EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 2, 2019, 590-604 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global On a nonsingular equation of length 9 over torsion free groups M. Fazeel Anwar1,∗, Mairaj Bibi2, M. Saeed Akram3 1 Department of Mathematics, Sukkur IBA University, Sukkur. 2 Department of Mathematics, COMSATS Institute of Information Technology, Islamabad. 3 Department of Mathematics, Khwaja Fareed University of Engineering & Information Technology, Rahim Yar Khan. Abstract. In [11], Levin conjectured that every equation is solvable over a torsion free group. In this paper we consider a nonsingular equation g1tg2tg3tg4tg5tg6t −1g7tg8tg9t −1 = 1 of length 9 and show that it is solvable over torsion free groups modulo some exceptional cases. 2010 Mathematics Subject Classifications: 20F05, 57M05 Key Words and Phrases: Asphericity; relative group presentations; torsion-free groups; group equations. 1. Introduction Let G be a non-trivial group, t be an unknown and let F be a free group generated by t. An equation in t over G is an expression of the form s(t) = g1t ε1g2t ε2 · · · gntεn = 1 (gi ∈ G, εi = ±1) in which it is assumed that εi + εi+1 = 0 implies gi+1 6= 1 in G. We call the integer n the length of the equation. A solution of s(t) = 1 over G is an embedding φ of G into a group H and an element h ∈ H such that φ(g1)hε1φ(g2)hε2 · · ·φ(gn)hε1 = 1 in H. Equivalently s(t) = 1 is solvable over G if and only if the natural map from G to 〈G∗F |s(t)〉 is injective, where G ∗ F is the free product of G and F . If G is a torsion free group then by Levin’s conjecture every equation is solvable [11]. The conjecture is known to be true for n ≤ 7 [5, 7, 9, 12]. The authors have done significant work in [1, 2, 4] to establish the conjecture for n = 8. The equation of length 9 have been consider in [3]. It has been proved that there are only three equations of length 9 which are still open. In this paper we consider a nonsingular equation of length 9 (one of three) and show that the equation has a solution ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i2.3405 Email addresses: fazeel.anwar@iba-suk.edu.pk (M. Fazeel Anwar) http://www.ejpam.com 590 c© 2019 EJPAM All rights reserved. M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 591 over G modulo some exceptional cases. This paper is the first step in proving Levin’s conjecture for equations of length 9. We first give some basic definitions. A relative group presentation is a presentation of the form P = 〈G, x | r〉 where r is a set of cyclically reduced words in G∗〈x〉. If the relative presentation is orientable and aspherical then the natural map from G to 〈G, x | r〉 is injective. In our case x and r consist of the single element t and s(t) respectively, therefore P is orientable and so asphericity implies s(t) = 1 is solvable. In this paper we use the weight test and the curvature distribution method to show that P is aspherical [6]. The star graph Γ of P has vertex set x∪x−1 and edge set r∗, where r∗ is the set of all cyclic permutations of the elements of r∪r−1 which begin with an element of x∪x−1. For R ∈ r∗ write R = Sg where g ∈ G and S begins and ends with x symbols. Then i(R) is the inverse of the last symbol of S, τ(R) the first symbol of S and λ(R) = g. A weight function θ on Γ is a real valued function on the set of edges of Γ which satisfies θ(Sh) = θ(S−1h−1). A weight function θ is called aspherical if the following three conditions are satisfied (W1) Let R ∈ r∗ with R = xε11 g1 · · · xεnn gn. Then n∑ i=1 (1− θ(xεii gi · · · x εn n gnx ε1 1 g1 · · · xεi−1 i−1 gi−1)) ≥ 2. (W2) Each admissible cycle in Γ has weight at least 2 (where admissible means having a label trivial in G). (W3) Each edge of Γ has a non-negative weight. If Γ admits an aspherical weight function then P is aspherical [6]. The following lemma [10] tells us that we can apply asphericity test in k−steps. Lemma 1. Let the relative presentation P = 〈H,x : r〉 define a group G and let Q = 〈G, t : s〉 be another relative presentation. If Q and P are both aspherical, then the relative presentation R = 〈H,x ∪ t : r ∪ s̃〉 is aspherical, where s̃ is an element of H ∗ F (x) ∗ F (t) obtained from s by lifting. For a detailed account on the curvature distribution method see [3]. It is clear from our definition of a group equation that if gi is a coefficient between a negative and a positive power of t than gi is not trivial in G. This fact will be used in all subsequent proofs without reference. 2. Main Results We now turn our attention to length 9 equations. A list of these equations is given in [3]. Consider the nonsingular equation of length 9 given by atbtctdtetft−1gthtit−1 = 1 . We write this as P = 〈G, t|s(t)〉, where s(t) = atbtctdtetft−1gthtit−1. Here b, c, d, e, h ∈ G and a, f, g, i ∈ G\{1}. The star graph Γ for P is given in Figure 2. We apply the transformation x = tb to get that b = 1 in G. By using the methods M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 592 x t t̄ β3 x̄ α4 β1 α3 α5 α2 β2 α1 x t t̄ α4 x̄ α2 α3 β1 β3 α5 α1 β2 x t t̄ β3 x̄ α4 α2 β1 α3 α7 α5 α1 β2 (a) (c)(b) α6 β4 β4 Figure 1: Star graph Γ given in [5, 8] we conclude that possible vertices of degree 2 (in the diagram associated to P) are (upto cyclic permutation and inversion) S = {ag, ag−1, fi, fi−1, bc−1, bd−1 be−1, bh−1, cd−1, ce−1, ch−1, de−1, dh−1, eh−1}. Since G is torsion free therefore it is clear that ag and ag−1 can not both hold at the same time. Similarly fi and fi−1 can not both hold at the same time. The following lemma gives some general results that will greatly simplify the proofs. This is an application of the results given in [1, 2]. t t̄ e b a h i c g f d Figure 2: Star graph Γ Lemma 2. The presentation P = 〈G, t|s(t)〉, where s(t) = atbtctdtetft−1gthtit−1 is as- pherical if any one of the following holds: (i) a = g−1 (ii) a = g, f = i M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 593 (iii) a = g, b = h Proof. A new generator x will be introduced to obtain the presentationQ = 〈G, t, x|r1, r2〉. (i) Let a = g−1. The relator s(t) is given by s(t) = atbtctdtetft−1a−1thtit−1. We substitute x = t−1a−1t to get r1 = x−1btctdtetfxhti and r2 = t−1a−1tx−1. The star graph Γ for Q is given by Figure 1 (a) in which (using r1) α1 = e, α2 = d, α3 = c, α4 = b, α5 = f, α6 = i, α7 = h; and (using r2) β1 = a−1, β2 = 1, β3 = 1. We assign a weight function θ such that θ(α4) = θ(α6) = θ(β1) = θ(β2) = 0. All other edges are assigned a weight 1. Then Σ(1 − θ(αi)) = Σ(1 − θ(βj)) = 2 shows that (W1) is satisfied. Also each cycle in Γ of weight less than 2 has label am or im, (m 6= 0) and (a, i 6= 1) and since G is torsion free (W2) is satisfied. Moreover (W3) clearly holds. (ii) We have s(t) = atbtctdtetit−1athtit−1, r1 = xbtctdtexh, r2 = tit−1atx−1. The star graph Γ is given by Figure 1 (b) in which α1 = c, α2 = d, α3 = e, α4 = h, α5 = b; and β1 = a, β2 = i, β3 = 1, β4 = 1. We assign a weight function θ such that θ(α3) = θ(α5) = θ(β1) = θ(β2) = 0. All other edges are assigned a weight 1. Then Σ(1− θ(αi)) = Σ(1− θ(βj)) = 2 shows that (W1) is satisfied. Also each cycle in Γ of weight less than 2 has label am or im, (m 6= 0) and (a, i 6= 1) and since G is torsion free (W2) is satisfied. Moreover (W3) clearly holds. (iii) We have s(t) = atbtctdtetft−1atbtit−1, r1 = xctdtetfxi, r2 = t−1atbtx−1. The star graph Γ is given by Figure 1 (c) in which α1 = e, α2 = d, α3 = f, α4 = i, α5 = c; and β1 = a, β2 = b, β3 = 1, β4 = 1. We assign a weight function θ such that θ(α3) = θ(α5) = θ(β1) = θ(β3). All other edges are assigned a weight 1. We have the desired result. Corollary 1. The presentation P = 〈G, t|s(t)〉, is aspherical if any one of the following holds: (i) a = g−1 and R ∈ {fi, fi−1, bc−1, bd−1 be−1, bh−1, cd−1, ce−1, ch−1, de−1, dh−1, eh−1} (ii) a = g, f = i and R ∈ {bc−1, bd−1 be−1, bh−1, cd−1, ce−1, ch−1, de−1, dh−1, eh−1} (iii) a = g, b = h and R ∈ {cd−1, ce−1, ch−1, de−1, dh−1, eh−1} Proof. The result is clear from lemma 3 by taking the weight function as given in lemma 3, part 1, 2 and 3 respectively. M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 594 Lemma 3. The presentation P = 〈G, t|s(t)〉, where s(t) = atbtctdtetft−1gthtit−1 is as- pherical if any one of the following holds: (i) a = g, b = c (ii) a = g, b = d (iii) a = g, b = e (iv) a = g, c = d (v) a = g, c = e (vi) a = g, c = h (vii) a = g, d = e (viii) a = g, d = h (ix) a = g, e = h (x) b = c, d = h (xi) b = c, e = h (xii) b = d, c = e (xiii) b = d, c = h (xiv) b = d, e = h (xv) b = e, c = h (xvi) b = h, c = d (xvii) b = c, d = e Proof. (i) In this case ∆ is shown in Figure 3. Since d∆(va) = d∆(vb) = 2 or d∆(vb) = d∆(vc) = 2 can not occur therefore it can be assumed that d∆(va) = d∆(vc) = d∆(vg) = 2 as shown in Figure 3. In this case c(∆) ≤ 0. M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 595 a b c d ef g h i c g a b a b c d ef g h i g a b Figure 3: Region ∆ (ii) In this case ∆ is shown in Figure 4. Since degree of vertices va and vb can not be 2 together so there are the following two cases to consider: (a) d∆(va) = d∆(vd) = d∆(vg) = 2; (b) d∆(vb) = d∆(vd) = d∆(vg) = 2. as shown in Figure 4. In both of these cases c(∆) ≤ 0. a b c d ef g h i d g a b a b c d ef g h i g a b a b c d ef g h i d a b Figure 4: Region ∆ (iii) In this case ∆ is shown in Figure 5. Since degree of vertices va and vb can not be 2 together so there are the following two cases to consider: M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 596 (a) d∆(va) = d∆(ve) = d∆(vg) = 2; (b) d∆(vb) = d∆(ve) = d∆(vg) = 2. as shown in Figure 5. In both of these cases c(∆) ≤ 0. a b c d ef g h i e g a b a b c d ef g h i g a b a b c d ef g h i e a b Figure 5: Region ∆ (iv) In this case ∆ is shown in Figure 6. Since degree of vertices va and vb can not be 2 together so there are the following two cases to consider: (a) d∆(va) = d∆(vc) = d∆(vg) = 2; (b) d∆(va) = d∆(vd) = d∆(vg) = 2. as shown in Figure 6. In both of these cases c(∆) ≤ 0. M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 597 a b c d ef g h i d g a c a b c d ef g h i g a a b c d ef g h i a d c g Figure 6: Region ∆ (v) In this case ∆ is shown in Figure 7. Here d∆(va) = d∆(vc) = d∆(ve) = d∆(vg) = 2 which implies l∆(va) = ag−1, l∆(vc) = ce−1, l∆(ve) = ec−1 ans l∆(vg) = ga−1 as shown in Figure 7. In order to have positive curvature the remaining vertices must be of degree 3. Observe that l∆(va) = ag−1 and l∆(vc) = ce−1 implies that l∆(vb) = h−1bd−1w where w ∈ {b, c, d, e, h} which implies d∆(vb) > 3. Notice that l∆(ve) = ec−1 and l∆(vg) = ga−1 implies that l∆(vf ) = d−1fi−1w where w ∈ {b, c, d, e, h} which implies d∆(vf ) > 3. Since d∆(vb) > 3 and d∆(vf ) > 3 so c(∆) ≤ 0. a b c d ef g h i c g a e a b c d ef g h i g a e c d i d h Figure 7: Region ∆ (vi) In this case ∆ is shown in Figure 8. Since degree of vertices vg and vh can not be 2 together so there are the following two cases to consider: M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 598 (a) d∆(va) = d∆(vc) = d∆(vg) = 2; (b) d∆(va) = d∆(vc) = d∆(vh) = 2. as shown in Figure 8. In both of these cases c(∆) ≤ 0. c a b c d ef g h i h g a a b c d ef g h i g a a b c d ef g h i h h g c Figure 8: Region ∆ (vii) In this case ∆ is shown in Figure 9. Since degree of vertices vd and ve can not be 2 together so there are the following two cases to consider: (a) d∆(va) = d∆(vd) = d∆(vg) = 2; (b) d∆(va) = d∆(ve) = d∆(vg) = 2. as shown in Figure 9. In both of these cases c(∆) ≤ 0. a b c d ef g h i d g a e a b c d ef g h i g a a b c d ef g h i a d e g Figure 9: Region ∆ (viii) In this case ∆ is shown in Figure 10. Since degree of vertices vg and vh can not be 2 together so there are the following two cases to consider: (a) d∆(va) = d∆(vd) = d∆(vg) = 2; M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 599 (b) d∆(va) = d∆(vd) = d∆(vh) = 2. as shown in Figure 10. In both of these cases c(∆) ≤ 0. d a b c d ef g h i h g a a b c d ef g h i g a a b c d ef g h i h h g d Figure 10: Region ∆ (ix) In this case ∆ is shown in Figure 11. Since degree of vertices vg and vh can not be 2 together so there are the following two cases to consider: (a) d∆(va) = d∆(ve) = d∆(vg) = 2; (b) d∆(va) = d∆(ve) = d∆(vh) = 2. as shown in Figure 11. In both of these cases c(∆) ≤ 0. e a b c d ef g h i h g a a b c d ef g h i g a a b c d ef g h i h h g e Figure 11: Region ∆ (x) In this case ∆ is shown in Figure 12. Since d∆(vb) = d∆(vc) = 2 or d∆(vc) = d∆(vd) = 2 can not occur therefore it can be assumed that d∆(vb) = d∆(vd) = d∆(vh) = 2 as shown in Figure 12. In this case c(∆) ≤ 0. M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 600 a b c d ef g h i c h d b a b c d ef g h i c h d Figure 12: Region ∆ (xi) In this case ∆ is shown in Figure 13. Since degree of vertices vb and vc can not be 2 together so there are the following two cases to consider: (a) d∆(vb) = d∆(ve) = d∆(vh) = 2; (b) d∆(vc) = d∆(ve) = d∆(vh) = 2. as shown in Figure 13. In both of these cases c(∆) ≤ 0. e a b c d ef g h i h c b a b c d ef g h i a b c d ef g h i h h e e c b Figure 13: Region ∆ (xii) In this case ∆ is shown in Figure 14. Since d∆(vb) = d∆(vc) = 2 or d∆(vc) = d∆(vd) = 2 or d∆(vd) = d∆(ve) = 2 can not occur therefore c(∆) ≤ 0. M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 601 a b c d ef g h i c h d b Figure 14: Region ∆ (xiii) In this case ∆ is shown in Figure 15. Since d∆(vb) = d∆(vc) = 2 or d∆(vc) = d∆(vd) = 2 can not occur therefore it can be assumed that d∆(vb) = d∆(vd) = d∆(vh) = 2 as shown in Figure 15. In this case c(∆) ≤ 0. a b c d ef g h i d b c h a b c d ef g h i d b c Figure 15: Region ∆ (xiv) In this case ∆ is shown in Figure 16. Since degree of vertices vd and ve can not be 2 together so there are the following two cases to consider: (a) d∆(vb) = d∆(vd) = d∆(vh) = 2; (b) d∆(vb) = d∆(ve) = d∆(vh) = 2. as shown in Figure 16. In both of these cases c(∆) ≤ 0. M. Fazeel Anwar, Mairaj Bibi, M. Saeed Akram / Eur. J. Pure Appl. Math, 12 (2) (2019), 590-604 602 e a b c d ef g h i h d b a b c d ef g h i a b c d ef g h i b h e e d d Figure 16: Region ∆ (xv) In this case ∆ is shown in Figure 17. Since degree of vertices vb and vc can not be 2 together so there are the following two cases to consider: (a) d∆(vb) = d∆(ve) = d∆(vh) = 2; (b) d∆(vc) = d∆(ve) = d∆(vh) = 2. as shown in Figure 17. In both of these cases c(∆) ≤ 0. c a b c d ef g h i b e h a b c d ef g h i a b c d ef g h i b b c c e h Figure 17: Region ∆ (xvi) In this case ∆ is shown in Figure 18. Since d∆(vb) = d∆(vc) = 2 or d∆(vc) = d∆(vd) = 2 can not occur therefore it can be assumed that d∆(vb) = d∆(vd) = d∆(vh) = 2 as shown in Figure 18. In this case c(∆) ≤ 0. REFERENCES 603 a b c d ef g h i h c b d a b c d ef g h i h c b Figure 18: Region ∆ (xvii) In this case ∆ is shown in Figure 19. Since d∆(vb) = d∆(vc) = 2 or d∆(vc) = d∆(vd) = 2 or d∆(vd) = d∆(ve) = 2 can not occur therefore c(∆) ≤ 0. a b c d ef g h i d e c b Figure 19: Region ∆ Remark 1. It is worth mentioning here that a few of the cases still remain open for this equation. These cases are extremely technical in detail and will be considered in a different article. References [1] M F Anwar, M Bibi, and M S Akram. On solvability of certain equations of arbitrary length over torsion-free groups. Preprint, 2019. [2] M F Anwar, M Bibi, and S Iqbal. On certain equations of arbitrary length over torsion-free groups. Preprint, 2019. [3] M Bibi. Equations of length seven over torsion free groups. PhD thesis, University of Notingham, 2013. REFERENCES 604 [4] M Bibi, M F Anwar, S Iqbal, and M S Akram. Solution of a non-singular equation of length 8 over torsion free groups. Preprint, 2019. [5] M Bibi and M Edjvet. Solving equations of length seven over torsion-free groups. Journal of Group Theory, 21(1):147–164, 2018. 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