1.1. Discussion of results

In previous papers [8] and [9] we considered orientable Seifert manifolds and the possible finite groups that can act fiber- and orientation-preservingly.

The main results in those papers established that firstly:

Theorem 1.1. Let M be a closed, compact, and orientable Seifert 3-manifold that fibers over an orientable base space. Let φ:G→Diff+fp(M) be a finite group action on M such that the obstruction class can expressed as

b=∑ i=1m(bi⋅#Orbφ(αi))

for a collection of fibers {α1,…,αm} and integers {b1,…,bm}. Then φ is an extended product action.

Where an extended product action is intuitively a product action on an orientable surface with boundary cross S¹ extended across Dehn fillings of the boundary tori.

Secondly, it was shown that:

Corollary 1.2. Suppose that φ:G→Diff(M) is a finite group action on an orientable Seifert manifold with a non-orientable base space. Then provided that the unique lifted group action φ˜:G→Diff(M˜) satisfies the obstruction condition, G is isomorphic to a subgroup of ℤ2×H where H is a finite group that acts orientation-preservingly on the orientable base space of M˜.

These two results will allow us to consider the elliptic 3-manifolds in particular and present the possible finite, fiber- and orientation-preserving groups that can act on them.

We then present a proof that all finite, fiber-preserving actions on Seifert manifolds with non-zero Euler class must be orientation-preserving by using our particular construction and in particular apply this to elliptic manifolds.

Finally, we consider the quotient orbifolds that will arise under the given actions and present a thorough example of one such action.

1.2. Preliminary definitions

Throughout we consider M to be an oriented smooth manifold of dimension 3 and without boundary. G is always assumed to be a finite group. We denote Diff(M) as the group of self-diffeomorphisms of M, and then define a G-action on M as an injection φ:G→Diff(M). We use the notation Diff+(M) for the group of orientation-preserving self-diffeomorphisms of M.

M is further assumed to be an orientable Seifert-fibered manifold. That is, M can be decomposed into disjoint fibers where each fiber is a simple closed curve and each fiber has a fibered neighborhood which can be mapped under a fiber-preserving map onto a solid fibered torus.

A Seifert bundle is a Seifert manifold M along with a continuous map p:M→B where p identifies each fiber to a point. For clarity, we denote the underlying space of B as B_U.

We use the normalized notation (g,ϵ|(q1,p1),…,(qn,pn),(1,b)) to indicate an orientable Seifert manifold with normalized Seifert invariants (q1,p1),…,(qn,pn), obstruction class b, and ϵ=o1 if the base space is orientable and ϵ=n2 is the base space is not orientable.

The Euler class of a Seifert manifold with normalized Seifert manifold is given by e=−(b+∑ i=1npiqi) and an elliptic Seifert manifold is such that e=0 and the base orbifold has positive Euler characteristic.

We say a G-action is fiber-preserving if for any fiber γ and any g∈G,φ(g)(γ) is some fiber of M. We use the notation Difffp(M) for the group of fiber-preserving self-diffeomorphisms of M (given some Seifert fibration). Given a fiber-preserving G-action, there is an induced action φBU:GBU→Diff(BU) on the underlying space B_U of the base space B.

Given a finite action φ:G→ Difffp(M), we define the orbit number of a fiber γ under the action to be #Orbφ(γ)=#{α|φ(g)(γ)=α for some g∈G}.

If we have a manifold M, then a product structure on M is a diffeomorphism k:A×B→M for some manifolds A and B. [6] If a Seifert-fibered manifold M has a product structure k:S1×F→M for some surface F and k(S1×{x}) are the fibers of M for each x∈ F, then we say that k:S1×F→M is a fibering product structure of M.

Given that the first homology group (equivalently the first fundamental group) of a torus is ℤ×ℤ generated by two elements represented by any two nontrivial loops that cross at a single point, we can use the meridian-longitude framing from a product structure as representatives of two generators. If we have a diffeomorphism f:T1→T2 and product structures ki:S1×S1→Ti, then we can express the induced map on the first homology groups by a matrix that uses bases for H1(Ti) derived from the meridian-longitude framings that arise from ki:S1×S1→Ti.

We denote this matrix as a11a12a21a22k2k1.

We say that a G-action φ:G→Diff(A×B) is a product action if for each g∈ G, the diffeomorphism φ(g):A×B→A×B can be expressed as (φ1(g),φ2(g)) where φ1(g):A→A and φ2(g):B→B. Here φ1:G→Diff(A) and φ2:G→Diff(B) are not necessarily injections.

Given an action φ:G→Diff(M) and a product structure k:A×B→M, we say that φ leaves the product structure k:A×B→M invariant if ψ(g)=k−1∘φ(g)∘k defines a product action ψ:G→Diff(A×B).

Suppose that we now have a fibering product structure k:S1×F→M. We then say that each boundary torus is positively oriented if the fibers are given an arbitrary orientation and then each boundary component of k({u}×F) is oriented by taking the normal vector to the surface according the orientation of the fibers.

We consider two particular types of 3-orbifold. We define the solid torus with exceptional core V(k) to be a solid torus with an exceptional set of order k running along the core loop of the solid torus. We define the Conway ball B(k) to be a ball with exceptional set consisting of two arcs of order two joined by an arc of order k according to Figure 1 below:

Figure 1. A solid torus V(k) and Conway ball B(k)

We begin with some preliminary results that we will use in the next section regarding orientation-reversing diffeomorphisms.

Lemma 2.1. Let F be an orientable surface with boundary. Let the boundary be positively oriented according to some orientation of F and f:F→F be a diffeomorphism. Then f is orientation-preserving on F if and only if f is orientation-preserving between some pair of boundary components.

Proof. In a regular neighborhood of two exchanged boundary components (they may be the same), the diffeomorphism is either a reflection or a rotation (given parameterizations of the annuli). If it is a reflection, the orientation on the boundary is reversed and and the orientation on F is reversed. If it is a rotation, the orientation on the boundary is preserved and and the orientation on F is preserved.

Corollary 2.2. Let M^ be an oriented trivially Seifert fibered 3-manifold with positively oriented boundary ∂M^=T1∪…∪Tn. Then a fiber-preserving diffeomorphism f:M^→M^ is orientation-preserving if and only if f is orientation-preserving between some pair of boundary tori.

Proof. Firstly, there is a fibering product structure k:S1×F→M^. Suppose that the diffeomorphism preserves the orientation of the fibers. Then the projected diffeomorphism on F must be orientation-preserving. By Lemma 2.1, this is if and only if it is orientation-preserving between some pair of boundary components. As the diffeomorphism preserves the orientation of a fiber, this is equivalent to f being orientation-preserving between some pair of boundary tori.

If now we suppose that the diffeomorphism reverses the orientation of the fibers. Then the projected diffeomorphism on F must be orientation-reversing. By Lemma 2.1, this is if and only if it is orientation-reversing between some pair of boundary components. As the diffeomorphism reverses the orientation of a fiber, this is equivalent to f being orientation-preserving between some pair of boundary tori.

We now use the previous section to establish some results about the conditions under which an orientation-reversing action is possible.

Firstly, a condition on the order of critical fibers:

Proposition 3.1. All finite, fiber-preserving actions on an orientable Seifert 3-manifold fibering over an orientable base space with at least one critical fiber of order greater than two are orientation-preserving.

Proof. Suppose for contradiction that there exists a periodic, fiber-preserving and orientation-reversing diffeomorphism f:M→M.

We begin with normalized invariants for M=(g,o1|(q1,p1),…,(qn,pn),(1,b)).

We then take a regular fiber γ with #Orbf(γ)=l for some l. Then adjust the invariants to yield M=(g,o1|(q1,p1),…,(qn,pn),(1,b1),…,(1,bl)) where each (1,bi) refer to a fiber in Orbf(γ). Necessarily, ∑ i=1lbi=b.

We can then proceed as in [8] to yield a manifold M^ with fibering product structure kM^:S1×F→M^ and a collection of solid tori X with product structure kX:S1×(D1∪…∪Dn+l)→X.

We can also now define a restricted map f^∈Diff(M^). Suppose that the filling of T_i yields a critical fiber of order greater than 2.

Suppose that f^(Ti)=Tj. It could be that i=j.

According to the given product structures (with positively oriented restrictions on the boundary) we then have the following homological diagram:

(d|∂Vi)*H1(Ti)←H1(∂Vi)f^*↓↓(d|∂Vj−1° f ^ °d|∂Vi)*H1(Tj)←H1(∂Vj)(d|∂Vj)*

Now, f:M→M is orientation-reversing and extends into the solid tori Vi,Vj, hence (d|∂Vj−1°f^°d|∂Vi)*=±10a1k∂Vjk∂Vi or ±−10a1k∂Vjk∂Vi. By Corollary 2.2, we must have the second case.

Then according to the framings on Vi,Vj, the fibrations are given by a (−qi,yi)=(−qj,yj) curve where qi=qj>2. f^ must preserve the fibration hence:

±−10a1−qiyi=±qi−yi

But this implies that −aqi+yi=−yi, and so aqi=2yi. This further implies that q_i divides 2 which is a contradiction.

Secondly, we establish that if the Euler class of the manifold is non-zero, then there are no orientation-reversing actions. This can be seen by noting that fiber-preserving diffeomorphisms induce isomorphisms on Euler class. In particular, orientation-preserving maps induce the identity and orientation-reversing induce the negative identity map. For more details, see Theorem 2.4 in [1]. The following result then follows straightforwardly, but we present a constructive proof to further illustrate our methods:

Proposition 3.2. All finite, fiber-preserving actions on an orientable Seifert 3-manifold fibering over an orientable base space with nonzero Euler class are orientation-preserving.

Proof. Again suppose for contradiction that there exists a periodic, fiber-preserving and orientation-reversing diffeomorphism f:M→M. We proceed as in the previous proposition to yield a manifold M^ with fibering product structure kM^:S1×F→M^, a collection of solid tori X with product structure kX:S1×(D1∪…∪Dn+l)→X, and a restricted diffeomorphism f^:M^→M^.

We now consider the first homology group of M^. We have the presentation:

H1(M^)=α1,…,αn+l,a1,b1,…,ag,bg,t|α1⋯αn+l=1,all commute

Where t represents an oriented fiber and α1,…,αn+l represent positively oriented loops kTi({u}×S1) on each boundary torus.

So we must have:

f^*(αi)=αj(i)±1tci

For some integer c_i and some permutation j∈perm{1,…n+l}.

Here the sign is the same for each αi. So then:

1=f^*(α1⋯αl)=t∑ i=1 n+lci

Hence,

∑ i=1 n+lci=0

Case 1: There are no critical fibers. That is, n=0.

Hence the obstruction is nonzero. We then consider the diagram:

(d|∂Vi)*H1(Ti)←H1(∂Vi)(f^|Ti)*↓↓(d|∂Vj(i)−1∘f^|Ti∘d|∂Vi)*H1(Tj(i))←H1(∂Vj(i))(d|∂Vj(i))*

So now (d|∂Vi)*=−1 bi 01kTik∂Vi and (d|∂Vj(i)−1∘f^|Ti∘d|∂Vi)*=±100−1k∂Vj(i)k∂Vi This is as the diffeomorphism extends, is fiber-preserving, and orientation-reversing as well as each V_i being trivially fibered. Here we again use Corollary 2.2.

Hence: (f^|Ti)*=±1−(bi +bj(i) )0−1kTj(i)kTi . So then from above, ci=∓(bi+bj). Hence, we have:

∑ i=1l(bi+bj(i))=2∑ i=1lbi=0

But by Theorem 1.1 of [7], ∑ i=1lbi is the obstruction term and by assumption is nonzero. Hence there can be no such f.

Case 2: There are critical fibers.

Let the fillings of T1,…,Tn be by nontrivially fibered solid tori and the fillings of Tn+1,…,Tl be by trivially fibered solid tori.

Firstly, for T1,…,Tn we have the diagram:

(d|∂Vi)*H1(Ti)←H1(∂Vi)(f^|Ti)*↓↓(d|∂Vj(i)−1°f^|Ti°d|∂Vi)*H1(Tj(i))←H1(∂Vj(i))(d|∂Vj(i))*

Now, in order for (d|∂Vj(i)−1∘f^|Ti∘d|∂Vi)* to extend into the solid torus, preserve a nontrivial fibration, and be orientation-reversing, according to Corollary 2.2 we must have:

(d|∂Vj(i)−1∘f^|Ti∘d|∂Vi)*=±10−1−1k∂Vj(i)k∂Vi

As the fibration on both V_i and Vj(i) is a (-2,1) fibration by Proposition 3.1. Hence, we have:

(d|∂Vi)*=0112kTik∂Vi

and

(d|∂Vj(i))*=0112kTik∂Vj(i)

So that:

(f^|Ti)*=±1−10−1kTj(i)kTi

That is, for those V_i that are nontrivially fibered, c_i=∓1. Here again the sign is the same for all.

For Tn+1,…,Tl we proceed as in Case 1, to yield ci=∓(bi+bj(i)) for i=n+1,…,l.

So now,

0=∑ i=1 n+lci=∑ i=1n∓1+∑ i=n+1l∓(bi+bj(i))=∓n∓2b=±2e

This is twice the Euler class of the bundle which is nonzero. This yields our contradiction

This proposition establishes the fact that there are no orientation-reversing actions on elliptic manifolds as these have nonzero Euler class.

We apply the results of [8] in the case where the base space of the fibration on the Seifert manifold M has underlying space S². Recall for an action φ:G→Difffp(M), there is an induced action φS2:GS2→Diff(S2). We first consider these possible actions.

4.1. Finite group actions on S²

By [11], the possible branching data of a quotient space of S² acted on by a finite group is given by Table 10.1.1. The semidirect product °− is defined so that for H∘−ℤ2, the ℤ2 generator anti-commutes with each element of H. Indeed, throughout, this will be the only semidirect product used. If H happens to be abelian, we use Dih(H) instead.

The notation here is such that rotnx is a rotation of order n about the x-axis when S² is embedded about the origin in ℝ3, similarly with rotny,rotnz. Then refxy is a reflection in the x-y plane, and again similarly with other reflections. Lastly rotL1,rotL2,rotL3 refer to rotations about lines regarding the rotational symmetry of a tetrahedron, an octahedron, and an icosahedron when inscribed inside S². For more details see [3]. Note that the groups may be given by different names in other sources. For example, A4∘−ℤ2 is really S₄, but we write as a semidirect product for convenience.

These groups form partially ordered sets. We do not expressly show these, but they can be worked out by referring to the generators given.

Remark 4.1. By reference to the generators, it is clear is that any finite group that acts on S² is a subgroup of a finite group that is a semidirect product of a group of orientation-preserving diffeomorphisms and a ℤ2 generated by an orientation-reversing element. Again, the semidirect product is such that the ℤ2 generator anti-commutes with all elements of the group of orientation-preserving diffeomorphisms.

This leads us to consider which of these will satisfy the obstruction condition in Table 1:

Remark 4.2. Note that for all actions with induced actions as above, the obstruction condition will be satisfied if the obstruction term is even, but there could actions that will not satisfy the obstruction condition if the obstruction term is odd. One such action is exhibited in [8].

4.2. Manifolds fibering over S²

We now prove a general result that will set up the group structure for the groups acting on manifolds fibering over an orbifold with underlying space S². Throughout this section we assume normalized form for Seifert invariants.

Proposition 4.3. Let M=(0,o1|(q1,p1),…,(qn,pn),(1,b)) and φ:G→Diff+fp(M) be a finite action that satisfies the obstruction condition. Then G is isomorphic to a subgroup of (ℤm×GS2+)∘−ℤ2 for some m∈ℕ and GS2+ is the orientation-preserving subgroup of the induced action φS2:GS2→Diff(S2).

Proof. As φ:G→Diff+fp(M) satisfies the obstruction condition, we can restrict to the action φ^:G→Diff(S1)×Diff(F).

Now, consider φ^S1(G), the projection onto Diff(S1). So then φ^S1(G) is a subgroup of Dih(ℤm)≅ℤm∘−ℤ2 for some m.

Also, φ^F(G) the projection onto Diff(F) will be a subgroup of φ^F(G)+∘−ℤ2 by the remark above, where φ^F+(G)+ is the orientation-preserving subgroup.

So now, φ^(G)⊂ φ ^ S1(G)× φ ^ F(G)⊂(ℤm∘−ℤ2)×( φ ^ F(G)+∘−ℤ2).

But, φ^(G) is orientation-preserving. Hence, we consider the orientation-preserving subgroup of (ℤm∘−ℤ2)×(φ^F(G)+∘−ℤ2).

Note that g∈((ℤm∘−ℤ2)×(φ^F(G)°−ℤ2))+ if and only if g=(g1,g2) or g=(g1z1,g2z2) for (g1,g2)∈ℤm×φ^F(G)+ and z1,z2 are respective generators of the two ℤ2 components. It therefore follows that ((ℤm∘−ℤ2)×(φ^F(G)+∘−ℤ2))+=(ℤm×φ^F(G)+)∘−ℤ2 where the ℤ2 is generated by z=(z1,z2), and the semidirect product is defined by z(g1,g2)z−1=(g1−1,g2−1).

Now, φ^F(G)+≅GS2+ and φ^(G)≅G so that G⊂(ℤm×GS2+)∘−ℤ2.

This result essentially states that we need only check that the obstruction condition is satisfied and calculate the possible orientation-preserving subgroup of the induced action φS2:GS2→Diff(S2). This we can do by reference to the Table 1 and 2.

Table 1 . Orbit numbers of finite group actions on S².

Number	Underlying Space	G	Data	Generators
1	S2	Trivial	()	id
2		ℤn	(n,n)	rotnz
3		Dih(ℤ2n)	(2,2,2n)	rot2nz,rot2y
4		Dih(ℤ2n+1)	(2,2,2n+1)	rot2n+1z,rot2y
5		A4	(2,3,3)	rot2z,rot3L1
6		S4	(2,3,4)	rot2z,rot3L2
7		A5	(2,3,5)	rot2z,rot3L3
8	D	ℤ2	(;)	refxy
9		ℤ2n×ℤ2	(2n;)	rot2nz,refxy
10		ℤ4n+2	(2n+1;)	rot2n+1z∘refxy
11		Dih(ℤn)	(;n,n)	rotnz,refyz
12		Dih(ℤ2n)∘−ℤ2	(;2,2,2n)	rot2nz,rot2y,refyz
13		Dih(ℤ2n+1)∘−ℤ2	(;2,2,2n+1)	rot2n+1z,rot2y,refyz
14		A4°−ℤ2	(;2,3,3)	rot2z,rot3L1,refyz
15		S4×ℤ2	(;2,3,4)	rot2z,rot3L2,refxy
16		A5×ℤ2	(;2,3,5)	rot2z,rot3L3,refxy
17		Dih(ℤ2n)∘−ℤ2	(2;2n)	rot2nz,rot2y,refxz
18		Dih(ℤ2n+1)∘−ℤ2	(2;2n+1)	rot2n+1z,rot2y,refxz
19		A4×ℤ2	(3;2)	rot2z,rot3L1,refxy
20	ℙ2	ℤ2	()	rot2z∘refxy
21		ℤ2n	(n)	rot2nz∘refxy

Table 2 . Orbit numbers of finite group actions on S²..

Number	G	Orbit Numbers of non-regular points	LCM	|G|/LCM	OC Satisfied?
1	Trivial	1	1	1	all b
2	ℤn	1, 1	n	1	all b
3	Dih(ℤ2n)	2, 2n, 2n	2n	2	b even
4	Dih(ℤ2n+1)	2, 2n + 1, 2n + 1	4n + 2	1	all b
5	A4	4, 4, 6	6	2	b even
6	S4	6, 8, 12	12	2	b even
7	A5	6, 10, 15	30	1	all b
8	ℤ2	1	2	1	all b
9	ℤ2n×ℤ2	2, n	2n	2	b even
10	ℤ4n+2	2, 2n + 1	4n + 2	1	all b
11	Dih(ℤn)	1, 1	2n	1	all b
12	Dih(ℤ2n)∘−ℤ2	2, 2n, 2n	4n	2	b even
13	Dih(ℤ2n+1)∘−ℤ2	2, 2n + 1, 2n + 1	8n + 4	1	all b
14	A4∘ℤ2	4, 4, 6	12	2	b even
15	S4×ℤ2	6, 8, 12	24	2	b even
16	A5×ℤ2	6, 10, 15	60	2	b even
17	Dih(ℤ2n)∘−ℤ2	2, 4n	4n	2	b even
18	Dih(ℤ2n+1)∘−ℤ2	2, 4n + 2	4n + 2	2	b even
19	A4×ℤ2	6, 8	12	2	b even
20	ℤ2	−	1	2	b even
21	ℤ2n	2	n	2	b even

We now proceed to consider the individual cases for the number of critical fibers. For each proof the construction set out in [8] provides the converse.

4.3. One critical fiber

We now consider the case where there is only one critical fiber.

Corollary 4.4. Let M=(0,o1|(q,p),(1,b)). There exists a finite action φ:G→Diff+fp(M) if and only if G is isomorphic to a subgroup of Dih(ℤm×ℤn) for some m,n∈ℕ.

Proof. Note that the induced action φS2:GS2→Diff(S2) must fix a point. By Table 1 and 2, we can assume that this is of the form of action 11. This action satisfies the obstruction condition for any b. Hence, GS2+≅ℤn for some n∈ℕ. Then by Proposition 4.1, G is isomorphic to a subgroup of (ℤm×ℤn)∘−ℤ2=Dih(ℤm×ℤn).

4.4. Two critical fibers

Now consider two critical fibers. Firstly, when the respective normalized fillings are not equal.

Corollary 4.5. Let M=(0,o1|(q1,p1),(q2,p2),(1,b)) with (q1,p1)≠(q2,p2). There exists a finite action φ:G→Diff+fp(M) if and only if G is isomorphic to a subgroup of Dih(ℤm×ℤn) for some m,n∈ℕ.

Proof. Note that the induced action φS2:GS2→Diff(S2) must fix two points. By Table 1 and 2, we again assume the form of action 11. This action satisfies the obstruction condition for any b. Hence, GS2+≅ℤn for some n∈ℕ. Then by Proposition 4.1, G is isomorphic to a subgroup of (ℤm×ℤn)∘−ℤ2=Dih(ℤm×ℤn).

Now consider when the fillings of the two critical fibers are equal.

Corollary 4.6. Let M=(0,o1|(q,p),(q,p),(1,b)) with b even. There exists a finite action φ:G→Diff+fp(M) if and only if G is isomorphic to a subgroup of (ℤm×Dih(ℤn))∘−ℤ2 for some m,n∈ℕ.

Proof. We assume that the induced action φS2:GS2→Diff(S2) exchanges two points referring to the critical fibers. Otherwise, we apply Corollary 4.3. Given that two points are exchanged, we consult the Tables to note that we can assume that φS2 is in the form of actions 12/13 or 17/18. The obstruction condition will be satisfied for each of these as we assume that b is even. Then in either case, GS2+≅Dih(ℤn) and by Proposition 4.1, G is isomorphic to a subgroup of (ℤm×GS2+)∘−ℤ2≅(ℤm×Dih(ℤn))∘−ℤ2.

Remark 4.7. Note that M=(0,o1|(q,p),(q,p),(1,b)) with b even is simply S2×S1. M=(0,o1|(q,p),(q,p),(1,b)) with b odd is a Lens space and so as an exception to our results, we refer the reader to [4] for a classification of finite actions on these manifolds.

4.5. Three critical fibers

We now move on to having three critical fibers and break into the three possible scenarios: that they all have different fillings; that two have the same fillings; and that they all have the same filling.

Corollary 4.8. Let M=(0,o1|(q1,p1),(q2,p2),(q3,p3),(1,b)) with (qi,pi) all different. There exists a finite action φ:G→Diff+fp(M) if and only if G is isomorphic to a subgroup of Dih(ℤm) for some m∈ℕ.

Proof. Note that the induced action φS2:GS2→Diff(S2) must fix three points. By Table 1 and 2, the only such induced action is the trivial action 1, that is GS2 is the trivial group. This action trivially satisfies the obstruction condition for any b.

Hence, by Proposition 4.1, G is a subgroup of (ℤm×GS2+)∘−ℤ2≅ℤm∘−ℤ2=Dih(ℤm).

Corollary 4.9. Let M=(0,o1|(q1,p1),(q,p),(q,p),(1,b)) with (q1,p1)≠(q,p). There exists a finite action φ:G→Diff+fp(M) if and only if G is isomorphic to a subgroup of Dih(ℤm×ℤ2) for some m∈ℕ.

Proof. Note that the induced action φS2:GS2→Diff(S2) must fix a point and at most exchange two others. By Table 1 and 2, the only such action is of the form of action 11 with n=2. This action satisfies the obstruction condition for any b. So GS2+≅ℤ2. Hence by Proposition 4.1, G is a subgroup of (ℤm×ℤ2)∘−ℤ2=Dih(ℤm×ℤ2).

Corollary 4.10. Let M=(0,o1|(q,p),(q,p),(q,p),(1,b)). There exists a finite action φ:G→Diff+fp(M) if and only if G is isomorphic to a subgroup of (ℤm×Dih(ℤ3))∘−ℤ2.

Proof. We assume that φS2:GS2→Diff(S2) exchanges three points, else apply Corollary 4.5. or Corollary 4.6. So now by Table 1 and 2 we can assume that φS2:GS2→Diff(S2) is of the form of action 13 with n=1. This action satisfies the obstruction condition for any b and GS2+≅Dih(ℤ3). Hence, by Proposition 4.1, G is isomorphic to a subgroup of (ℤm×Dih(ℤ3))∘−ℤ2.

4.6. No critical fibers

In the case where there are no critical fibers, we note that there are no restrictions on φS2:GS2→Diff(S2). Hence we cannot guarantee that the obstruction condition will be satisfied unless b is even. In such a case the group will be a subgroup of a group of the form (ℤm×H)∘−ℤ2 where H is a group from the list of groups that act orientation-preservingly on S². Note, however that once again, these manifolds are Lens spaces of the form L(b,1) and we again refer the reader to [4].

4.7. Manifolds fibering over ℙ2

We here apply the results of [9] to yield the following result:

Corollary 4.11. Let M=(1,n2|(q,p),(1,b)). There exists a finite action φ:G→Diff+fp(M) if and only if G is isomorphic to a subgroup of ℤ2×Dih(ℤn) for some n∈ℕ.

Proof. Let M˜=(0,o1|(q,p),(q,p),(1,2b)) be the orientable base space double cover of M. According to [9], we consider a corresponding finite action φ˜:G→Diff+fop(M˜) that commutes with the covering translation τ:M˜→M˜.

Now note that the induced action φ˜S2:GS2→Diff(S2) can exchange two points but must be orientation-preserving as φ˜:G→Diff+fop(M˜) is fiber-orientation-preserving. We can then assume that φS2 is in the form of actions 3/4. Then, GS2≅Dih(ℤn) for some n∈ℕ. Again by Table 2, it will satisfy the obstruction condition as 2b is even.

So now apply the results of [9] to note that there is a restricted action φ˜^:G→Diff( M˜^) and product structure k:S1×F→M˜^ (F is in fact an annulus) so such that (k−1∘φ˜^(g)∘k)(u,x)=(ϵ(g)u, φ ˜ ^2(g)(x)) for ϵ(g)=±1. We then note that φ˜^2(G)≅GS2≅Dih(ℤn). Hence, φ˜^(G)≅G is isomorphic to a subgroup of φ˜^1(G)× φ˜^2(G)≅ℤ2×Dih(ℤn). Again, our construction set out in [8] provides the converse.

Recall that elliptic 3-manifolds are Seifert manifolds where χorb(B)>0 and the Euler class of the Seifert bundle is nonzero. [10] By [11], the orbifolds without boundary that have positive orbifold Euler characteristic are:

S2,S2(q1),S2(q1,q2),S2(2,2,q),ℙ2(q),S2(2,3,3),S2(2,3,4),S2(2,3,5)

We note that by Proposition 3.2, all fiber-preserving actions on elliptic manifolds are orientation-preserving as the Euler class must be nonzero. Hence we can break down the possible base spaces and apply the results of the previous sections. In each subsection, suppose that we have a finite action φ:G→Difffp(M).

5.1. Base space S²

These manifolds are lens spaces fibered without critical fibers. By [4], these are of the form L(p,q) where q=±1(modp).

By Remark 2, we can only certainly work with even obstruction condition and in which case the lens space is constructed by two (b,1) fillings of S1×A. We then calculate:

−1b01100−1−1b01=1−2b0−1

Thus we have the lens spaces L(2b,1) for nonzero b∈ℤ.

So now we apply Section 4 to state that the group G will be a subgroup of a group of the form (ℤm×H)∘−ℤ2 where H is a group from the list of groups that act orientation-preservingly on S² and m∈ℕ.

5.2. Base space S²(q)

These manifolds are again lens spaces, but fibered with one critical fiber. All lens spaces can be given such a fibration except those of the form L(p,q) where q=±1(modp) mentioned above. This follows from fibering one solid torus side of the Heegaard torus trivially and inducing a fibration on the other side.

We can now apply Corollary 4.2. to find that the group G is a subgroup of Dih(ℤm×ℤn) for m,n∈ℕ.

5.3. Base space S2(q1,q2)

So M=(0,o1|(p1,q1),(p2,q2),(1,b)). Once again, these manifolds are lens spaces, but now fibered with two critical fibers. All lens spaces can be fibered in this way.

We first consider (p1,q1)≠(p2,q2). Then G is a subgroup of Dih(ℤm×ℤn) for m,n∈ℕ by Corollary 4.3.

If (p1,q1)=(p2,q2) then our results only apply in the case where b is even, in which case the manifold is not elliptic by Remark 3.

5.4. Base space S2(2,2,q)

So M=(0,o1|(q,p),(2,1),(2,1),(1,b)). These manifolds are now prism manifolds fibered longitudinally. We split into the two cases:

Case 1: (q,p)=(2,1)

In this case we apply Corollary 4.7 to yield that G is a subgroup of (ℤm×Dih(ℤ3))∘−ℤ2 for m∈ℕ.

Case 2: (q,p)≠(2,1)

In this case we instead apply Corollary 4.6 to yield that G is a subgroup of Dih(ℤm×ℤ2) for m∈ℕ.

5.5. Base space ℙ2(p)

These manifolds are again prism manifolds but fibered meridianally.

We apply Corollary 4.8 to yield that the group G is a subgroup of ℤ2×Dih(ℤn) for some n∈ℕ.

5.6. Base space S2(2,3,3)

In this case, M=(0,o1|(2,1),(3,p1),(3,p2),(1,b)) for p1=1,2 and p2=1,2. We hence break into the two possible cases:

Case 1: p1=p2

In this case we apply Corollary 4.6 to yield that G is a subgroup of (ℤm×Dih(ℤ3))∘−ℤ2 for m∈ℕ.

Case 2: p1≠p2

In this case we instead apply Corollary 4.5 to yield that G is a subgroup of Dih(ℤm) for some m∈ℕ.

5.7. Base space S2(2,3,4) and S2(2,3,5)

In both of these cases M=(0,o1|(2,1),(3,p1),(q2,p2),(1,b)) for q2≠2,3. Hence we apply apply Corollary 4.7 to yield that G is a subgroup of Dih(ℤm) for some m∈ℕ.

We now consider the quotient spaces under these constructed actions.

6.1. General outline of construction

We first note that an orientation and fiber-preserving action on a fibered torus will have quotient type either a torus or a S2(2,2,2,2). This follows from [10] and the fact that S2(2,3,6), S2(3,3,3), and S2(2,4,4) cannot be Seifert fibered. We then consider the quotient of M^≅S1×F under a product action and the stabilizers of the boundary tori. Here, F will be a disc with holes. There will be then glued in either a solid torus with exceptional core or a Conway ball.

The main part is to establish what form the quotients of M^ and each Vi will be, and then how the gluing maps look under the projection.

Formally, for a representation M=(0,o2|(q1,p1),…,(qn,pn)), we take an action φ:G→Diff+fp(M) that restricts to an action φ^:G→Diff+fp(M^) which leaves some fibering product structure k:S1×F→M^ invariant and which extends over the fillings of a collection of fibered solid tori X=V1∪…∪Vn. We denote φX:G→Diff+fp(X) to be the restricted action on X.

We then let p^:M^→M^/φ^ and $pX:X→X/φX be the quotient maps.

We then have the diagram:

d|∂ViTi←∂Vip^|Ti↓↓(pX)|∂ViTi/Stab φ ^ (Ti)←∂Vi/Stab φX(∂Vi)d′|∂Vi/Stab φX(∂Vi)

We hence need to find the following:

•M^/φ^

•Vi/StabφX(Vi)

•d′|∂Vi/StabφX(∂Vi)

6.2. M^/φ^

We first consider actions constructed via the method of [8] that are fiber-orientation-preserving. For this section we consider F in the more general setting as any orientable surface with boundary.

Lemma 6.1. Let φ^S1×F:G→Diff+(S1)× Diff+(F) be a finite group action such that no element leaves an isolated fiber invariant. Then (S1×F)/φ^S1×F is a trivially fibered Seifert 3-manifold with fibering product structure S1×(F/(φ^S1×F)F).

Proof. It is clear that (S1×F)/φ^S1×F is a trivially fibered Seifert 3-manifold. It remains to show that it has the fibering product structure S1×(F/(φ^S1×F)F). We examine the diagram:

projFS1×F→FpφS1×F↓(S1×F)/φ^S1×F=S1×F′→ F′proj F ′

Now, pφS1×F:S1×F→S1×F′ can be chosen so that pφS1×F(u,x)=(p1(u,x),p2(x)).

So now, we have projF′(p(u,x))=p2(x) and p₂ is the covering map for (φ^S1×F)F. Hence F′=F/(φ^S1×F)F.

Remark 6.2. As φ^S1×F:G→Diff+(S1)×Diff+(F) will be extended over some fillings, if it does happen to leave some isolated fibers invariant, then we can simply drill out these fibers and restrict to φ^′ S1 ×F:G→Diff+(S1)×Diff+(F′) and then consider the resultant torus boundaries to be filled according to a (1,0) filling. Therefore, for our purposes, we can without loss of generality assume that our action φ^S1×F:G→Diff+(S1)×Diff+(F) does not leave any isolated fibers invariant and so the previous lemma holds.

We now allow the fibers to be reversed.

Lemma 6.3. Let φ^S1×F:G→Diff(S1)×Diff(F) be a finite, orientation-preserving group action so that (φ^S1×F)+:G+→Diff+(S1)× Diff+(F) is such that no element leaves an isolated fiber invariant. Then any element that reverses the orientation on both components will induce some product involution f=(f1,f2) of S1×(F/((φ^S1×F)+F) that also reverses the orientation on both components. Then (S1×F)/φ^S1×F is found by taking I×(F/(φ^S1×F)+F) and identifying (i,x) with (i,f2(x)) for i=0,1 and leaving exceptional sets of order 2 as properly embedded arcs or circles according to the fixed point set of f₂.

Proof. If g− is an element of G, so that φ^S1×F(g−) reverses the orientation on both components, then we have some f:S1×(F/((φ^S1 ×F)+F)→S1×(F/((φ^S1 ×F)+F) so that p(φ^S1×F)+∘φ(g−)=f∘p(φ^S1×F)+.

To see that f is an involution requires only the observation that G+ is an index two subgroup of G.

To see that it is a product, we note that if it does not preserve the product structure S1×(F/((φ^S1×F)+F), then g− cannot preserve the product structure S1×F. Hence f is a product reversing the orientation on both components.

The result therefore follows.

Corollary 6.4. Let F be a genus 0 surface with boundary. Let φ^S1×F:G→Diff(S1)×Diff(F) be a finite, orientation-preserving group action so that (φ^S1×F)+:G+→Diff+(S1)× Diff+(F) is such that no element leaves an isolated fiber invariant. Then any element that reverses the orientation on both components will induce some product involution f=(f1,f2) of S1×(F/((φ^S1×F)+F) that also reverses the orientation on both components. Then (S1×F)/φ^S1×F is a ball B less a disjoint collection of balls and solid tori in the interior of B with exceptional sets of order 2 as properly embedded arcs or circles according to the fixed point set of f₂.

Proof. From Lemma 6.2, we note that S1×(F/((φ^S1×F)+F) is simply another genus 0 surface with boundary cross I. It then follows that the boundary identification will fold S1×(F/((φ^S1×F)+F) up to a ball with removed interior balls and solid tori and exceptional sets of order 2 as properly embedded arcs or circles according to the fixed point set of f₂.

Example 6.5. Consider F×S1 where F is a disc with three discs removed. Then take a Dih(ℤ2)-action on F×S1 generated by g₁: an order 2 rotation on F fixing two of the boundary components and exchanging the other two with no rotation in the S1 component, and g₂: the antipodal map on F and a reflection on S1. Then (F×S1)/〈g1〉 is an annulus cross S1. g₂ induces an involution on this space consisting of the antipodal map on the annulus and a reflection in the S1 component. This quotients to a ball with no interior balls removed and no exceptional set.

Example 6.6. Now consider again F×S1 where F is a disc with three discs removed. This time take a Dih(ℤ2)-action on F×S1 generated by g₁: an order 2 rotation in the S1 component and the identity on F and g₂: a reflection on F that leaves two boundary components invariant, exchanging the other two, and a reflection in the S1 component.

Here (F×S1)/〈g1〉 is homeomorphic to F×S1 and g₂ induces the same map on the quotient space. This then quotients to the following space:

Figure 2. Quotient under the action of Example 6.6

6.3. Vi/StabφX(Vi)

We begin by assuming that the action preserves the orientation of the fibers and note that the filling is of a fibered solid torus where the critical fiber is also an exceptional set. By [5] the action of the stabilizer on V_i will be a ℤm×ℤl-action where m divides l with generators φ(g1)(u,v)=(e2aπiu,e2bπiv) and φ(g2)(u,v)=(e2cπiu,e2dπiv) where a=a1a2,b=b1b2,c=c1c2,d=d1d2 are rational numbers. The quotient will then be a solid torus with an exceptional core of order k, where m|k|l. This follows again from [5].

Lemma 6.7. The quotient of a solid torus under a ℤm-action with generator φ(g1)(u,v)=(e2aπiu,e2bπiv) is a solid torus with exceptional core of order:

k=b2gcd(a2,b2)

Proof So φ(g1)a2(u,v)=(u,e2a2bπiv) and φ(g1)a2 is an order lcm(a2,b2)a2=b2gcd(a2,b2) element that fixes the core. The quotient space then has an exceptional core of order k=b2gcd(a2,b2).

Lemma 6.8. The quotient of a solid torus under a ℤm×ℤl-action where m divides l with generators φ(g1)(u,v)=(e2aπiu,e2bπiv) and φ(g2)(u,v)=(e2cπiu,e2dπiv) is a solid torus with exceptional core of order:

k=b2d2gcd(a2,c2)gcd(d2gcd(a2,c2)gcd(a2,b2),a2b1c1d2z+b2c2d1)

Where z is such that a1z+1a2∈ℤ.

Proof We begin by noting that the quotient of the solid torus under the normal group generated by g₁ is V(k') where k′=b2gcd(a2,b2). We then claim the projection under the restricted action of g1, is pg1(u,v)=(ua2,uz′ vk′ ) where z′=a2b1gcd(a2,b2)z for z such that a1z+a2y=−1.

To prove this, we first note that:

pg1(e2aπiu,e2bπiv)=((e2aπiu)a2,(e2aπiu)z′ (e2bπiv)k′ )=(ua2,e2πi(a z ′+b k ′)uz′ vk′ )

So then:

az′+bk′=b1gcd(a2,b2)a1z+b1gcd(a2,b2)   =a1b1gcd(a2,b2)(−1−a2y)+b1gcd(a2,b2)   =−a2ygcd(a2,b2)∈ℤ

So that:

pg1(e2aπiu,e2bπiv)=(ua2,uz′ vk′ )

Also, if we solve pg1(u,v)=(1,1), we yield ua2=1 and u z′v k′=1. So that there are a2k′=a2b2gcd(a2,b2) possible solutions. This is the order of g1.

Now, there is an induced map φ(g2)¯ such that φ(g2)¯∘pg1=pg1∘φ(g2).

We compute:

φ(g2)¯(ua2,uz′ vk′ )=(φ(g2)¯∘pg1)(u,v)      =(pg1∘φ(g2))(u,v)      =pg1(e2cπiu,e2dπiv)      =(e2a2cπiua2,e2πi(c z ′+d k ′)uz′ vk′ )

It follows that φ(g2)¯(u,v)=(e2πa2ciu,e2π(cz′+dk′)iv).

So then φ(g2)¯c2gcd(a2,c2)(u,v)=(u,e2πc2gcd(a2,c2)(cz′+dk′)iv).

Now φ(g2)¯c2gcd(a2,c2) is an element that fixes the core of V(k') and is of order the denominator of c2gcd(a2,c2)(cz′+dk′) when in reduced form. We calculate:

c2gcd(a2,c2)(cz′+dk′)=c1d2z′+d1c2k′d2gcd(a2,c2)

Hence φ(g2)¯c2gcd(a2,c2) has order d2gcd(a2,c2)gcd(d2gcd(a2,c2),c1d2z′+c2d1k′).

So finally, the order of the exceptional core of quotient space of the whole action is:

k=k′d2gcd(a2,c2)gcd(d2gcd(a2,c2),c1d2z′+c2d1k′)=b2gcd(a2,b2)d2gcd(a2,c2)gcd(d2gcd(a2,c2),c1d2z′+c2d1k′)=b2d2gcd(a2,c2)gcd(d2gcd(a2,b2)gcd(a2,c2),c1d2a2b1z+c2d1b2)

We now consider an action of the stabilizer that reverses the orientation of the fibers. By [5] the action will be a Dih(ℤm×ℤl)-action where m divides l with generators φ(g1)(u,v)=(e2aπiu,e2bπiv), φ(g2)(u,v)=(e2cπiu,e2dπiv), and φ(g3)(u,v)=(u−1,v−1). We here note that similar to the proof of Lemma 6.2. we can consider the quotient of the ℤm×ℤl-action and then the induced involution upon it. The following lemma then holds:

Lemma 6.9. The quotient of a solid torus under a Dih(ℤm×ℤl)-action where m divides l with generators φ(g1)(u,v)=(e2aπiu,e2bπiv), φ(g2)(u,v)=(e2cπiu,e2dπiv), and φ(g3)(u,v)=(u−1,v−1) is a Conway ball with exceptional core of order:

k=b2d2gcd(a2,c2)gcd(d2gcd(a2,c2)gcd(a2,b2),a2b1c1d2z+b2c2d1)

Where z is such that a1z+1a2∈ℤ.

Proof. This follows from considering an orientation-preserving involution on V(k) the quotient of the ℤm×ℤl-action generated by φ(g1)(u,v)=(e2aπiu,e2bπiv) and φ(g2)(u,v)=(e2cπiu,e2dπiv).

6.4. d′|∂Vi/StabφX(∂Vi)

We again begin by assuming that the action preserves the orientation of the fibers. So now M^/φ^ has a collection of boundary tori. These will be filled by solid tori with a possible exceptional core. It remains to show how the gluing map from the boundary of the solid tori into M^/φ^ will look.

By using product structures k:S1×F→M^ and k′:S1×F/φ^F→M^/φ^ that restrict to positively oriented product structures kTi:S1×S1→Ti and k′T′i:S1×S1→T′i, we can consider:

H1(M^)=t,x1,…,xs,a1,b1,…,ag,bg|x1⋯xs=1,allcommute
H1(M^/φ^)=t′, x ′ 1,…, x ′ s ′ , a ′ 1, b ′ 1,…, a ′ g ′ , b ′ g ′ | x ′ 1⋯ x ′ s ′ =1,allcommute

Here note that we again allow F to be more generally any orientable surface with boundary as it presents no extra complication to the calculations. t,t' represent a fiber of M^ and M^/φ^ respectively; x1,…,xs represent the boundary loops of k({1}×F); and similarly x′1,…,x′s′ represent the boundary loops of k′({1}×F/φ^F).

We then we have that:

(pφ^)*(t)=t′a,(pφ^)*(xi)=xj(i)′mj(i)t′lj(i)

Here j:{1,…,s}→{1,…,s′} is a surjection.

Note that this is well-defined as if j(i1)=j(i2) then (pφ^)*(xi1)=(pφ^)*(φ(g)*(xi2)) for some g∈G.

Now, 1=(pφ^)*(x1⋯xs)=x1′m1#j−1(1)⋯x s′′m s′ #j−1(s ′)t′l1#j−1(1)+…+l s′ #j−1(s ′).

So that 0=l1#j−1(1)+…+ls′ #j−1(s′) and necessarily mj(i1)#j−1(i1)=mj(i2)#j−1(i2) for any i1,i2∈{1,…,s}.

Now, for any torus (either on the boundary of M^ or on the boundary of one of the solid tori) we have that Stab(T)≅ℤm or ℤm×ℤl where m divides l. [5]

So we consider the diagram:

d|∂ViTi←∂Vipφ^|Ti↓↓p φX|∂Vi T′ i′=Ti/Stab(Ti)←∂ V′ i′d′|∂ V′ i′

We begin with the cyclic case. We can then choose the product structure such that (kTi−1∘φ^(g)∘kTi)(u,v)=(e2aπiu,e2bπiv) for some a,b∈ℚ and g a generator of Stab(Ti). Letting a=a1a2,b=b1b2 be fully reduced, this has order lcm(a2,b2).

Lemma 6.10. If Stab(Ti)≅ℤm, then (kT′i′−1∘pφ^∘kTi)(u,v)=(ulcm(a2,b2)b2vlj(i),vb2) where lj(i) is an integer such that a1lcm(a2,b2)a2b2+lj(i)b1b2=a1gcd(a2,b2)+lj(i)b1b2 is integer valued.

Proof. The projection pφ^|Ti will need to send a fiber to a fiber, hence (kT′i′−1∘pφ^∘kTi)(u,v)=(urvs,vt). But now:

(urvs,vt)=(kT′i′−1∘pφ^∘kTi)(e2aπiu,e2bπiv))=(e2(ar+bs)πiurvs,e2btπivt)

So then take t=b2.

Now consider (kT′i′−1∘pφ^∘kTi)(u,v)=(urvs,vt)=(1,1). This should have lcm(a2,b2) solutions. So ur=1 has lcm(a2,b2)b2 solutions and r=lcm(a2,b2)b2.

Now, ar+bs=alcm(a2,b2)b2+bs∈ℤ. So let s=lj(i) be a solution to this. This exists as gcd(a2,b2) divides b₂ by [2]. There are however an infinite number of choices depending upon the product structure k′T′i:S1×S1→T′i.

The projection pφX|∂Vi will need to extend over the entire solid torus and so will need to send a meridian to a meridian. We can again choose the product structure such that (k∂Vi−1∘φX(g)∘k∂Vi)(u,v)=(e2aπiu,e2bπiv). Hence pφX|∂Vi will similarly give (k∂V′i′−1∘pφ^∘k∂Vi)(u,v)(ua2,uzvlcm(a2,b2)a2). Here the choice of z will not affect the filling but depends upon the product structure k′∂V′i′:S1×S1→∂V′i′.

We proceed with Stab(Ti)≅ℤm×ℤl where m divides l. Then (kTi−1∘φ^(g1)∘kTi)(u,v)=(e2aπiu,e2bπiv) and (kTi−1∘φ^(g1)∘kTi)(u,v)=(e2cπiu,e2dπiv) for some a,b,c,d∈ℚ and g1,g2 generators of Stab(Ti). Here m=lcm(a2,b2) and l=lcm(c2,d2).

Lemma 6.11. If Stab(Ti)≅ℤm×ℤl, then (kT′i′−1∘pφ^∘kTi)(u,v)=(umllcm(b2,d2)vlj(i),vlcm(b2,d2)), where lcm(b2,d2) divides lj(i).

Proof. The projection pφ^|Ti will again need to send a fiber to a fiber. Hence it will again be of the form (kT′i′−1∘pφ^∘kTi)(u,v)=(urvs,vt). Now:

(urvs,vt)=(kT′i′−1∘pφ^∘kTi)(e2aπiu,e2bπiv))=(e2(ar+bs)πiurvs,e2btπivt)
(urvs,vt)=(kT′i′−1∘pφ^∘kTi)(e2cπiu,e2dπiv))=(e2(cr+ds)πiurvs,e2dtπivt)

Hence we take t=lcm(b2,d2).

Now consider (kT′i′−1∘pφ^∘kTi)(u,v)=(urvs,vt)=(1,1). This should have ml solutions. So ur=1 has mllcm(b2,d2) solutions and r=mllcm(b2,d2).

Finally, s is such that amllcm(b2,d2)+bs∈ℤ and cmllcm(b2,d2)+ds∈ℤ. We now calculate:

lcm(a2,b2)lcm(c2,d2)=lcm(lcm(a2,b2),lcm(c2,d2))gcd(lcm(a2,b2),lcm(c2,d2))=lcm(lcm(a2,b2),lcm(c2,d2))lcm(a2,b2)

So that:

lcm(c2,d2)=lcm(lcm(a2,b2),lcm(c2,d2))=lcm(lcm(a2,c2),lcm(b2,d2))

Then:

amllcm(b2,d2)=a1lcm(a2,b2)lcm(c2,d2)a2lcm(b2,d2)    =a1lcm(a2,b2)lcm(lcm(a2,c2),lcm(b2,d2))a2lcm(b2,d2)∈ℤ

Similarly, cmllcm(b2,d2)∈ℤ. So then we require that bs,ds∈ℤ. Hence, b₂ and d₂ must divide s and we take s=lj(i) to be a multiple of lcm(b2,d2).

The projection pφX|∂Vi will need to extend over the entire solid torus and so will need to send a meridian to a meridian. We can again choose the product structure such that (k∂Vi−1∘φX(g1)∘k∂Vi)(u,v)=(e2aπiu,e2bπiv) and (k∂Vi−1∘φX(g1)∘k∂Vi)(u,v)=(e2cπiu,e2dπiv). Hence pφX|∂Vi will similarly give (k∂V′i′−1∘pφ^∘k∂Vi)(u,v)=(ulcm(a2,c2),uzvmllcm(a2,c2)). Here the choice of z will not again affect the filling but depends upon the product structure k′∂V′i′:S1×S1→∂V′i′.

So now we have from above that 0=l1#j−1(1)+…+ls′ #j−1(s′). Hence we have the degree of freedom to choose c1,…,cs′−1 (according to the conditions), but then cs′ will be uniquely determined.

Each filling d′|∂ V ′ i ′ will now be determined be solving:

(pφ^| Ti)*(d|∂Vi)*=(d′|∂ V ′ i ′ )*(p φX|∂Vi)*

Example 6.12. We consider a Dih(ℤ6×ℤ12)-action on the lens space M=(0,o1|(3,2),(1,5)) constructed by:

f1:S1×A→S1×A,f1(u,ρv)=(e2πi6u,ρe2πi3v)
f2:S1×A→S1×A,f1(u,ρv)=(u,ρe2πi12v)
f2:S1×A→S1×A,f1(u,ρv)=(u−1,ρv−1)

Here we parameterize A={ρv|1≤ρ≤2,v∈S1}. Note that according to the product structure S1×A one boundary torus is positively oriented and the other negatively depending on the orientation on the fiber. We take S1×S1 to be positively oriented and S1×2S1 to be negatively oriented.

We calculate first (S1×A)/Dih(ℤ6×ℤ12). This will be I×A where (0,ρv) is identified with (0,ρv−1) and (0,ρv) is identified with (0,ρv−1) with four arcs of order 2. It will be S2×S1 with four properly embedded arcs looking as shown in Figure 3:

Figure 3. Quotient space (S1×A)/Dih(ℤ6×ℤ12)

Next we compute the orders of the exceptional sets of the two Conway balls that fill the two boundary components. We first calculate the generators of the induced action on the solid tori V₁ and V₂ that correspond to the fillings (3,2) and (1,5).

Firstly, for V₁, we compute:

(d|∂V1−1∘f1∘d|∂V1)(u,v)=(d|∂V1−1∘f1)(u−1v2,u−1v3)=d|∂V1−1(e2πi6u−1v2,e2πi3u−1v3)=(e2πi(−36+23)u,e2πi(−16+13)v)=(e2πi6u,e2πi6v)(d|∂V1−1∘f2∘d|∂V1)(u,v)=(d|∂V1−1∘f2)(u−1v2,u−1v3)=d|∂V1−1(u−1v2,e2πi12u−1v3)=(e2πi(212)u,e2πi(312)v)=(e2πi6u,e2πi4v)

So then by Lemma 6.6, the exceptional set will have order:

k=(6)(12)gcd(6,6)gcd(12gcd(6,6)gcd(6,6),(6)(1)(1)(12)z+(6)(6)(11))=432gcd(432,72z+396)

Here z is such that z+16∈ℤ. So take z=-1 and then k=432gcd(432,324)=432108=4.

Secondly, for V₂, we compute:

(d|∂V2−1∘f1∘d|∂V2)(u,v)=(d|∂V2−1∘f1)(u−1v5,v)=d|∂V2−1(e2πi6u−1v5,e−2πi3v)=(e2πi(−16−53)u,e−2πi3v)=(e2πi6u,e4πi3v)(d|∂V2−1∘f2∘d|∂V2)(u,v)=(d|∂V2−1∘f2)(u−1v5,v)=d|∂V2−1(u−1v5,e−2πi12v)=(e2πi(−512)u,e−2πi12v)=(e14πi12u,e10πi12v)

So then again using Lemma 6.6, the exceptional set will have order:

k=(3)(12)gcd(2,12)gcd(12gcd(2,12)gcd(2,3),(2)(2)(7)(12)z+(3)(12)(5))=72gcd(24,288z+180)

Here z is such that z+12∈ℤ. So take z=-1 and then:

k=72gcd(24,108)=7212=6

We now compute the projection maps. By section 6.1, the projection map from both S1×S1 and S1×2S1 will have the matrix:

mllcm(b2,d2)c0lcm(b2,d2)=6c012

Here lcm(b2,d2)=12 divides c, so we take c=12.

The projection map from ∂V1 will have matrix:

lcm(a2,c2)0zmllcm(a2,c2)=60612

The projection map from ∂V2 will have matrix:

lcm(a2,c2)0zmllcm(a2,c2)=12066

We now calculate the projected filling of S1×S1 with V₁ by solving:

(pφ^|S1×S1)*(d|∂V1)*=(d′|∂ V ′ 1)*(p φX|∂V1)*
612012−12−13=x′p′y′q′60612

This yields:

−1848−1236=6x′+6p′12p′6y′+6q′12q′

So then p′=4,q′=3,x′=−7, and y'=-5.

We now calculate the projected filling of S1×2S1 with V₂ by solving: (pφ^|S1×2S1)*(d|∂V2)*=(d′|∂ V ′ 2)*(p φX|∂V2)*

612012−1501=x′p′y′q′12066

This yields:

−642012=12x′+6p′6p′12y′+6q′6q′

So then p′=7,q′=2,x′=−4, and y'=-1.

This fully characterizes the quotient space. We visualize in Figure 4:

Figure 4. Full quotient space