On the asphericity of length five relative group presentations J Howie and V Metaftsis∗
Abstract We investigate asphericity of the relative group presentation hG, t | atbtctdtet = 1i and prove it aspherical provided the subgroup of G generated by {ab−1 , bc−1 , cd−1 , de−1 } is neither finite cyclic nor a finite triangle group. We also prove a similar result for the closely related relative group presentation hG, s, t | αsβsγt = 1 = δtεtζs−1 i.
1
Introduction
A relative group presentation is an expression of the form P = hG, x | ri, where G is a group, x a set disjoint from G, and r a set of cyclically reduced words in the free product G ∗ hxi. This notion was introduced by Bogley and Pride [2], who studied the relative group presentation hG, t | atbtct = 1i. Relative group presentations are closely connected with the study of equations over groups in the following way. If P = hG, x | ri is a relative group presentation, then each r ∈ r is a word in G ∗ hxi. The group defined by P is P = G∗hxi , where N N is the normal closure of r in G ∗ hxi. If the natural map G → P is injective, then the collection xN (x ∈ x) of elements of P is a solution to the system of equations r(G, x) = 1 (r ∈ r) in the overgroup P of G. In this case the relative group presentation P is said to be injective. Bogley and Pride [2] also introduce the notion of asphericity for relative group presentations. In the case where G is the trivial group, we have an ordinary group presentation, and asphericity means diagrammatic reducibility in the sense of Gersten [5]. More generally, a relative group presentation is aspherical if every spherical diagram over it contains a dipole. (See Section 2 for definitions). From a topological point of view, a relative group presentation hG, x | ri can be modelled by a relative 2-complex (Y, X), where X is a K(G, 1)-space. ∗ The second author wishes to thank the Department of Mathematics of Heriot-Watt University and especially J. Howie for their hospitality during the preparation of this paper. Also partial financial support from INTAS grant 94-0921 is acknowledged.
1
Asphericity in this context implies that π2 (Y, X) is generated by some particularly simple elements of π2 (Y ), arising whenever some r ∈ r is equal (as a word in G ∗ hxi) to a conjugate of s∓1 for some s ∈ r. In particular, every aspherical relative group presentation is injective. The converse is far from true, however. Even in cases where injectivity is known and relatively easy to prove, the asphericity of a relative group presentation hG, x | ri depends in a crucial way on the elements of G appearing in the words r ∈ r (the coefficients), and can be difficult to analyse. For example, a theorem of Levin [7] asserts that any relative group presentation of the form P = hG, t | g1 tg2 t . . . gk t = 1i is injective. It is trivial to analyse the asphericity of P for k ≤ 2, but less so as k increases. Bogley and Pride [2] obtain a complete classification of when P is aspherical for k = 3, and Baik, Bogley and Pride [1] obtain an almost complete classification when k = 4 (modulo a few open cases). In a similar way, a relative group presentation hG, t | atbtct−1 = 1i is injective by a theorem of Howie [6], while Edjvet [4] classifies the cases where it is aspherical (again, modulo a few open cases). In the present paper we extend the investigations of [1, 2, 4] to the case k = 5. In other words we study the relative presentation Q1 = hG, t | atbtctdtet = 1i. As expected, we are unable to obtain a complete classification. There are a number of open cases (discussed in Section 7 below). However, we are able to prove the following result in the positive direction. Theorem A With the above notation, let H be the subgroup of G generated by {ab−1 , bc−1 , cd−1 , de−1 }. If H is neither finite cyclic nor a finite triangle group, then Q1 is aspherical. The one-relator relative presentation Q1 is closely connected to the two-relator relative presentation Q2 = hG, s, t | αsβsγt = 1 = δtεtζs−1i. To see this, note that Q2 can be transformed to Q1 by using one of its relations to eliminate one of the generators s, t. Conversely, if appropriate coefficients in Q1 coincide (e.g. a = c) then we can transform Q1 to Q2 . Thus the asphericity of Q1 and Q2 is closely interlinked and we exploit this connection in our arguments. In particular, an immediate consequence of Theorem A is: Theorem B With the above notation, let H be the subgroup of G generated by {αδε−1, ζγε−1, ζβα−1}. If H is neither finite cyclic nor a finite triangle group, then Q2 is aspherical. The structure of the paper is as follows. In Section 2 below we recall some preliminary definitions and results from [1, 2]. In Section 3 we reduce the study of Q1 to a few special cases, studied in Sections 4 through 6. In Section 7 we pull these results together to get proofs of Theorems A and B, and give a discussion of the open cases we are unable to treat. 2
2
Preliminaries
The definitions of this section are taken from [2] and [1]. The reader should consider the above cited papers for more details. A picture P is a finite collection of pairwise disjoint discs {∆1 , . . . , ∆m } in the interior of a disc D2 , together with a finite collection of pairwise disjoint simple arcs {α1 , . . . , αn } embedded in the closure of D2 − ∪m i=1 ∆i in such a way that S each arc meets ∂D2 ∪ m ∆ transversely in its end points. The boundary of P i i=1 2 is the circle ∂D , denoted ∂P. For 1 ≤ i ≤ m, the corners of ∆i are the closures of the connected components of ∂∆i − ∪nj=i αj , where ∂∆i is the boundary of ∆i . The regions of P, denoted by Fj , are the closures of the connected components of D2 −
m [
n [
∆i ∪
i=1
αj .
j=1
An inner region of P is a simply connected region of P that does not meet ∂P. The picture P is nontrivial if m ≥ 1, is connected if m [
∆i ∪
i=1
n [
αj
j=1
is connected, and is spherical if it is nontrivial and if none of the arcs meets the boundary of D2 . The region F1 is a neighbour of F2 if they share a common edge. The number of edges in a region F is called the degree of F and is denoted by d(F ). If P is a spherical picture, the number of different discs to which a disc ∆i is connected is called the degree or the valence of ∆i , denoted by deg(∆i ). Introduce the following labelling: each arc αj is equipped with a normal orientation, indicated by a short arrow meeting the arc transversely, and labelled by an element of x ∪ x−1 . Each corner of P is oriented clockwise (with respect to D2 ) and labelled by an element of G. If κ is a corner of a disc ∆i of P, then W (κ) is the word obtained by reading in a clockwise order the labels on the arcs and corners meeting ∂∆i beginning with the label on the first arc we meet as we read the clockwise corner κ. If we cross an arc labelled x in the direction of its normal orientation, we read x, otherwise we read x−1 . A picture P is called a picture over the relative presentation P if the above labelling satisfies the following conditions 1. for each corner κ of P, W (κ) ∈ r∗ , the set of all cyclic permutations of the members of r ∪ r−1 which begin with a member of x; 2. if g1 , . . . , gl is the sequence of corner labels encountered in an anticlockwise traversal of the boundary of an inner region F of P, then the product g1 . . . gl = 1 in G. We say that g1 . . . gl is the label of F .
3
In the case where G = 1, we get the usual notion of picture which can be found in [8, 9]. A dipole in a labelled picture P over P consists of corners κ, κ0 of P together with an arc joining the two corners such that κ and κ0 belong to the same region and such that if W (κ) = Sg where g ∈ G and S begins and ends with a member of x ∪ x−1 , then W (κ0 ) = S −1 h−1 . The picture P is reduced if it does not contain a dipole. A relative presentation P is called aspherical if every connected spherical picture over P contains a dipole. If P is not aspherical, there is a reduced spherical picture over P. If P is a spherical picture over a relative presentation P and F is a region that contains just two corners then the two arcs in the boundary of F constitute a double bond . In the sequel, double bonds are treated as arcs and so two regions in P will be called neighbours if they share either a single or a double bond. Also by an n-region we mean a region of P consisting of n vertices and n boundary arcs either single or double bonds. Finally, the discs of a spherical picture P over a relative presentation P are also called vertices of P. The star graph P st of a relative presentation P is a graph whose vertex set is x ∪ x−1 and edge set is r∗ . For R ∈ r∗ , write R = Sg where g ∈ G and S begins and ends with a member of x ∪ x−1 . The initial and terminal functions are given by: ι(R) is the first symbol of S, τ (R) is the inverse of the last symbol of S. The labelling function on the edges is defined by λ(R) = g −1. This induces a homomorphism from the path groupoid of P st to G also called the labelling function. The subgroup H of G in Theorems A and B is the image under λ of st the fundamental group of Qst 1 and Q2 respectively. A non-empty cyclically reduced cycle (closed path) in P st will be called admissible if it has trivial label in G. Each inner region of a reduced picture over P supports an admissible cycle in P st . A weight function θ on P st is a real valued function on the set of edges of P st which satisfies θ(Sg) = θ(S −1 g −1 ) with Sg = R ∈ r∗ . A weight function θ on P st is weakly aspherical if the following two conditions are satisfied 1. Let R ∈ r∗ , with R = xε11 g1 . . . xεnn gn . Then n X
ε
i−1 (1 − θ(xεi i gi . . . xεnn gn xε11 g1 . . . xi−1 gi−1 )) ≥ 2
i=1
2. Each admissible cycle in P st has weight at least 2. If there is a weakly aspherical weight function on P st , then P is aspherical. A weight function satisfying condition 1 above is admissible. An angle function on a picture P is a real-valued function φ on the set of corners of P. Associated to φ is a curvature function c defined on the discs ∆ of 4
P by
X
c(∆) = 2π −
φ(κ)
κ⊆∂∆
and on the regions F of P by
c(F ) = 2π −
X
(π − φ(κ)) .
κ⊆∂F
If P is a connected spherical picture, then there is the fundamental curvature formula X X c(∆) + c(F ) = 2πχ(S 2 ) = 4π. ∆
F
The sums are taken over all discs and regions of P. It follows immediately that for any angle function on any connected spherical picture, some disc or region has positive curvature. An application of the curvature formula we use in this paper is due to Edjvet (see [4]) and is called curvature distribution. More specifically, let φ be an angle function on a connected spherical picture P with associated curvature function c. Suppose also that every disc ∆ of P is flat in the sense that c(∆) = 0. Then the curvature formula implies the existence of at least one region F such that c(F ) > 0. If F 0 is a region of P that neighbours F across an arc α in the boundary of F then we can subtract η (any real scalar) from one of the corners of F that touches α and then add η to the adjacent corner in F 0 . This results in a new angle function on P with associated curvature function c∗ . Obviously, c∗ (∆) = c(∆) for each disc ∆ of P, c∗ (F ) = c(F ) − η and c∗ (F 0 ) = c(F 0 ) + η. Other regions are unaffected. More generally, if F is the set of regions of P we can define a distribution scheme on P as the function η : F × F → R . In this paper we consider spherical pictures over Q1 . Each vertex (disc) in such a picture has one of the forms shown below.
a-1 e -1 -1 _ -1 b d -1 c
e a d + b c Figure 1
The − disc is obtained from the + disc by planar reflection. The coefficients labelling the corners are also inverted. In any picture P over Q1 or any other 5
equivalent relative presentation, each arc connectes a + vertex to a − vertex and so each region contains an even number of corners. Each region also supports a relation between the coefficients. Specifically, if we read the corner labels that are encountered in an anticlockwise traverse of the boundary of any simply connected region, then we find a word that represents the identity in G. Furthermore, if a connected spherical picture P over Q1 (or any equivalent relative presentation) is reduced, then each region supports a cyclically reduced word in the alphabet {a∓1 , b∓1 , c∓1 , d∓1 , e∓1 }.
3
Reduction to special cases
Our first observation is that the presentation Q1 is aspherical unless there is a picture over Q1 that contains a vertex connected to only three or two other vertices, in other words has valence two or three. Indeed, in the opposite case, every vertex is connected to at least four more vertices and so there are no reduced k-wheels for k < 4 in the sense of [2]. So the presentation satisfies C(4). On the other hand, if every vertex is connected to at least four other vertices this implies that all coefficients a, b, c, d, e are pairwise −1 different and the star graph Qst to 1 of Q1 consists of five edges oriented from t t with labels a, b, c, d, e. Hence, the smallest admissible cycle in Q1 is of length 4 and therefore Q1 satisfies T(4). Thus by theorem 2.2 in [2], Q1 is aspherical. Now assume that there is a reduced spherical picture over Q1 . Then all coefficients a, b, c, d, e cannot be different. So up to cyclic permutation we have the following cases. Case I. If a = b the using the substitutions x = ta, g1 = a−1 c, g2 = a−1 d g3 = a−1 e the relative presentation Q1 is transformed to the equivalent P = hG, x | x3 g1 xg2 xg3 = 1i. If all g1 , g2 , g3 are different (and different than 1) then one can easily see that any picture over P cannot contain a vertex of valence three or two, a contradiction to our assumptions. If g1 = g3 then by substituting s = xg1 x the relative presentation transforms to P = hG, x, s | xg1 xs−1 = 1 = sg2 sxi, a special case of Q2 . If g1 = g2 , then
P1 = hG, x | x3 g1 xg1 xg3 i
with a−1 c = a−1 d and hence c = d. Case g2 = g3 is equivalent. Case II. If a = c, using again the substitution x = ta, g1 = a−1 b, g2 = a−1 d and g3 = a−1 e the relative presentation Q1 is transformed to P2 = hG, x | x2 g1 x2 g2 xg3 = 1i 6
and by substituting x2 = s to P2 = hG, x, s | x2 s−1 = 1 = sg1 sg2 xg3 i. We note three subcases of P2 that will require particular attention in what follows: P3 = hG, x | x4 g1 xg2 = 1i P4 = hG, x | x2 g1 x3 g2 = 1i P5 = hG, x | x2 g 2x2 gxh = 1i which by using the substitution x2 = s transform to the following relative presentations: P3 = hG, x, s | x2 s−1 = 1 = s2 g1 xg2 i P4 = hG, x, s | x2 s−1 = 1 = sg1 xsg2 i P5 = hG, x, s | x2 s−1 = 1 = sg 2 sgxgi. Lemma 1 If a = b = c = d in Q1 then the relative presentation is aspherical if and only if a−1 e has infinite order. Proof. Let g = a−1 e. Then using also the substitution x = ta, Q1 transforms to the relative presentation P = hG, x | x5 g = 1i. If g has infinite order then Lemma 3 of [1] applies to give us the result. If, on the other hand, g has finite order n then P = G∗A G0 where G0 = hg, x | n g = 1 = x5 gi = hx | x5n = 1i and A = gp(x5 ). So G0 is finite and P contains a finite subgroup that is not conjugate to a subgroup of G so by Theorem 1-(2) in [1], P is not aspherical. 2 Lemma 2 The relative presentation Q2 is aspherical unless there is an admissible cycle of length 3 or 4 in Qst 2. Proof. One can easily see that the star graph of the relative presentation Q2 is a complete graph of 4 vertices.
7
t
ε
t -1 ζ α
s-1
γ
β
δ
s Figure 2 On Qst 2 assign a weight function θ with θ(i) = 1/3 for every i in {α, β, γ, δ, ε, ζ}. Then each admissible cycle with length greater than 4 has length at least 6 and hence satisfies the conditions for asphericity. 2 Lemma 3 Assume that Qst 2 has an admissible cycle of length 3. Then Q2 reduces to Q1 with a = c = d (hence to the relative presentation P4 ). Proof. Without loss of generality we can assume that the cycle ζ −1αβ −1 is admissible. Substitute s = β −1 u. Then the relations of the relative presentation become αβ −1 u2 γt = 1 δtεtζu−1β = 1. Solving the first relation for t we get t = γ −1 u−2 βα−1. Substitute in the second relation to get δγ −1 u−2 βα−1εγ −1 u−2 βα−1 ζu−1β = 1 or equivalently u(ζ −1αβ −1 )u2(γε−1 αβ −1 )u2 (γδ −1 β −1 ) = 1. Since ζ −1αβ −1 = 1 the relation becomes u3 (γε−1 αβ −1 )u2 (γδ −1 β −1 ) = 1 and so the relative presentation Q2 is equivalent to P = hG, u | u3 g1 u2 g2 = 1i with g1 = γε−1 αβ −1 and g2 = γδ −1 β −1 . From the discussion in the beginning of section 3, Q2 is equivalent to P4 . 2 Let P4 = hG, x, s | x2 s−1 = 1 = sg1 xsg2 i. For simplicity, let g1 = g and g2 = h and rewrite the relative presentation P4 as P4 = hG, x, s | x2 s−1 = 1 = sgxshi. We can now prove the following. 8
Lemma 4 Let H be the subgroup of G generated by {g, h} and assume that 1 6= g 6= h∓1 6= 1. Then P4 is aspherical unless one of the following holds 1. g = h2 or h = g 2 2. H is cyclic of order 6 generated by g or h 3. H is an infinite dihedral group 4. 1/o(g) + 1/o(h) + 1/o(gh−1) > 1 Proof. Assume that P4 is not aspherical. Then for any admissible weight function on P4st there exists an admissible cycle of weight less than 2. Now P4st is the complete graph with 4 vertices of Figure 1 with α = δ = ε = ζ = 1 and β = g, γ = h. Assign to P4st a weight function θ in the following way: θ(β) = θ(γ) = θ(ζ) = 0, θ(α) = 1 and θ(ε) = θ(δ) = 1/2. Then the possible admissible cycles in P4st with weight less than 2 which do not contradict our assumptions are labelled h2 , h3 , (gh−1 )2 , (gh−1 )3 , h−1 g 2 , gh−2, gh−3 and gh−1 gh−2. If h2 = 1 then assign θ in P4st with θ(β) = θ(γ) = θ(ζ) = 0, θ(α) = 1, θ(ε) = 1/3 and θ(δ) = 2/3. Here the only possible admissible cycles that do not contradict our assumptions are either (gh−1 )2 or h−1 g 2 . The first case gives us a dihedral group if the order of g is infinite, or 1/o(g) + 1/o(h) + 1/o(gh−1) > 1 if the order of g is finite. The second case gives us that H is a finite group of order 4 generated by g and h = g 2 . If h3 = 1 then assign θ in P4st with θ(β) = θ(γ) = θ(ζ) = 0, θ(α) = 1, θ(ε) = 1/3 and θ(δ) = 2/3. Again the only possible admissible cycles that do not contradict our assumptions are either (gh−1 )2 or h−1 g 2 . The latter implies that H is cyclic of order 6 generated by g with h = g 2. In the former case, we assign to P4st a new weight function θ with θ(α) = θ(ε) = 1/3, θ(γ) = θ(δ) = 2/3 and θ(ζ) = θ(β) = 0. Then the possible admissible cycles in P4st give that either o(g) = 2, 3, 4, 5 or g 2 h−1 = 1 or g 2 h = 1. Here, the first case implies that 1/o(g) + 1/o(h) + 1/o(gh−1) > 1 and the second and third cases that H is cyclic of order 6 generated by g and h = g 2 or g = h4 . For the rest of the proof assume that o(h) ≥ 4. If (gh−1)2 = 1 or (gh−1 )3 = 1 then assign θ in P4st with θ(β) = θ(γ) = θ(ζ) = 0, θ(α) = 1, θ(ε) = 2/3 and θ(δ) = 1/3. In this case there is no admissible cycle that does not contradict our assumption. If g = h3 then assign θ on P4st with θ(β) = θ(ζ) = 0, θ(α) = 2/3, θ(γ) = 1/3 and θ(δ) = θ(ε) = 1/2. The only possibility for H is to be of order 6 generated by h and g = h3 . Finally, if gh−1 gh−2 = 1 then assign θ with θ(β) = θ(ζ) = 0, θ(α) = 2/3, θ(γ) = 1/3 and θ(δ) = θ(ε) = 1/2. Then g 2 = 1. Here g = (gh−1)3 and h = (gh−1 )2 , hence H is cyclic generated by gh−1 . Assign θ with θ(β) = θ(γ) = θ(ζ) = 0, θ(α) = 1, θ(ε) = 2/3 and θ(δ) = 1/3. Then either o(gh−1 ) < 6 or h−1 (h−1 g)2 = 1 or g(h−1 g)3 = 1. Each possibility contradicts our assumptions. 9
In each case we have shown that one of the conditions 1-4 holds or derived a contradiction, so the lemma is proved. 2 Lemma 5 Assume that Qst 2 has an admissible cycle of length 4. Then either Q2 reduces to Q1 with a = b = c (hence to the relative presentation P3 ) or Q2 reduces to P5 . Proof. There are 3 cycles of length 4 in Qst 2 (up to cyclic permutation and inversion). Assume first that ζβδε−1 = 1. In Q2 substitute t = ε−1 u. Then the relations become αsβsγε−1 u = 1 = δε−1 u2 ζs−1 . By solving the second relation for s = δε−1 u2 ζ and substituting in the first we get (αδε−1)u2 (ζβδε−1)u2 (ζγε−1)u = 1 and since ζβδε−1 = 1 the relative presentation for Q2 is transformed to hG, u | u4 g1 ug2 = 1i with g1 = ζγε−1 and g2 = αδε−1 . Hence, the result follows from the discussion in the beginning of section 3. A similar argument holds in the case where αβ −1γε−1 = 1. Now suppose that αδγ −1 ζ −1 = 1. Then by using the substitution t = γ −1 u in Q2 the relations become αsβsu = 1 = δtεtζs−1. The second relation gives s = δtεtζ and substituting in the first we get (αδγ −1 )u(εγ −1)u(ζβδγ −1)u(εγ −1)uζu = 1. Since αδγ −1 = ζ the above is transformed into (ζu)2(εγ −1 )u(ζβδγ −1)u(εγ −1)u = 1 and substituting ζu = v to v 3 (εγ −1 ζ −1 )v(ζβδγ −1ζ −1 )v(εγ −1ζ −1 ) = 1. Let for simplicity εγ −1 ζ −1 = g and ζβδγ −1ζ −1 = h. Then Q2 is transformed to the relative presentation P = hG, v | v 3 gvhvg = 1i or equivalently to P = hG, x, s | xgxs−1 = 1 = shsxi. The star graph P st of the above presentation is the graph of Figure 2 with α = δ = γ = ζ = 1 and β = g, ε = h. Assign to P st a weight function θ in the following way: θ(β) = θ(ε) = 0 and θ(α) = θ(δ) = θ(γ) = θ(ζ) = 1/2. Then the possible admissible cycles in P st with weight less than 2 are labelled gh−1 and gh. 10
If gh−1 = 1 then g = h and P = hG, x | x3 gxgxg = 1i. Using the transformation xg = w, P trasnforms to hG, w | w 4 g −1wg −1 = 1i which is a special case of P3 . Finally, if gh = 1 then P = hG, x | x3 gxg −1xg = 1i. Using the substitution w = xg we get the relative presentation P 0 =< G, w | w 2 g −2 w 2g −1 wg −1 = 1i which transforms to P5 using the appropriate substitutions. 2 Let P3 = hG, x, s | x2 s−1 = 1 = s2 g1 xg2 i. For simplicity, let g1 = g and g2 = h. Hence the relative presentation P3 is transformed to P3 = hG, x, s | x2 s−1 = 1 = s2 gxhi. We now prove the following. Lemma 6 Let H be the subgroup of G generated by {g, h} and assume that 1 6= g 6= h∓1 6= 1. Then P3 is aspherical unless one of the following holds 1. g = h2 or h = g 2 2. H is cyclic of order 6 generated by g or h 3. H is an infinite dihedral group 4. 1/o(g) + 1/o(h) + 1/o(gh−1) > 1 Proof. The star graph of the relative presenation P3 , P3st is the complete graph with 4 vertices shown in Figure 2 with α = h, γ = g and β = δ = ε = ζ = 1. Let θ be the weight function on P3st with θ(α) = θ(γ) = θ(ε) = 0, θ(β) = 1 and θ(ζ) = θ(δ) = 1/2. Then θ is an aspherical weight function unless there is a word in {g, h} equal to 1 in H corresponding to an admissible cycle. Hence we can assume that one of g 2 , g 3 , h2 , h3 , gh∓2 , hg ∓2 is the identity in H. By symmetry, assume that either g 2 = 1 or g 3 = 1 or gh−2 = 1 or gh2 = 1. If gh2 = 1 then assign a new weight function on P3st with θ(α) = θ(γ) = 0, θ(β) = 1, θ(δ) = 1/2 and θ(ζ) = θ(ε) = 1/4. Then the admissible cycles give that either g 2 = 1 or g 3 = 1 or (gh−1)2 = 1. In all cases H is cyclic of order 4,6 or 6 respectively, generated by h. If g 2 = 1 then assign a weight function to P3st with θ(γ) = θ(δ) = θ(α) = 0, θ(β) = 1 and θ(ε) = θ(ζ) = 1/2. Assuming that gh∓2 6= 1, we have that one of the following holds in H: h2 = 1 or h3 = 1, (gh−1 )2 = 1 or (gh−1 )3 = 1 or gh−3 = 1 or gh−1 gh−2 = 1. If h2 = 1 or (gh−1 )2 = 1 then H is dihedral. If gh−3 = 1 then H is cyclic of order 6 generated by h. If gh−1 gh−2 = 1 then H is cyclic of order 6 generated by gh−1 . If h3 = 1 then assign a new weight function on P3st with θ(γ) = θ(δ) = θ(α) = 0, θ(β) = 1, θ(ε) = 2/3 and θ(ζ) = 1/3. Then either H is cyclic or (gh−1 )2 = 1 or (gh−1)3 = 1 or (gh−1 )4 = 1 or (gh−1 )5 = 1. Any one of the last four relations implies that 1/o(g) + 1/o(h) + 1/o(gh−1) > 1. 11
If g 2 = 1 = (gh−1 )3 then assign θ on P3st in the following way: θ(γ) = θ(δ) = θ(α) = 0, θ(β) = 1, θ(ε) = 1/3 and θ(ζ) = 2/3. Again, H is either cyclic or h2 = 1 or h3 = 1 or h4 = 1 or h5 = 1. Any one of the last four relations implies that 1/o(g) + 1/o(h) + 1/o(gh−1) > 1. Finally, if g 3 = 1, then we can assume that h2 6= 1 (otherwise we can apply the argument of the previous case, by symmetry) and that gh∓2 6= 1. Assign a weight function θ on P3st with θ(γ) = θ(α) = 0, θ(β) = 1 and θ(δ) = θ(ε) = θ(ζ) = 1/3. The only possible admissible cycle is then (gh−1 )2 = 1. In that case consider the weight function with θ(γ) = θ(α) = θ(e) = 0, θ(β) = 1, θ(δ) = 1/3 and θ(ζ) = 2/3. Then either H is cyclic of order 6 or one of the following hold: h3 = 1 or h4 = 1 or h5 = 1. All three relations imply that 1/o(g)+1/o(h)+1/o(gh−1) > 1.
2
4
The relative presentation P4
For the rest of this section we consider the relative presentation P4 = hG, x | x3 gx2 h = 1i. Lemma 7 Let H be the subgroup of G generated by {g, h}. If H is an infinite dihedral group then P4 is aspherical. Proof. Suppose first that g 2 = h2 = 1, w = x3 gx2 h. Then we lift to the 2-fold cover Pˆ4 corresponding to the kernel of the map hG, x | w = 1i → Z2 with g, h → 1, x → 0. Define z = gh. We now have a system of two equations in two unknowns y0 = x and y1 = gxg over hzi ∼ = Z, given by y03y12 z = 1 = y13y02 z −1 . Then we can tranform the presentation of Pˆ4 to the one-relator presentation hy0 , y1 | y05y15 = 1i which is aspherical. Hence our original presentation is also aspherical. If g 2 = (gh−1)2 = 1 then use the lift to the 2-fold cover Pˆ4 corresponding to the kernel of the map hG, x | w = 1i → Z2 with g, x → 1, h → 0. Here we obtain the system of equations over hhi y0 y1 y02 y1 h = 1 = y1 y0 y12 y0 h−1 with y0 = xg and y1 = gx. We can transform the presentation Pˆ4 to a one-relator presentation hy0 , y1 | y0 y1 y02 y12 y0 y10y0 = 1i which again is aspherical. The case h2 = (gh−1 )2 = 1 is similar. 12
2
Lemma 8 If g = h∓1 then P4 is aspherical if and only if g has finite order. Proof. If g = h∓1 and g has infinite order then Lemma 3 in [1] applies to show that P4 aspherical. If g = h and g k = 1, k < ∞ then we can construct the following spherical picture over P4
1 g
g -1
-1
g
1 1
1 g
g g
-1
1
1
...
1
g-1
Figure 3 Finally, if g = h−1 and g k = 1 then the equation x3 gx2 h = 1 transforms to x = gx2 g −1 . Hence, −3
k
k−1
x(−3) = gx2(−3)
g −1 = g 2x2
2 (−3)k−2
g −2 = . . . = g k x2 g −k = x2 . k
k
This last relations implies that x has finite order, hence Theorem 1 in [1] applies to show that P4 is not aspherical. 2 Lemma 9 Assume that g = h2 . If o(g) ≤ 5 then P4 is not aspherical. (There is an analogous result if the rˆ oles of g and h are interchanged.) Proof. A coset enumeration reveals that the group P4 = hx, h | hk = 1 = x3 h2 t2 hi k = 2, 3, 4, 5 is finite with orders 10, 105, 500, 2525 respectively. So by Theorem 1 in [1] P4 is not aspherical. 2 Lemma 10 If
1 o(g)
+
1 o(h)
+
1 o(gh−1 )
> 1 then P4 is not aspherical.
Proof. Since there is no loss in assuming that 0 ≤ o(g) ≤ o(h) and (by Lemma 8) that o(gh−1) ≥ 2 we have the following possible cases o(g) = 2, {o(gh−1), o(h)} = {2, n}, {3, 3}, {3, 4} or {3, 5} 13
o(g) = 3, o(gh−1 ) = 2 and o(h) = 3, 4, 5. Suppose o(g) = 2. Start with a tesselation T of the sphere in which each vertex has valence o(gh−1 ) and each face is a o(h)-gon. Surround each vertex −1 of T with an o(gh−1)-gon with label (gh−1 )o(gh ) in such a way that all edges are double bonds. Now the double bonds that face to neighbouring vertices of the tesselation are connected with a g −2 -region (in the sense of [1]). You have also created ho(h) -regions that complete the construction. As an example we have constructed the picture for the case (o(g), o(h), o(gh−1)) = (2, 4, 3). 1
h g
-1
g
h
-1
1 h
1 g
g
h 1 g -1 h 1
h
-1
1
-1
1 g
-1
h
g -1
-1
h
h 1
g
1
g
g
-1
1
-1
g
g
1
g
h
g
-1
1
-1
g
g -1
1
g
1
h
-1
1
g
h
h
-1
-1
g
-1
g
h -1
1
g
h
-1
1
h
h
h g
h
1
-1
1 g
g
1
h
1 -1
-1 -1
h
h
-1
-1
-1
h
g
-1
1
g
-1
h
-1
1
h
h
1
h
1
1
h
g
1 g
1
h
h
-1
h
-1
1
1 g
g 1
-1
g -1
1
h
h
g
1
1
1
-1
h 1
g
-1
h
h
g -1
h
-1
1
-1
g
1
-1
g 1
h
-1 -1
g
1
h
1
g
1
h
g
1
-1
-1
h
g h
-1
h
1
g
1
h
g
g
-1
g
g h
-1
g 1
1
g -1
g
h
-1
h 1
h
Figure 4 If o(g) = 3 and o(gh−1 ) = 2 we make a similar construction. Here each vertex of the tesselation has valence 3 and each face is an o(h)-gon. Place g 3 -regions around the vertices of the tesselation such that the edge that face neighbouring vertices of the tesselation are double bonds. Close those double bonds by double bonds and you also created h−o(h) -regions. This completes the construction. We leave it to the reader to carry out the constructions in individual cases. 2 14
Theorem 1 The relative presentation P4 is aspherical unless one of the following. 1. g = h∓1 and the order of g is finite 2. 1/o(g) + 1/o(h) + 1/o(gh−1) > 1, with o(g), o(h), o(gh−1) < ∞ 3. g = h2 and o(g) < 6 or h = g 2 and o(h) < 6 4. g = h2 and o(g) > 6 or h = g 2 and o(h) > 6 5. The group H generated is cyclic of order 6 generated by g or h 6. The group H generated is cyclic of order 6 generated by gh−1 In cases 1, 2, 3 and 6 the relative presentation P4 is not aspherical. Proof. The proof of the Theorem is an immediate consequence of Lemmas 4, 7, 10, 9, 8 and from the fact that the group hx, gh−1 | (gh−1 )6 = 1 = x3 (gh−1 )2 x2 (gh−1 )3 i is finite of order 12090.
5
2
The relative presentation P3
For the rest of this section we assume that the relative presentation P3 is P3 = hG, x | x4 gxh = 1i. Lemma 11 Let H by the subgroup of G generated by {g, h}. If H is an infinite dihedral group then P3 is aspherical. Proof. The argument is similar to the one used in Lemma 7. Again if h2 = g 2 = 1, lift to the 2-fold cover Pˆ3 corresponding to the kernel of the map hG, x | x4 gxh = 1i → Z2 with g, h → 1 and x → 0. If z = gh, the system of equations we get is y04y1 z = 1 = y14 y0 z −1 with y0 = x and y1 = gxg. This, again, transforms to the one-relator presentation hy0 , y1 | y05y15 = 1i which is aspherical. If g 2 = (gh−1)2 = 1 the map P3 → Z2 is given by g, x → 1 and h → 0. Here we obtain two equations over hhi, namely y0 y1 y0 y12 h = 1 = y1 y0 y1 y02 h−1 15
with y0 = xg and y1 = gx. The presentation Pˆ3 can transform to the one-relator presentation hy0 , y1 | y0 y1 y0 y13 y0 y1 y02 = 1i which is aspherical and hence P3 is aspherical. A similar argument applies if h2 = (gh−1 )2 = 1. 2 Lemma 12 If
1 o(g)
+
1 o(h)
+
1 o(gh−1 )
> 1 then P3 is not aspherical.
Proof. The proof is similar to that of Lemma 10 and is left to the reader.
2
Lemma 13 If g = h∓1 then the relative presentation P3 is aspherical if and only if g has finite order. Proof. If g = h∓1 and g has infinite order then Lemma 3 in [1] applies to show that P3 is aspherical. If g = h and g k = 1 then the following is a spherical picture over P3 and hence P3 is not aspherical.
1 -1
g
g
g-1
1
1 g
g 1
...
1
Figure 5 If g = h−1 and g k = 1 then the equation x4 gxg −1 = 1 can be written as k x−4 = gxg −1. Hence x(−4) = g k xg −k = x and hence x has finite order. Therefore, by Theorem 1 in [1], P3 is not aspherical. 2 Lemma 14 If g = h2 and o(g) < 6 then P3 is not aspherical. If g = h2 and o(g) > 6 then P3 is aspherical. Similarly if we interchange the rˆ oles of g and h. Proof. A coset enumeration reveals that the group with presentation hx, h | hk = 1 = x4 h2 xhi for k = 2, 3, 4, 5 is finite of order 10, 195, 820, 3775 respectively and by Theorem 1 in [1] P3 is not aspherical. On the other hand, if o(g) > 6 then the presentation hx, h | hk = 1 = x4 h2 xhi is a C(4)-T(4) presentation so π2 is generated by dipoles arising from the relation hk = 1, hence P3 is aspherical (see also [1]). 2 16
Theorem 2 The relative presentation P3 is aspherical unless one of the following holds: 1. g = h∓1 and g has finite order 2. g = h2 and o(g) ≤ 5 or h = g 2 and o(h) ≤ 5 3. 1/o(g) + 1/o(h) + 1/o(gh−1) > 1, o(g), o(h), o(gh−1) < ∞ 4. The group H is cyclic of order 6 generated by gh−1 5. The group H is cyclic of order 6 generated by g or h In cases 1, 2, 3 and 4, the relative presentation P3 is not aspherical. Proof. The proof of the Theorem is an immediate consequence of Lemmas 11, 12, 13, 14 and the fact that the groups hgh−1, x | x4 (gh−1 )3 x(gh−1 )2 = 1 = (gh−1 )6 i hgh−1, x | x4 (gh−1 )2 x(gh−1 )3 = 1 = (gh−1 )6 i are finite of order 38010.
6
2
The relative presentations P5 and P1
Lemma 15 The presentation P5 = hG, x | x2 g 2 x2 gxg = 1i is aspherical if and only if g is of finite order. Proof. If g has infinite order then Lemma 3 in [1] applies and P5 is aspherical. Now let o(g) < ∞. If o(g) = 2 then P5 has a reduced spherical picture shown in Lemma 13. On the other hand if o(g) = n > 2 then the following is a reduced spherical picture over P5 .
17
g
-2
g g
1 g-1 g
g2 g g-2
-2
g
1
g
g -2
g
g2
g -2
1 g2
g-1
g -1
1
g -1
g2
g2
g -1
1 g
g-2
g
g
... Figure 6 This completes the proof of the lemma. 2 For the rest of this section let P1 be the relative presentation P1 = hG, x | x3 gxgxh = 1i with g 6= 1 6= h and P1 the group defined by P1 Let also H be the group generated by {g, h} in G. Lemma 16 In P1 the rˆoles of h and gh−1 can be interchanged. Proof. In P1 we substitute x = zg −1 . Then the relation transforms to zg −1 zg −1 z 3 g −1h = 1 and by reversing and replacing z −1 by y we get P10 = hG, y | y 3gygyh−1g = 1i. Hence, all the results that apply to P1 also apply to P10 .
2
Lemma 17 If P1 is not aspherical then at least one of the following conditions holds 1. g = h∓1 2. g = h2 or h = g 2 3. 2 ∈ {o(g), o(h)} 4. o(gh−1) = 2 and 3 ∈ {o(g), o(h)}
18
Proof. Suppose that P1 with g 6= 1 6= h is not aspherical. Assume also that none of the conditions 1, 2, 3 hold. Thus g 6= h∓1 , g 6= h2 , h 6= g 2 and both g and h have order at least three. Since P1 is not aspherical, there is a nontrivial reduced connected spherical picture P over P1 . We define an angle function on P as follows. Each corner within a double bond has angle zero. Every other corner has angle 2π/deg(∆) where deg(∆) is the number of different vertices connected to vertex ∆. In other words every corner in a vertex with valence three has angle 2π/3, every angle in a vertex with valence four has angle π/2 and every corner in a vertex with valence five has angle 2π/5. Since we will make use of this scheme again, we will refer to it as the standard angle function. Denoting the associated curvature function by c, one has that each vertex ∆ (disc) of P is c-flat: c(∆) = 0. Any region within any double bond is also flat. Then using the curvature formula one finds that the only possibility for a positively curved n-region F is
6−n c(F ) ≥ π 3
and hence c(F ) is positive only if n = 4. By considering all cyclically reduced words of length at most four in the alphabet {g ∓1 , h∓1 } and having in mind our assumptions, we see that a 4-region can exist only if it is a (gh−1 )∓2 -region (in the sense of [1]) or all corners are labelled by 1 ∈ G. Since a possitively curved region needs to contain at least one double bond, the latter case is impossible to make without creating a dipole. Hence, o(gh−1 ) = 2. Now, consider the alternative angle function to be the following. All corners within a double bond have angle 0. All corners with labels g ∓1 or h∓1 have angle π/2. The corners labelled by 1 not in a double bond have angle π if they belong to a vertex of valence three, π/2 if they belong to a vertex of valence 4 and π/4 if they belong to a vertex of valence five. One can easily see again that c(∆) = 0, in words all vertices are flat. If F is a positively curved region in P, F can have at most half of its vertices labelled by 1 ∈ G with angles π. Hence the maximum curvature for a region F can be 8−n c(F ) = π 4 and so can only be a 4-region or a 6-region. If F is a positively curved 4-region, then it necessarily contains a corner labelled by 1 ∈ G belonging to a vertex of valence three. The labels of the other three corners cannot be 1 (or else you create a dipole) and any other choice contradicts our assumptions (that conditions 1,2,3 do not hold). Hence a 4-region is not possible. If F is a positively curved 6-region then F cannot have fewer than three corners labelled by 1 with angles π or it has negative or zero curvature. Also, 19
the above three corners cannot belong to vertices that are connected or else we create a dipole. Hence, there is no loss in assuming that all three corners belong to negative vertices. The other three corners, all belonging to positive vertices, are labelled by g or h (again a label 1 would lead to a dipole). So there are four different choices for the label of the 6-region. Either g 3 = 1 or h3 = 1 or g −1 = h2 or h−1 = g 2. The first two cases imply that either o(g) = 3 or o(h) = 3. The last two cases give us that the group H generated by {g, h} is cyclic of order 6 either generated by h and g = h4 or generated by g and h = g 4 . In the first case o(gh−1 ) = 2 and o(g) = 3 and in the second case o(gh−1 ) = 2 and o(h) = 3. 2 Lemma 18 If g = h then P1 is aspherical if and only if g has infinite order. If g = h−1 then P1 is not aspherical if o(g) < 4 and is aspherical if g has infinite order. Proof. If g = h∓1 and o(g) = ∞ then asphericity of P1 follows from Lemma 3 in [1]. If g = h and g k = 1 then the following spherical picture over P1 shows that P1 not aspherical.
1 -1
g
g
g-1
1
1 g
g 1
...
1
Figure 7 Finally, if g = h−1 and o(g) = 2, 3 then a coset enumeration reveals that the group hx, g | g k = 1 = x3 gxgxg −1i is of finite order 30 or 960 when k = 2 or 3 respectively. 2 Lemma 19 The relative presentation P1 is not aspherical if 1/o(g) + 1/o(h) + 1/o(gh−1) > 1 with o(g), o(h), o(gh−1) < ∞. Proof. The proof, similar to that of Lemmas 10 and 12, is left to the reader.
2
Lemma 20 Let H be the subgroup of G generated by {g, h}. If H is an infinite dihedral group then P1 is aspherical. 20
Proof. The argument used is similar to that in Lemma 7. If P1 is the group defined by P1 and g 2 = (gh−1 )2 = 1 then the map P1 → Z2 is given by g → 1, h, x → 0. Lifting to the 2-fold cover that corresponds to the kernel of the map P1 → Z2 we obtain two equations in y0 = x, y1 = gxg over hhi ∼ = Z namely y03 y1 y0 h = 1 = y13y0 y1 h−1 . The above presentation can transform to Pˆ1 = hy0, y1 | y03 y1 y0 y13y0 y1 = 1i. This last presentation is aspherical as a two-generator, one-relator presentation and so is P1 . If h2 = g 2 = 1 then define the map P1 → Z2 by h, x, g → 1. Here we obtain a system of two equations over an infinite cyclic group generated by z = hg, y0 y1 y0 z −1 y0 z −1 y0 = 1 = y1 y0 y1 zy1 zy1 with y0 = hx and y1 = xh. Use the first equation to eliminate y1 , obtaining an aspherical one-relator presentation: hy0 , z | y0−2 zy0−1 (zy0−2 (zy0−1 )2 )3 = 1i. If h2 = (gh−1 )2 = 1 then define the map P1 → Z2 by h, x → 1, g → 0. Then we obtain a system of equations over hgi y0 y1 y0 gy1 g −1y0 = 1 = y1 y0 y1 g −1y0 gy1 with y0 = hx and y1 = xh. Note here that this is a standard HNN-presentation of a semi-direct product hy0 , y1 i o hgi, which is known to be aspherical by standard results (see e.g. [3]). 2 Lemma 21 If o(g) = 3, o(gh−1) = 2 and o(h) ≥ 6 then P1 is aspherical unless H is cyclic of order 6 generated by h with g = h4 . There is an analogous result if the rˆ oles of g and h are interchanged. Proof. Let us assume that P1 is not aspherical and that P is a spherical picture over the relative presentation P1 . Assign to the vertices (discs) of P the alternative angle function defined in the proof of Lemma 17. Again using the curvature formula for a region we can see that a positively curved region can only be a 4-region or a 6-region. More specifically, arguing as in Lemma 17 we see that the only positively curved region can be a 6-region with label g 3 or gh2 . The second case implies that H is cyclic of order 6 generated by h with g = h4 . Excluding this case, assume that all positively curved regions are 6-regions with label g 3 .
21
The maximum curvature for such a region is π/2. Define a distribution scheme on P as follows.
π/6 if c(F ) > 0 and F is separated from F 0 by a double bond with η(F, F 0) = corner labels 1 and g 0 otherwise If η(F, F 0) > 0 then at least one corner of F 0 has label h∓1 . Let F 0 be a positively curved n-region after the curvature distribution. In F 0 , two double bonds each belonging also to a positively curved region cannot be adjacent. In other words, only every second boundary arc of F 0 can belong to a positively curved region. So the maximum curvature for F 0 is
24 − 2n c(F ) = π . 12 0
This last equation implies that F 0 has positive curvature only if n = 4, 6, 8 or 10. If a 10-region F 0 exists then its maximum curvature before the distribution was −π/2 and hence F 0 needs to have at least four double bonds belonging also to positively curved regions and five corners labelled by 1 belonging to vertices with valence three. This implies that the label of the region is either h∓5 or (gh4 )∓1 . Using our hypothesis we arrive in contradiction in both cases. If an 8-region F 0 exists then its maximum curvature before the distribution was 0 and so we have the following cases Case I. Four vertices with valence three and corner labels 1 and at least one double bond belonging also to a positively curved region and so having corner labelled h∓1 . A simple investigation shows that all possible labels for such a region contradict hypothesis. Case II. Three vertices with valence three and cornel labels 1 and four double bonds belonging also to positively curved regions. There is only one possible such region with label h∓4 , contradiction. If a 6-region F 0 exists, then this region had before the distribution negative or zero curvature, hence it has at least one double bond belonging also to a positively curved region. Moreover, it has at least two corners labelled 1, belonging to vertices of valence three. Case I. The two corners with labels 1 and angles π belong to positive vertices (the case both negative is equivalent). Then the only possible labels for region F 0 are h−1 g −2, h−2 g −1 , gh−3 , h−2 , h−3 . Using our hypothesis we either get that H is cyclic of order 6 generated by h and g = h4 or a contradiction. Case II. The two corners with labels 1 and angles π belong to a positive and a negative vertex. Then the only possible labels for region F 0 are h−1 g, h−2 g, h−1 g 2 and h−2 g 2 . Again, using the conditions in hypothesis we get either a contradiction or that H is cyclic of order 6 generated by h and g = h4 . 22
Finally, if a 4-region F 0 exists, then this region has at least one double bond belonging also to a positively curved region. This means that two corners of F 0 are labelled by 1 and h−1 . A simple investigation shows that all possible labels of such a region contradict to our assumptions. Therefore, after the distribution, all regions have negative or zero curvature which contradicts the existence of a spherical picture. Along to the same lines is the case where the rˆoles of g and h are interchanged. The details of the proof are left to the reader. 2 Lemma 22 If o(g) = 2, o(h) ≥ 4 and o(gh−1) ≥ 4 then P1 is aspherical. Proof. Let us assume that there is spherical picture P over P1 . Assign to the vertices of P the standard angle function defined in the proof of Lemma 17. Then the only positively curved region F of P is -1
1 h
g
-1
h
-1
g
g
1
1
g
-1
h
h
1 Figure 8
The maximum curvature for F is 2π/3. Define a distribution scheme on P as follows. π/6 if c(F ) > 0 and F is separated 0 from F 0 by a double bond η(F, F ) = 0 otherwise Notice that in the above region F if you “break” any of the double bonds (replacing it by a single bond) you decrease the curvature of the picture by at least π/6 since one of the vertices connected by the broken bond will have valence at least 4. So the above distribution scheme still applies. If η(F, F 0) > 0 then at least one corner of F 0 has label h∓1 . Also notice that if F 0 is a region after the distribution, F 0 cannot have two consequtive double bonds belonging also to positively curved regions. Hence, the maximum curvature for an n-region F 0 after the distribution is 24 − 3n π 12 23
and that is positive only if F 0 is a 4-region or a 6-region. If a 6-region F 0 exists then we have the following cases. Case I. At least five of the six vertices of F 0 have valence three. Then an analysis shows that the label of F 0 is (g −1 h)∓3 contrary to hypothesis. Case II. F 0 has four vertices of valence three then at least three double bonds of F 0 belong also to positively curved region giving possible labels for F 0 either (gh−1 )3 or h−3 , contradiction. If a 4-region F 0 exists then two of the four corners have labels either 1, h−1 or g −1 , h. A simple investigation shows that every possible combination of labels for F 0 contradicts to hypothesis. Hence a positively curved F 0 cannot exist, and that is a contradiction to the assumption that a spherical picture P exists. 2 Theorem 3 The relative presentation P1 is aspherical unless one of the following holds 1. g = h and g has finite order 2. g = h−1 and g has finite order 3. 1/o(g) + 1/o(h) + 1/o(gh−1) > 1 with o(g), o(h), o(gh−1) < ∞ 4. H is cyclic of order 6 generated by g or h 5. g = h2 or h = g 2 In cases 1 and 3, P1 is not aspherical. In case 2, P1 is not aspherical if o(h) < 4. Proof. A consequence of Lemmas 17, 18, 19, 20, 21, 22 combined with Lemma 16 when necessary. 2
7
Conclusions
Combining the results of the previous sections, we are now able to prove the results stated in the Introduction. Proof of Theorem A. Suppose a, b, c, d, e ∈ G such that Q1 is not aspherical. Then by the results of Section 3 we can assume that we are in one of the subcases P1 , P3 , P4 or P5 . The result then follows form Theorems 1, 2 or 3 and Lemma 15 respectively. 2 Proof of Theorem B. The relative presentation Q2 transforms to one of type Q1 by a Tietze transformation that respects asphericity and (up to conjugacy) the subgroup H. Then Theorem B follows from Theorem A. 2
24
Theorem A and B give strong restrictions on when the relative presentations Q1 and Q2 can fail to be aspherical. However these are not the only restrictions, and more details can be read off from the other results in the paper. For example, Lemma 2 says that Q2 is aspherical unless there is an admissible cycle in the star graph of length three or four. From this it can be deduced that Q1 is aspherical if the coefficients a, b, c, d, e ∈ G take at least four different values – except possibly in the case where (up to cyclic permutation) a = c and a−1 eb−1 d = 1. In this latter case we follow through the proof of Lemma 5 to transform Q1 into P3 = hG, x | x4 g1 xg2 = 1i with g1 = a−1 db−1 a and g2 = d−1 a, so Theorem 2 yields further restrictions. Our results also show in many cases that Q1 and/or Q2 is not aspherical. For example, if a, b, c, d, e take only two values g, h ∈ G and g −1h has finite order, then Q1 is not aspherical. However, they also leave open some intriguing questions. Question 1 Suppose G = hg | g 6 = 1i and h = g k for k = 2, 3 or 4. Are (any of) hG, t | t4 gth = 1i, hG, t | t3 gt2 h = 1i, hG, t | t3 gtgth = 1i or hG, t | t3 hthtg = 1i aspherical ? Question 2 Suppose G = hg | g k = 1i for k ≥ 7. Are (any of) hG, t | t3 gt2 g 2 = 1i, hG, t | t3 gtgtg 2 = 1i or hG, t | t3 g 2 tg 2 tg = 1i aspherical ? Question 3 Suppose G = hg | g k = 1i for k = 4 or 5. Is hG, t | t3 gtgtg −1 = 1i aspherical ? We have been unable to answer these questions with the existing methods.
References [1] Y.G. Baik, W.A. Bogley and S.J. Pride, On the asphericity of length four relative group presentations, Int. J. Algebra and Comp. 7 (3), 227–312 (1997). [2] W.A. Bogley and S.J. Pride, Aspherical relative presentations, Proc. Edinburgh Math. Soc. 35, 1–39 (1992). [3] I.M. Chiswell, D.J. Collins and J. Huebschmann, Aspherical group presentations, Math. Z. 178, 1–36 (1981). [4] M. Edjvet, On the asphericity of one-relator relative presentations, Proc. Royal Soc. Edinburgh 124A, 713–728 (1994). [5] S. M. Gersten, Reducible diagrams and equations over groups. In : S.M. Gersten (ed), Essays in group theory, 15–73, Math. Sci. Res. Inst. Publ., 8, Springer, New York-Berlin, 1987. 25
[6] J. Howie, The solution of length three equations over groups, Proc. Edinburgh Math. Soc. 26, 89–96 (1983). [7] F. Levin, Solutions of equations over groups, Bull. Amer. Math. Soc. 68, 603–604 (1962). [8] S.J. Pride, Identities among relations of group presentations. In: E. Ghys, A. Haefliger and A. Verjovsky (eds), Group theory from a geometrical viewpoint (Trieste, 1990), 687–717. [9] C.P. Rourke, Presentations and the trivial group. In: R. Fenn (ed), Topology of low-dimensional manifolds (Proc. Second Sussex Conf., Chelwood Gate, 1977), pp. 134–143, Lecture Notes in Math. 722, Springer, Berlin, 1979.
Department of Mathematics, Heriot-Watt University, Edinburgh EH14 4AS, Scotland. Email:
[email protected] Department of Mathematics, University of the Aegean, Karlovassi, GR83 200 Samos, Greece. Email:
[email protected]
26
