AUTOMORPHISMS OF THE MAPPING CLASS GROUP OF A ...

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Mar 13, 2014 - Since CMod(Nn h )(PMod+(Nn h )) is contained in the centralizer of the subgroup generated by all Dehn twits, it is trivial by [11, Theorem 6.2].
arXiv:1403.2774v1 [math.GT] 11 Mar 2014

AUTOMORPHISMS OF THE MAPPING CLASS GROUP OF A NONORIENTABLE SURFACE ˙ FERIHE ATALAN AND BLAZEJ SZEPIETOWSKI Abstract. Let Mod(Ngn ) be the mapping class group of a nonorientable surface of genus g ≥ 5 with n ≥ 0 punctures. We prove that the outer automorphism group of Mod(Ngn ) is trivial. This is an analogue of Ivanov’s theorem on automorphisms of the mapping class groups of an orientable surface, and also an extension and improvement of the first author’s previous result.

1. Introduction n ) denote the orientable (resp. nonorientable) surface Let Σng,b (resp. Ng,b of genus g with b boundary components and n punctures (or distinguished n ) denote the mapping class group of N n , which is points). Let Mod(Ng,b g,b n , where diffeomorthe group of isotopy classes of all diffeomorphisms of Ng,b phisms and isotopies are the identity on the boundary. The mapping class group Mod(Σng,b ) is defined analogously, but we consider only orientation pren ) serving maps. The pure mapping class groups PMod(Σng,b ) and PMod(Ng,b n n are the subgroups of Mod(Σg,b ) and Mod(Ng,b ) respectively, consisting of the isotopy classes of diffeomorphisms fixing each puncture. We denote by n ) the subgroup of PMod(N n ) consisting of the isotopy classes PMod+ (Ng,b g,b of diffeomorphisms preserving local orientation at each puncture. If b or n equals 0, then we drop it from the notation. The first author proved in [2] that the outer automorphism group of Mod(Ng ) is cyclic for g ≥ 5. In this paper we improve this result and also extend it to the case of surfaces with punctures.

Theorem 1.1. The outer automorphism group Out(Mod(Ngn )) is trivial for g ≥ 5 and n ≥ 0. The analogous theorem for the mapping class group of an orientable surface is due to Ivanov [4], who proved that if Σ is an orientable surface of genus g ≥ 3, then every automorphism of Mod(Σ) is induced by a diffeomorphism of Σ, not necessarily orientation preserving one. Later, Ivanov and Date: March 13, 2014. 2010 Mathematics Subject Classification. 20F38, 57N05. Key words and phrases. Nonorientable surface, mapping class group, outer automorphism. The first author is supported by TUBITAK-110T665. The second author is supported by NCN 2012/05/B/ST1/02171. 1

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McCarthy [5] proved (among other things) that any injective endomorphism of Mod(Σ) must be an isomorphism. Finally, by recent results of Castel [3] and Aramayona-Souto [1], any nontrivial endomorphism of Mod(Σ) must be an isomorphism. Similarly as in [4] and [5], the main ingredient of our proof of Theorem 1.1 is an algebraic characterization of Dehn twists (Theorem 2.3), from which we conclude that any automorphism of Mod(Ngn ) maps Dehn twists on Dehn twists. However, unlike for orientable surfaces, Mod(Ngn ) is not generated by Dehn twists (and neither are PMod(Ngn ) and PMod+ (Ngn ), see [6] and [12]). It seems reasonable to expect that Theorem 1.1 is true also for g < 5, provided that n is sufficiently big. On the other hand, it is false for g = 2 or 3 and n = 0 [2, Proposition 4.5]. Finally, we note that Theorem 1.1 together with the fact that the center of Mod(Ngn ) is trivial [11, Corollary 6.3], imply that Aut(Mod(Ngn )) is isomorphic to Mod(Ngn ) for g ≥ 5. 2. Preliminaries Let G be a group, X ⊆ G a subset and x ∈ G an element of G. Then C(G), CG (X) and CG (x) will denote the center of G, the centralizer of X in G and the centralizer of x in G, respectively. 2.1. Circles and Dehn twists. By a circle a on a surface S we understand in this paper an unoriented simple closed curve. According to weather a regular neighborhood of a is an annulus or a M¨obius strip, we call a twosided or one-sided respectively. If a bounds a disc with at most one puncture or a M¨obius band, then it is called trivial. Otherwise, we say that it is nontrivial. Let S a denote the surface obtained by cutting S along a. If S a is connected, then we say that a is nonseparating. Otherwise, a is called separating. If a is two-sided, then we denote by ta a Dehn twist about a. On a nonorientable surface it is impossible to distinguish between right and left twists, so the direction of a twist ta has to be specified for each circle a. Equivalently we may choose an orientation of a regular neighborhood of a. Then ta denotes the right Dehn twist with respect to the chosen orientation. Unless we specify which of the two twists we mean, ta denotes any of the two possible twists. It is proved in [11] that many well known properties of Dehn twists on orientable surfaces are also satisfied in the nonorientable case. We will use these properties in this paper. For two circles a and b we denote by i(a, b) their geometric intersection number (see [11] for definition and properties). We say that a and b are equivalent if there exists a diffeomorphism h : S → S such that h(a) = b. We say that a collection of circles C = {a1 , . . . , ak } is a generic family of disjoint circles if the circles ai are nontrivial, pairwise disjoint, pairwise nonisotopic, and none is isotopic to a boundary component of S. We denote by S C the surface obtained by cutting S along all circles of C.

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2.2. Pants and skirts. We will use some properties of pants and skirts (PS) decompositions defined in [11, Section 5]. We say that a generic family of disjoint circles C is a P-S decomposition if each a ∈ C is two-sided and each component of S C is diffeomorphic to one of the following surfaces: • disc with 2 punctures (pantalon of type 1), • annulus with 1 puncture (pantalon of type 2), • sphere with 3 holes (pantalon of type 3), • M¨obius strip with 1 puncture (skirt of type 1), • M¨obius strip with 1 hole (skirt of type 2). A P-S decomposition C is called separating if each a ∈ C is a boundary of two different connected components of S C . Lemma 2.1. Let S = Ngn for g ≥ 3 and s = 3g−7 2 + n if g is odd, or s = 3g−8 2 +n if g is even. Suppose that a is a two-sided circle on S. There exits a P-S decomposition C = {a1 , . . . , as } of S, such that each ai is equivalent to a, if and only if S a is connected and nonorientable. Furthermore, if g + n > 3 then such P-S decomposition must be separating. Proof. The “if” part is left to the reader. Suppose that a is separating. Then all ai are separating. Furthermore, either each ai separates a pantalon of type 1, or each ai separates a skirt of type 1. It follows that s ≤ n, a contradiction. Now suppose that S ai is connected and orientable (this is possible only for even g). Then every component of S C is a pantalon of type either 2 or 3. Note, however, that for i 6= j the circles ai , aj together separate S (there can be no curve on S disjoint from ai and intersecting aj once; such a curve would be two-sided and one-sided at the same time). It follows that no component of S C is a pantalon of type 3, hence all components are pantalons of type 2. We have s ≤ n, a contradiction. Now suppose that g + k > 3 and ai is a boundary of only one connected component P of S C . Because −χ(S) = g + n − 2 > 1, S C has more then one component. It follows that P must be a pantalon of type 3 and the third boundary component of P must separate S. This is a contradiction, because all ai are nonseparating. Thus C is a separating P-S decomposition of S.  Lemma 2.2. Let S = Ngn for g ≥ 5 and suppose that C = {a1 , . . . , as } is a P-S decomposition as in Lemma 2.1. For k ≥ 1 let TCk be the subgroup of Mod(S) generated by tkai for 1 ≤ i ≤ s. Then, for each k ≥ 1: (a) TCk is a free abelian group of rank s; (b) CMod(S) (TCk ) = TC1 . Proof. The assertion (a) follows from [11, Proposition 4.4]. To prove (b) we use an idea from the proof of [11, Theorem 6.2]. Suppose f ∈ CMod(S) (TCk ). Then tkai = f tkai f −1 = tkf (ai ) for all i. It follows that f fixes each circle ai , hence it permutes the connected components S C . Suppose that f interchanges some two components P1 and P2 of S C . By the proof of Lemma 2.1,

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there are no pantalons of type 1 and no skirts of type 1 in the decomposition. Suppose that P1 and P2 are skirts of type 2 glued along a circle ai . Then the remaining boundary circles aj ⊂ P1 and al ⊂ P2 must be glued together (al = f (aj ) = aj ), hence S is the closed nonorientable surface of genus 4. Similarly, if P1 and P2 are pantalons of type 2 or 3, then S must be a Klein bottle with two punctures, or a closed nonorientable surface of genus 4 respectively. Since g ≥ 5 by assumption, f fixes each component of S C . Furthermore, since f centralizes the boundary twists of each pantalon, it preserves its orientation. Because the mapping class groups of a pantalon of type 2 or 3, and that of the skirt of type 2 are generated by boundary twists, f is a product of some powers of tai for 1 ≤ i ≤ s. Thus CMod(S) (TCk ) ⊆ TC1 and the opposite inclusion is obvious.  Note that (b) of Lemma 2.2 implies that TC1 is a maximal abelian subgroup of Mod(S). 2.3. Pure subgroups. Let S denote the surface Ngn for g ≥ 3 and n ≥ 0. We recall from [2] the construction of finite index pure subgroups Γ′ (m) of Mod(S) (see Section 2 of [2] for more details). Fix an orientable double cover Σ = Σ2n g−1 of S. Then Mod(S) can be identified with the subgroup of Mod(Σ), consisting of the isotopy classes of diffeomorphisms commuting with the covering involution. Consequently, Mod(S) acts on H1 (Σ, Z/mZ) for all m ≥ 0. We define Γ′ (m) to be the subgroup of Mod(S) consisting of all elements inducing the identity on H1 (Σ, Z/mZ). If m ≥ 3, then Γ′ (m) is a pure subgroup of Mod(S). Fix m ≥ 3 and suppose that f ∈ Γ′ (m) preserves a generic family of disjoint circles C. Then f fixes each circle of C and, furthermore, it can be represented by a diffeomorphism equal to the identity on a regular neighborhood of each circle of C. If the restriction of f to any connected component of S C is isotopic (by an isotopy that does not have to fix pointwise the boundary components of S C ) either to the identity or to a pseudo-Anosov map, then C is called a reduction system for f . The intersection of all reduction systems for f is called the canonical reduction system for f . 2.4. Algebraic characterization of a Dehn twist. The key ingredient of the proof of our main result is an algebraic characterization of a Dehn twist about a nonseparating circle in the mapping class group. Theorem 2.3 below is an extension of Theorem 3.1 of [2] to punctured surfaces. The proof closely follows Ivanov’s ideas [4]. Theorem 2.3. Let S = Ngn be a connected nonorientable surface of genus g ≥ 5 with n punctures and let Γ′ be a finite index subgroup of Γ′ (m) for m ≥ 3. An element f ∈ Mod(S) is a Dehn twist about a nonseparating simple closed curve with nonorientable complement if and only if the following conditions are satisfied:

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(i) C(CΓ′ (f k )) ∼ = Z, for any integer k 6= 0 such that f k ∈ Γ′ . 3g−7 (ii) Set s = 2 + n if g is odd, or s = 3g−8 2 + n if g is even. There exist elements f2 , . . . , fs ∈ Mod(S), each conjugate to f1 = f , such that f1 , . . . , fs generate a free abelian group K of rank s. (iii) For k ≥ 1 let Kk be the subgroup of Mod(S) generated by fik for 1 ≤ i ≤ s. Then CMod(S) (Kk ) = K. Proof. Assume that the above conditions are satisfied, then we have to show that f is a Dehn twist about a nonseparating circle. Choose any integer k 6= 0 such that f k ∈ Γ′ . Because f has infinite order by (ii), f k is not the identity element. Let C be the canonical reduction system for f k . Let G denote the subgroup generated by the twists about the two-sided circles in C. Set G′ = G ∩ Γ′ . Then G and G′ are free abelian groups. Firstly, we will show that G′ ⊂ C(CΓ′ (f k )). Let g ∈ CΓ′ (f k ). Since g commutes with f k , it preserves the canonical reduction system C. Because g is pure, it fixes each circle of C and also preserves orientation of a regular neighborhood of each twosided circle of C. It follows that g commutes with each generator G, hence G ⊆ CM(S) (CΓ′ (f k )). So, G′ ⊂ CΓ′ (CΓ′ (f k )) = C(CΓ′ (f k )). For the last equality observe that, since f k ∈ CΓ′ (f k ), CΓ′ (CΓ′ (f k )) ⊆ CΓ′ (f k ), hence CΓ′ (CΓ′ (f k )) ⊆ C(CΓ′ (f k )) and the opposite inclusion is obvious. The assumption C(CΓ′ (f k )) = Z implies that C contains at most one two-sided circle. Assume that C has no two-sided circle, so that C = {c1 , · · · , cl }, where each ci is a one-sided circle. Then S C is connected. Let Stab+ (C) be the subgroup of Mod(S) consisting of elements fixing each circle of C and preserving its orientation. Note that CΓ′ (f k ) ⊆ Stab+ (C). The mapping h 7→ h|S C defines an isomorphism Stab+ (C) → Mod(S C )/Zl , where Zl is the subgroup generated by the boundary twists of S C (see [10, Section 4]). We also have an isomorphism Mod(S C )/Zl → PMod+ (S ′ ), where S ′ is the surface obtained by collapsing each boundary component to a puncture. By composing these two maps we obtain an isomorphism θ : Stab+ (C) → PMod+ (S ′ ). Because C is the canonical reduction system for f k , θ(f k ) is either the identity or pseudo-Anosov. In the former case f k must be the identity, a contradiction. Suppose θ(f k ) is pseudo-Anosov. Set H = Γ′ ∩ Kk , where Kk is the group from condition (iii). We have H ⊆ CΓ′ (f k ) ⊆ Stab+ (C) and θ(H) is a free abelian subgroup of PMod+ (S ′ ) containing θ(f k ). Since θ(f k ) is pseudoAnosov, θ(H) must have rank 1. This is a contradiction, as H has rank s > 1. We have C = {c1 , · · · , cl , a}, where a is a two-sided circle and each ci is one-sided. Let D be the subgroup generated by f k and the twist about a and denote the intersection D ∩ Γ′ by D ′ . Hence, D ′ ⊂ C(CΓ′ (f k )) and hence D ′ is isomorphic to Z. It follows that f n1 = tm a for some integers m and k1 (possibly greater than k).

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Let f1 , . . . , fs be the elements from condition (ii). For 1 ≤ i ≤ s we have fik1 = tm ai for some circle ai equivalent to a1 = a. We claim that C = {a1 , . . . , as } is a P-S decomposition of S. If not, then we can complete C to a P-S decomposition C ′ . Let TC ′ be the free abelian group generated by twists about the circles of C ′ . We have TC ′ ⊆ CMod(S) (Kk1 ) = K. It follows that rank(TC ′ ) ≤ s, hence C ′ = C. By (iii) and (b) of Lemma 2.2 we have K = CMod(S) (Kk1 ) = CMod(S) (TCm ) = TC1 . By (ii) f is a primitive element of K = TC1 , hence f = ta1 . It follows from Lemma 2.1 that a1 is nonseparating and has nonorientable complement. The proof of the opposite implication is straightforward and left to the reader (see [4]).  Remark 2.4. It follows from condition (iii) that the group K from (ii) is a maximal abelian subgroup of Mod(S). However, if g is even, then the rank of K, which equals 3g−8 2 + n, is not the maximal rank of an abelian subgroup of Mod(S), as there exist subgroups of rank 3g−6 2 +n. We illustrate this on an example for g = 8. In Figure 1 (a) the abelian subgroup K of rank 8, which is generated by the Dehn twists about the curves a1 , · · · , a8 , is maximal. However, in Figure 1 (b) the abelian subgroup K of rank 8, which is again generated by the Dehn twists about the curves a1 , · · · , a8 , is not maximal. It becomes maximal only if the Dehn twist about the curve a9 is also added to the subgroup. Note that a9 has orientable component. It is not difficult to show that 3g−8 2 + n is the maximal rank of an abelian subgroup of Mod(S) generated by Dehn twists about nonseparating curves with nonorientable complements.

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Figure 1. (a) The abelian subgroup K of rank 8 is maximal (b) The abelian subgroup K of rank 8 is not maximal. Corollary 2.5. Let g ≥ 5 and suppose that ϕ : Mod(Ngn ) → Mod(Ngn ) is an isomorphism. If f ∈ Mod(Ngn ) is a Dehn twists about a nonseparating circle with nonorientable complement, then so is ϕ(f ). Proof. Fix m ≥ 3. Because f satisfies the conditions (i), (ii), (iii) of Theorem 2.3 with Γ′ (m) as Γ′ , it follows that ϕ(f ) also satisfies (i), (i), (iii) of Theorem 2.3 with Γ′ = ϕ(Γ′ (m)) ∩ Γ′ (m). 

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2.5. Chains. A sequence (a1 , . . . , ak ) of circles is called a chain if i(ai , ai+1 ) = 1 for 1 ≤ i ≤ k − 1 and i(ai , aj ) = 0 for |i − j| > 1. The integer k ≥ 1 is called the length of the chain. If all circles in a chain are two-sided, then a regular neighborhood of the union of these circles is orientable. Let tai be right Dehn twists with respect to some orientation of such a neighborhood for 1 ≤ i ≤ k. Then (a) tai tai+1 tai = tai+1 tai tai+1 for 1 ≤ i ≤ k − 1 (b) tai taj = taj tai for |i − j| > 1. Conversely, if a sequence of Dehn twists (ta1 , · · · , tak ) satisfies (a) and (b), then (a1 , . . . , ak ) is a chain, and the twists are right with respect to some orientation of a regular neighborhood of the union of the circles of the chain (see [11, Section 4]). A sequence of Dehn twists satisfying (a) and (b) will also be called a chain. Observe that if (a1 , a2 ) is a 2-chain of two-sided circles, then S ai must be connected and nonorientable for i = 1, 2. 2.6. Trees. We will now define a tree of circles (and Dehn twists) as a generalization of a chain. Suppose that C is a collection of circles, such that i(a, b) ∈ {0, 1} for all a, b ∈ C. Let Γ(C) be a graph with C as the set of vertices, and where a and b are connected by an edge if and only if i(a, b) = 1. We will call C a tree if and only if Γ(C) is a tree (connected and acyclic). If all circles in a tree are two-sided, then a regular neighborhood of the union of these circles is orientable. Let ta be right Dehn twists with respect to some orientation of such a neighborhood for a ∈ C. Then (a’) ta tb ta = tb ta tb if a and b are connected by an edge, (b’) ta tb = tb ta otherwise. Conversely, suppose that T = {ta : a ∈ C} is a set of Dehn twists for some set of circles C, where each two twists of T either commute, or satisfy the braid relation. Then the geometric intersection number of the underlying circles is, respectively, either 0 or 1. We say that T is a tree of twists if and only if C is a tree. Note that all twists of a tree are right with respect to some orientation of a regular neighborhood of the union of the underlying circles. The following corollary follows immediately from Corollary 2.5 Corollary 2.6. Let g ≥ 5 and suppose that ϕ : Mod(Ngn ) → Mod(Ngn ) is an isomorphism. If T = {ta : a ∈ C} is a tree of Dehn twists of cardinality at least 2, then ϕ(T ) is also a tree of Dehn twists for some set of circles C ′ , such that Γ(C) and Γ(C ′ ) are isomorphic (as abstract graphs). 2.7. Nonorientable triangle. Suppose that ta , tb , tc are Dehn twists about circles a, b, c. We say that (ta , tb , tc ) is a nonorientable triangle of Dehn twists if and only if the following braid relations hold in the mapping class group: ta tb ta = tb ta tb

ta tc ta = tc ta tc

−1 −1 t−1 c tb tc = tb tc tb

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Lemma 2.7. Suppose that (ta , tb , tc ) is a nonorientable triangle of Dehn twists, where the circles a, b, c intersect each other minimally. Then a regular neighborhood of a ∪ b ∪ c is N4,1 , a genus four nonorientable surface with one boundary component. Proof. It follows from the braid relations and [11, Proposition 4.8] that the intersection number is 1 for every pair of the circles a, b, c. Let M be a regular neighborhood of a ∪ b ∪ c. Suppose that M is orientable and fix an orientation such that ta is right Dehn twist. It follows from the first two braid relations that tb and tc are also right Dehn twists. However, because of the third braid relation, tb and tc can not be both right. Therefore M must be nonorientable. Note that M has Euler characteristic −3 and 1 boundary component, hence it is diffeomorphic to N4,1 .  Conversely, it is not difficult to find a nonorientable triangle of twists in Mod(N4,1 ). This implies the following corollary. Corollary 2.8. Let S = Ngn and suppose that C is a collection of pairwise disjoint two-sided circles on S. Let K be the subgroup of Mod(S) generated by Dehn twists about the circles of C. Then CMod(S) (K) contains a nonorientable triangle of Dehn twists if and only if S C has a nonorientable connected component of genus at least 4, where S C is the surface obtained by cutting S along C. 3. Automorphisms of Mod(Ngn ) The aim of this section is to prove Theorem 1.1. We will use the following lemma proved in [4]. Lemma 3.1 (Ivanov). Let H be a normal subgroup of a group G such that CG (H) is trivial. If ϕ : G → G is an automorphism fixed on H, then ϕ = idG . Let D and E be the trees of circles from Figures 2 and 3. We will abuse notation and denote by the same symbols the corresponding trees of Dehn n n ), (resp. N2g+2 twists. Let Σg,n+1 (resp. Σg,n+2 ) be a subsurface of N2g+1 n n ) supporting D (resp. E), obtained by removing from N2g+1 (resp. N2g+2 n open discs, each containing one puncture, and a M¨obius band (resp. an n annulus with core a). For i = 1, 2 the inclusion Σg,n+i ⊂ N2g+i induces a n homomorphism Mod(Σg,n+i ) → Mod(N2g+i ). We define sub-trees Γ ⊂ D and Λ ⊂ E as Γ = {ta1 , ta3 , ta5 } ∪ {ta2i : 1 ≤ i ≤ 2g − 1} ∪ {tbj : 1 ≤ j ≤ n − 1}, Λ = {ta1 , ta3 , ta5 } ∪ {ta2i : 0 ≤ i ≤ 2g − 1} ∪ {tbj : 1 ≤ j ≤ n − 1}. Lemma 3.2. Suppose that h = 2g + 1 for g ≥ 2. Then CMod(Nhn ) (Γ) = 1. Proof. Let H denote the image of Mod(Σg,n+1 ) in Mod(Nhn ). It can be easily deduced from the main result of [7] that H is generated by twists of Γ. Thus

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CMod(Nhn ) (Γ) = CMod(Nhn ) (H). Set D ′ = D\{ta4i−2 : 1 ≤ i ≤ g}. The curves supporting the twists of D ′ form a separating pants and skirts decomposition of Nhn (see Subsection 2.2 for the definition). Let h ∈ CMod(Nhn ) (H). Since Q i D ′ ⊂ H, h ∈ CMod(Nhn ) (D ′ ). By the proof of (b) of Lemma 2.2, h = tm ai ′ for some integers mi , where the product is taken over all tai ∈ D . By [8, Proposition 3.4], for every tai ∈ D ′ there exists a simple closed curve c on Σg,n+1 , such that i(c, ai ) > 0 and tc commutes with all twists in D ′ \{tai }. Since tc ∈ H, it also commutes with h. It follows that tc commutes with i tm ai , which is possible only for mi = 0, hence h = 1 and CMod(Nhn ) (H) is trivial. 

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Lemma 3.3. Suppose that h = 2g + 2 for g ≥ 2. Then CMod(Nhn ) (Λ) is the infinite cyclic group generated by ta , where a is the circle from Figure 3. Proof. Let H denote the image of Mod(Σg,n+2 ) in Mod(Nhn ). Similarly as in the odd genus case, H is generated by twists of Λ, thus CMod(Nhn ) (Λ) = CMod(Nhn ) (H). Note that ta ∈ H, because a is isotopic to a boundary component of Σg,n+2 . Set E ′ = E ∪ {ta }\{ta4i−2 : 1 ≤ i ≤ g}. The curves supporting the twists of E ′ form a separating P-S decomposition of Nhn . Let h ∈ CMod(Nhn ) (H). By a similar argument as in the proof of (b) of Lemma Q mi 2.2, h = tm tai for some integers mi and m, where the product is taken a over all tai ∈ E ′ \{ta }. By the same argument as in the proof for odd genus, all mi = 0, hence h = tm  a . The following lemma is well-known (see [7, Proposition 2.12]). Lemma 3.4. Suppose that (tc1 , . . . , tc2k+1 ) is a chain of twists, where the circles ci intersect each other minimally. Let S be a regular neighborhood of the union of the circles ci , oriented so that tci are right twist (Figure 4). Then (tc1 · · · tc2k+1 )2k+2 is equal to the product of 2 right twists about the boundary components of S.

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Figure 4. A chain of two-sided circles of odd length and its regular neighborhood Lemma 3.5. Suppose that (tc1 , . . . , tc5 ) is a chain of twists, and tc0 is a twist such that i(c0 , c2 ) = 1, i(c0 , ci ) = 0 for i 6= 2, and tc0 6= tc1 . Let S be a regular neighborhood of the union of the circles ci , oriented so that tci are right twist, and let u be the component of ∂S bounding a pair of pants with c0 and c1 (Figure 5). Suppose that S is embedded in a surface M , so that c5 , c3 and c0 bound a pair of pants in M . Then tu = (tc0 tc1 tc2 tc3 tc4 tc5 )5 (tc0 tc2 tc3 tc4 tc5 )−6 in the mapping class group of M . Proof. The boundary of S consists of three circles: u, v and w. Suppose that w bounds a 4-holed sphere with c5 , c3 and c0 . Then, by assumption, w bounds a disc in M . By [7, Proposition 2.12] we have (tc0 tc1 tc2 tc3 tc4 tc5 )5 = tw tv t2u = tv t2u and by Lemma 3.4 (tc0 tc2 tc3 tc4 tc5 )6 = tv tu . 

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w

Figure 5. The circles from Lemma 3.5 Lemma 3.6. Suppose that h = 2g + 1 (resp. h = 2g + 2) for g ≥ 2 and ϕ : Mod(Nhn ) → Mod(Nhn ) is an automorphism. Then there exits f ∈ Mod(Nhn ) such that ϕ(t) = f tf −1 for each t ∈ D (resp. t ∈ E ∪ {ta }). Proof. Suppose h = 2g + 1. By Corollary 2.6, ϕ(Γ) is a tree of Dehn twists for which the underlying tree of circles is isomorphic (as abstract graphs) to that of Γ. For tai , tbj ∈ Γ choose circles ci , dj such that tci = ϕ(tai ), tdj = ϕ(tbj ). These circles may be chosen to intersect each other minimally. Let M be a closed regular neighborhood of the union of ci and dj for tci , tdj ∈ ϕ(Γ). Note that M is an orientable surface of genus g with n + 2 (or 3 if n = 0) boundary components. v d'1

c5 c4g-2

c4g-6

c'1 c6

c2

u

... ...

c4g-4 d'n-1

c4 c'3

Figure 6. The neighborhood M supporting ϕ(Γ). Similarly, let M ′ ⊂ Σg,n+1 be a closed regular neighborhood of the union of the curves supporting Γ. Orient M and M ′ so that tai , tbj and tci , tdj are right Dehn twists. Fix an orientation preserving diffeomorphism f0 : M ′ → M such that f0 (a2i ) = c2i for 1 ≤ 1 ≤ 2g − 1 , f0 (a5 ) = c5 , {f0 (a1 ), f0 (a3 )} = {c1 , c3 } and {f0 (bj ) : 1 ≤ j ≤ n − 1} = {dj : 1 ≤ j ≤ n − 1}. If (g, n) = (2, 0) then we can also assume f0 (ai ) = ci for i = 1, 3. Set c′i = f0 (ai ) for i = 1, 3 and d′j = f0 (bj ) for 1 ≤ j ≤ n − 1. Either (c′1 , c′3 ) = (c1 , c3 ) or (c′1 , c′3 ) = (c3 , c1 ). Analogously, (d′1 , . . . , d′n−1 ) is some (possibly nontrivial) permutation of (d1 , . . . , dn−1 ). The neighborhood M and the curves supporting ϕ(Γ) are shown on Figure 6. By Lemma 3.2, CMod(Nhn ) (ϕ(Γ)) = ϕ(CMod(Nhn ) (Γ)) = 1. It follows that Dehn twists about the boundary components of M are trivial, hence each

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˙ FERIHE ATALAN AND BLAZEJ SZEPIETOWSKI

component of ∂M bounds either a M¨obius band or a disc with 0 or 1 puncture. It is clear that exactly 1 component bounds a M¨obius strip, and exactly n components bound once-punctured discs. Consider the component u of ∂M which bounds a pair of pants together with c′1 and c′3 . By Corollary 2.8, CMod(Nhn ) {ta1 , ta3 } contains no orientation reversing triangle, hence neither does CMod(Nhn ) {tc1 , tc3 }. It follows that u bounds a M¨obius strip, for otherwise c1 and c3 would separate a nonorientable subsurface of genus ≥ 4. Suppose (g, n) 6= (2, 0) and consider the component v of ∂M which bounds a 4-holed sphere together with c5 , c4 and c′1 . For i = 1, 3 set xi = (ta5 ta6 ta4 ta2 tai )6

and

yi = (tc5 tc6 tc4 tc2 tc′i )6 .

Suppose that (c′1 , c′3 ) = (c3 , c1 ). Then ϕ(x3 ) = y1 . By Lemma 3.4, x3 is a product of 2 twists commuting with ta1 , whereas y1 does not commute with tc′3 , a contradiction. Hence c′i = ci for i = 1, 3. It also follows that y3 commutes with tc1 , which implies that v bounds a non-punctured disc. It is now clear that f0 can be extended to f : Nhn → Nhn . We have ϕ(tai ) = f tai f −1 for all tai ∈ Γ. Since each taj ∈ D can be expressed in terms of tai ∈ Γ, we have ϕ(taj ) = f taj f −1 for all taj ∈ D. It remains to prove that d′i = di for 1 ≤ i ≤ n − 1. We proceed by induction. Consider the once-punctured annulus A1 , whose boundary is the union of b1 and a4g−3 . Let u1 be the boundary of a small disc contained in A1 and containing the puncture. Let w1 be the expression for tu1 in terms of taj ∈ D, given by relation (R5) of [7, Theorem 3.1]. Now ϕ(w1 ) is equal to a twist about the circle bounding a disc containing all punctures of the annulus A′1 , whose boundary is the union of d1 and f (a4g−3 ). Since tu1 = 1, we have ϕ(w1 ) = 1. It follows that A′1 contains only 1 puncture, hence d1 = d′1 . Now suppose that d′i = di for 1 ≤ i ≤ k − 1 for some k < n. Consider the once-punctured annulus Ak , whose boundary is the union of bk−1 and bk . Let uk be the boundary of a small disc contained in Ak and containing the puncture. Let wk be the expression for tuk in terms of taj ∈ D and tbk−1 , given by relation (R6) of [7, Theorem 3.1]. Now ϕ(wk ) is equal to a twist about the circle bounding a disc containing all punctures of the annulus A′k , whose boundary is the union of dk and dk−1 . As above, it follows that dk = d′k . For g = 2g +2 we proceed as above, to obtain a diffeomorphism f0 : M ′ → M , where M (resp. M ′ ) is a regular neighborhood of the union of the circles supporting ϕ(Λ) (resp. Λ), such that f0 (a2i ) = c2i for 1 ≤ 1 ≤ 2g − 1, f0 (a5 ) = c5 , {f0 (a0 ), f0 (a1 ), f0 (a3 )} = {c0 , c1 , c3 } and {f0 (bj ) : 1 ≤ j ≤ n − 1} = {dj : 1 ≤ j ≤ n − 1}, where ϕ(tai ) = tci and ϕ(tbj ) = tdj . Set c′i = f0 (ai ) for i = 0, 1, 3 and d′j = f0 (bj ) for 1 ≤ j ≤ n−1. Note that M and M ′ are orientable of genus g with n + 3 (or 4 if n = 0) boundary components (see Figure 7). By Lemma 3.3, CMod(Nhn ) (ϕ(Λ)) = ϕ(CMod(Nhn ) (Λ)) ≈ Z.

AUTOMORPHISMS OF THE MAPPING CLASS GROUP

13

v d'1

c5 c4g-2

c4g-6

c'1 c6

u1

c2

... c4

...

c4g-4 d'n-1

c'0 c'3

u2

Figure 7. The neighborhood M supporting ϕ(Λ). Consider components u1 , u2 of ∂M , each bounding a pair of pants together with two circles from {c′0 , c′1 , c′3 }. By Corollary 2.8, CMod(Nhn ) {tai , taj } contains no orientation-reversing triangle for all i, j ∈ {0, 1, 3}, hence neither does CMod(Nhn ) {tci , tcj }. It follows that u1 and u2 bound an annulus (exterior to M ) such that the union of M and that annulus is a nonorientable surface of genus 2g + 2 = h. Otherwise the complement of some ci ∪ cj for i, j ∈ {0, 1, 3} would have a nonorientable component of genus ≥ 4. For i ∈ {0, 1, 3} set xi = (ta5 ta6 ta4 ta2 tai )6

and

yi = (tc5 tc6 tc4 tc2 tc′i )6 .

Suppose (g, n) 6= (2, 0) and consider the component v of ∂M which bounds a 4-holed sphere together with c5 , c4 and c′1 . It follows from Lemma 3.4 that {ta0 , ta1 , ta3 } ∩ CMod(Nhn ) {x0 , x1 , x3 } = {ta1 }, hence {tc0 , tc1 , tc3 } ∩ CMod(Nhn ) {y0 , y1 , y3 } = {tc1 }. Since neither tc′0 nor tc′3 commute with y1 , we have c1 = c′1 . Furthermore, since tc1 commutes with y0 and y3 , v bounds a non-punctured disc. Now for i = 0, 3 set wi = (tai ta1 ta2 ta4 ta6 ta5 )5 (tai ta2 ta4 ta6 ta5 )−6 zi = (tc′i tc1 tc2 tc4 tc6 tc5 )5 (tc′i tc2 tc4 tc6 tc5 )−6 Suppose that (c′0 , c′3 ) = (c3 , c0 ). Then ϕ(w0 ) = z3 . By Lemma 3.4, w0 is a Dehn twist commuting with ta3 , whereas z3 does not commute with tc′0 , a contradiction. It follows that c′i = ci for i ∈ {0, 1, 3}. If (g, n) = (2, 0) then we have {ta0 , ta1 , ta3 } ∩ CMod(Nhn ) {x0 , x1 , x3 } = {ta1 , ta3 }. It follows that c′0 = c0 , and by composing f0 by a suitable diffeomorphism if necessary we may assume c′i = ci for i = 1, 3. It is clear that f0 can be extended to f : Nhn → Nhn . The rest of the proof follows as in the odd genus case.  For k ∈ {5, 6} let Nk,1 be a nonorientable surface of genus k with one boundary component, represented on Figure 8 as disc with k crosscaps. This

14

˙ FERIHE ATALAN AND BLAZEJ SZEPIETOWSKI

means that interiors of the k shaded discs should be removed from the disc, and then antipodal points in each of the resulting boundary components should be identified. Let us arrange the crosscaps as shown on Figure 8 and number them from 1 to k.

...

... 1

i

... j

k

Figure 8. The surface Nk,1 and the curve ci,j . For i ≤ j let ci,j denote the simple closed curve on Nk,1 from Figure 8. Note that ci,j is two-sided if and only if j −i is odd. In such case tci,j denotes the twist about ci,j in the direction indicated by the small arrows on Figure 8. We denote by u the crosscap transposition defined to be the isotopy class of the diffeomorphism of Nk,1 interchanging the (k − 1)’st and k’th crosscaps as shown on Figure 9, and equal to the identity outside a disc containing these crosscaps.

u

−→ k−1

k

Figure 9. The crosscap transposition

n and Lemma 3.7. For g ≥ 2 there are embeddings θ1 : N5,1 → N2g+1 n θ2 : N6,1 → N2g+2 , such that: n (a) for i = 1, 2, N2g+i \θi (N4+i,1 ) is an orientable surface of genus g − 2 with n punctures containing the curves ai for all i > 8; (b) for i = 1, 2, a5 = θi (c1,2 ), a6 = θi (c2,3 ), a4 = θi (c3,4 ), a2 = θi (c4,5 ), a1 = θi (c1,4 ); (c) a3 = θ1 (tc4,5 u(c1,4 )); (d) a0 = θ2 (c5,6 ), a = θ2 (c1,6 ); (e) θ2 maps boundary circles of a regular neighborhood of c1,6 ∪ c5,6 ∪ c6,6 on a1 and a3 .

AUTOMORPHISMS OF THE MAPPING CLASS GROUP

15

Proof. Suppose h = 2g + 1. Set c5 = c1,2 , c6 = c2,3 , c4 = c3,4 , c2 = c4,5 , c1 = c1,4 and c3 = tc4,5 u(c1,4 ). By changing these curves by small isotopy, we may assume that they intersect each other minimally. Then we have |ci ∩ cj | = |ai ∩ aj | for all i, j ∈ {1, . . . , 6}. Let M (resp. M ′ ) be a regular neighborhood of the union of ci (resp. ai ) for i ∈ {1, . . . , 6}. Observe that M and M ′ are both diffeomorphic to Σ2,3 . There is a diffeomorphism θ ′ : M → M ′ such that θ ′ (ci ) = ai for i ∈ {1, . . . , 6}. To see that θ ′ can be extended to an embedding θ1 : N5,1 → Nhn observe that (1) c1 , c4 and c5 (resp. a1 , a4 and a5 ) bound a pair of pants on N5,1 (resp. Nhn ); (2) c3 , c4 , c5 and ∂N5,1 bound a 4-holed sphere; (3) c1 and c3 (resp. a1 and a3 ) bound a subsurface of N5,1 (resp. Nhn ) diffeomorphic to N1,2 . Suppose h = 2g + 2. Set c5 = c1,2 , c6 = c2,3 , c4 = c3,4 , c2 = c4,5 , c1 = c1,4 , c0 = c5,6 . Let K be a regular neighborhood of c1,6 ∪ c5,6 ∪ c6,6 . Observe that K is a Klein bottle with two holes, whose one boundary component is isotopic to c1 = c1,4 . Let c3 denote the other component of ∂K. We have |ci ∩ cj | = |ai ∩ aj | for all i, j ∈ {0, . . . , 6}. Let M (resp. M ′ ) be a regular neighborhood of the union of ci (resp. ai ) for i ∈ {0, . . . , 6}. Observe that M and M ′ are both diffeomorphic to Σ2,4 . There is a diffeomorphism θ ′ : M → M ′ such that θ ′ (ci ) = ai for i ∈ {0, . . . , 6}. To see that θ ′ can be extended to an embedding θ2 : N6,1 → Nhn observe that (1) c1 , c4 and c5 (resp. a1 , a4 and a5 ) bound a pair of pants on N6,1 (resp. Nhn ); (2) c3 , c4 , c5 and ∂N6,1 bound a 4-holed sphere; (3) two boundary circles of M (resp. M ′ ) bound an annulus with core c1,6 (resp. a). The conditions (a, b, d, e)  follow immediately from the construction of θ2 : N6,1 → Nhn . n Via these embeddings, we will treat N4+i,1 as a subsurface of N2g+i for i = 1, 2. Consequently, we will identify circles on N4+i,1 with their images n n on N2g+i , and also treat Mod(N4+i,1 ) as a subgroup of Mod(N2g+i ) (in particular ta5 = tc1,2 etc.). Since the composition tck−1,k u is a crosscap slide, the following proposition is an immediate corollary from [12, Theorem 4.1] n n )) is generated by ) (resp. PMod+ (N2g+2 Proposition 3.8. PMod+ (N2g+1 u and D (resp. E ∪ {ta }).

Lemma 3.9. Let h = 2g + 1 for g ≥ 2, D ′ = D\{tai : i = 1, 2, 3, 4} and H = CMod(Nhn ) (D ′ ). Let c be the nontrivial boundary component of a regular neighborhood of the union of the circles supporting D ′ . Then CH {ta1 , ta2 } is the free abelian group of rank 2 generated by (ta1 ta2 )3 and either tc if (g, n) 6= (2, 0), or (ta5 ta6 )3 if (g, n) = (2, 0). Proof. Let d be the boundary of a regular neighborhood of a1 ∪ a2 (torus with one hole) and set ρ = (ta1 ta2 )3 . Then ρ2 = td . Since tc can be expressed in terms of twists of D ′ , we have CH {ta1 , ta2 } ⊂ CMod(Nhn ) {tc , td }. It follows that any x ∈ CH {ta1 , ta2 } can be represented by a diffeomorphism, also denoted by x, equal to the identity on regular neighborhoods of c and d. The complement of the union of such neighborhoods has three connected components S ′ , S ′′ and N , where S ′ is diffeomorphic to Σng−1,1 (containing

˙ FERIHE ATALAN AND BLAZEJ SZEPIETOWSKI

16

a5 and a6 ), S ′′ is diffeomorphic to Σ1,1 (containing a1 and a2 ), and N is diffeomorphic to N1,2 . Clearly x preserves each of these components. Furthermore, x restricts to a diffeomorphism x′ of S ′ , which commutes with all twists of D ′ up to isotopy. Since PMod(S ′ ) is generated by twists of D ′ , x′ ∈ CMod(S ′ ) (PMod(S ′ )). By [8, Proposition 5.5 and Theorem 5.6], CMod(S ′ ) (PMod(S ′ )) = C(Mod(S ′ )) is the infinite cyclic group generated either by tc if (g, n) 6= (2, 0), or by (ta5 ta6 )3 if (g, n) = (2, 0) (note that tc = (ta5 ta6 )6 if (g, n) = (2, 0)). Thus x′ is isotopic on S ′ to some power of tc (or (ta5 ta6 )3 ). Analogously, x restricts to a diffeomorphism x′′ of S ′′ , isotopic on S ′′ to some power of ρ. Finally, since Mod(N ) is generated by the boundary twists, the restriction of x to N is isotopic to the product of some power of tc and some power of td .  Lemma 3.10. Let h = 2g + 1 for g ≥ 2 and suppose that ϕ is an automorphism of Mod(Nhn ) such that ϕ(t) = t for all t ∈ D. Then ϕ restricts to the identity on PMod+ (Nhn ). Proof. By Proposition 3.8, it suffices to prove ϕ(u) = u. Let M be the subgroup of Mod(Nhn ) generated by u, ta1 and ta2 . By [9, Theorem 4.1], M is isomorphic to Mod(N3,1 ). More specifically, it is the mapping class group of the nonorientable subsurface of Nhn bounded by the circle c from −1 −1 Lemma 3.9. Set u2 = u and u1 = t−1 a2 ta1 u ta1 ta2 . The following relations are satisfied in M (see [9]). (1) ta2 ta1 ta2 = ta1 ta2 ta1 (3) u2 u1 ta2 = ta1 u2 u1 (5)

ui tai u−1 i

=

t−1 ai

(2) u2 u1 u2 = u1 u2 u1 (4) ta2 u1 u2 = u1 u2 ta1

for i = 1, 2

(6) u2 ta1 ta2 u1 = ta1 ta2

−1 Set e = ta2 u2 ta1 u−1 2 ta2 . Note that e is a Dehn twist about the circle ta2 u2 (a1 ) = a3 (see (b) and (c) of Lemma 3.7). In particular ϕ(e) = e. Set v = eu−1 1 . We have (5)

(6)

−1 −1 −1 −1 e = ta2 u2 ta1 u−1 2 ta2 = ta2 u2 ta1 ta2 u2 = ta2 ta1 ta2 u1 u2

v = ta2 ta1 ta2 (u1 u2 u1 )−1 2 It follows from relations (1,3,4,5) that vtai v −1 = t−1 ai for i = 1, 2, and v = −2 −1 (u1 u2 u1 ) = tc (for the last equality see [9, Subsection 3.2]). Observe that ϕ(v)v −1 commutes with all twists of D ′ , where D ′ is as in Lemma 3.9, and also with tai for i = 1, 2. Suppose that (g, n) 6= (2, 0). By Lemma 3.9, 2 2k−1 , hence ϕ(v) = vtkc (ta1 ta2 )3m for some k, m ∈ Z. We have t−1 c = ϕ(v ) = tc k = 0. If (g, n) = (2, 0), then by Lemma 3.9, ϕ(v) = v(ta5 ta6 )3k (ta1 ta2 )3m , 2 k−1 , hence k = 0. We have ϕ(v) = v(t t )3m . and because t−1 a1 a2 c = ϕ(v ) = tc −1 −1 It follows that ϕ(u1 ) = u1 (ta1 ta2 )3m and ϕ(u2 ) = u2 (ta1 ta2 )3m . Set td = (ta1 ta2 )6 (d bounds a regular neighborhood of a1 ∪ a2 ) and y = ta2 u2 . Observe that the circles y(a4 ) and a4 are disjoint up to isotopy (recall a4 = c3,4 ), hence yta4 y −1 commutes with ta4 . By applying ϕ2 we

AUTOMORPHISMS OF THE MAPPING CLASS GROUP

17

−m −1 commutes with ta4 . By [11, Proposition 4.7] it obtain that ytm d ta4 td y m follows that i(td (a4 ), y −1 (a4 )) = 0. Set a′4 = y −1 (a4 ). We have i(a4 , a′4 ) = 0, i(a4 , d) = i(a′4 , d) = 2, i(a4 , a1 ) = i(a′4 , a1 ) = 0 and i(a4 , a2 ) = i(a′4 , a2 ) = 1. From these data, it is not difficult to deduce, using the bigon criterion [11, ′  Poposition 2.1], that i(tm d (a4 ), a4 ) = 4|m|, hence m = 0.

Lemma 3.11. Let h = 2g + 2 for g ≥ 2 and suppose that ϕ is an automorphism of Mod(Nhn ) such that ϕ(t) = t for all t ∈ E ∪ {ta }. Then ϕ restricts to an inner automorphism of PMod+ (Nhn ). Proof. Let K denote the nonorientable connected component of the surface obtained by removing from Nhn an open regular neighborhood of a1 ∪ a3 . Thus, K is a Klein bottle with two holes, and the other connected component is diffeomorphic to Σng−1,2 . Furthermore, by (e) of Lemma 3.7, K is a regular neighborhood of c1,6 ∪ c5,6 ∪ c6,6 . We will treat Mod(K) as a subgroup of Mod(Nhn ). Set u′ = ϕ(u). Since u′ commutes with all twists of E supported on Nhn \K, it can be represented by a diffeomorphism supported on K, by a similar argument as in the proof of Lemma 3.9. Hence u′ ∈ Mod(K). Since ′ u′ ta0 u′ −1 = t−1 a0 , u preserves the isotopy class of a0 by [11, Proposition 4.6]. Let S denote the subgroup of Mod(K) consisting of elements fixing the isotopy class of a0 , and let S + be the subgroup of index 2 of S consisting of elements preserving orientation of a regular neighbourhood of a0 . Note that every element of S + can be represented by a diffeomorphism equal to the identity on a neighborhood of a0 . By cutting K along a0 we obtain a four-holed sphere, and it follows from the structure of the mapping class group of this surface, that S + is isomorphic to Z3 × F2 , where the factor Z3 is generated by ta1 , ta3 and ta0 , and F2 is the free group of rank 2 generated by ta and uta u−1 . Set v = ta u. By [10, Lemma 7.8] we have v 2 = ta1 ta3 . Note that v ∈ S\S + . It follows from the previous paragraph, that S admits a presentation with generators ta1 , ta0 , ta and v, and the defining relations ta0 ta = ta ta0 , ta1 v = vta1 ,

vta0 = t−1 a0 v, ta1 ta0 = ta0 ta1 ,

v 2 ta = ta v 2 ta1 ta = ta ta1

Let H denote the subgroup generated by ta0 , ta1 and v 2 = ta1 ta3 . It follows from above presentation that H is normal in S and S/H is isomorphic to the free product Z ∗ Z2 . More specifically, denoting by A and V the images in S/H of respectively ta and v, we see that S/H has the presentation hA, V | V 2 = 1i. Since ϕ(tai ) = tai for i = 0, 1, 3 and ϕ(ta ) = ta and u′ = ϕ(u) ∈ S, ϕ preserves S and, by the same argument, ϕ−1 also preserves S, hence ϕ|S is an automorphism of S. Since ϕ is equal to the identity on H, it induces φ ∈ Aut(S/H). We have φ(A) = A. Note that every element of order 2 in S/H is conjugate to V . In particular φ(V ) is conjugate to V . It is an easy exercise to check, using the normal form of elements of the free product,

18

˙ FERIHE ATALAN AND BLAZEJ SZEPIETOWSKI

that in order for φ to be surjective, we must have φ(V ) = An V A−n for some n ∈ Z. n −n It follows that ϕ(v) = tka1 tla3 tm a0 ta vta for some integers l, k and m. We have ta1 ta3 = ϕ(v 2 ) = t2k+1 t2l+1 a1 a3 , hence k = l = 0. By composing ϕ with n the inner automorphism x 7→ t−n a xta we may assume n = 0 (note that ta commutes with all tai ). Thus ϕ(u) = tm a0 u. −1 As in the proof of Lemma 3.10, we have i(t−m a0 (a2 ), y (a2 )) = 0, where −1 y = ta0 u. It can be verified, by the bigon criterion, that i(t−m a0 (a2 ), y (a2 )) = |m|, hence m = 0. 

Proof of Theorem 1.1. Suppose that ϕ is any automorphism of Mod(Nhn ) for h ≥ 5. By Lemma 3.6, there exists f ∈ Mod(Nhn ) such that ϕ′ defined by ϕ′ (x) = f −1 ϕ(x)f for x ∈ Mod(Nhn ) is the identity on D (if h is odd) or E ∪ {ta } (if h is even). By Lemma 3.10 or Lemma 3.11, ϕ′ restricts to an inner automorphism of PMod+ (Nhn ). Thus, by composing ϕ′ with an inner automorphism we obtain ϕ′′ , which restricts to the identity on PMod+ (Nhn ). Since CMod(Nhn ) (PMod+ (Nhn )) is contained in the centralizer of the subgroup generated by all Dehn twits, it is trivial by [11, Theorem 6.2]. Lemma 3.1 implies that ϕ′′ is trivial, hence ϕ is inner. 

References [1] J. Aramayona and J. Souto, Homomorphisms between mapping class groups, Geom. Topol. 16 (2012), 2285–2341. [2] F. Atalan, Outer automorphisms of mapping class groups of nonorientable surfaces, Internat. J. Algebra Comput. 20(3) (2010) 437–456. [3] F. Castel, Geometric representations of the braid groups, Preprint 2011, arXiv:1104.3698 [4] N. V. Ivanov, Automorphisms of Teichmuller modular groups, in Lecture Notes in Math. 1346 (Springer, Berlin, 1988) pp. 199–270. [5] N. V. Ivanov and J. D. McCarthy, On injective homomorphisms between Teichm¨ uller modular groups I, Invent. Math. 135 (1999), 425–486. [6] M. Korkmaz, Mapping class groups of nonorientable surfaces, Geom. Dedicata 89 (2002) 109-133. [7] C. Labru`ere and L. Paris, Presentations for the punctured mapping class groups in terms of Artin groups, Algebr. Geom. Topol. 1 (2001), 73–114. [8] L. Paris and D. Rolfsen, Geometric subgroups of mapping class groups, J. Reine. Angew. Math. 521 (2000), 47–83. [9] L. Paris and B. Szepietowski, A presentation for the mapping class group of a nonorientable surface, Preprint 2013, arXiv:1308.5856 [10] B. Szepietowski, A presentation for the mapping class group of a non-orientable surface from the action on the complex of curves, Osaka J. Math. 45 (2008), 283–326. [11] M. Stukow, Dehn twists on nonorientable surfaces, Fund. Math. 189 (2006), 117–147. [12] M. Stukow, The twist subgroup of the mapping class group of a nonorientable surface, Osaka J. Math. 46 (2009) 717–738.

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19

Department of Mathematics, Atilim University, 06836 Ankara, TURKEY E-mail address: [email protected] ´ sk University, Wita Stwosza 57, 80-952 Gdan ´ sk, Institute of Mathematics, Gdan Poland E-mail address: [email protected]