ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 15 (2018) 297-321 https://doi.org/10.26493/1855-3974.806.c9d (Also available at http://amc-journal.eu) Finite actions on the 2-sphere, the projective plane and I-bundles over the projective plane John Kalliongis Department of Mathematics and Statistics, Saint Louis University, 220 North Grand Boulevard, Saint Louis, MO 63103 Ryo Ohashi Department of Mathematics and Computer Science, King's College, 133 North River Street, Wilkes-Barre, PA 18711 Received 5 February 2015, accepted 6 February 2018, published online 25 June 2018 In this paper, we consider the finite groups which act on the 2-sphere S2 and the projective plane P2, and show how to visualize these actions which are explicitly defined. We obtain their quotient types by distinguishing a fundamental domain for each action and identifying its boundary. If G is an action on P2, then G is isomorphic to one of the following groups: S4, A5, A4, Zm or Dih(Zm). For each group, there is only one equivalence class (conjugation), and G leaves an orientation reversing loop invariant if and only if G is isomorphic to either Zm or Dih(Zm). Using these preliminary results, we classify and enumerate the finite groups, up to equivalence, which act on P2 x I and the twisted I-bundle over P2. As an example, if m > 2 is an even integer and m/2 is odd, there are three equivalence classes of orientation reversing Dih(Zm)-actions on the twisted I-bundle over P2. However if m/2 is even, then there are two equivalence classes. Keywords: Achiral symmetry, chiral symmetry, equivalence of actions, finite group action, isometry, orbifold, symmetry. Math. Subj. Class.: 57S25, 05E18, 57M60, 57R18, 58D19, 57M20 1 Introduction The finite orientation preserving groups which act effectively on S2 are known. (See for example Gross and Tucker [5] and Zimmermann [9].) They are the octahedral symmetric group S4, the dodecahedral/icosahedral alternating group A5, the tetrahedral alternating E-mail addresses: kalliongisje@slu.edu (John Kalliongis), ryoohashi@kings.edu (Ryo Ohashi) Abstract ©® This work is licensed under https://creativecommons.Org/licenses/by/3.0/ 297 Ars Math. Contemp. 15 (2018) 441-466 group A4, the cyclic group Zm or the dihedral group Dih(Zm). Using this classification, the actions on the projective plane P2 are also known as folklore, and one can easily compute them by this theorem of Singerman [7] and Tucker [8]. Theorem. Let F be a closed non-orientable surface and let p: F ^ F be the orientable double cover with covering translation t: F ^ F. Then any finite group G acting on F, lifts to an orientation preserving action of G on F that commutes with t. Moreover, the action of G on F is determined by the action of G x (t) on F. If t: S2 ^ S2 is the covering translation such that S2 / (t) = P2, one checks that any rotation of S2 commutes with t. Therefore since these groups consists of rotations, it follows that the orientation preserving actions on S2 project to P2, giving the following corollary. Corollary. Any finite group acting on P2 is isomorphic to one of the following groups: S4, As, A4, Zm or Dih(Zm). A finite G-action on a manifold M is a monomorphism p: G ^ Homeo(M), where G is a finite group, and Homeo(M) is the group of homeomorphisms of M. Two actions p1 and p2 are equivalent if there exists a homeomorphism h of M such that hp1 (G)h-1 = p2 (G). For an action p, the quotient space M/p is an orbifold which is referred to as the quotient type of the action. In this paper, we describe how to visualize the finite groups which act on the 2-sphere S2 and the projective plane P2, and show how to obtain their quotient types. Our approach, for the groups which are not cyclic or dihedral, is to view these groups as subgroups of the symmetric group Sn for an appropriate n, tiling the 2-sphere with appropriate polygons with n vertices for each group, and explicitly defining each action. As for the cyclic and dihedral groups, we use spherical coordinates to precisely describe their actions on S2. For all these groups, we can easily identify an explicit fundamental region for each action and see its quotient type, which is obtained by identifying the boundary of the fundamental region. In this way, it is easy to see the actions on S2, P2 and their quotient types. This part of the paper may be considered expository, and we obtain the following theorem where the description of these quotient types may be found in Figure 1. Theorem 7.1. Let p: G ^ Homeo(P2) be a finite group action on P2. Then G is isomorphic to one of the following groups: S4, A5, A4, Zm or Dih(Zm). The orbifold quotient P2/p is an orbifold homeomorphic to one of the following orbifolds: Oh, Ih, Tv, Zm, S2m, Dm or Dim. There is only one equivalence class for each group. (1) G ~ S4 if and only if P2/p = Oh. (2) G ~ As if and only if P2/p = Ih. (3) G ~ A4 if and only if P2/p = Tv. (4) G ~ Zm and m is even if and only if P2/p = Zm. (5) G ~ Zm and m is odd if and only if P2/p = S2m. (6) G ~ Dih(Zm) and m odd if and only if P2/p = Dm. (7) G ~ Dih(Zm) and m even if and only if P2/p = D^j m J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles. 299 This approach relates to topics in topological graph theory found in Gross and Tucker [5]. There, graphs are embedded on surfaces and finite groups act on these spaces with quotient spaces, branch covering maps and branch points, relating to orbifold covering maps and cone points. Using the above result, we classify in Theorem 7.4 the finite group actions, up to equivalence, on P2 x I for I = [0,1]. If G is an action on P2 x I, then G is isomorphic to one of the following groups: S4, S4 x Z2, As, As x Z2, A4, A4 x Z2, Zm, Zm x Z2, Dih(Zm) or Dih(Zm) x Z2. We indicate the number of equivalence classes for each group in Theorem 7.4. If W is the twisted I-bundle over the projective plane P2, then we obtain the following results: Corollary 8.12. Let <: G ^ Homeo(W) be a finite orientation preserving G-action on W. Then G is isomorphic to one of the following groups: S4, A5, A4, Zm or Dih(Zm). The orbifold quotient for each action is a twisted I-bundle orbifold over the following 2-orbifolds: Oh (for S4), Ih (for As), Tv (for A4), Zh (for Zm and m even), S2m (for Zm and m odd), Dvm (for Dih(Zm) and m odd) and Dh (for Dih(Zm) and m even). There is one equivalence class for each quotient type. Theorem 9.4. Let <: G ^ Homeo(W) be an orientation reversing G-action. Then G is isomorphic to one of the following groups: S4, Zm with m even, Dih(Zm), S4 x Z2, As x Z2, A4 x Z2, Zm x Z2 or Dih(Zm) x Z2. (1) If G is either S4, S4 x Z2, As x Z2, A4 x Z2, Zm x Z2 with m even or Dih(Zm) with m odd, there is only one equivalence class. (2) If G is Zm with m > 2 even and m/2 odd, then there are two equivalence classes of Zm = Zm/2 x Z2-actions on W. (3) If G is Zm with either m/2 even or m = 2, then there is only one equivalence class. (4) If G is Dih(Zm) with m > 2 and m/2 even, there are two equivalence classes of Dih(Zm)-actions on W. (5) If G is Dih(Zm) with m > 2 and m/2 odd, there are three equivalence classes of Dih(Zm)-actions on W. (6) If G is Dih(Zm) x Z2 with m even, there is only one equivalence class. (7) If G is Dih(Zm) x Z2 with m odd, then Dih(Zm) x Z2 ~ Dih(Z2m) and there are three equivalence classes of Dih(Z2m)-actions on W. We list all the closed 2-orbifolds with positive Euler number, of which there are 14. (See Figure 1.) In referring to these orbifolds, we use Schonflies notation found in Coxeter and Moser [1], and Dunbar [3]. There are five orientable 2-orbifolds with positive Euler number which have as their underlying space a 2-sphere with the cone points indicated in the notation. They are E(2, 2, n) = Dn, E(2, 3, 3) = T, E(2, 3, 4) = O, E(2, 3, 5) = I, and £(n,Z) = Cn,;. These double cover the following nine non-orientable 2-orbifolds where the double lines are reflector lines. The superscripts h and v stand for horizontal and vertical reflections in their orientable double covers. Except for E(n, Z) = Cn,; where the cone points are at the north and south poles, all the cone points are located on the equator. In this article, where appropriate and depending on the context, we use the same symbol to denote the quotient space and the group acting on S2. For example, O = E(2, 3,4) and O also denote the octahedral group. 300 Ars Math. Contemp. 15 (2018) 441-466 Here is a brief outline of the paper. We consider each of these orbifolds in Sections 2 through 6, and give model maps which we consider as standard actions, to obtain each quotient type. Summarizing we give the main results for finite actions on P2 and P2 x I in Section 7. Sections 8 and 9 are devoted to classifying the finite actions on the twisted I-bundle W over P2. The authors wish to thank the referees for many helpful comments and suggestions. 2 Chiral octahedral symmetry O and achiral octahedral symmetry Oh We describe the groups O = S4 and Oh = S4 x Z2 acting on the 2-sphere S2, and show how O acts on the projective plane P2. We view S2 as an octahedron which has eight triangles (faces): A125, A145, A126, A146, A235, A236, A345 and A346. (See Figure 2.) Consider elements of S6 where a = (1,2)(3,4)(5,6) and b = (1, 2, 5)(3,4,6). The two elements act on the octahedron. We can see that a is a 180° rotation about the axis passing through the midpoint of edges 1,2 and 3,4. On the other hand, b is a 120° rotation about the axis passing through the barycenter of A125 and A346 respectively. Further, ab = (2, 6,4,5) where ab is a 90° rotation about the axis passing through vertices 1 and 3. As a result, the two elements a and b generate a group isomorphic to S4, and we denote this group by O = (a,b | a2 = b3 = (ab)4 = 1), the octahedral group. Next, we use E to denote the quotient space of S2 by O, and we will find a fundamental region for E on S2. We first claim that A125 will tile the whole octahedron S2 by the action of O. Observe that the action by a sends A125 to A216. Further, b2 (ab)b-2 = (1,4, 3,2) is a 90° rotation about the axis passing through vertices 5 and 6, which shows our claim. Note that the number of fundamental regions for E on S2 must be 24 as the number is the order of the octahedral group O = S4. Since the S2 currently has eight faces, we will have to triangulate them further. Our approach is that we will add one more vertex on the barycenter on each triangle. For instance, one of the triangulations on A125 is shown in the Figure 2. We now show that A12y becomes a fundamental region for E. Since a rotational axis of b passes the vertex y, the barycenter of A125, one can see that b permutes those three triangles A12y, A51y and A25y. In the meantime, edges 1,xi and 2, xi are identified J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles. 301 5 5 2 2 xi 1 6 (a) Front of S2. 6 (b) Back of S2. (c) Triangle 125. Figure 2: S2 as an octahedron. by a. Likewise, edges 1, y and 2, y are identified by b. Points 1, x1 and y will be cone points under the action. Each has an order 4, 2 and 3 respectively. Consequently, we obtain £ = S2/O = £(2,3,4). In order to obtain Oh, we consider an action i = (1,3)(2,4)(5, 6) which is the antipodal map on S2. Notice that the antipodal map commutes with the elements in O, hence it induces the reflection map on S2/O = £(2, 3,4). Now, we choose a triangle whose vertices are 1, x1 and y. Apply (ab)2b(ab)i on the triangle gives us the triangle with vertices 2, x1 and y. Notice that segments 1x1 and 2x1; 1y and 2y have been identified under O-action and the segment x1y has been fixed under the map (ab)2b(ab)i. This argument shows that A1x1y is a fundamental region for O x (i)-action on S2. The vertices of A1x1y become corner reflectors, and the edges minus the vertices become the reflector lines. As a result, S2/[O x (i)] = Oh, where O x (i) = S4 x Z2 = n1 (Oh). We remark that S2/(i) = P2 is the projective plane. Since the antipodal map i commutes with O, the octahedral action on S2 induces the action generated by a and b on P2, which is isomorphic to the octahedral group O. As aresult, we also obtain P2/(a, b) = Oh. We will now describe the octahedral action O on P2. The left diagram in Figure 3 illustrates a fundamental region on S2 used to obtain P2 under the antipodal map i = (1,3)(2,4)(5, 6). For any arc x, y, z, we let [x, y, z] be its projection in P2. The arc 1, 2, 3 (or 3,4,1 etc) on S2 projects to an orientation reversing loop [172,3] on P2. The generator a maps the loop [172,3] onto [2714] = [2TT][174] = [2,1][3, 2], which traces the same loop as [1,2,3] starting at a different point. Thus a leaves the loop [1, 2,3] invariant and restricted to this loop is a rotation. On the other hand, the map b maps the loop [1, 2,3] onto [2, 5,4] whose image is shown as a bold line in the middle _2 _ _ diagram in Figure 3 above. Moreover, b maps the loop [1,2,3] onto the loop [1, 3, 5]. Thus the Z3-action generated by b does not leave the orientation reversing loop [1,2,3] on P2 invariant. However, it is important to emphasize that this does not imply the Z3-action leaves no orientation reversing loops invariant. In fact, we can find another orientation reversing loop on P2 which is left invariant under the map b. It can be found by looking at the octahedron S2 which double covers P2. Consider the circle on S2 which contains the vertices consisting of the midpoints of 4, 5, 5,3, 3, 2, 2, 6, 6,1 and 1,4. One can check that this circle is left 302 Ars Math. Contemp. 15 (2018) 441-466 invariant under b and the covering translation i, hence it projects to an orientation reversing loop on P2 left invariant under b. It follows that the entire S4 -action on P2 does not leave any orientation reversing loop invariant. Lemma 2.1. Let Z2 be a subgroup of n1(Oh) such that P2 ^ Oh is the covering corresponding to Z2. Then Z2 = (i). Proof. We will show that there is only one element of order two in Oh = S4 x Z2 acting on S2 which is fixed point free and orientation reversing, and that element is i. Since the elements in S4 and Z2 commute, we will first look at all elements of order two in S4. In this group, there are nine such elements. Six of them are a rotation of 180° where their rotational axes are on midpoints of edges. For example, one rotational axis passes the midpoint of 1,4 and 2,3. Another one passes the midpoint of 2, 5 and 4,6. Notice that all six types of these rotations are conjugate in S4. Moreover, there are three types of 90° rotations, call them r1, r2 and r3, where r1 = (1, 2,3,4), r2 = (1, 6,3,5) and r3 = (2, 5,4,6) respectively. Clearly, they generate three kinds of 180° rotations which are conjugate in S4. As a result, S4 has two conjugacy classes of order two elements, and we will choose a and (ab)2 from the group to represent each class. There is an easy way to verify if two elements in Sn are conjugate for n G N by checking their cycle types. (See [2, Chapter 4].) Now, we compose them with the antipodal map i to obtain ai = (1,4)(2,3) and (ab)2i = (1,3). Since both maps have a fixed point, if P2 ^ Oh is the covering corresponding to any Z2, then Z2 = (i). □ Proposition 2.2. Let <: G ^ Homeo(P2) be a finite action such that P2/< is homeomor-phic to Oh. Then G ~ S4 and < is conjugate to the standard action S4 = (a, b). Moreover, no orientation reversing loop is left invariant by the G-action. Proof. Let v: P2 ^ P2/ (a, b) and vv : P2 ^ P2/< be the orbifold covering maps. By assumption there exists a homeomorphism h: P2/(a, b) ^ P2/<. By Lemma 2.1, the Z2 subgroup of ni(P2/<) giving rise to a covering P2 ^ P2/< is unique. Hence h lifts to a homeomorphism h: P2 ^ P2 and we obtain the following commutative diagram: P2 h ) P2 P2/(a, b) —4 P2/< This implies that G ~ S4 and h conjugates < to the standard action (a, 6). □ J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles. 303 3 Chiral dodecahedral/icosahedral symmetry I and achiral dodeca-hedral/icosahedral symmetry Ih We describe the groups I = A5 and Ih = I x Z2 = A5 x Z2 acting on the 2-sphere S2, and show how A5 acts on P2. We view S2 as a dodecahedron consisting of 12 pentagons as shown in the first two figures from the left in Figure 4. We also consider two elements a and b in S20 where a = (1,2)(3, 7)(4,13)(5,8)(6,14)(9,12)(10,19)(11, 20)(15,18)(16,17) and the element b = (2, 5,7)(3,6,13)(4,12, 8)(9,11,19)(10,18,14)(15,17,20). The two elements act on the dodecahedron S2, and we can see that a is a 180° rotation about the axis passing through the midpoint of edges 1,2 and 16,17. On the other hand, b is a 120° rotation about the axis passing through the vertices 1 and 16. Moreover, ab-1 is a 72° rotation about the axis passing through the barycenter of the pentagon whose vertices are 1, 2, 3, 4, 5 and 16, 17,18, 19, 20 respectively since ab-1 = (1,2,3,4, 5)(6, 7,8,9,10)(11,12,13,14,15) (16,17,18,19, 20). Consequently, a and b generate a group isomorphic to I = A5 written by I = (a, b | a2 = b3 = (ab-1)5 = 1). We use E to denote the quotient space of S2 by I, and we will look for a fundamental region for E on S2. We will first observe that one of the pentagons consists of vertices 1, 2, 3, 4 and 5 tiles the remaining pentagons on S2. This pentagon is sent to the pentagon with the vertices 1, 5, 6, 12, 7 by b. Then, ab-1 permutes the remaining pentagons in the front of S2. On the other hand, (ab-1)2 sends the vertices 1, 5, 6,12, 7 to the vertices 4, 3, 9,15, 10. Then, the map a sends them to the vertices 13, 7, 12, 18, 19 on the back of S2. At this stage, one can see that ab-1 permutes all pentagons on the back of S2 except the one on the center whose vertices are 16, 17, 18, 19, 20. However, it can be obtained by applying the map b-1(ab-1)2 on the vertices 4, 3, 9, 15,10. 8 8 14 14' (a) Front of S2. (b) Back of S2. (c) Triangulated face. 5 2 Figure 4: S2 as a dodecahedron. Next, we will add a vertex on the barycenter of the pentagon 1, 2, 3, 4, 5 (see Figure 4), which we denote by y. We also add vertices x (1 < i < 5) on this pentagon. We can see that A12y tiles the remaining triangles on the pentagon 1, 2, 3, 4 and 5 (see Figure 4) by the map ab-1. By the argument above, this proves that A12y is a fundamental region for E. Now the edges 1,x1 and 2, x1 are identified by a G I. Likewise, ab-1 G I identifies edges 1, y and 2, y. The vertices 1, x1 and y are fixed by the elements b, a and ab-1 respectively. Thus, the vertices project to the cone points on E of orders 3, 2 and 5 respectively. Consequently, E = S2/I = E(2, 3, 5). 304 Ars Math. Contemp. 15 (2018) 441-466 In order to obtain Ih, we consider an antipodal map on S2 defined by i = (1,16)(2,17) (3,18)(4,19)(5,20)(6,14)(7,15)(8,11)(9,12)(10,13). It is easy to check that the antipodal map commutes with the elements of I, hence I x Z2 = (a, b, i | a2 = b3 = (ab-1)5 = i2 = 1, [a, i] = [b, i] = 1). As a result, i induces the reflection map on S2/I = E(2,3, 5). We choose a triangle whose vertices are 1, x1 and y on the fundamental region A12y. When the map a(ab-1)3b-1(ab)i is applied on A1x1y, its image is A4x3y. Since A4x3y is identified to A2x1y by (ab-1)-2 G I, this shows the antipodal map i on S2 induces the reflection map on E(2,3,5) at its equator line. Consequently, S2/[1 x (i)] = Ih, where I x (i) = A5 x Z2. The quotient space S2/[1 x (i)] is a mirrored disk where cone points of order 2, 3 and 5 are on the mirror. Note that S2/ (i) = P2 is the projective plane and the map i commutes with I. Thus, the icosahedral action on S2 induces the action generated by a and b on P2, which is isomorphic to the icosahedral group I. Hence, we obtain P2/ (a, b) = Ih. We will now describe the I = A5-action on P2, and show that it is unique up to conjugation. The front and back of S2 in Figure 4 describe a fundamental region used to obtain P2 = S2/(i) where i is the antipodal map on S2. Note that the boundary of each region in the diagram is left invariant and the interior of each region is exchanged under i. The arc 7,13, 8,14,9,15 (or 15,10,11,6,12,7 etc) projects to an orientation reversing loop [7,13, 8,14, 9,15] on P2. The map ab-1 leaves the outer most loop containing the arc invariant up to the covering translation i. Thus, the induced map ab-1 in P2 leaves this orientation reversing loop invariant. On one hand, a leaves the circle containing vertices 1, 2, 3, 9,15,16,17,18,12, 7, 1 in S2 invariant which double covers an orientation reversing loop on P2. Note that a leaves this orientation reversing loop invariant. However, the orientation reversing loops [7,13, 8,14,9,15] and [3,4,5, 6,12,18] = [3,4,5,6,9, 3] in P2 are exchanged by a. Finally, b will induce a map b on P2. One can see this since three orientation reversing loops in P2, namely [7,13, 8,14, 9,15], [2,3,4,10,11,17] = [2,3,4,10,8, 2] and [5,6,12,18,4,5] = [5, 6, 9,3,4, 5], are permuted under b. Note that although we can find an orientation reversing loop left invariant under b, no common orientation reversing loop exists which is left invariant by both a and b since the two maps generate an A5-action on P2. Lemma 3.1. Let Z2 be a subgroup of n1(1h) such that P2 ^ Ih is the covering corresponding to Z2. Then Z2 = (i). Proof. We claim that there is only one element of order two in Ih = A5 x Z2 acting on S2 which is fixed point free and orientation reversing up to a conjugacy. Notice that all elements in Ih have the form of al bmin for some l, m, n G Z where a, b G A5 and i G Z2. Since a corresponding covering space must be regular, the group generated by al bmin must be a normal subgroup in Ih. In particular, albm generates a normal subgroup of A5 which is impossible unless l = m = 0. Therefore, a covering space of the orbifold Ih corresponding to a Z2 subgroup in n1(/h) = A5 x Z2 is S2/(i) = P2. Therefore, an A5-action on P2 with quotient type Ih is unique up to conjugacy. □ Proposition 3.2. Let f: G ^ Homeo(P2) be a finite action such that P2/f is homeomor-phic to Ih. Then G ~ A5 and f is conjugate to the standard action I = (a, b). Moreover, no orientation reversing loop is left invariant by the G-action. Proof. The proof is similar to that of Proposition 2.2 and uses the above Lemma 3.1. □ J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles... 305 4 Chiral tetrahedral symmetry T and pyritohedral symmetry Tv We consider the groups T = A4 and Tv = T x Z2 = A4 x Z2 acting on the 2-sphere S2 and describe how T acts on the projective plane. 4 4 4 1 2 9 1 1 13 3 3 14 2 2 14 3 Figure 5: S2 as a tetrahedron. We view S2 as a tetrahedron which has four faces: A124, A134, A234 and A123 (see Figure 5). We add a total of 14 vertices on the faces to triangulate the tetrahedron, and A123 in the Figure 5 illustrates a "bottom" of the tetrahedron. Consider elements of S14 where a = (1,2, 3)(5, 6,7)(9,14,13)(10,12,11) and b = (1,2)(3,4)(5, 8)(6, 7)(10,13)(11,14). These two generators act on the tetrahedron S2. For instance, a is a 120° rotation about the axis passing throught the vertices 4 and 8; and b is a 180° rotation about the axis passing throught the vertices 9 and 12. It is easy to see that ab = (1, 3,4)(5,8,6)(9,14,10)(11,13,12) is a 120° rotation about the axis passing through the vertices 2 and 7. Hence, the two elements a and b generate a group isomorphic to A4, and we call this group by T = (a, b | a3 = b2 = (ab)3 = 1), which is the tetrahedral group. Let E be the quotient space of S2 by the group T. We will observe that the face A123 on the "bottom" of this tetrahedron tiles the rest of its faces. To understand this, we look at the map b which sends from A123 to A124. Then, the map a permutes A124 by 120° each time to tile the whole tetrahedron. However, this argument shows that we may choose A128 for a fundamental region for E since the map a permutes within the three triangles A128, A238 and A318 on the "bottom" face A123 of S2. Notice that b, which has the order two, fixes the vertex 9. Hence, it becomes an exceptional point of order two. Further, b identifies edges 1, 9 and 2, 9. On the other hand, a fixes the vertex 8, hence this vertex becomes a cone point of order three. Also, a identifies edges 1,8 and 2,8. Moreover, the map ab, which has an order three, fixes the vertex 2 to obtain an additional cone point of order three. Consequently, E = S2/T =E(2,3,3). Next, we will discuss how to obtain Tv. An antipodal map defined by i = (1, 6)(2, 7) (3,5)(4,8)(9,12)(10,13)(11,14) on S2 commutes with the elements in T, hence we have T x Z2 = (a, b, i | a3 = b2 = (ab)3 = i2 = 1, [a, i] = [b, i] = 1) and i induces a map on S2/T = E(2,3,3). However, it requires some work to analyze what map i induces on the orbifold E(2,3, 3). First, let x and y be the mid-point of the edge 1,8 and 2, 8 respectively. Since the points x and y are identified in E(2, 3, 3) by a G T, we may view the union of x, 9 and 9, y as the vertical equator line on E(2,3,3). Notice that the induced map i on E(2,3,3) fixes all points on the vertical equator line. It can be checked by observing that a-1(ab)-1(a2b)i fixes the points on the line x, 9; and (ab)(a2b)i fixes the points on the line 9, y. Secondly, we will show that the induced map i on E(2, 3, 3) is a reflection on the vertical equator line x, 9 U 9,y. To see this, consider A189 lying on our fundamental 306 Ars Math. Contemp. 15 (2018) 441-466 region A128. Apply the map a(ab)2i on A189 gives us A819 which is a reflection on x, 9. Likewise, A289 is reflected on 9, y by the map a(ab)2bi to get A829. As a result, the induced map i on E(2, 3, 3) is a reflection at the vertical equator line on the orbifold. By the argument above, S2/[T x (i)j = Tv, where T x (i) = A4 x Z2 and the quotient space S2/[T x (i)] is a mirrored disk containing a corner reflector containing one exceptional points of order 2 and 3 on it and one exceptional point of order 3 in its interior. Recall that the antipodal map i commutes with T on S2, hence a, b e T induce maps a and b on S2/(i) = P2. Moreover, P2/(a, b) = S2/[T x (i)] = Tv, where (a, b) is isomorphic to T. We will now describe the T = A4-action on P2 and show that it is unique up to conju-gacy. 2 14 11 3 V r/ /A 4 _____> 2 14 11 3 >To\ /12\ 4 2 14 11 3 r// yTff\ /12N. 4 Figure 6: Fundamental region on tetrahedron. The left diagram in Figure 6 above illustrates a fundamental region on S2 used to obtain P2 under the antipodal map i, where S2 is viewed as in Figure 5 and S2/ (i) = P2. This can be seen by observing the circle containing vertices, 5, 2, 14, 3, 7,11 is left invariant by i, and the vertices 4, 10, 6, 12 are sent to 8, 13, 1, 9 respectively. The projective plane is obtained by identifying the opposite side in this polygon. Recall a and b are generators of the tetrahedral group T = A4 operating on S2. Furthermore i e T = A4. Thus T induces an action on P2 and the elements a, b e T induce maps a, b on P2. Notice that the generator a maps the loop [5,2,14, 3] = [5~2][2,14][14, 3] in P2 onto [6T3][37T3][T37T] = [6T3][5;i0][106] = [6,3,5,10,6] = [5,10,6, 3], and a2 maps this loop onto [2,6,12,7]. Each image is expressed as a bold line in the Figure 6 above. Thus, the map a does not leave this orientation reversing loop invariant in P2. Likewise, ab and Jab)2 map the loop [5, 2,14, 3] = [2,14,3,7] onto [2,10,4,7] and [2,6,12, 7] respectively. Furthermore, b maps the loop [5,2,14, 3] onto [11,4,6,14]. The loop consists of vertices 2, 6, 12, 7,1 and 9 on the tetrahedron S2 is left invariant by the map b e T and the covering translation i, hence the arc having vertices 2, 6,12 and 7 projects to an orientation reversing loop on P2. There is no orientation reversing loop in P2 which is left invariant by both a and b. Lemma 4.1. Let Z2 be a subgroup of ni(Tv) such that'. sponding to Z2. Then Z2 = (i). ^ Tv is the covering corre- Proof. We will show that the orbifold Tv has only one P2 covering space up to a conjugacy. 22 Notice that A4 has three elements of order two. These elements are b, aba 1 and a2ba which are all equivalent. Thus, A4 x Z2 has two conjugacy classes of order two elements 5 5 5 7 7 7 J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles... 307 which reverse orientation, namely bi and i. Since bi = (1,7)(2, 6)(3, 8)(4,5)(9,12) fixing vertices 10, 11, 13 and 14, we have a desired conclusion. □ Proposition 4.2. Let p: G ^ Homeo(P2) be a finite action such that P2/p = Tv. Then G ~ A4 and p is conjugate to the standard action generated by (a, b). Moreover, no orientation reversing loop is left invariant by the G-action. Proof. The proof follows as in Proposition 2.2 and uses Lemma 4.1. □ We remark that [4] contains excellent figures to show us how each element in A4 acts on a tetrahedron. 5 Achiral tetrahedral symmetry Th In Section 1, we have seen the O = S4-action on S2 where S2/O is £(2, 3,4), which is an orientable orbifold. In this section, we will investigate another O = S4-action on S2. However, the resulting quotient space S2/O = Th will be non-orientable this time. More specifically, it will be a mirrored disk which contains two cone points of order three and one cone point of order two on the mirror. Note that we will triangulate S2 as shown in Section 4 which is a tetrahedron. First, we will begin by providing generators to define a group isomorphic to S4. Consider two elements a = (1, 2)(6, 7)(10,11)(13,14) and b = (2,4, 3)(5, 7, 8)(9,11,13) (10,12,14) in Si4. We can see that a is a reflection on the circle containing vertices 4, 5, 9, 8, 3 and 12 in S2. On the other hand, b is a 120° rotation about the axis passing through vertices 1 and 6. It is easy to check ab = (1,2,4, 3)(5, 6,7,8)(9,10,12,13)(11,14). Although ab reverses an orientation, it is called improper rotation. As a result, S4 = (a,b | a2 = b3 = (ab)4 = 1). Secondly, A4 is an index two subgroup of S4 and the subgroup can be expressed by using the two generators for S4. In order to get a presentation for A4, consider (ab)2 = (1,4)(2,3)(5, 7)(6, 8)(9,12)(10,13) which is a 180° rotation about the axis passing through vertices 11 and 14. Then, b(ab)2 = (1, 3,4)(5,8, 6)(9,14,10)(11,13,12) is a 120° rotation about the axis passing through vertices 2 and 7. Consequently, we obtain a desired subgroup A4 = (b, (ab)2 | [(ab)2]2 = b3 = [b(ab)2]3 = 1). Thirdly, we will look for a fundamental region for S2/A4. It is easy to compute that the map b permutes A134, A123 and A142. Further, b(ab)2 maps from A123 to A432. Thus, we will look at A134. However, A137 tiles A134 using the element b(ab)2. Then, the vertices 1 and 7 become order 3 cone points since they are fixed by b and b(ab)2 respectively. Thus, we may choose A137 for our fundamental region. Notice that the vertex 11 is fixed under (ab)2, which becomes the order 2 cone point, and it is identified to the vertex 13 by b(ab)2 € A4. Now, b(ab)2 identifies 17 and 3/7; b(ab)2b-1 identifies 1,13 and 3713. As a result, the quotient space S2/A4 is indeed E(2, 3, 3). Finally, we will discuss how to obtain the orbifold Th. Recall the map a € S4 reflects on the circle containing vertices 4, 5, 9, 8, 3 and 12 in S2. We compose this map by a covering translation (ab)-1[(ab)2b]2(ab) = (1,2,3)(5, 6, 7)(9,14,13)(10,12,11) € A4, which is a 120° rotation about the axis passing through vertices 4 and 8. Then, (ab)-1[(ab)2b]2(ab)a sends the triangle containing vertices 1, 13 and 7 to the triangle containing vertices 3, 7 and 13. Notice that 1, 7 and 3,7 are identified in E(2,3,3). Likewise, 1,13, and 3,13 are identified. Thus, the circle containing vertices 1, 7, 13 becomes 308 Ars Math. Contemp. 15 (2018) 441-466 the line of reflection under the map induced by a on E(2,3, 3). Consequently, we obtain S2/S4 = Th. Unlike the previous orbifolds, Th is not covered by a projective plane. Notice that ni(Th) = S4 contains six elements of order two which are orientation reversing. All of them are a reflection at a plane whose intersection with the tetrahedron is a triangle containing either vertices 2, 11, 3; vertices 1, 12, 2; vertices 3, 10, 1; vertices 4, 9, 3; vertices 4,13, 2; or vertices 4,14,1. Clearly, none of them give a fixed point free action on the tetrahedron S2, and hence this yields the following lemma. Lemma 5.1. The orbifold Th is not covered by a projective plane. 6 Cyclic and dihedral actions We describe the cyclic and dihedral actions on S2 and the projective plane P2. In describing these actions, it is convenient to use spherical coordinates. Therefore for any point (x, y, z) € S2, we let x = sin ^ • cos 0, y = sin ^ • sin 0 and z = cos We begin by defining a rotation of order m on S2 as follows: r(x,y, z) = (sin^ • cos(0 + m), sin^ • sin(0 + m), cos Note that r fixes only the points (0,0,1) and (0,0, -1). A spinning map s which rotates through an angle of n about the y-axis is defined by s(x, y, z) = (—x, y, -z). In terms of the spherical coordinate system, the map is defined by s(x, y, z) = (sin(^ + n) • cos(-0), sin(^ + n) • sin(-0), cos(^ + n)). One can check that s o r o s-1 = r-1, and therefore (r, s) generates a dihedral group Dih(Zm) acting on S2. Finally we define the antipodal map i on S2 by i(x, y, z) = (—x, -y, -z). In terms of the spherical coordinate system, i(x, y, z) = (sin(^ + n) • cos 0, sin(^ + n) • sin 0, cos(^ + n)). We have S2/ (i) = P2. Observe that i o s o i-1 = s and i o r o i-1 = r. Hence i commutes with r and s which implies the following lemma: Lemma 6.1. The maps r and s induce homeomorphisms r and s on P2 respectively. Let k(x,y,z) = (sin^• cos(0 + ^), sin^• sin(0 + ^), cos. A computation shows that k o s o k-1 = r o s, k o r o k-1 = r and k o i = i o k. This implies that the induced map s on P2 conjugates s to s o s and commutes with s. Notice that we can express the three maps above in terms of a PL-category. Let m € N. We assume that vertices from 1 to 2m are located on the equator line of S2. The vertices 2m +1 and 2m + 2 are on the poles. As a result, we obtain 4m many faces (triangles) from these vertices on S2. If m > 1 is odd, then the rotation r is expressed by r = (1,3,..., 2m-1)(2,4,..., 2m) whose order is m. On one hand, if m is even, then r = (1, 2, 3,..., 2m) whose order is 2m. In each case, the vertices 2m + 1 and 2m + 2 are fixed under r since they are the north and the south poles. The spinning map for m > 1 passing through the y-axis is defined by s = (2, 2m)(3, 2m - 1) • • • (m, m + 2)(2m + 1, 2m + 2). The J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles... 309 vertices 1 and m + 1 are fixed under s. The antipodal map for m > 1 is defined by i = (1, m + 1)(2, m + 2) • • • (m, 2m)(2m + 1, 2m + 2). For the case when r has order two, we place vertices 1 to 4 on the equator of S2 and vertices 5 and 6 are on the poles (see the Figure 2 in Section 2). Then r = (1,3)(2,4), s = (2,4)(5,6), and i = (1,3)(2,4)(5, 6). Define a map j = (1,6,3,5). A computation shows that j o s o j _1 = r, j o r o j _1 = s and j o i o j _1 = i. Therefore j is conjugate to j on P2. Summarizing we have the following lemma: Lemma 6.2. There exists a homeomorphism k on P2 which conjugates s to j o j and commutes with r. When j has order two, there exists a homeomorphism j on P2 which conjugates rs to s and s to rs. 6.1 Quotient types S(0, m, m), S(2, 2, m), D^ and D^ The space E(0, m, m) is an orbifold whose underlying space is a 2-sphere with two cone points each of order m. Similarly E(2, 2, m) is an orbifold whose underlying space is a 2-sphere with three cone points, two of order 2 and one of order m. The orbifold Dm is a mirrored disk containing a cone point of order m and 2 on the mirror and its interior respectively. The orbifold Dm is a mirrored disk with three cone points on the mirror, one of order m and two of order 2. Observe that we obtain S2/(r) = E(0, m, m), which double covers S2/(r, s) = E(2,2, m). Since i commutes with r and s, we have Dih(Zm) x Z2 = [(r) o_1 (s)] x (i) acting on S2. Now r and s acting on S2 induce a Dih(Zm)-action on S2/(i) = P2. Furthermore, i operating on S2 induces an orientation reversing involution i on E(2,2, m), and we have E(2,2, m)/(i) = P2/(r, s) = S2/(Dih(Zm) x Z2). Thus the fundamental group of the quotient space P2/(r, s) is Dih(Zm) x Z2 = [(r) o_1 (s)] x (i). Let p: S2 ^ S2/(r, s) = E(2, 2, m) be the orbifold covering map and note that p(0,0,1) = p(0,0, -1) is the cone point of order m. Since i(0,0,1) = (0,0, -1) and s(0,0, -1) = (0,0,1), it follows that_i(p(0,0,1)) = p(0,0,1), and thus i fixes the cone point of order m in E(2,2, m). Hence i is a reflection. If m is odd, rk(0,1,0) = (0, -1,0) for any k. Thus p(0,1,0) and p(0, -1,0) are the two distinct cone points of order 2 in E(2,2, m). If m is even, then p(0,1,0) = p(0, -1,0) is a cone point of order 2 since rm (0,1,0) = (0, -1,0). We will consider the cases m odd and m even separately. Suppose m is odd. Then since i(0,1,0) = (0, -1,0), it follows that i(p(0,1,0)) = p(0, -1,0) and thus i exchanges the two cone points of order 2. Since i fixes the cone point of order m, it follows that P2/(r, s) = Dm. The order two elements in Dih(Zm) x Z2 are: i, s, rjsi. One can check that rjsi(x, y, z) = ( sin ^ • cos(-0 + ), sin ^ • sin(-0 + ), cos . By choosing ^ = 0 or n, the map fixes the points (0,0, ±1) on S2. Note that E(0, 2, 2) = S2/(s) is not a regular covering space of Dm since (s) is not a normal subgroup of n1(Dm) = Dih(Zm) x Z2. Thus i is the only orientation reversing element which is fixed-point free. This implies that when m is odd, n1(Dm) has a unique normal Z2 subgroup generated by a fixed-point free orientation reversing element, and the covering of Dm corresponding to this subgroup is P2. Next we suppose m is even and show how to obtain Dm. Write m = 2n and observe that the rotation r of order 2n on S2 is defined as follows: r(x, y, z) = (sin ^ • cos(0 + n), sin ^ • sin(0 + n), cos . 310 Ars Math. Contemp. 15 (2018) 441-466 Note that r fixes only the points (0,0, ±1), and since rns(1,0,0) = (1,0,0) it follows that p(1,0,0) is one of the cone points of order 2 in E(2, 2,2n). Consider the point (sin(f) • cos(), sin(f) • sin(), cos(f)). We see that rn+1 s fixes (sin(72) • cos(), sin(f) • sin(), cos(f)), and so it follows that the point p((sin(72) • cos(), sin(f) • sin(2n), cos(f))) is the other cone point of order 2 in E(2, 2, rn). Define a reflection /: S2 ^ S2 by /(x,y, z) = (sin(-^) • cos(—0 + ), sin(-^) • sin(—0 + ), cos(—. A calculation shows that /s/-1 = rs and /r/-1 = r-1. Thus we have Dih(Z2n) o Z2 = [(r) o-1 (s)j o (/) acting on S2 and an induced map f acting on E(2, 2, m) = S2/(r, s). A further computation shows that /(1,0,0) = /(sin(n) • cos(0), sin(|) • sin(0), cos(|)) = (sin(-n) • cos(), sin(-n) • sin(), cos( —r)). Applying rn to this element, we see that rn(sin(-n) • cos(), sin() • sin(n), cos(-n)) = (sin(2) • cos(), sin(n) • sin(n), cos(f)). Hence the induced map / exchanges the two cone points of order two. In addition, consider a set F C S2 defined by F = {(sinf • cos(+ n), sinf • sin( in + n), cosf) | f G R}. Notice that /(sin f • cos( Jn + n), sin f • sin( + n), cos f) = (sin(—f •cos(-in - n + 2n), sin(—f •sin(-in - n + 2n)>cos(-^)) = ( - sin f • cos(in + n), - sin f • sin( jn + n), cos(f)) = rm (sin f • cos( Jn + n), sin f • sin( Jn + n), cos f). Therefore, p(F) = fixj!} in E(2, 2, m) where p denotes the covering map. Consequently, / is a reflection exchanging the cone points of order 2. Thus E(2,2, m)/(f) = D2n, and n1(D2n) = Dih(Z2n) o Z2 = [(r) o-1 (s)] o (/) where /s/-1 = rs and /r/-1 = r^1. The elements of order two are: rn, rks, rk/ (for any integer k = 0,1,..., 2n — 1). The only orientation reversing elements of order two are rk/, and they all fix the points (0,0,1) and (0,0, —1). Thus there is no orbifold covering P2 ^ D2n. We summarize the above in the following theorem. Theorem 6.3. Let f: G ^ Homeo(P2) be a finite action such that P2/f = Dm. Then m is odd, G ~ Dih(Zm) and f is conjugate to the standard action generated by (f, f). Proof. By the above m is odd. Let v: P2 ^ Dm = P2/ (f, f) be the covering map corresponding to the standard action. For the action f: G ^ Homeo(P2) with P2/f = Dm, let ^: P2 ^ P2/f be the covering map and h: Dm ^ P2/f be a homeomorphism. By the above the subgroup (n1(P2)) in n1(P2/f) is unique. Thus h lifts to a homeomorphism h of P2 such that hv = ^h. This implies that the two actions are conjugate by h. □ J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles... 311 Suppose m is even. Since r2 (0,1,0) = (0,-1,0) and ¿(0,1,0) = (0,-1,0), it follows that i fixes the cone point p(0,1,0). Since i also fixes the cone point of order m, we have that i is a reflection leaving each cone point fixed and P2/ (r, s) = . The order two elements in Dih(Zm) x Z2 are: i, s, rjsi or r2¿. Since rjsi(0,0,1) = (0,0,1), we only need to consider r2 i(x, y, z) = ( sin(^ + n) • cos(0 + n), sin(^ + n) • sin(0 + n), cos(^ + n)). Letting ^ = n and 0 = 0, we see that the point (1,0,0) is fixed by r2 i. Thus has a unique P2 covering up to conjugation. This implies that when m is even, ) has a unique normal Z2 subgroup generated by a fixed-point free orientation reversing element, and the covering of corresponding to this subgroup is P2. We now suppose m is odd and show how to obtain . Define a reflection /0 : S2 ^ S2 by /0(x, y, z) = (sin(-• cos(—0), sin(—• sin(-0), cos(—= (—x, y, z). One can check that 10s1-1 = s and /0 r/-1 = r-1. Hence Dih(Zm)oZ2 = [(r)o_1(s)]o(/0) acting on S2 and an induced map T0 acting on E(2,2, m) = S2/(r, s). Clearly /0 fixes the points (0,1,0) and (0, —1,0). Recall p(0,1,0) = (0, —1,0). Hence the induced map Z"o on E(2, 2, m) is a reflection which fixes each cone point. Thus E(2, 2, m)/(Z0) = and ni(Dm) = Dih(Zm) o Z2 = [(r) o_i (s)] o (/0). The elements of order two are: s, rk s, rk/0 (any integer k = 0,1,..., m — 1), and s/0. The only orientation reversing elements of order two are rk/0 and s/0, but they all have fix-points. Thus there is no orbifold covering P2 ^ . We summarize the above in the following theorem whose proof is similar to Theorem 6.3. Theorem 6.4. Let p: G ^ Homeo(P2) be a finite action such that P2/p = . Then m is even, G — Dih(Zm) and p is conjugate to the standard action generated by (Z, ZZ). 6.2 Quotient types S2m and Z^ We use S2m and Z^ to denote a projective plane that has one cone point of order m and a mirrored disk containing an order m cone point in its interior respectively. The orbifold Zq denotes a mirrored disk without an exceptional point within its interior, and if m = 1, then S2(1) = P2. Recall (r) x (i) = Zm x Z2 acts on S2. Hence, the involutions in this group are either i, r 2 or ir 2 for an even number m. If m is even, then ir2(x, y, z) = (sin^ • cos0, sin^ • sin0, — cos. The fixed point set of this map is the circle at the equator on S2 and occurs when ^ = 2. Thus, S2/(ir2) = S2/Z2 = Zq. Furthermore, r on S2 induces a rotation r on Zq fixing a point not on the mirror, and inducing an action r acts on P2 = S2/ (i). In the meantime, i on S2 induces a reflection i on E(0, m, m) = S2/(r) since r 2 ( — 1,0,0) = (1,0,0) and i(1,0,0) = ( —1,0,0). As a result, we obtain Z^ = ZQV(r) = P2/(r) = S(0,m,m)/(i) for m is even. Note that n1(Z^) — Zm x Z2 is generated by r and i, where i is the only fixed-point free orientation reversing element. This implies that when m is even, n1(Z^) has a unique normal Z2 subgroup generated by a fixed-point free orientation reversing element, and the covering of Zh corresponding to this subgroup is P2. We now show how to obtain Zh when m is odd. Let p be a homeomorphism of S2 defined by p(x, y, z) = (sin ^ • cos 0, sin ^ • sin 0, — cos . A computation shows that 312 Ars Math. Contemp. 15 (2018) 441-466 p and r commute. We obtain an orbifold covering map S2 ^ E(0, m, m) = S2 / (r) with p inducing a reflection p on E(0, m, m). The quotient space E(0, m, m)/(p) = and n1(Zm) — Zm x Z2 is generated by r and p. Since m is odd, the only element of order 2 in n1 (Zm) is p which has fixed points. Thus there is no orbifold covering P2 ^ Z^ when m is odd. Consequently, the following theorem is obtained: Theorem 6.5. Let p: G ^ Homeo(P2) be a finite action. If P2/p = Zm, then m is even, G — Zm and p is conjugate to the standard action generated by (r). If m is odd, we again have r inducing a map r on P2 = S2/ (i), and one can check that the induced map i on E(0, m, m) = S2/(r) is the antipodal map. Consequently, we obtain S2m = P2/(r) = E(0, m, m)/(l). Furthermore n1(S2m) — Zm x Z2 is generated by r and i, where the only order two fixed-point free orientation reversing element is i. Hence when m is odd, n1(S2m) has a unique normal Z2 subgroup generated by a fixed-point free orientation reversing element, and the covering of S2m corresponding to this subgroup is P2. To obtain S2m when m is even, we write m = 2n and define a homeomorphism h of S2 by h(x,y,z) = (sin(^ + n) • cos(0 + ), sin(^ + n) • sin(0 + ), cos(^ + n)). Observe that h is a composition of the antipodal map and a rotation through n/2n, and h generates a Z2(2n)-action on S2 and S2/(h2) = E(0,2n, 2n). It follows that the induced map h on E(0,2n, 2n) is the antipodal map and E(0, 2n, 2n)/(h) = S2(2n). Furthermore n1(S2(2n() — Z2(2n) is generated by h. The only element of order 2 is h2n, and h2n(x, y, z) = (sin^ • cos(0 + n), sin^ • sin(0 + n), coshas fixed-points. Thus there is no orbifold covering P2 ^ S2m when m is even. Summarizing these results we obtain the following theorem: Theorem 6.6. Let p: G ^ Homeo(P2) be a finite action. If P2/p = S2m, then m is odd, G — Zm and p is conjugate to the standard action generated by (r). 6.3 Nonexistence of quotient type Cm m The orbifold Cm,m is a mirrored disk with two cone points on the mirror of order m. We will show that the orbifold Cm m is obtained by some covering translations on S2. Recall the reflection map on the yz-plane defined on R3 by Zo(x, y, z) = (—x, y, z) and the rotation r(x, y, z) = (sin^ • cos(0 + m), sin^ • sin(0 + m), cos. It is easy to check that Dih(Zm) = (r) o_1 (Z0). Then, we obtain E(0, m, m) = S2/(r) and the reflection on S2 induces a reflection Z0 on E(0, m, m). As a result, Cm,m = E(0, m, m)/(Z0) where n1(Cm,m) = Dih(Zm). The order two elements in n1(Cm,m) are rjZ0 for 0 < j < m, or r ^r for m even. A calculation shows that rjZ0(x,y,z) = (sin(-<£) • cos(-0 + m), sin(-<£) • sin(-0 + ^), cos(-^)), which has fixed points at (0,0, ±1) G S2 when ^ = 0 or n. Since Z0 and rtt when m is even, have fixed points, Cm,m is not covered by P2. We therefore have shown the following proposition: Proposition 6.7. The projective plane does not cover CV m. J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles... 313 7 Finite group actions on P2 and P2 x I In this section, we summarize the above results and classify the finite group actions on P2 and P2 x I. Theorem 7.1. Let f: G ^ Homeo(P2) be a finite group action on P2. Then G is isomorphic to one of the following groups: S4, A5, A4, Zm or Dih(Zm). The orbifold quotient P2/f is an orbifold homeomorphic to one of the following orbifolds: Oh, Ih, Tv, , S2m, Dm or Dm. There is only one equivalence class for each group. (1) G ~ S4 if and only if P2/f = Oh. (2) G ~ A5 if and only if P2/f = Ih (3) G ~ A4 if and only if P2/f = Tv. (4) G ~ Zm and m is even if and only if P2/f = Zm. (5) G ~ Zm and m is odd if and only if P2/f = S2m. (6) G ~ Dih(Zm) and m odd if and only if P2/f = Dm. (7) G ~ Dih(Zm) and m even if and only if P2/f = Dm. Proof. Let f: G ^ Homeo(P2) be a finite group action. Then P2 /f is a non-orientable 2-orbifold with positive euler number x(P2 /f). The non-orientable good orbifolds (orbifolds which have manifolds for their universal covering spaces) with positive euler numbers are the following: Cm,m, S2m, Zm, Dm, Dm, Th, Oh, Ih and Tv. The result then follows by the above. □ Theorem 7.2. Lei f: G ^ Homeo(P2) be a finite group action. The action f (G) does not leave any orientation reversing loop in P2 invariant if and only if G is isomorphic to S4, A5 or A4. Furthermore, f is equivalent to one ofthese standard actions. Proof. This follows from Sections 2 through 6. □ Theorem 7.3. Let f: G ^ Homeo(P2 x I) be a finite group action. If for every g G G f(g)(P2 x {0}) = P2 x {0}, then G is isomorphic to one of the following groups: S4, A5, A4, Zm or Dih(Zm). Furthermore, there is only one equivalence class for each group which is represented by one of the standard actions. Proof. By the comment following Theorem 8.1 in [6], we may conjugate f (G) so that it is a product action. This implies that there is a G-action f 1: G ^ Homeo(P2) such that for any g G G, we have f (g)(z, t) = (f1 (g)(z),t). By Theorem 7.1, there exists a homeomorphism k of P2 such that kf 1(G)k-1 is one of the standard actions (1) through (7) listed there. Conjugating this action further by k x id proves the result. □ Theorem 7.4. Let f: G ^ Homeo(P2 x I) be a finite group action. Then G is isomorphic to one of the following groups: S4, S4 x Z2, A5, A5 x Z2, A4, A4 x Z2, Zm, Zm x Z2, Dih(Zm) or Dih(Zm) x Z2. (1) If G is isomorphic to S4, then there are two equivalence classes. (2) If G is isomorphic to either S4 x Z2, A5, A5 x Z2, A4, A4 x Z2, Zm x Z2 or Dih(Zm) x Z2, then there is one equivalence class. 314 Ars Math. Contemp. 15 (2018) 441-466 (3) Suppose G is isomorphic to Zm. If m is odd, then there is one equivalence class. If m is even, then there are two equivalence classes. (4) Suppose G ~ Dih(Zm). If m = 2 or if m is odd, then there are two equivalence classes. If m > 2 and m is even, there are 3 equivalence classes. Proof. Again by [6], we may assume <: G ^ Homeo(P2 x I) is a product action. Thus there exists a homomorphism <1: G ^ Homeo(P2) such that for any g G G, <(g)(z,t) = (<1(g)(z),t) or <(g)(z,t) = (<\(g)(z), 1 - t). There exists a Z2-action on P2 x I generated by a map R defined by R(z, t) = (z, 1 - t). Suppose first that <1 : G ^ Homeo(P2) is not one-to-one. This implies there exists an element g0 = 1 G G such that 2. When m = 2, Lemma 6.2 also shows that H2 and H3 are conjugate, and so we have only two equivalence classes in this case. When m is odd, the only two possibilities are ((a x id), ( a x id)) and ((a x id), ( a x id) o R). □ J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles... 315 8 Finite actions on twisted I-bundle over P2 For S2 x I, define a fixed-point free orientation preserving involution a: S2 x I ^ S2 x I by a(z, t) = (i(z), 1 - t). The manifold S2 x I/(a) = W is a twisted I-bundle over the one-sided projective plane P2. Let v: S2 x I ^ W be the covering map and note that v(S2 x {1/2}) = P2 is a one-sided projective plane. The levels of W are v(S2 x {t}),and a homeomorphism h of W is level preserving if h(v(S2 x {t})) = v(S2 x {t}). We may view W as the set of equivalence classes {[z, t] | (z, t) is equivalent to (i(z), 1 - t)}. Let Homeo( W, P2) be the group of homeomorphisms which leave the projective plane P2 invariant. Denote by Centp(a) the subgroup of the centralizer of a which leaves S2 x {1 / 2 } invariant and preserves the sides of S2 x{ 1 / 2 }. Every homeomorphism which leaves P2 invariant lifts to two homeomorphisms of S2 x I, one of which preserves the sides of S2 x {1/2} while the other doesn't. Thus for any homeomorphism f G Homeo(W, P2) there is a unique lift f G Centp(a), and we obtain an isomorphism L: Homeo(W, P2) ^ Centp(a). Note that since v|S2x{0}: S2 x {0} ^ dW is a homeomorphism, it follows that f is orientation preserving if and only if f is orientation preserving. We obtain the following proposition. Proposition 8.1. L: Homeo(W, P2) ^ Centp(a) is an isomorphism. There exists a map R: Homeo(W, P2) ^ Homeo(P2) defined by restricting any homeomorphism to P2. Proposition 8.2. Let p: G ^ Homeo( W, P2) be an effective orientation preserving enaction. Then the restriction Rp: G ^ Homeo(P2) is an effective G-action. Proof. Let p = L o p: G ^ Centp(a) be an orientation preserving G-action on S2 x I. Suppose there exists an element g G G such that Rp(g) = id, and thus p(g) |S2 x{1/2> = id or i. Since p(g) does not reverse the sides of S2 x {1/2} and p(g) is orientation preserving, we have that p(g) |S2 x{1/2} = id implying p(g) = id. This implies that p(g) = id proving the result. □ Remark 8.3. The assumption that the G-action in Proposition 8.2 be orientation preserving is necessary. For if we define an involution p of W by p[z, t] = [z, 1 - t] = [i(z), t], then Rp = id|P2 but p = id on W. Proposition 8.4. Let p1,p2: G ^ Homeo(W, P2) be two orientation preserving G-actions such that Rp1 and Rp1 are effective G-actions on P2 with Rp1(G) = Rp2 (G). Then there exists a homeomorphism k of W isotopic to the identity such that kp1 (G)k-1 = p2(G). Proof. Let Rp1 = p1 and Rp2 = p2 be the effective G-actions on P2. Replacing p2 by p2p-1p1, we may assume p1 = p2. Let Lp1 = p1 and Lp2 = p2. Since p1 and p2 are both orientation preserving, it follows that p51(g)|S2 x{1/2} = p2(g)|S2x{1/2} for any g G G. Consider pi(G)|S2x[0i1/2]: S2 x [0,1/2] ^ S2 x [0,1/2]. Again using [1], both these actions are conjugate to product actions, and hence there exist homeomorphisms kj such that kip5i(G)k-1 is a product action. The conjugating maps kj may be chosen to be the identity on S2 x {1/2}. Since both actions agree on S2 x {1/2}, we have k1p1(G)k-1 = k2p2(G)k-1. Letting h = k-1k1, we obtain a homeomorphism 316 Ars Math. Contemp. 15 (2018) 441-466 h: S2 x [0,1/2] ^ S2 x [0,1/2], isotopic to the identity relative to S2 x {1/2}, such that h((i5i(G)|§2X[0,1/2])h-1 = <^i(G)|s2X[o,i/2]. Extend h to S2 x [1/2,1] by letting h|s2x [1/2,1] = (a|s2x [0,1/21) ◦ (h|s2x [0,1/2]) ◦ (a|s2x [1/2,1]) and note that h and a commute. Let g G G and z G S2 x [1/2,1]. Now we have h(1(g)h-1(z) = h(1(g)(a o h-1|S2x[o,1/2] ◦ a)(z) = a(h(^1(g)h-1)(a(z)) = ac^^Mz) for some g/ G G. Since a(i52(g/)a(z) = (52(g/)(z), we have shown h(i51(G)h-1 = (2(G). Letting L-1(h) = k, we have h(1(G)h-1 = (2(G) proving the result. □ Remark 8.5. The assumption in Proposition 8.4 that the actions are both orientation preserving is necessary. For example, let f: S2 ^ S2 be an orientation preserving home-omorphism commuting with i such that f2n = id. Define two Z2n-actions on W by F[z,t] = [f (z),t] and G[z,t] = [if (z),t]. Clearly G|P2 = F|P2 since i projects to the identity on P2, however they are not conjugate as F is orientation preserving, and G is orientation reversing. Corollary 8.6. Let (1,(2: G ^ Homeo(W, P2) be effective orientation preserving enactions such that R(1(G) = R(2(G). Then there exists a homeomorphism k of W isotopic to the identity such that k(1(G)k-1 = (2(G). Proof. By Proposition 8.2 R(1 and R(1 are effective G-actions on P2, and so the result follows by Proposition 8.4. □ Proposition 8.7. Let (: G ^ Homeo(P2) be an effective G-action. Then ( extends to an effective level preserving orientation preserving G-action ( on W. Proof. By [7] and [8], there exists an action (: G ^ Cent+(i) where Cent+(i) consists of orientation preserving elements in the centralizer Cent(i) of i in Homeo(S2). Define an action 0: G ^ Centp(a) by 0(g)(x,t) = (((g)(z), t). Then : G ^ Homeo(W, P2) is the extension. □ Let E (P2, G) be the set of equivalence classes of effective G-actions on P2, and let E+(W, G) be the set of equivalence classes of effective orientation preserving G-actions on W. Denote by E+ ((W, P2), G) the subset of E+ (W, G) which have a representative that leaves a one-sided projective plane invariant. Proposition 8.8. Let (: G ^ Homeo(W) be a finite action on W. Then there exists a one-sided projective plane P such that ((g)(P) = P for all g G G. Proof. Let Homeo(S2 x I, S2 x {0}) be the group of homeomorphisms which leave S2 x {0} invariant. There exists an injection L0: Homeo(W) ^ Homeo(S2 x I, S2 x {0}) n Cent(a) defined by lifting any homeomorphism to a homeomorphism of S2 x I leaving S2 x {0} invariant. Letting L0( = (: G ^ Homeo(S2 x I, S2 x {0}) n Cent(a), we obtain a G x Z2 action on S2 x I where the Z2-action is generated by a, which projects to the (-action on W. This action is equivalent to a product action by [1], and thus there is an G x Z2-invariant 2-sphere S in int(S2 x I). Furthermore, a(S) = S and v(S) is a ((G)-invariant projective plane in W. □ Corollary 8.9. E+((W, P2), G) = E+(W, G). J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles... 317 Proposition 8.10. Let P be a one-sided projective plane in W. Then there exists a homeo-morphism k of W, isotopic to the identity, such that k(P) = P2. Proof. Isotope P to intersect P2 in simple closed curves. We may assume the number of curves in P n P2 is minimal. We will show that the number of simple closed curves in P n P2 is one. Note first that P n P2 = 0, for otherwise P C W - P2 which is isomorphic to S2 x [0,1), and this is impossible. If the number of intersections of P n P2 exceeds one, and hence the number of simple closed curves in P exceeds one, then there is a simple closed curve S G P n P2 which bounds a disk A in P. We may assume A is innermost in P, in the sense that int(A) n P2 — 0. Since S bounds a disk in WV, it follows that S is an orientation preserving loop in P2, and thus bounds a disk D in P2. Now D U A is a separating 2-sphere. If D U A bounds a ball, then we may isotope P to eliminate S. We therefore assume D U A does not bound a ball, and is therefore parallel to the sphere boundary dW = v(S2 x {0}). Lift P to an a-invariant 2-sphere S in S2 x I, let A1 and A2 be the two lifts of A in S, and let D1 and D2 be the two lifts of D in S2 x {1/2}. Denote dAi by Si. We may assume D1U A1 C S2 x [0,1 /2] and D2 U A2 C S2 x [1 /2,1]. Furthermore, there is an a-invariant simple closed curve 7 G S n S2 x {1/2}, separating S1 and S2. Note that (S2 x {1/2} - int(D1)) is a disk in S2 x {1/2} whose boundary is the boundary of the disk A1 in S. Now D1 U A1 is parallel to S2 x {0}, which implies that A1 U (S2 x {1/2} - int(D1)) bounds a ball in S2 x [0,1/2]. Thus we may construct an a-equivariant isotopy, relative to 7, which eliminates the intersections S1 and S2. Projecting this isotopy to W eliminates the S-intersection of P n P2. Thus we have shown that S n S2 x{1/2} is a single simple closed curve 7 which projects to a non-contractable simple closed curve 7 in P n P2. By an argument similar to the one above, there is an a-equivariant isotopy, relative to 7, which isotopes S to S2 x {1/2}. Projecting this isotopy to W, we obtain an isotopy taking P to P2. □ Theorem 8.11. The map r: E(P2, G) ^ E+((W, P2), G) defined by extending G-actions from P2 to W is a bijection. Proof. Let [(] G E(P2, G). By Proposition 8.7, ( can be extended to a G-action (p on W. Define r([(]) = [(]. Suppose - is a G-action on P2 such that [-] = [( G E(P2, G). Then there exists a homeomorphism h of P2 such that h((G)h-1 = -0(G). Lift h to an orientation preserving homeomorphism k of S2 and note that ik = ki. Extend k to a homeomorphism p by p(x, t) = (k(x),t). Letting h = L-1p. we see that h is an extension of h. Since R(h(p(G)h-1) = h((G)h-1 = -(G) = R(-p(G)), it follows by Proposition 8.4 that [(p] = [-], and thus r is well defined. Let [S] g E((W, P2),G). Thus there is a one-sided projective plane P such that S(g)(P) = P for all g G G. By Proposition 8.10, there exists a homeomorphism of W taking P to P2. This implies that we may choose a representative S' in [S] such that S'(g)(P2) = P2 for all g G G. By Proposition 8.2, the restriction RS' is an effective G-action on P2 and therefore represents an element in E(P2, G). Let r([RS']) = [RS']. Since R(RS') = RS', it follows by Proposition 8.4 that [RS7] = [S'], and thus r([RS']) = [S'] showing r is a surjection. To show r is one-to-one, suppose that [(], [0] G E (P2, G) are such that their level preserving extensions [(p] = [p] in E+((W, P2), G). Now W/( and W/0 are homeomorphic twisted I-bundle orbifolds over one of the following 2-orbifolds: Oh, Ih, Tv, , S2m, Dm or Dm. Since by Theorem 7.1, there is only one equivalence class for each action 318 Ars Math. Contemp. 15 (2018) 441-466 on P2 which determines a unique quotient type, it follows that [<] = [0] showing r is one-to-one. □ Corollary 8.12. Let <: G ^ Homeo(W) be a finite orientation preserving G-action on W. Then G is isomorphic to one of the following groups: S4, A5, A4, Zm or Dih(Zm). The orbifold quotient for each action is a twisted I-bundle orbifold over the following 2-orbifolds: Oh (for S4), Ih (for A5), Tv (for A4), (for Zm and m even), S2m (for Zm and m odd), Dm (for Dih(Zm) and m odd) and Dm (for Dih(Zm) and m even). There is one equivalence class for each quotient type. 9 Orientation reversing finite actions on twisted I-bundle over P2 Recall that W = {[z, t] | (z,t) is equivalent to (i(z), 1 - t)} with P2 = {[z, 1/2] G W}, and L : Homeo(W, P2) ^ Centp(a) is an isomorphism. Let / be a homeomorphism of P2, and let / be a lift of /1 to S2. We remark that / commutes with i. A homeomorphism / : W ^ W is a product homeomorphism if /[z,t] = [f1(z),t]. Note that /|P2 = /1. Let Homeo(S2 x I, S2 x {1/2} ) be the group of homeomorphisms of S2 x I which leave S2 x {1/2} invariant. Define the map R : Homeo(S2 x I, S2 x {1/2}) ^ Homeo(S2) by restricting any homeomorphism to S2 x {1/2}. Lemma 9.1. Let f : G ^ Homeo(W, P2) be an effective G-action and let f = Lf : G ^ Centp(a) C Homeo(S2 x I, S2 x {1/2}). Then Rf : G ^ Homeo(S2) is an effective G-action. Proof. Suppose there exists an element g G G such that Rf (g) — ¿d|§2x{1/2}. Since Rf (g) does not reverse the sides of S2 x {1/2}, it follows that f (g) = id. □ Remark 9.2. Note that the involution p of W defined by p[z, t] = [i(z), t] = [z, 1 - t], has the property that Rp = id|P2, but RLp(z, t) = (i(z), t) and thus does not restrict to the identity on S2 x {1/2}. Theorem 9.3. Let f : G ^ Homeo(W) be an effective G-action. Then f is conjugate to a product action on W. Proof. By Propositions 8.8 and 8.10, we may assume f (g)(P2) = P2 for every g G G. Let f = Lf : G ^ Centp(a). By Lemma 9.1, Rf : G ^ Homeo(S2) is an effective G-action which commutes with i. Define an action 0: G ^ Centp (a) C Homeo(S2 x I, S2 x{1/2}) by 0(g) = Rf (g) x id. Thus %)|§2x{1/2} = f (g)|s2x{1/2} for any g G G. Projecting this action to W, we obtain an effective product action 0 : G ^ Homeo(W, P2). We now use the proof in Proposition 8.4 to construct a homeomorphism h which commutes with a and conjugates 0(G) to f (G). The homeomorphism h projects to a homeomorphism of W which conjugates 0(G) to f (G), thus completing the proof. □ We will now define the standard actions S4 x Z2, A4 x Z2, A4 x Z2, Zm x Z2 or Dih(Zm) x Z2 on W. Consider first the group S4 = (a, b | a2 = b3 = (ab)4 = 1} acting on S2 commuting with i, and its projection S4 = (a, b | a2 = b3 = (ab)4 = 1} to P2. Define the product maps A, B : W ^ W by A[z,t] = [a(z),t] and B[z,t] = [b(z),t]. Note that (A, B, p} = S4 x Z2. The other standard group actions on W are defined in a similar fashion. J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles... 319 Theorem 9.4. Let f: G ^ Homeo(W) be an orientation reversing G-action. Then G is isomorphic to one of the following groups: S4, Zm with m even, Dih(Zm), S4 x Z2, A5 x Z2, A4 x Z2, Zm x Z2 or Dih(Zm) x Z2. (1) If G is either S4, S4 x Z2, A5 x Z2, A4 x Z2, Zm x Z2 with m even or Dih(Zm) with m odd, there is only one equivalence class. (2) If G is Zm with m > 2 even and m/2 odd, then there are two equivalence classes of Zm = Zm/2 x Z2-actions on W. (3) If G is Zm with either m/2 even or m = 2, then there is only one equivalence class. (4) If G is Dih(Zm) with m > 2 and m/2 even, there are two equivalence classes of Dih(Zm)-actions on W. (5) If G is Dih(Zm) with m > 2 and m/2 odd, there are three equivalence classes of Dih(Zm)-actions on W. (6) If G is Dih(Zm) x Z2 with m even, there is only one equivalence class. (7) If G is Dih(Zm) x Z2 with m odd, then Dih(Zm) x Z2 ~ Dih(Z2m) and there are three equivalence classes of Dih(Z2m)-actions on W. Proof. Let f: G ^ Homeo( W) be an effective orientation reversing G-action. We may assume by Theorem 9.3, that there exists G-actions f 1: G ^ Homeo(S2) and f 1: G ^ Homeo(P2), such that f(g)[z,t] = [ 2, or to (Rp, S) when m = 2. If m > 2, then any generator of Zm in (Rp, S) is an odd power of Rp relatively prime to m, and thus orientation reversing. On the other hand, any generator of Zm in (R, Sp) is orientation preserving. Hence these groups cannot be conjugate. This implies that if m/2 is even, there are two equivalence classes of Dih(Zm)-actions. We note that when m/2 is odd and not equal to one, there are three equivalence classes of Dih(Zm)-actions. They are (Rp) 0-1 (S), (R) 0-1 (Sp) and (R2p) 0-1 (S). The last group has an element (R2p)m/2 = p restricting to the identity on P2, and the group may be viewed as ((R2) o-1 (S)) x (p) = Dih(Zm/2) x Z2.This group was identified in the second paragraph of this proof when we assumed p1: G ^ Homeo(P2) was not an effective G-action. The proof is completed by noting that if G is isomorphic to Dih(Zm) x Z2 and m is odd, then Dih(Zm) x Z2 is isomorphic to Dih(Z2m), and this case has already been dealt with. □ References [1] H. S. M. Coxeter and W. O. J. Moser, Generators and Relations for Discrete Groups, volume 14 of Ergebnisse der Mathematik und ihrer Grenzgebiete, Springer-Verlag, New York, 4th edition, 1980, doi:10.1007/978-3-662-21943-0. [2] D. S. Dummit and R. M. Foote, Abstract Algebra, John Wiley & Sons, Hoboken, New Jersey, 3rd edition, 2004. J. Kalliongis andR. Ohashi: Finite actions on the 2-sphere, the projective plane andI-bundles... 80 [3] W. D. Dunbar, Geometric orbifolds, Rev. Mat. Univ. Complut. 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