Generating the surface mapping class group by two elements

Wajnryb proved that the mapping class group of an orientable surface is generated by two elements. We prove that one of these generators can be taken as a Dehn twist. We also prove that the extended mapping class group is generated by two elements, again one of which is a Dehn twist. Another result we prove is that the mapping class groups are also generated by two elements of finite order.


Introduction
Let Σ be a compact connected oriented surface of genus g with one boundary component. We denote by Mod 1 g the mapping class group of Σ, the group of isotopy classes of orientation-preserving diffeomorphisms Σ → Σ which restrict to the identity on the boundary. The isotopies are also required to fix the points on the boundary. If the diffeomorphisms and the isotopies are allowed to permute the points on the boundary of Σ, then we get the group Mod g,1 . The extended mapping class group Mod * g,1 of Σ is defined to be the group of isotopy classes of all (including orientation-reversing) diffeomorphisms of Σ. These three groups are related to each other as follows: Mod g,1 is contained in Mod * g,1 as a subgroup of index two and the groups Mod 1 g and Mod g,1 fit into a short exact sequence 1 → Z → Mod 1 g → Mod g,1 → 1, where Z is the subgroup of Mod 1 g generated by the Dehn twist about a simple closed curve parallel to the boundary component of Σ.
In this paper, we will be interested in the groups Mod 1 g , Mod g,1 and Mod * g,1 . The mapping class group and the extended mapping class group of the closed surface of genus g obtained from Σ by gluing a disc along the boundary component are denoted by Mod g and Mod * g . The mapping class group is a central object in low-dimensional topology. Therefore, its algebraic structures are of interest. The problem of finding the generators for the mapping class group of a closed orientable surface was first considered by Dehn. He proved in [D] that Mod g is generated by a finite set of Dehn twists. About quarter century later, Lickorish [L1, L2] also proved the same result, showing that 3g−1 Dehn twists generate Mod g . This Date: March 29, 2022. * Supported by TÜBA-GEBİP. number was improved to 2g + 1 by Humphries [Hu]. These 2g + 1 generators are the Dehn twists about the curves b, a 1 , a 2 , . . . , a 2g in Figure 1, where the closed surface is obtained from Σ by gluing a disc along the boundary component. Humphries proved, moreover, that in fact the number 2g + 1 is minimal; i.e. Mod g cannot be generated by 2g (or less) Dehn twists. Johnson [J] proved that the 2g + 1 Dehn twists about b, a 1 , a 2 , . . . , a 2g on Σ also generate Mod 1 g . Finally, the minimal number of generators for the mapping class group is determined by Wajnryb [W2]. He showed that Mod 1 g , and hence Mod g , can be generated by two elements; one is a product of two Dehn twists (one is right and one is left) and the other is a product of 2g Dehn twists. Since the mapping class group is not abelian, the number two is minimal. Recently, it was shown by Brendle and Farb in [BF] that the mapping class group is generated by three torsion elements and by seven involutions.
Since Mod g,1 is a quotient of Mod 1 g , it is generated by the corresponding 2g + 1 Dehn twists. In order to generate the extended mapping class group Mod * g,1 , it suffices to add one more generator, namely the isotopy class of any orientation-reversing diffeomorphism.
In this paper we have three main results. First, we improve Wajnryb's result. We show that one of the two generators of Mod 1 g can be taken as a Dehn twist. All Dehn twists involved in our generators are Dehn-Lickorish-Humphries generators. We also prove that the extended mapping class group Mod * g,1 is generated by two elements, again one of which is a Dehn twist. Our proof is independent from that of Wajnryb [W2]. Next, we prove that the mapping class groups Mod g,1 and Mod g are also generated by two torsion elements. Of course, this is not true for Mod 1 g since it is torsion-free. In the last section of the paper, we transform the presentation of the mapping class group in [W3] into a presentation on our two generators.

Preliminaries
Recall that if a is a simple closed curve on the oriented surface Σ, then the (right) Dehn twist A about a is the isotopy class of the diffeomorphism obtained by cutting Σ along a, twisting one of the side by 2π to the right and gluing two sides of a back to each other. We denote the curves by the letters a, b, c, d (possibly with subscripts) and the Dehn twists about them by the corresponding capital letters A, B, C, D. Notationally we do not distinguish a diffeomorphism/curve and its isotopy class. We use the functional notation for the composition of two diffeomorphisms; if F and G are two diffeomorphisms, then the composition F G means that G is applied first.
We define the curves c i , d i ,c i andd i on Σ as shown in Figure 2, so that c i (resp.d i ) is obtained from c i (resp. d i ) by rotating the surface Σ about the y-axis by π. These curves will be used through out the paper. Here, we assume that the surface of the paper is the yz-plane, the positive side of the x-axis is pointing above the page and the surface is invariant under the rotation by π about the y-axis. Let G be a subgroup of Mod 1 g or Mod * g,1 . Then G acts on the set of the isotopy classes of simple closed curves. If c is a simple closed curve, then we denote by c G the G−orbit of c; We record the following lemmas.
Lemma 1. Let c be a simple closed curve on Σ, let F be a self-diffeomorphism of Σ and let F (c) = d. Then F CF −1 = D r , where r = ±1 depending on whether F is orientation-preserving or orientation-reverving.
Lemma 2. Let c and d be two simple closed curves on Σ. If c is disjoint from d, then CD = DC.
Lemma 3. Let c and d be two simple closed curves on Σ. Suppose that C ∈ G and the curve d is contained in c G . Then D is also contained in G.
Proof. Since d ∈ c G , F (c) = d for some F ∈ G. By Lemma 6, F CF −1 = D r . This shows that D is contained in G.
3. The mapping class group Mod 1 g Let S denote the product A 2g A 2g−1 · · · A 2 A 1 of 2g Dehn twists in Mod 1 g and let G be the subgroup of Mod 1 g generated by B and S. We prove in this section that G = Mod 1 g . It follows that the mapping class groups Mod g,1 and Mod g are also generated by B and S. The main idea of the proof is to show that the G-orbit b G of the curve b contains the simple closed curves a 1 , a 2 , . . . , a 2g .
Lemma 4. The curves c 1 , c 2 , . . . , c g−1 and d 1 , d 2 , . . . , d g−2 of Figure 2 Proof. This follows from these easily verified facts: These facts will be used in Section 5 in the proof of the fact that the mapping class groups Mod g,1 and Mod g are generated by two torsion elements Figure 3. g is odd.
Theorem 5. Suppose that g ≥ 2. The subgroup G generated by B and S is equal to the mapping class group Mod 1 g .

Proof. It can easily be shown that S(a
Suppose that the curve a i 0 is contained in b G for some i 0 . Since B ∈ G, by Lemma 3 we get that A i 0 is also contained in G. Therefore, all A i are contained in G. Since the mapping class group Mod 1 g is generated by the 2g+1 Dehn twists A 1 , A 2 , . . . , A 2g and B, we conclude that the subgroup G is in fact equal to Mod 1 g . Therefore, in order to finish the proof of the theorem it suffices to show that a i is contained in b G for some i with 1 ≤ i ≤ 2g.
Suppose first that g is odd. It is easy to see that the diffeomorphism maps the curve S(b) to a curve α and C 2 D 1 maps α to a 3 (cf. Figure 3).
Suppose now that g is even. In this case the diffeomorphism This concludes the proof of the theorem.
4. The extended mapping class group Mod * g,1 In this section we prove that the extended mapping class group Mod * g,1 is also generated by two elements, one of which is a Dehn twist.
Consider the surface Σ embedded in the 3-space as shown in Figure 1. Let R denote the reflection across the xy-plane and let T denote the product A 2g A 2g−1 · · · A 2 A 1 R. Let H denote the subgroup of Mod * g,1 generated by B and T . We prove that H = Mod * g,1 . Again, it follows that Mod * g is also generated by B and T . Recall that the H-orbit of the simple closed curve b is denoted by b H .
Lemma 6. (i) If g is even, then the curves c 1 ,c 2 , c 3 ,c 4 , . . . ,c g−2 , c g−1 and Proof. It can be shown easily that the diffeomorphism T −1 maps The lemma follows from these and Lemma 3.
is contained in H and it maps the curve b to e 1 (c.f. Figure 5). This proves that e 1 ∈ b H . For the proof of e 2 ∈ b H let U denote the diffeomorphism Now it suffices to show that C 1 UC g−1 T −1 maps the curvec g−1 to e 2 . This can be seen from Figure 6. The curve e 3 is the image of the curve e 2 under the diffeomorphism E −1 1 (C 2 ) −1 (c.f. Figure 7). This proves that e 3 ∈ b H . Finally, T 3 (e 3 ) = e 4 proving that e 4 ∈ b H (c.f. Figure 8). Proof. We prove this theorem in the same way as Theorem 5; we show that the H-orbit b H of the curve b contains simple closed curves a 1 , a 2 , . . . , a 2g . It is easy to show that T (a i ) = a i−1 for i = 2, 3, . . . , 2g − 1. Hence, T A i T −1 = A i−1 , and thus A i ∈ H if and only if A i−1 ∈ H.
Suppose that a i 0 ∈ b H for some i 0 . Since B is contained in H, by Lemma 3 we get that A i 0 is also contained in H. Therefore, all A i are contained in H.
Since T ∈ H, the reflection R is also contained in H. The extended mapping class group Mod * g,1 is generated by the 2g + 1 Dehn twists B, A 1 , A 2 , . . . , A 2g and the reflection R. We conclude that the subgroup H is in fact equal to Mod * g,1 . Therefore, in order to finish the proof of the theorem it suffices to show that b H contains a i for some i with 1 ≤ i ≤ 2g.
Suppose first that g is even. It follows from Lemma 3 and Lemma 6 that the diffeomorphism Figure 9 shows that V maps the curve c g−1 to a 1 . Since c g−1 ∈ b H , a 1 is also in b H . This finishes the proof in this case.
Suppose now that g is odd. It easy to verify that (C 1 ) −1 E 4 (e 2 ) = a 1 . Since e 2 ∈ b H , and C 1 and E 4 are in H, we conclude that a 1 ∈ b H . This completes the proof of the theorem.  Figure 9. The proof of a 1 ∈ b H for even g.

Generating mapping class groups by two elements of finite order
In this section we prove that the mapping class groups Mod g,1 and Mod g are generated by two elements of finite order. Clearly, this will be the minimum number of such generators. In their paper [BF], Brendle and Farb proved that these mapping class groups are generated by three torsion elements and asked if they can be generated by two. Therefore our result gives a positive answer to their question.
Let Σ be a surface with one boundary component as in Figure 1. In Mod g,1 , let S denote the product A 2g A 2g−1 · · · A 2 A 1 . Note that S is of order 4g + 2. Throughout this section, let G denote the subgroup of Mod g,1 generated by the two torsion elements S and BSB −1 . We prove that G = Mod g,1 for g ≥ 3.
b 2 e Figure 10. The curves e and b 2 .
In the proof of the main result of this section, we use the celebrated lantern relation, which was discovered by Dehn and redisovered by Johnson. This relation is read as follows; where the B 2 and E are the Dehn twists about the curves b 2 and e in Figure 10. We rewrite the relation as The strategy of the proof is to show that the statements inside each paranthesis are in G. Then the rest of the proof is easy.
Let us define B as the subset of nonseparating simple closed curves consisting of those curves x such that XB −1 ∈ G. That is, x is a nonseparating simple closed curve and XB −1 ∈ G}.
We first state the following easy to prove, but very useful, lemmas.
Lemma 9. Suppose that two simple closed curves x and y are contained in B. Then XY −1 is contained in the subgroup G.
Lemma 10. Suppose that y ∈ B and XY −1 ∈ G. Then x ∈ B.
Lemma 11. Suppose that a curve y is contained in B and x is in the Sorbit of y. Then x is contained in B.
Proof. By assumption, there is an integer k such that x = S k (y). Since the element is contained in G, the lemma is proved.
Proof. The fact that c i , d j are in the S -orbit of b is shown in Section 3. It can be shown that S 2g+1 is isotopic to the rotation by π about the yaxis. Sincec i = S 2g+1 (c i ) andd j = S 2g+1 (d j ), the corollary follows from Lemma 11. Lemma 13. Suppose that g ≥ 3. Then for each i = 1, 2, . . . , g, the curves a i , b 2 and e are contained in B.
Proof. We prove first that the curves a i are contained in B. Since all a i are contained in the same S -orbit, it is enough to prove that any one of a i is in B.
Suppose that g is even, so that g ≥ 4. Let V denote the product We have shown in the proof of Theorem 5 that the diffeomorphism BD −1 2 V S maps the curve b to a 4 . Since the curvē c 1 is disjoint from each c i and d j for i ≥ 1, j ≥ 3, the Dehn twistC 1 about c 1 commutes with each C i and D j . Therefore it commutes with V . Let By the above lemmas, we also have SBS −1C −1 1 ∈ G. Therefore is contained in G. This shows that a 4 , and hence, all a i , is in B. Suppose next that g is odd and g ≥ 5. Now again let V denote the diffeomorphism D −1 3 C −1 4 D −1 5 C −1 6 · · · D −1 g−2 C −1 g−1 . We have shown in the proof of Theorem 5 that the diffeomorphism C 2 D 1 BV S maps the the curve b to a 3 . In this case, let x denote the curve V S(b), as above. The equation (3) shows that x ∈ B. We now use the equality to conclude that a 3 is in B. Suppose now that g = 3. It can easily be shown that C −1 1 D −1 1 S 2 BS(b) = a 5 . We use this fact to prove that a 5 is contained in B. Notice that shows that b 2 ∈ B. Let y denote the curve BS(b). From the equation , we conclude that y ∈ B. By Lemma 11, z = S 2 (y) = S 2 BS(b) is also contained in B. Finally, the fact that a 5 is in B follows from the equation This concludes the proof of a i ∈ B. In order to finish the proof of the lemma, it remains to prove that b 2 and e are contained in B.
It is easy to see that C −1 2 A 6 A 5 A 4 (b) = b 2 and A 2 A 1 A −1 4 C 1 (a 5 ) = e. It can be shown that the diffeomorphisms A −2 1 C −1 2 A 6 A 5 A 4 and B −2 This completes the proof.
Theorem 14. The mapping class group Mod g,1 (and hence Mod g ) is generated by two elements of finite order.
Proof. If g = 1 then A 2 A 1 and A 1 A 2 A 1 are of orders 6 and 4, respectively, and they generate Mod 1,1 . If g = 2 then A 4 A 3 A 2 A 1 and A 5 A 4 A 3 A 2 A 1 are of orders 10 and 6, respectively, and they generate Mod 2,1 . Suppose that g ≥ 3 and let G be the subgroup of Mod g,1 generated by the elements S = A 2g A 2g−1 · · · A 2 A 1 and BSB −1 , which are both of order 4g + 2. Since the curves a 3 , a 5 , b, b 2 , d 1 and e are all contained in B, the elements BA −1 3 , D 1 A −1 5 and EB −1 2 are contained in G. By the lantern relation (2), Finally, the element C −1 1 A −1 1 A 4 A 3 is in G and maps a 2 to b. Since A 2 ∈ G, this shows that B ∈ G. Consequently, G = Mod g,1 .
This finishes the proof of the theorem.

A presentation of the mapping class group on two generators
In this last section we transform the Wajnryb presentation of the mapping class group Mod 1 g to a presentatoin on the two generators B and S. It turns out that the number of relations in the new presentation depends linearly on g, whereas it is quadratic in the original presentation.
Theorem 15. ( [W1, W3]) Let g ≥ 3. The mapping class group Mod 1 g admits a presentation with generators B, A 1 , A 2 , . . . , A 2g and with defining relations Suppose that g ≥ 3. We have shown above that the curve b is mapped to a 3 by the diffeomorphism if g is odd, and to a 4 by if g is even.
From the equalities S −2k+1 (b) = c k and S −2k (b) = d k we get that S −2k+1 BS 2k−1 = C k and S −2k BS 2k = D k , which should replace C k and D k in the equations (5) and (6). Also note that S(a k ) = a k−1 and hence SA k S −1 = A k−1 . Let us now define so that S −k X(b) = a k , and therefore for every k = 1, 2, . . . , 2g.
The presentation of the mapping class group Mod 1 g in Theorem 15 has 2g+ 1 generators and 2g 2 +g+2 relations. By replacing each A k in Theorem 15 by S −k XBX −1 S k , we get another presentation of Mod 1 g on the two generators B and S. Now these relations are not as nice as those of Wajnryb's. But, the number of relations in the new presentation reduces to 4g + 1.
For any i, j with |i − j| > 1 and for any k, all the relations A i+k A j+k = A j+k A i+k reduce to the single relation B X B S i−j X = B S i−j X B X . This shows that the relations in (i) gives rise to (2g − 1) + (2g − 2) = 4g − 3 relations.
Each braid relation A k A k+1 A k = A k+1 A k A k+1 reduces to the single relation B X B S −1 X B X = B S −1 X B X B S −1 X . Thus the relations in (ii) reduces to two relations.
Consequently the new presentation of Mod 1 g is given as follows: Let P denote B X and let U denote S −1 , where X is defined as in (7).
Theorem 16. Let g ≥ 3. The mapping class group Mod 1 g admits a presentation with generators B and U , and with defining relations (i) BP U i = P U i B, for 1 ≤ i ≤ 2g and i = 4, (ii) P P U i = P U i P for 2 ≤ i ≤ 2g − 1, (iii) BP U 4 B = P U 4 BP U 4 and P P U P = P U P P U , (iv) (P U P U 2 P U 3 ) 4 = BB V , and (v) P U P U 3 P U 5 B W = BB t −1 1 t −1 2 B t −1 2 , where V = P U 4 P U 3 P U 2 P U 1 P U 1 P U 2 P U 3 P U 4 . t 1 = P U 2 P U P U 3 P U 2 , t 2 = P U 4 P U 3 P U 5 P U 4 and W = P U 6 P U 5 P U 4 P U 3 P U 2 (t 2 P U 6 P U 5 ) −1 B(t 2 P U 6 P U 5 )(P U 4 P U 3 P U 2 P U ) −1 .
The presentation of the mapping class group Mod g is obtained from that of Mod 1 g by adding one more relation (c.f. [W3]). Hence, in the new presentation on the generators B and S the number of relation reduces to 4g + 2.