}. By Proposition 3.4, there exists a generic simple closed curve b : is the Dehn twist about b, we see that B commutes with A j , j = i. B also commutes with all the C j and with ξ since they are central. It follows that B commutes with A r i i . By Proposition 3.7 we conclude that r i = 0. Since i was arbitrary, ξ = C s 1 1 · · · C sq q .
Case 2: m + q = 1. We may now suppose that a 1 is not separating, but a 2 , . . . , a p are separating curves, and the pantalon (M 1 , P 1 ) is of type III with two boundary curves identified to a 1 , so that φ 1 (M 1 ) is a surface of genus one with one boundary component, which we may take to be a 2 (see Figure 16). Moreover we for integers k, r 1 , r 3 , . . . , r p , s 1 . Then Employing the same argument as in Case 1, we conclude that 2r 1 = k = 2r 3 = · · · = 2r p = 0, and so ξ = C s 1 1 if q = 1. If q = 0 and m = 1 we similarly conclude that ξ = 1. Case 3: m + q = 0. In particular, q = 0. Then we need two singular pantalons in the decomposition of M , ξ has an expression as a product of two half-twists together with powers of the A i and we conclude by examining ξ 2 , exactly as in case 2 that the powers are all zero and therefore ξ = 1.

Commensurability
If G is a group, then two subgroups H, H ′ < G are said to be commensurable if H ∩ H ′ has finite index in both H and H ′ . Commensurability is an equivalence relation on the set of all subgroups of G, really of interest only for infinite groups. Following is an elementary property of commensurable subgroups, which we shall find useful. Proof: We may assume b transverse to all the a i and argue by induction on the cardinality of b ∩ (a 1 ∪ · · · ∪ a p ). If b ∩ (a 1 ∪ · · · ∪ a p ) is empty, there is nothing to prove; suppose it is nonempty. Choose i so that b ∩ a i = ∅. Proposition 3.2 implies that b and a i cobound a bigon D. There may be other intersections of a j with D, but there is always an outermost bigon D ′ ⊂ D cobounded by b and some a k , and otherwise disjoint from a 1 ∪ · · · ∪ a p (see Figure 17). Now we can push b across D ′ to obtain a curve b ′ isotopic with b and |b ′ ∩ (a 1 ∪ · · · ∪ a p )| = |b ∩ (a 1 ∪ · · · ∪ a p )| − 2.
By inductive hypothesis, there exists an isotopy from b ′ to an essential simple closed curve c in M \ P such that a i ∩ c = ∅ for all i = 1, . . . , p. Definitions: Recall that a subsurface is essential if none of its exterior components is a disk with zero marked points. By a marked subsurface we mean a pair (N, Q) where N is a subsurface of M and Q = P ∩ N . We will say that an essential marked subsurface ( If (N, Q) and (N ′ , Q ′ ) are marked subsurfaces of (M, P ), we say they are isotopic provided there is a continuous family of homeomorphisms h t ∈ H(M, P ), t ∈ [0, 1] such that h 0 = identity, and h 1 (N, Q) = (N ′ , Q ′ ). In particular, Q = Q ′ .
We would like to be able to say that geometric subgroups are commensurable if and only if they are equal, if and only if their defining subsurfaces are isotopic. However, just as with centres, there are some exceptions to the general principle. Our first family are the infinite cyclic geometric subgroups. The only marked surfaces with mapping class Z are (S 1 × I, 0) and (D 2 , 2); we note for future reference that these are the only mapping class groups which contain Z as a finiteindex subgroup. In case (1), N ′ is isotopic with N . In case (2), (N ′ , Q ′ ) = (D 2 , 2), then one of the components (N 1 , Q 1 ) of the exterior of (N, Q) in (M, P ) is a (D 2 , 2), and (N 1 , Q 1 ) is isotopic with (N ′ , Q ′ ) (see Figure 18).
Proposition 3.6 implies that a ′ is isotopic with a or with a −1 . Since N and N ′ are regular neighborhoods, respectively, of a and a ′ , it follows that N and N ′ are isotopic in this case. Case 2: (N ′ , Q ′ ) is (D 2 , 2). Choose notation as in Case 1, except that a ′ now denotes the boundary curve of N ′ . Again we have nonzero k and l so that A k = A ′l , and we conclude that a ′ is isotopic with a ±1 . After an isotopy, we may assume a ′ equals a boundary component of N . It then follows that N ′ is a component of the exterior of N .
Definition: A doubled pantalon is the marked surface obtained by pasting together two pantalons of the same type along their boundaries (see Figure 19). One reason to be interested in doubled pantalons is that they provide examples of nonisotopic subsurfaces inducing commensurable geometric subgroups, which we will see to be an exceptional case.  Proof: In the case of doubled pantalons of Type II or III, the two geometric subgroups both equal the free abelian group generated by twists along the common boundary of N and N ′ . This has rank 2 or 3, respectively. For Type I, the generator of M(N, N ∩ P ) is a half-twist interchanging the two points N ∩ P , which clearly does not belong to M(N ′ , N ′ ∩ P ), all of whose elements fix N ∩ P pointwise. Likewise the generator of M(N ′ , N ′ ∩ P ) is not in M(N, N ∩ P ). But the squares of the generators, being a Dehn twist along the common boundary, coincide. The question of isotopy is clear for types I and II, because the surfaces N and N ′ enclose different marked points. In type III, the two pantalons, N and N ′ are nonisotopic, too. For an isotopy taking N to N ′ would take the boundary to itself, but with orientation reversed. But a simple homological calculation shows that if a, b, c are the (oriented) boundary curves of N , a cannot be isotopic with a −1 , b −1 or c −1 .
In the following, consider two connected marked injective subsurfaces (N, Q), (N ′ , Q ′ ) ⊂ (M, P ), it being understood that Q = N ∩ P and Q ′ = N ′ ∩ P . We will refer to the following four statements Proof: It is obvious that (c) ⇒ (b) ⇒ (a) in all three cases. Before breaking into cases, let d 1 , . . . , d l : S 1 → M \ P denote the boundary components of N and similarly d ′ 1 , . . . , d ′ l ′ the components of ∂N ′ . Let {a i } be curves in N defining a pantalon decomposition of N (see Figure 20). Let N j denote the components of M \ N . Choose curves b jk giving a pantalon decomposition of (N j , N j ∩ P ) for (N j , N j ∩ P ) = (S 1 × I, ∅). We wish to show that in all cases (a) ⇒ (d). So we assume commensurability of the geometric subgroups M(N, Q) and M(N ′ , Q ′ ); the strategy is to show that after an isotopy, the boundaries of N and N ′ can be made to coincide. We break the argument into two steps.
Step I(d i , d ′ ) = 0 for all i. As in the previous case, a power of D ′ belongs to M(N, Q) and so commutes with D i , a twist of ∂N . Now Proposition 6.2 implies that d ′ may be assumed, after an isotopy, to be disjoint from the a i , b jk , d i , i. e. d ′ lies entirely inside one of the pantalons in the decomposition of M corresponding to those curves. Being essential, d ′ is isotopic with one of the boundary curves of that pantalon (or its inverse). We have concluded that, up to isotopy of M rel P , we have one (and only one) of the following possibilities, for some i, j, k: In summary, we have shown that in case (i), (c) ⇒ (b) ⇒ (a) ⇒ (c). Cases (ii) and (iii) also follow from the above arguments and Proposition 6.4.

Normalizers and Commensurators
The commensurator of a subgroup H of a group G is In general we have For any subsurface N of M , we define the stabilizer Stab(N ) as , we see this is a well-defined subgroup of M(M, P ).
As in the previous section, the doubled pantalons are exceptional cases, so we make the following constructions. Let (M, N, P ) be a doubled pantalon of type I, II or III, embedded in the xyz-space symmetrically with respect to 180 degree rotation s : M → M about the x-axis, and so that (N, N ∩ P ) is the intersection of (M, P ) with the half-space y ≥ 0 (see Figure 21). The map s interchanges N and M \ N , and reverses the orientation of ∂N . We will call σ = [s] : M(M, P ) → M(M, P ), the exchange map, for each of the three types of doubled pantalon. Here σ denotes the cyclic subgroup of order 2 generated by the exchange σ of (M, N, P ) and Stab(N ) is normal in the semidirect product. Next suppose that N is a cylinder and P ∩ N is empty. Then h(N ) is also a cylinder and by Proposition 6.3 M(N, P ∩ N ) = M(h(N ), P ∩ h(N )) and h(N ) is isotopic to N rel P . Again we conclude Com G (H) = N G (H) = Stab(N ). Now assume that (M, N, P ) is a doubled pantalon of type II or III. By Theorem 6.5, M(N, P ∩ N ) = M(h(N ), P ∩ h(N )) and h(N ) is isotopic to N or M \ N rel P . Therefore Com G (H) = N G (H), but this group is bigger than Stab(N ); for example σ belongs to the normalizer, but does not stabilize N .
We can define an epimorphism f : The kernel of f is Stab (N ). The homomorphism f has a section s : Z/2Z → Com G (H) given by s(0) = [id], s(1) = σ, where σ is the exchange of (M, N, P ). It follows that Com G (H) = Stab(N ) × σ . By Propositions 5.3 and 5.5, σ is in the centre of M(M, P ), so the product is, in fact, a direct product.
Finally, consider the case that (M, N, P ) is a doubled pantalon of type I. By Theorem 6.5, h(N ) is isotopic with N or M \ N . Moreover, if ξ ∈ N G (H), then h(N ) is isotopic with N , so we conclude that N G (H) = Stab(N ). By defining the homomorphism f and its section s exactly as above, we also conclude that We proceed now to study the group Stab(N ) which, as seen before, determines the commensurator and the normalizer of M(N, N ∩ P ) in M(M, P ).
We first state the following proposition which can be proved in the same manner as Theorem 4.1.
and is isomorphic with Z l . Figure 22: Marked surfaces whose union is M.
Let (M, P ) be a marked surface, and let N be an injective subsurface of M . Let c 1 , . . . , c l denote the boundary components of N and Σ l the symmetric group of {1, . . . , l}. There is a natural homomorphism τ : Stab(N ) → Σ l which associates to [h] ∈ Stab(N ) the unique σ ∈ Σ l such that h(c i ) is isotopic to c σ(i) . Here we assume that the curves c i are provided with the orientation induced by the one of N . So, since h(N ) is isotopic to N , such a σ exists and is unique, even if N is an annulus and N ∩ P = ∅. The homomorphism τ is not surjective in general. However, one can explicitly describe its image which depends on the topology of the complement of N \ N ∩ P in M \ P . This image is long and tedious to describe and we let the reader off this description.
Let iii) Suppose M \ N has two connected components, N 1 and N 2 (see Figure  23.ii).Then we have an exact sequence Proof: First, suppose that M is a torus and P is empty. By the structure of the mapping class group of an annulus, the kernel of τ is M(N ) ≃ Z. Moreover, σ ∈ Stab(N ) and σ permutes (up to isotopy) the boundary components of N , thus τ is surjective and the map (1, 2) → σ gives a section of τ . Since σ is central in M(M ), it follows that Stab(N ) = M(N ) × σ .
Assume now that M \ N has a unique connected component, N 1 , and (M, P ) = (T 2 , ∅). Let c 1 , c 2 denote the boundary components of N . We can cut N 1 along some closed essential curve d into two subsurfaces such that one of them is a pantalon of type III having d, c 1 and c 2 as boundary components (see Figure  23.i). Pasting this pantalon with N one obtains a genus one surface with one boundary component, d. Let ρ be the half-twist of this surface of genus one along d relative to c 1 , as defined in Section 5. Then ρ ∈ Stab(N ) and τ (ρ) = (1,2). This shows that τ is surjective. Let 2 ). The equality Imι = Kerτ and the inclusion Imθ ⊂ Kerι are obvious, and the injectivity of θ follows from Corollary 3.9. So, it remains to prove the inclusion Kerι ⊂ Imθ. If (N 1 , N 1 ∩ P ) = (S 1 × I, ∅), then M(N 1 , N 1 ∩P ) = M(N ) is the infinite cyclic subgroup generated by C 1 = C 2 , and therefore, since M(N 2 , N 2 ∩ P ) injects in M(M, P ) by Theorem 4.1, Kerι ⊂ Imθ. If (N 1 , N 1 ∩ P ) = (S 1 × I, ∅) = (N 2 , N 2 ∩ P ), then the inclusion Kerι ⊂ Imθ follows from Proposition 7.2.