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Forum Mathematicum

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Ed. by Blomer, Valentin / Cohen, Frederick R. / Droste, Manfred / Duzaar, Frank / Echterhoff, Siegfried / Frahm, Jan / Gordina, Maria / Shahidi, Freydoon / Sogge, Christopher D. / Takayama, Shigeharu / Wienhard, Anna


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Volume 31, Issue 1

Issues

Asphericity of positive free product length 4 relative group presentations

Suzana Aldwaik / Martin Edjvet
  • Corresponding author
  • School of Mathematical Sciences, University of Nottingham, Nottingham NG7 2RD, United Kingdom
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/ Arye Juhász
Published Online: 2018-09-10 | DOI: https://doi.org/10.1515/forum-2017-0141

Abstract

Excluding some exceptional cases, we determine the asphericity of the relative presentation 𝒫=G,xaxmbxn, where a,bG{1} and 1mn. If H=a,bG, the exceptional cases occur when a=b2 or when H is isomorphic to C6.

Keywords: Relative group presentation; picture; relative diagram; curvature; asphericity

MSC 2010: 20F05; 57M05

1 Introduction

A relative group presentation is an expression of the form 𝒫=G,𝒙𝒓, where G is a group, and 𝒙 is a set that is disjoint from G. Denoting the free group on 𝒙 by 𝒙, 𝒓 is a set of cyclically reduced words in the free product G𝒙. The group defined by 𝒫 is G(𝒫)=(G𝒙)/N, where N is the normal closure of 𝒓 in G𝒙.

A relative group presentation is said to be orientable if no element of 𝒓 is a cyclic permutation of its inverse; and is said to be aspherical if every spherical picture over it contains a dipole (see Section 2.1). If G is the trivial group, 𝒫 is then an ordinary group presentation, and asphericity means diagrammatic reducibility as defined in [13] under the assumption that 𝒫 has no proper power relators. These notions were introduced and developed in [6], where it is shown that if 𝒫 is orientable and aspherical, then group theoretic information about G(𝒫) can be deduced. In particular, the natural homomorphism GG(𝒫) is injective [6]; and torsion in G(𝒫) can be described [6, 15].

There has been some progress in the problem of determining the asphericity of 𝒫 when both 𝒙 and 𝒓 consist of a single element (see, for example, [3, 1, 2, 6, 10, 11, 12, 14, 18]). Our particular interest is when 𝒙={x} and 𝒓={axmbxεn} where a,bG{1}, 1mn and ε=±1. Recent studies and applications of this case can be found in [7, 8, 12]. When ε=-1 and m=1, the asphericity of 𝒫 has been determined (modulo some exceptional cases) when n=2 in [11], when n=3 in [3] and when n4 in [10]. When ε=+1 and m=1, the asphericity of 𝒫 has been determined when n=2 in [6], and (again modulo some exceptional cases) when n=3 in [2], when n=4 in [14] and when n5 in [1]. In the present paper, we consider the case ε=+1. Before stating our main theorem, observe that axmbxn=1 if and only if b-1x-ma-1x-n=1, and it follows that there is no loss of generality in working modulo ab. Note that this allows us to assume that |a||b|. For example, suppose that |a|=3, |b|=2, [a,b]=1 and mn. Then axmbxn=1 if and only if xnaxmb=1 if and only if b-1x-ma-1x-n=1. Replacing x by x-1 yields a relator of the form axmbxn, where |a|=|b-1|=2, |b|=|a-1|=3 and [a,b]=[b-1,a-1]=1.

We list the following exceptional cases (in which |b| denotes the order of b in G):

  • (E1)

    a=b2, |b|{5,6} and n{m,2m},

  • (E2)

    a{b3,b4}, |b|=6 and nm,

  • (E3)

    a=b2, 6<|b|< and m<n<2m.

Theorem 1.1.

Let P be the relative group presentation P=G,xaxmbxn, where 1mn, xG and a,bG{1}. Suppose that none of the conditions in (E1), (E2) or (E3) holds. Then P is aspherical if and only if (modulo ab) none of the following holds:

  • (a)

    ( m=n ), 1<|ab-1|<.

  • (b)

    ( mn )

    • (i)

      a=b±1 has finite order.

    • (ii)

      a=b2 and |b|=4.

    • (iii)

      a=b2, 4<|b|< and n=2m.

    • (iv)

      |a|=2, |b|=3 and [a,b]=1.

    • (v)

      1|a|+1|b|+1|ab-1|>1 , where 1:=0.

Moreover, if P is aspherical, then x has infinite order in G(P).

Clearly, the element x is not conjugate in G(𝒫) to any element of G and |x|m+n>1 in G(𝒫). The final statement in Theorem 1.1 then follows immediately from, for example, statement (0.4) in the introduction of [6] and the fact that 𝒫 is orientable.

An example is given by the following: Let F be the free group with basis u0,,ul-1, and let θ be the automorphism of F such that uiθ=ui+1 (mod l). For ωF, recall that the cyclically presented group Gl(ω) is given by the group presentation

Gl(ω)=u0,,ul-1ω,ωθ,,ωθl-1.

For integers A0, r1, s1 and f1, consider

ω=u0ufu2fu(r-1)fuAuA+fuA+2fuA+(s-1)f.

Then Gl(ω) belongs to the class of groups of type 𝔉, as defined in [8]. Now the automorphism θ of F induces an automorphism of Gl(ω) and the resulting split extension El(ω) of Gl(ω) by the cyclic group of order l has presentation El(ω)=u,ttl,w(u,t), where w(u,t) is obtained from ω by the rewrite uit-iuti [16]. In our case, ω rewrites to (ut-f)rtrf-A(ut-f)stsf+A, so letting x=ut-f, we obtain El(ω)=G,xaxrbxs, where G=ttl, a=trf-A and b=tsf+A. More generally, it follows from [8, Lemma 6] and [4, Theorem 4.1] that Theorem 1.1 can be applied to obtain asphericity classifications for group of type 𝔉.

In Section 2, we give the method of proof and introduce the concepts of pictures and curvature distribution. In Section 3, some preliminary results are proved. The proof of Theorem 1.1 is completed in Section 4.

2 Method of proof

2.1 Pictures

The definitions in this subsection are taken from [6]. The reader is referred to [6] and [2] for more details.

A picture is a finite collection of pairwise disjoint discs {D1,,Dm} 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-i=1mDi in such a way that each arc meets D2i=1mDi transversely at its end points. The boundary of is the circle D2, denoted . For 1im, the corners of Di are the closures of the connected components of Di-j=1nαj, where Di is the boundary of Di. The regions of are the closures of the connected components of D2-(i=1mDij=1nαj). An inner region of is a simply connected region of that does not meet . The picture is non-trivial if m1, is connected if i=1mDij=1nαj is connected, and is spherical if it is non-trivial and if none of the arcs meets the boundary of D2. Thus the set of regions of a connected spherical picture consists of the simply connected inner regions together with a single annular region that meets . The number of edges in Δ is called the degree of the region Δ and is denoted by d(Δ). A region of degree n will be called an n-gon. If is a spherical picture, the number of different discs to which a disc Di is connected is called the degree of Di, denoted by d(Di). The discs of a spherical picture are also called vertices of .

Suppose that the picture is labeled in the following sense: Each arc αj is equipped with a normal orientation, indicated by a short arrow meeting the arc transversely, and labeled with an element of 𝒙𝒙-1. Each corner of is oriented clockwise (with respect to D2) and labeled with an element of G. If κ is a corner of a disc Di of , then W(κ) will be the word obtained by reading in a clockwise order the labels on the arcs and corners meeting Di beginning with the label on the first arc we meet as we read the clockwise corner κ.

If we cross an arc labeled x in the direction of its normal orientation, we read x, else we read x-1.

A picture over the relative presentation 𝒫=G,𝒙𝒓 is a picture labeled in such a way the following are satisfied:

  • (1)

    For each corner κ of , W(κ)𝒓, the set of all cyclic permutations of 𝒓𝒓-1 which begin with a member of 𝒙.

  • (2)

    If g1,,gl is the sequence of corner labels encountered in anticlockwise traversal of the boundary of an inner region Δ of , then the product g1g2gl=1 in G. We say that g1g2gl is the label of Δ, denoted by l(Δ)=g1g2gl.

A dipole in a labeled picture over 𝒫 consists of corners κ and κ of together with an arc joining the two corners such that κ and κ belong to the same region and such that if W(κ)=Sg where gG and S begins and ends with a member of 𝒙𝒙-1, then W(κ)=S-1h-1. The picture is reduced if it is non-empty and 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, then there is a reduced spherical picture over 𝒫.

A connected spherical picture over 𝒫 is defined to be strictly spherical if the product of the corner labels in the annular region taken in anticlockwise order defines the identity in G. The relative presentation 𝒫 is weakly aspherical if each strictly spherical connected picture over contains a dipole. Let G(𝒫) denote the group defined by 𝒫. It is shown in [6] that if 𝒫 is weakly aspherical and if the natural map of G into G(𝒫) is injective, then 𝒫 is aspherical.

Now let 𝒫=G,xaxmbxn. Then the natural map GG(𝒫) is injective [17], and so it suffices to show that 𝒫 is weakly aspherical. Let be a reduced connected strictly spherical picture over 𝒫. Then the vertices (discs) of are given by Figure 2.1 (i), (ii). It is clear from the orientation of the edges that a positive vertex can only be connected to a negative vertex, in particular, the degree of each region of is even.

Vertices of 𝒫{\mathcal{P}} and the star graph Γ.
Figure 2.1

Vertices of 𝒫 and the star graph Γ.

2.2 The star graph Γ

The star graph Γ of a relative presentation 𝒫 is a graph whose vertex set is 𝒙𝒙-1 and edge set is 𝒓. For R𝒓, write R=Sg, where gG and S begins and ends with a member of 𝒙𝒙-1. The initial and terminal functions are given as follows: ι(R) is the first symbol of S, and τ(R) is the inverse of the last symbol of S. The labeling function on the edges is defined by λ(R)=g-1 and is extended to paths in the usual way. A non-empty cyclically reduced cycle (closed path) in Γ will be called admissible if it has trivial label in G.

In general, we have that only each inner region of a reduced spherical picture over 𝒫 supports an admissible cycle in Γ. However, since we are only considering strictly connected spherical , the same holds for the annular region as well.

The star graph Γ of 𝒫=G,xaxmbxn is given by Figure 2.1 (iii). In particular, a word obtained from a cyclically reduced closed path in Γ does not contain aa-1,a-1a,bb-1,b-1b up to cyclic permutation although it can contain the subwords 11-1,1-11 provided that different edges of Γ labeled by 1 are used. (Note that the structure of Γ also implies that the degree of a region must be even.)

Using Γ, we see that the possible labels or regions of degree 2 or 4 are (up to cyclic permutation and inversion) as follows:

d(Δ)=2:l(Δ){ab-1,11-1},d(Δ)=4:l(Δ){(ab-1)2,ab-1a1-1,ab-11b-1,ab-111-1,a1-11b-1,a1-1a1-1,a1-1b1-1,b1-1b1-1,11-111-1}.

2.3 Curvature

Our aim is to show that, given certain conditions on m, n, a and b, 𝒫 is aspherical. To this end, assume by way of contradiction that is a reduced connected strictly spherical picture over 𝒫. Our method is curvature distribution (see, for example, [1]). Proceed as follows: Contract the boundary to a point which is then deleted. This way all regions Δ of the amended picture, also called , are simply connected and form a tessellation of the 2-sphere. An angle function on is a real-valued function on the set of corners of . Given this, the curvature of a vertex of is 2π less the sum of all the angles at that vertex; and then the curvature c(Δ) of a region Δ of is (2-d(Δ))π plus the sum of all angles of the corners of Δ. The angle functions we will define results in each vertex having zero curvature, and so the total curvature c() of is given by c()=Δc(Δ). Given this, it is a consequence of Euler’s formula, for example, that c()=4π, and so must contain regions of positive curvature. Our strategy will be to show that the positive curvature that exists in can be sufficiently compensated by the negative curvature. To this end, we locate each Δ satisfying c(Δ)>0 and distribute c(Δ) to near regions Δ^ of Δ. For such regions Δ^, define c(Δ^) to equal c(Δ^) plus all the positive curvature Δ^ receives minus all the curvature Δ^ distributes during this distribution procedure. We prove that c(Δ^)0 and, since the total curvature of is at most c(Δ^), this yields the desired contradiction.

The standard angle function assigns the angle 0 to each corner of v which forms part of a region of degree 2 and assigns 2πd(v) to the remaining corners. Therefore if Δ is a region of degree k>2 with vertices v1,,vk such that d(vi)=di (1ik), then c(Δ)=c(d1,,dk)=(2-k)π+i=1k2πdi; or if d(Δ)=2, then c(Δ)=0. Since c(3,3,3,3,3,3)=0 this shows, for example, that must contain a region of degree 4 (since d(Δ) is even); and since c(4,4,4,4)=0, must contain a vertex of degree <4.

3 Preliminary results

Recall that 𝒫=G,xaxmbxn (1mn) and denotes a reduced connected strictly spherical picture over 𝒫 amended as described in Section 2.3. We can make the following assumptions without any loss of generality:

  • (P1)

    is minimal with respect to number of vertices.

  • (P2)

    Subject to (P1), is maximal with respect to number of 2-gons.

Bridge move at the 4-gon Δ.
Figure 3.1

Bridge move at the 4-gon Δ.

Lemma 3.1.

If Δ is a region of P, then l(Δ)(11-1)k for k2.

Proof.

Consider the 4-gon Δ of Figure 3.1 (i) having label (11-1)2. Observe that there are ki 2-gons between vi and vi+1 (1i4, subscripts mod 4). Apply r=min(k2+1,k4+1) bridge moves [9] of the type shown in Figure 3.1 (ii), (iii). Then each of the first r-1 bridge moves will create and destroy two 2-gons leaving the total number unchanged. The r-th bridge move however will create two 2-gons but destroy at most one. Since bridge moves do not alter the number of vertices, we obtain a contradiction to (P2). The proof now proceeds by induction on k. Indeed if l(Δ)=(11-1)k where k>2, then a sequence of bridge moves can produce a new picture with at least the same number of 2-gons but with a region having label (11-1)k-1. ∎

Lemma 3.2.

The following conditions hold:

  • (i)

    If K=a,bG is infinite cyclic, then 𝒫 is aspherical.

  • (ii)

    If |ab-1|= , then 𝒫 is aspherical.

Proof.

(i) The result follows immediately from [5, Lemma 3.8]. (ii) If a,b is infinite cyclic, the result follows by (i); otherwise the proof of [2, Theorem 3] which uses a weight test [6] on the star graph shows that 𝒫 is aspherical. ∎

Lemma 3.3.

If m=n, then P is aspherical if and only if |ab-1|{1,}.

Proof.

If |ab-1|=, the result follows from Lemma 3.2. Let |ab-1|=1, and so the relator is axmaxm. If m=1, it is clear that every picture contains a dipole, so let m>1. Then the degree of each vertex of any given picture is at least 4, and it follows from the last paragraph in Section 2.3 that 𝒫 is aspherical. Finally, if 1<|ab-1|<, then since (axm)2=ab-1, it follows that |axm|< and non-asphericity follows, for example, from [6, statement (0.4)]. ∎

Lemma 3.4.

Suppose that mn. Then the following hold:

  • (i)

    If a=b±1 and |a|< , then 𝒫 is not aspherical.

  • (ii)

    If a=b2 and |b|=4 , then 𝒫 is not aspherical.

  • (iii)

    If a=b2, 4<|b|< and n=2m , then 𝒫 is not aspherical.

  • (iv)

    If a=b2, n>2m and |b|7 , then 𝒫 is aspherical.

Proof.

(i) If a=b-1, then |x|< and so 𝒫 is not aspherical; or if a=b, then, for example, a spherical picture can easily be constructed (Figure 3.2 (i) for the case |a|=3).

(ii) In this case, there is the spherical picture of Figure 3.2 (ii). (This also follows from [7, Lemma 2.1].)

(iii) Since b2xmbx2m=1 if and only if b-1(xmb)b=(xmb)-2, it follows that |xmb|< and 𝒫 is not aspherical.

Spherical pictures and vertices.
Figure 3.2

Spherical pictures and vertices.

(iv) Let 𝕂 be a connected spherical picture over 𝒫. If 𝕂 contains a vertex of degree 2, then, since n>2m, 𝕂 contains a dipole and the result follows, so assume otherwise. Now fill in the regions of 𝕂 using, if necessary, b-vertices with label b±k to obtain a connected spherical picture 𝕃 over the ordinary presentation b,x;bk,b2xmbxn. The vertices of 𝕃 are given (up to inversion) in Figure 3.2 (iii). Clearly, each region of 𝕃 not of degree 2 has degree 4; and, since k7, each b-vertex has degree 4. Curvature considerations now tells us that there must be a non-b-vertex v such that d(v)<4, and so v must form a dipole in 𝕃. But d(v)3 in 𝕂, so it follows that d(v)=3 in both 𝕂 and 𝕃 which forces the dipole v forms in 𝕃 to be a dipole in 𝕂, as required. ∎

It will be assumed from now on that none of the following exceptional cases holds:

  • (E1)

    a=b2, |b|{5,6} and n{m,2m},

  • (E1´)

    b=a2, |a|{5,6} and n{m,2m},

  • (E2)

    a{b3,b4}, |b|=6 and nm,

  • (E2´)

    b{a3,a4}, |a|=6 and nm,

  • (E3)

    a=b2, 6<b< and m<n<2m,

  • (E3´)

    b=a2, 6<a< and m<n<2m.

It follows from Lemmas 3.23.4, the exceptional cases and [1] that from now on the following assumptions can be made:

  • (A1)

    1<m<n,

  • (A2)

    a,b is not infinite cyclic,

  • (A3)

    |ab-1|<,

  • (A4)

    ab±1,

  • (A5)

    ab2 and ba2.

Given this, we have the following lemmas:

Lemma 3.5.

The following conditions hold:

  • (i)

    If none of (ab-1)2, a2 or b2 are trivial in a,b , then 𝒫 is aspherical.

  • (ii)

    If 1|a|+1|b|+1|ab-1|>1 , then 𝒫 is not aspherical.

Proof.

(i) Any reduced closed path in the star graph Γ of length at most 4 involving a or b yields (see Section 2.2) one of the relators ab±1, a2b-1, ab-2, (ab-1)2, a2 or b2, and so (A4) and (A5) together with assuming that none of (ab-1)2, a2, b2 are trivial forces the degree of each region (not of degree 2) to be at least 6, and so 𝒫 is aspherical (see Section 2.3).

A spherical picture.
Figure 3.3

A spherical picture.

(ii) If the condition holds, then reduced spherical pictures over 𝒫 can be constructed. For example, if (|a|,|ab-1|,|b|)=(2,3,4), then is given by Figure 3.3. The other spheres can be obtained in similar fashion, we omit the details. ∎

Lemma 3.6.

If any of the following sets of conditions holds, then P is aspherical:

  • (i)

    |a|=2, |b|4 and |ab-1|4.

  • (ii)

    |a|4, |b|4 and |ab-1|=2.

  • (iii)

    |a|=2, |b|= and |ab-1|=2.

Proof.

Assign the following angle function to the vertices of : If d(v)>3, then assign 2πd(v) to each corner of v not part of a 2-gon and 0 otherwise; and if d(v)=3, then for (i), (ii), (iii) respectively, assign angles as shown in Figure 3.4 (i), (ii), (iii) respectively. In each case, it follows that if Δ is a region of , then Δ has at most 12d(Δ) corners having angle π. Therefore c(Δ)(2-d(Δ))π+12d(Δ).π+12d(Δ).π2=π(8-d(Δ)2), and so c(Δ)0 for d(Δ)8.

Assignment of angles at vertices of degree 3.
Figure 3.4

Assignment of angles at vertices of degree 3.

(i) If d(Δ)=4, then l(Δ)=a1-1a1-1 and c(Δ)-2π+4.π2=0; and if d(Δ)=6, then checking the star graph Γ shows that (up to cyclic permutation and inversion)

l(Δ){ab-1a1-1b1-1,ab-11a-1b1-1,a1-1a1-111-1},

and so c(Δ)-4π+2.π+4.π2=0.

(ii) If d(Δ)=4, then l(Δ)=(ab-1)2 and c(Δ)-2π+4.π2=0; and if d(Δ)=6, then

l(Δ){ab-1ab-111-1,b-1ab-1a1-11,ab-11a-1b1-1,ab-11b-1a1-1},

and so c(Δ)-4π+2.π+4.π2=0.

In (i) and (ii), there are no regions of positive curvature, a contradiction which completes the proof in these cases.

(iii) If d(Δ)=4 and c(Δ)>0, then l(Δ)=(a1-1)2 is given by Figure 3.5 (i) and (ii). In Figure 3.5 (i), c(Δ)(2-4)π+π+3.π2=π2 is distributed from Δ to Δ^ as shown; and in Figure 3.5 (ii), 12c(Δ)π2 is distributed from Δ to each of Δ^1 and Δ^2 as shown. If d(Δ)=6 and c(Δ)>0, then Δ is given by Figure 3.5 (iii), and hence 12c(Δ)12[(2-6)π+3.π+3.π2]=π4 is distributed from Δ to each of Δ^1 and Δ^2 as shown. Note that positive curvature is distributed across a (b-1,1)-edge as shown in Figure 3.5 (i)–(iii) or the inverse (1-1,b)-edge as in Figure 3.5 (iv), where the maximum amount of π2 is indicated.

Distribution of curvature for Lemma 3.6 (iii).
Figure 3.5

Distribution of curvature for Lemma 3.6 (iii).

Distribution of curvature for aεi⁢wi⁢aεi+1{a^{\varepsilon_{i}}w_{i}a^{\varepsilon_{i+1}}} segments.
Figure 3.6

Distribution of curvature for aεiwiaεi+1 segments.

Let Δ^ be a region that receives positive curvature. Then l(Δ^) involves b and, since |b|= and ab, must also involve a at least twice. Therefore l(Δ^)=aε1w1aε2w2aεrwr, where εi=±1 (1ir), r>1 and each wi (1ir) does not involve a. It follows that c(Δ^)2π+r.π2+α, where the r.π2 is the contribution from the r vertices with corner label a±1, and α is the total curvature contributed to c(Δ^) by the r segments aεiwiaεi+1 (subscripts mod r). First observe that it can be assumed without any loss that 1-11 is not a sublabel of l(Δ) since the contribution to c(Δ^) made by the three edges and two intermediate vertices of Figure 3.6 (i) is at most -3π+3π2=-3π2, whereas the contribution made by the single edge shown is -π. Given this, it can be further assumed that 11-1 is not a sublabel since the possibilities a-111-1b and b-111-1a each contribute at most -3π+3π2+π2=-π which equals the contribution made by the edge of a-1b or b-1a (Figure 3.6 (i)). Given these two assumptions, it follows that, up to inversion, there are four types of segment aεiwiaεi+1, and these are

ab-11b-11b-11b-1a,a1-1b1-1b1-1b1-1a,ab-11b-11b-11a-1,a-11b-11b-11b-1a.

These are shown in Figure 3.6 (ii)–(v).

The contribution from each segment is made up from the edges ei (1il), the vertices vj (1jl-1) and any positive curvature Δ^ receives across the ei. It follows that

  • the segment of Figure 3.6 (ii) contributes at most

    -l.π+(l2-1)π+l2.π2+(l2-1)π2=-3π2;

  • the segment of Figure 3.6 (iii) contributes at most

    -l.π+l2.π+(l2-1)π2+(l2-1)π2=-π;

  • the segment of Figure 3.6 (iv) contributes at most

    -l.π+(l-12)π+(l-12)π2+(l-12)π2=-π;

  • the segment of Figure 3.6 (v) contributes at most

    -l.π+(l-12)π+(l-12)π2+(l-32)π2=-3π2.

Examples of when the maximum can be obtained are given by Figure 3.6 (vi)–(ix). Note that d(v1)=3 in Figure 3.6 (vii) and d(vl-1)=3 in Figure 3.6 (viii) for otherwise the maximum would be -3π2. For this reason we have the additional rule that π2 is distributed from Δ^ to Δ^1 in Figure 3.6 (iii) when d(v1)=3 and in Figure 3.6 (iv) when d(vl-1)=3. Note that the curvature is distributed as before across a (b-1,1)-edge or (1-1,b)-edge and so the above calculations remain unchanged. Therefore the net contribution to Δ^ of the segments of Figure 3.6 (ii)–(v) are each at most -3π2 and so c(Δ^)2π+r.π2-r.3π20 for r2, a contradiction that completes the proof. ∎

4 Proof of Theorem 1.1

The statements in Lemma 3.5 imply that we need only consider 1|a|+1|b|+1|ab-1|1, where at least one of |a|, |b| or |ab-1| equals 2. Working modulo ab, it can be assumed that |a||b|, and so there are seven possibilities, namely |a|=2, |b|=2 and |ab-1|= which is not allowed by (A3), conditions (i)–(iii) of Lemma 3.6, and the following three cases that remain to be considered:

|a|=2,|b|=3and|ab-1|6,|a|=2,|b|6and|ab-1|=3,|a|=3,|b|6and|ab-1|=2.

As before suppose by way of contradiction that is a reduced connected strictly spherical amended picture over 𝒫. For the remaining cases, we have found it easier to work with the dual of . This yields a so-called strictly spherical relative diagram, 𝑫 say, which is connected and simply connected. The regions of 𝑫 are given (up to inversion) by the region Δ of Figure 4.1 (i). The product of the corner and edge labels of Δ read clockwise gives axmbxn (up to cyclic permutation). If v is a vertex of 𝑫, then the label l(v) of v is (up to cyclic permutation) the product of the corner labels at v read in an anti-clockwise direction. Each vertex label corresponds to an admissible cycle in the star graph Γ of Figure 2.1 (iii). Note that when listing possible labels of a vertex of a given degree, we do so up to cyclic permutation and inversion. If d(Δ)=3, then Δ is given (up to inversion) by Figure 4.1 (ii), where from now on, for ease of presentation, vertices of degree 2 and the labels and arrows on x-edges are omitted. If d(Δ^i)=3 (1i3) in Figure 4.1 (ii), then we fix notation for the neighboring regions in Figure 4.1 (iii). From now on, we will use only the standard angle function. Each corner at a vertex v will be assigned the angle 2πd(v). For regions of degree 3, the following curvature identities will be useful: c(4,4,k)=2πk, c(4,6,6)=π6, c(4,6,8)=π12, c(4,6,10)=π30, c(4,6,12)=c(4,8,8)=0.

Case 1: |a|=2, |b|=3 and |ab-1|6

If ab=ba, then 𝒫 is not aspherical by [7, Theorem A] and noting that, although more general, non-asphericity in [7] implies the existence of a reduced spherical picture, that is, implies non-asphericity in the sense used here, so assume otherwise.

Checking Γ (and using the assumption abba together with the hypothesis on |a|, |b|, |ab-1|) shows that if d(v)8, then

l(v){a1-1a1-1,a1-1a1-111-1,b1-1b1-1b1-1,a1-1ab-11b-11b-1,(a1-1)4,a1-1a1-111-111-1,a1-111-1a1-111-1,b1-1b1-1b1-111-1}.

In particular, as shown in Figure 4.1 (iv), the vertices of two adjacent corners each with label 1 cannot both have vertex label a1-1a1-1, that is, be of degree 4.

If c(Δ)>0, then Δ is given by Figure 4.1 (ii) in which d(v1)4, d(v2)6 and d(v3)4.

Regions of 𝑫{\boldsymbol{D}} and adjacent vertices of degree 4.
Figure 4.1

Regions of 𝑫 and adjacent vertices of degree 4.

First let d(v1)=d(v3)=4. Then Δ is given by Figure 4.2 (i) in which d(Δ^3)>3, since, more generally, it can be seen from Figure 4.1 (ii) that if d(Δ)=3, then the only allowable subwords of l(Δ) of length 2 are ab, b1, 1a, b-1a-1, 1-1b-1 and a-11-1. If d(Δ^1)>3 and d(Δ^2)>3, then distribute 34c(Δ)34c(4,4,6)=π4 to c(Δ^3) and 18c(Δ)π24 to each of c(Δ^1) and c(Δ^2) as shown in Figure 4.2 (i); if d(Δ^1)=3 and d(Δ^2)>3, then add c(Δ)c(4,4,8)=π4 to c(Δ^3) as in Figure 4.2 (ii); if d(Δ^1)>3 and d(Δ^2)=3, then either d(v2)8 in which case add c(Δ)π4 to c(Δ^3) as in Figure 4.2 (iii), or d(v2)=6 in which case distribute π4 of c(Δ)=π3 to c(Δ^3) and the remaining π12 to c(Δ^1) as in Figure 4.2 (iv); and if d(Δ^1)=d(Δ^2)=3, then add c(Δ)π4 to c(Δ^3) as in Figure 4.2 (v).

Distribution from degree 3 regions: |a|=2{\lvert a\rvert=2}, |b|=3{\lvert b\rvert=3}, |a⁢b-1|≥6{\lvert ab^{-1}\rvert\geq 6}.
Figure 4.2

Distribution from degree 3 regions: |a|=2, |b|=3, |ab-1|6.

Suppose now that exactly one of d(v1) or d(v3) equals 4. If at least one Δ^i (1i3) has >3, then distribute c(Δ) as shown in Figure 4.3. The details are as follows: If d(Δ^i)>3 (1i3), then add 13c(Δ)π18 to c(Δ^i) as in Figure 4.3 (i); if d(Δ^i)=3 for exactly one of the Δ^i, then add 12c(Δ)π12 to c(Δ^j) (ji) as in (ii)–(iv); if d(v1)=4 and d(Δ^i)>3 for exactly one Δ^i, then the possibilities are shown in (v)–(ix) and in each case add c(Δ) to c(Δ^i); or if d(v3)=4 and d(Δ^i)>3 for exactly one Δ^i, then the possibilities are shown in (x)–(xiv) and in each case add c(Δ) to c(Δ^i). Observe that if Δ^ receives positive curvature in Figure 4.3, then d(Δ^)>3.

Distribution from degree 3 regions: |a|=2{\lvert a\rvert=2}, |b|=3{\lvert b\rvert=3}, |a⁢b-1|≥6{\lvert ab^{-1}\rvert\geq 6}.
Figure 4.3

Distribution from degree 3 regions: |a|=2, |b|=3, |ab-1|6.

Suppose that d(Δ^i)=3 (1i3) and d(v1)=4 only. Then Δ^1 and Δ^3 are given by Figure 4.4 (i). If Δ^2 is given by Figure 4.4 (i) also, then c(Δ)c(4,8,10)<0, so assume that Δ^2 is given by Figure 4.4 (ii), in which case c(Δ)>0 implies d(v2)=10 and d(v3)=6. If the vertex v of Figure 4.4 (ii) has degree 6 as indicated, then add c(Δ)=c(4,6,10)=π30 to c(Δ^2)c(6,6,10)=-2π15 as shown; if d(v)=4 and the corner label x equals 1 in Figure 4.4 (ii), then d(Δ^5)>3 and so c(Δ)+c(Δ^2)=2.π30 is added to c(Δ^5) as shown in Figure 4.4 (iii); if d(v)=4 and x=b in Figure 4.4 (ii), then (the inverse of) l(v2)=b-1ab-1aw and d(v2)=10 implies

l(v2){b-1ab-1a1-1ba-1ba-11,b-1ab-1a1-1ba-1b1-1a}

and either d(Δ^4)>3 in which case add c(Δ)+c(Δ^1)2.π30 to c(Δ^4) as in Figure 4.4 (iv), or d(Δ^4)=3 in which case add c(Δ)=π30 to c(Δ^1)c(4,8,10)=-π20. We note that in order to obtain the two possible l(v2) for d(v2)=10 and l(v2)=b-1ab-1aw, our method, here and elsewhere, is to enumerate all closed paths in Γ modulo cyclic permutation and inversion that do not contain the subwords aa-1, a-1a, bb-1 or b-1b. We then, often with the use of GAP [19], delete all those words in our list that contradict either abba or our hypothesis on |a|, |b| and |ab-1|. This is a routine but lengthy procedure and we omit the details for reasons of space.

Distribution from degree 3 regions: |a|=2{\lvert a\rvert=2}, |b|=3{\lvert b\rvert=3}, |a⁢b-1|≥6{\lvert ab^{-1}\rvert\geq 6}.
Figure 4.4

Distribution from degree 3 regions: |a|=2, |b|=3, |ab-1|6.

Suppose now that d(Δ^i)=3 (1i3) and d(v3)=4 only. Then Δ^2 and Δ^3 are given by Figure 4.5 (i). If Δ^1 is also given by Figure 4.5 (i), then c(Δ)0, so let Δ^1 be given by Figure 4.5 (ii) in which case c(Δ)>0 implies that d(v1)=10 and d(v2)=6. If the vertex v of Figure 4.5 (ii) has degree 6 as indicated, then add c(Δ)=π30 to c(Δ^1)c(6,6,10)=-2π15 as shown. Let d(v)=4. If d(Δ^9)>3, then add c(Δ)+c(Δ^1)=2.π30 to c(Δ^9) as shown in Figure 4.5 (iii). Let d(Δ^9)=3, in which case l(v1)=b-1ab-1aw and as before

l(v1){b-1ab-1a1-1ba-1ba-11,b-1ab-1a1-1ba-1b1-1a}.

If d(Δ^10)>3 in Figure 4.5 (iv), then add c(Δ)+c(Δ^1)+c(Δ^9)3.π30 to c(Δ^10) as shown; if d(Δ^10)=3 and d(u)10 for the vertex u of Figure 4.5 (iv), then add c(Δ)+c(Δ^1)=2.π30 to c(Δ^4)c(4,10,10)=-π10 as in Figure 4.5 (v); or if d(Δ^10)=3 and d(u)=8, then add c(Δ)+c(Δ^1)+c(Δ^9)=2.π30-π20=π60 to c(Δ^10)c(6,8,10)=-13π60 as shown in Figure 4.5 (vi).

Distribution from degree 3 regions: |a|=2{\lvert a\rvert=2}, |b|=3{\lvert b\rvert=3}, |a⁢b-1|≥6{\lvert ab^{-1}\rvert\geq 6}.
Figure 4.5

Distribution from degree 3 regions: |a|=2, |b|=3, |ab-1|6.

This completes the description of distribution of curvature from regions of degree 3. It follows that if Δ^ receives π4, then, see Figure 4.2, it does so across exactly one edge; if Δ^ receives π10, then, see Figure 4.5 (iv), Δ^ contains at least one vertex of degree 10; if d(Δ^)=3, then checking Figures 4.4 and 4.5 shows that Δ^ receives curvature across at most one edge; if Δ receives curvature from Δ^2 in Figure 4.4 (iv) or (v), then this situation is described by Figure 4.5 (v) and (vi) in which Δ plays the rôle of Δ^2; Δ does not receive any curvature from Δ^3 in Figure 4.5 (ii)–(v) for otherwise l(v1)=b-1ab-1ab-1w and d(v1)>10, a contradiction; and the maximum amount of curvature to cross an edge is either π4 or π10 or π12. It follows in turn from these observations that if d(Δ^)=3, then c(Δ^)0.

There remains to be described distribution of curvature from a region of degree 4. Let Δ^ be the region of degree 4 as in Figure 4.6 (i) in which d(Δ^11)>3 and d(Δ^12)>3. If d(Δ)>3, then add c(Δ^)c(4,4,4,6)+π4=π12 to c(Δ^11) as shown; or if d(Δ)=3, then add 12c(Δ^)c(4,4,4,6)+π4+π12=π12 to each of c(Δ^11) and c(Δ^12) as in Figure 4.6 (ii). Note that these final distribution rules do not alter the fact that Δ^ receives no curvature from Δ, Δ^11 or Δ^12 in Figure 4.6 (i) or from Δ^11 or Δ^12 in Figure 4.6 (ii) nor does it alter any of the consequences listed above as a result of distribution of curvature from regions of degree 3.

Distribution from degree 4 regions: |a|=2{\lvert a\rvert=2}, |b|=3{\lvert b\rvert=3}, |a⁢b-1|≥6{\lvert ab^{-1}\rvert\geq 6}.
Figure 4.6

Distribution from degree 4 regions: |a|=2, |b|=3, |ab-1|6.

The description of curvature distribution is now complete so let d(Δ^)5. As noted earlier, the vertices of two adjacent corners each with label 1 cannot both be of degree 4 and so Δ^ must contain at least two vertices of degree 6, therefore c(Δ^)c(4,4,4,6,6)+π4+4.π10<0. Finally, let d(Δ^)=4. If Δ^ contains at most one vertex of degree 4, then c(Δ^)c(4,6,6,6)+4.π10<0. The case when Δ^ contains three vertices of degree 4 is dealt with by Figure 4.6 (i), (ii) and so this leaves Δ^ having exactly two vertices of degree 4. But then either c(Δ^)c(4,4,6,6)+4.π12=0 or c(Δ^)c(4,4,6,10)+4.π10<0 or c(Δ^)c(4,4,6,6)+π4+π12=0 or c(Δ^)c(4,4,6,10)+π4+π10<0 or Δ^ is given by Figure 4.6 (iii), (iv) in which case either c(Δ^)c(4,4,6,8)+π4+2.π12=0 or, see Figure 4.5 (iv), c(Δ^)c(4,4,6,10)+π4+π10+π12<0.

Case 2: |a|=2, |b|6 and |ab-1|=3

If ab=ba, then we obtain the exceptional case (E2), so assume otherwise.

Checking Γ (and using the assumption abba together with the hypotheses on |a|, |b| and |ab-1|) shows that if d(v)8, then

l(v){a1-1a1-1,a1-1a1-111-1,(ab-1)3,(ab-1)311-1,a1-11b-1ab-1ab-1,ab-1ab-11a-11b-1,(a1-1)4,a1-1a1-111-111-1,a1-111-1a1-111-1}.

If c(Δ)>0, then Δ is given by Figure 4.1 (ii) in which d(v1)4, d(v2)6 and d(v3)4.

First let d(v1)=d(v3)=4. Then Δ is given by Figure 4.7 (i) in which d(Δ^3)>3. If d(Δ^1)>3 and d(Δ^2)>3, then distribute 34c(Δ)13c(4,4,6)=π4 to c(Δ^3) and 18c(Δ)π24 to each of c(Δ^1) and c(Δ^2) as shown in Figure 4.7 (i); if d(Δ^1)=3 and d(Δ^2)>3, then distribute 34c(Δ)π4 to c(Δ^3) and 14c(Δ)π12 to c(Δ^2) as shown in Figure 4.7 (ii); if d(Δ^1)>3 and d(Δ^2)=3, then add c(Δ)c(4,4,8)=π4 to c(Δ^3) as in Figure 4.7 (iii); and if d(Δ^1)=d(Δ^2)=3, then add c(Δ)π4 to c(Δ^3) as in Figure 4.7 (iv).

Distribution from degree 3 regions: |a|=2{\lvert a\rvert=2}, |b|≥6{\lvert b\rvert\geq 6}, |a⁢b-1|=3{\lvert ab^{-1}\rvert=3}.
Figure 4.7

Distribution from degree 3 regions: |a|=2, |b|6, |ab-1|=3.

Suppose now that exactly one of d(v1) or d(v3) equals 4. If at least one Δ^i (1i3) has degree >3, then distribute c(Δ) as shown in Figure 4.8. The details are as follows. If d(Δ^i)>3 (1i3), then add 13c(Δ)13c(4,6,6)=π18 to c(Δ^i) (1i3) as in Figure 4.8 (i); if d(Δ^i)=3 for exactly one of the Δ^i, then add 12c(Δ)π12 to c(Δ^j) (ji) as in (ii)–(iv); if d(v1)=4 and d(Δ^i)>3 for exactly one of the Δ^i, then the possibilities are shown in (v)–(ix) and in each case add c(Δ) to c(Δ^i); or if d(v3)=4 and d(Δ^i)>3 for exactly one of the Δ^i, then the possibilities are shown in (x)–(xiv) and in each case add c(Δ) to c(Δ^i). Observe that if Δ^ receives positive curvature in Figure 4.8, then d(Δ^)>3.

Distribution from degree 3 regions: |a|=2{\lvert a\rvert=2}, |b|≥6{\lvert b\rvert\geq 6}, |a⁢b-1|=3{\lvert ab^{-1}\rvert=3}.
Figure 4.8

Distribution from degree 3 regions: |a|=2, |b|6, |ab-1|=3.

Suppose that d(Δ^i)=3 (1i3) and d(v1)=4 only. Then Δ^1 and Δ^3 are given by Figure 4.9 (i). If Δ^2 is also given by Figure 4.9 (i), then c(Δ)c(4,8,8)=0, so assume that Δ^2 is given by Figure 4.9 (ii), in which case c(Δ)>0 forces d(v2)=6 and d(v3)=10. If the vertex v of Figure 4.9 (ii) has degree 6 as indicated, then add c(Δ)=c(4,6,10)=π30 to c(Δ^2)c(6,6,10)=-2π15 as shown; so let d(v)=4. We then see from Figure 4.9 (iii) that (the inverse of) l(v3)=b1-1b1-1w and d(v3)=10 implies

l(v3){b1-1b1-1ab-11b-1a1-1,b1-1b1-1ab-11b-11a-1}.

If d(Δ^6)>3, then add c(Δ)+c(Δ^2)=2.π30 to c(Δ^6) as shown in Figure 4.9 (iii); if d(Δ^6)=3, then either c(Δ^6)c(4,10,10)=-π10 and so add c(Δ)+c(Δ^2)=2.π30 to c(Δ^6) as in Figure 4.9 (iv), or c(Δ^6)=c(4,8,10)=-π20 in which case add c(Δ)+c(Δ^2)+c(Δ^6)=π60 to the region Δ^ of Figure 4.9 (v) in which d(Δ^)>3 or Δ^ of Figure 4.9 (vi) in which d(Δ^)=3 and c(Δ^)c(6,8,10)=-13π60.

Distribution from degree 3 regions: |a|=2{\lvert a\rvert=2}, |b|≥6{\lvert b\rvert\geq 6}, |a⁢b-1|=3{\lvert ab^{-1}\rvert=3}.
Figure 4.9

Distribution from degree 3 regions: |a|=2, |b|6, |ab-1|=3.

Suppose now that d(Δ^i)=3 (1i3) and d(v3)=4 only. Then Δ^2 and Δ^3 are given by Figure 4.10 (i). If Δ^1 is also given by Figure 4.10 (i), then c(Δ)c(4,8,8)=0, so let Δ^1 be given by Figure 4.10 (ii), in which case c(Δ)>0 implies that d(v1)=6 and d(v2)=10. If the vertex v of Figure 4.10 (ii) has degree 6 as indicated, then add c(Δ)=π30 to c(Δ^1)c(6,6,10)=-2π15 as shown. Let d(v)=4. If d(Δ^4)>3, then add c(Δ)+c(Δ^1)=2.π30 to c(Δ^4) as shown in Figure 4.10 (iii). Let d(Δ^4)=3, in which case l(v2)=b1-1b1-1w and so

l(v2){b1-1b1-1ab-11b-1a1-1,b1-1b1-1ab-11b-11a-1}.

If d(Δ^5)>3, then add c(Δ)+c(Δ^2)2.π30 to c(Δ^5) as in Figure 4.10 (iv); otherwise d(Δ^5)=3 and so add c(Δ)=π30 to c(Δ^2)c(4,8,10)=-π20 as in Figure 4.10 (v).

Distribution from degree 3 regions: |a|=2{\lvert a\rvert=2}, |b|≥6{\lvert b\rvert\geq 6}, |a⁢b-1|=3{\lvert ab^{-1}\rvert=3}.
Figure 4.10

Distribution from degree 3 regions: |a|=2, |b|6, |ab-1|=3.

This completes the description of distribution of curvature from regions of degree 3. It follows that if Δ^ receives π4, then, see Figure 4.7, it does so across exactly one edge; if d(Δ^)=3, then checking Figures 4.9 and 4.10 shows that Δ^ receives curvature across at most one edge; if Δ receives positive curvature from Δ^1 in Figure 4.10 (iv) or (v), then this situation is described in Figure 4.9 (iv)–(vi) in which Δ plays the rôle of Δ^1; Δ does not receive any curvature from Δ^3 in Figure 4.9 (ii)–(vi) for otherwise l(v3)=b1-1b1-1bw and d(v3)>10, a contradiction; and that the maximum amount of curvature to cross any given edge is π4 or π12. Given all of this, it now follows that if d(Δ^)=3, then c(Δ^)0.

There remains to be described distribution of curvature from a region of degree 4. Let Δ^ be the region of degree 4 shown in Figure 4.11 (i) in which d(Δ^11)>3 and d(Δ^12)>3. If d(Δ)>3 as shown, then c(Δ^)c(4,4,4,8)+π4=0; or if d(Δ)=3, then add c(Δ^)c(4,4,4,10)+π4+-π12=π30 to c(Δ^11) as in Figure 4.11 (ii). Note that this distribution rule does not alter the fact that Δ^ receives no curvature from Δ, Δ^11 or Δ^12 in Figure 4.11 (i) or from Δ^11 or Δ^12 in Figure 4.11 (ii) nor does it alter any of the consequences given above as a result of curvature distribution from regions of degree 3.

Distribution from degree 4 regions: |a|=2{\lvert a\rvert=2}, |b|≥6{\lvert b\rvert\geq 6}, |a⁢b-1|=3{\lvert ab^{-1}\rvert=3}.
Figure 4.11

Distribution from degree 4 regions: |a|=2, |b|6, |ab-1|=3.

The description of curvature distribution is now complete so let d(Δ^)5. Once more it can be seen from Figure 4.1 (iv) that the vertices of two adjacent corners each with label 1 cannot both be of degree 4 and so Δ^ must contain at least two vertices of degree 6 therefore c(Δ^)c(4,4,4,6,6)+π4+4.π12<0. Finally, let d(Δ^)=4. If Δ^ contains at most one vertex of degree 4, then c(Δ^)c(4,6,6,6)+4.π12<0. The case when Δ^ contains three vertices of degree 4 is dealt with by Figure 4.11 (i), (ii) so this leaves Δ^ having exactly two vertices of degree 4. But then either c(Δ^)c(4,4,6,6)+4.π12=0 or Δ^ is given by Figure 4.11 (iii), in which case c(Δ^)c(4,4,6,8)+π4+2.π12=0.

Case 3: |a|=3, |b|6 and |ab-1|=2

If ab=ba, we obtain the exceptional case (E2), so assume otherwise.

Checking Γ (and using the assumption abba together with the hypotheses on |a|, |b| and |ab-1|) shows that if d(v)8, then

l(v){(ab-1)2,ab-1ab-111-1,ab-1a1-11b-1,(a1-1)3,(ab-1)4,(ab-1)2(11-1)2,ab-1a1-11b-111-1,ab-1a1-111-11b-1,(ab-111-1)2,b-1ab-111-1a1-11,(b-1a1-11)2,ab-11a-11a-11b-1,(a1-1)311-1}.

In particular, any given region has at most two vertices of degree 4, and Figure 4.11 (iv) shows that two such vertices cannot be adjacent.

If c(Δ)>0, then Δ is given by Figure 4.1 (ii) in which d(v1)4, d(v2)4 and d(v3)6. Moreover, d(v1)=4 implies d(v2)6 and so c(Δ)c(4,6,6)=π6. If at least one Δi (1i3) has degree >3, then distribute c(Δ) as shown in Figure 4.12. The details are as follows: If d(Δ^i)>3 (1i3), then add 13c(Δ)13c(4,6,6)=π18 to c(Δ^i) (1i3) as in Figure 4.12 (i); if d(Δ^i)=3 for exactly one of the Δ^i, then add 12c(Δ)π12 to c(Δ^j) (ji) as in (ii)–(iv); if d(v1)=4 and d(Δ^i)>3 for exactly one Δ^i, then the possibilities are shown in (v)–(xi) and in each case add c(Δ) to c(Δ^i); or if d(v2)=4 and d(Δ^i)>3 for exactly one Δ^i, then the possibilities are shown in (xii)–(xx) and in each case add c(Δ) to c(Δ^i). Note that if Δ^ receives positive curvature in Figure 4.12, then d(Δ^)>3.

Distribution from degree 3 regions: |a|=3{\lvert a\rvert=3}, |b|≥6{\lvert b\rvert\geq 6}, |a⁢b-1|=2{\lvert ab^{-1}\rvert=2}.
Figure 4.12

Distribution from degree 3 regions: |a|=3, |b|6, |ab-1|=2.

Suppose that d(Δ^i)=3 (1i3) and d(v1)=4. Then Δ^1 and Δ^3 are given by Figure 4.13 (i). If Δ^2 is also given by Figure 4.13 (i), then add c(Δ)c(4,6,8)=π12 to c(Δ^2)c(6,6,8)=-π12, so let Δ^2 be given by Figure 4.13 (ii). If the vertex v of Figure 4.13 (ii) has degree 6 as indicated, then add c(Δ)π12 to c(Δ^)-π12 as shown. Let d(v)=4. Observe in Figure 4.13 (iii) that length considerations force l(v2)=1-1b1-1bw and so c(Δ)>0 implies d(v2)=10 and d(v3)=6. If d(Δ^5)>3, then add c(Δ)+c(Δ^2)=2.π30 to c(Δ^5) as in Figure 4.13 (iii); or if d(Δ^5)=3, then add c(Δ)+c(Δ^2)+c(Δ^5)3.π30 to the region Δ^ as in Figure 4.13 (iv) when d(Δ^)>3 or as in Figure 4.13 (v) when d(Δ^)=3 and c(Δ^)c(6,6,10)=-2π15.

Distribution from degree 3 regions: |a|=3{\lvert a\rvert=3}, |b|≥6{\lvert b\rvert\geq 6}, |a⁢b-1|=2{\lvert ab^{-1}\rvert=2}.
Figure 4.13

Distribution from degree 3 regions: |a|=3, |b|6, |ab-1|=2.

Now suppose that d(Δ^i)=3 (1i3) and d(v2)=4. Then Δ^1 and Δ^2 are given by Figure 4.14 (i). If Δ^3 is also given by Figure 4.14 (i), then add c(Δ)c(4,6,8)=π12 to c(Δ^3)c(6,6,8)=π12 as shown. Let Δ^3 be given by Figure 4.14 (ii). If the vertex v of Figure 4.14 (ii) has degree 6, then again add c(Δ)π12 to c(Δ^3)-π12. Let d(v)=4 as in Figure 4.14 (iii). Then (the inverse of) l(v3)=1-1b1-1bw together with c(Δ)>0 implies d(v1)=6, d(v2)=10 and l(v2){1-1b1-1ba-11b-11a-1b,1-1b1-1ba-11b-11b-1a}. If d(Δ^6)>3, then add c(Δ)+c(Δ^2)2.π30 to c(Δ^6) as in Figure 4.14 (iii); or if d(Δ^6)=3, then again add c(Δ)+c(Δ^2)2.π30 to c(Δ^6)c(6,6,10)=-2π15 as in Figure 4.14 (iv).

Distribution from degree 3 regions: |a|=3{\lvert a\rvert=3}, |b|≥6{\lvert b\rvert\geq 6}, |a⁢b-1|=2{\lvert ab^{-1}\rvert=2}.
Figure 4.14

Distribution from degree 3 regions: |a|=3, |b|6, |ab-1|=2.

This completes the description of curvature distribution. Observe that if Δ receives curvature from Δ^3 in Figure 4.14 (iii) or (iv), then this situation is described by Figure 4.13 (iv) or (v) in which Δ plays the rôle of Δ^3. Moreover, Δ does not receive any curvature from Δ^1 in Figure 4.13 (ii)–(v) for if so, then Δ^1 of Figure 4.13 (ii)–(v) would have to coincide with the inverse of Δ of Figure 4.14 (iii), (iv); in particular, the vertex v2 of Figure 4.13 (ii)–(v) would coincide with the inverse of the vertex v3 of Figure 4.14 (iii), (iv), but the inverse of the corner label a of Δ^6 in Figure 4.14 (iii), (iv) differs from the corresponding corner label of Δ^2 in Figure 4.13 (ii)–(v) which is 1-1, a contradiction. Given this it follows that the maximum amount of curvature to cross any edge is π6 or π10 or π12. A further observation is the following: Checking the degrees and labels of the vertices involved in Figure 4.12 in the six instances where Δ^ receives π6 shows that Δ^ cannot receive π6 across consecutive edges.

Checking Figures 4.13 and 4.14 shows that if d(Δ^)=3, then Δ^ cannot receive curvature across three edges and there are exactly two cases when Δ^ can receive positive curvature across two edges. The first case is when Δ^2 of Figure 4.13 (i) coincides with Δ^3 of Figure 4.14 (ii). But this forces l(v3)=b-11b-11a-1w in Figure 4.13 (i) which implies d(v3)=10 and so c(Δ^2)=c(Δ^3)c(6,6,10)+2.π30<0. The second case is when Δ^ of Figure 4.13 (v) coincides with Δ^3 of Figure 4.14 (ii), but then c(Δ^3)=c(Δ^)c(6,6,10)+3.π30+π12<0.

It can be deduced from above that if d(Δ^)=3, then c(Δ^)0. Let d(Δ^)6. Then c(Δ^)c(4,4,6,6,6,6)+3.π6+3.π10<0. Let d(Δ^)=5. Then c(Δ^)c(4,4,6,6,6)+2.π6+3.π10<0. Finally, let d(Δ^)=4. Then either c(Δ^)c(6,6,6,6)+2.π6+2.π10<0 or c(Δ^)c(4,6,6,6)+2.π6+2.π12=0 or, see Figure 4.13 (iv), (v), c(Δ^)c(4,6,6,10)+2.π6+2.π10<0, or Δ^ has exactly two vertices of degree 4 and is given by Figure 4.15 in which d(u)8. But again checking the six instances in Figure 4.12 when π6 crosses an edge shows that in fact Δ^ cannot receive π6 across any of its edges and so either c(Δ^)c(4,4,6,6)+4.π12=0 or c(Δ^)c(4,4,6,10)+4.π10<0.

Distribution into a degree 4 region: |a|=3{\lvert a\rvert=3}, |b|≤6{\lvert b\rvert\leq 6}, |a⁢b-1|=2{\lvert ab^{-1}\rvert=2}.
Figure 4.15

Distribution into a degree 4 region: |a|=3, |b|6, |ab-1|=2.

In each of the three cases above, we have shown that if Δ^ receives positive curvature, then c(Δ^)0. This contradiction to c()=4π completes the proof of Theorem 1.1.

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About the article


Received: 2017-07-03

Revised: 2018-06-11

Published Online: 2018-09-10

Published in Print: 2019-01-01


Citation Information: Forum Mathematicum, Volume 31, Issue 1, Pages 49–68, ISSN (Online) 1435-5337, ISSN (Print) 0933-7741, DOI: https://doi.org/10.1515/forum-2017-0141.

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