EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 4, 2020, 914-938 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Levin conjecture for group equations of length 9 Muhammad Saeed Akram1,∗, Maira Amjid1, Sohail Iqbal 2 1 Department of Mathematics, Khwaja Fareed University of Engineering and Information Technology, Rahim Yar Khan, 64200, Pakistan 2 Department of Mathematics, CUI, Islamabad, Pakistan Abstract. Levin conjecture states that every group equation is solvable over any torsion free group. The conjecture is shown to hold true for group equation of length seven using weight test and curvature distribution method. Recently, these methods are used to show that Levin conjecture is true for some group equations of length eight and nine modulo some exceptional cases. In this paper, we show that Levin conjecture holds true for a group equation of length nine modulo 2 exceptional cases. In addition, we allude the list of cases that are still open for two more equations of length nine. 2020 Mathematics Subject Classifications: 20F05, 20E06, 57M05 Key Words and Phrases: Group equations, torsion-free groups, relative group presentations, asphericity, weight test, curvature distribution 1. Introduction: Let G be a non-trivial group and t an element not in G. A group equation over G is an equation of the form s(t) = g1t l1g2t l2 ...gnt ln = 1 (gi ∈ G, li = ±1) such that li + li+1 = 0 implies gi+1 6= 1 ∈ G (subscripts modulo n). The non-negative integer n is known as the length of equation s(t) = 1. The equation s(t) = 1 is said to be solvable over G if s(h) = 1 for some element h of a group H which contain G. Equivalently, s(t) = 1 is solvable over G if and only if the natural homomorphism from G to G∗(t) N is injective, in which N is the normal closure of s(t) = 1 in the free product G∗(t). The equation s(t) = 1 is called singular if ∑n i=1 li = 0 and non-singular otherwise. The study of group equations was initiated by Neumann [17] who solved an equation t−1g1tg2 = 1 over any torsion free group. Motivated by the solvability of the polynomial ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i4.3786 Email addresses: mrsaeedakram@gmail.com (Muhammad Saeed Akram), mairaamjad450@gmail.com (Maira Amjid), soh.iqbal@gmail.com (Sohail Iqbal) https://www.ejpam.com 914 c© 2020 EJPAM All rights reserved. Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 915 equations over fields, Levin [16] studied the analogous problem for group equations and proved that the equation s(t) = 1 (li non-negative and not necessarily 1) is solvable over any group G, for l1 + l2 + ... + ln = n . These findings of Neumanm and Levin gave the hope for the conjecture that every equation is solvable over a torsion free group, which is known as Levin conjecture. There has been significant work to verify the Levin conjecture [11, 12, 14, 15] for group equations of length less than or equal to six. Recently, Mairaj and Edjvet [8] proved the Levin conjecture for all group equations of length seven by using weight test and curvature distribution method. By employing the methods used in [8], Mairaj et al. [7] have proved the conjecture for a non-singular equation of length eight modulo one exceptional case. The authors have done some significant work in [4, 5] using weight test which establishes the conjecture up to great extent for length eight. The equations of length nine are considered in [6], where it is proved that there are only three equations of length nine which are open. More recently, Fazeel et al. [3] have investigated the conjecture for a non-singular equation of length nine (one of three) given by s1(t) = atbtctdtetft−1gthtit−1 = 1 by applying these methods. Fazeel et al. [4] solved 41 cases of this equation. In this paper we have continued our study of exploring the validity of Levin conjecture for the group equation s1(t) = 1 initiated in [4] and found that the total cases of s1(t) = 1 are 245 in which 183 cases are solved by weight test, 60 cases are solved by curvature distribution method and 2 cases are still open. These findings are formulated in the form of the following Theorem which is the main result of the paper. Theorem. The group equation s1(t) = 1 is solvable modulo two exceptional cases: (i) a = g, h = e, d = b, c = b, d = c; (ii) a = g, e = b, e = c, c = b, e = d, d = b, d = c. The authors in [2] and [1] have explored the validity of Levin Conjecture by applying weight test and curvature distribution to the remaining two equations of length nine given by s2(t) = atbtct−1dtetft−1gthtit−1 = 1 and s3(t) = atbtctdtet−1ftgthtit−1 = 1. They found that Levin conjecture holds true for these equations modulo some exceptional cases. The authors in [2] have found that the total cases of s2(t) = 1 are 318 in which 147 cases are solved by weight test, 117 cases are solved by curvature distribution method and 55 cases are still open. The authors in [1] have found that the total cases of s3(t) = 1 are 245 in which 136 cases are solved by weight test, 70 cases are solved by curvature distribution method and 39 cases are still open. Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 916 2. Methodology: A presentation is said to be relative (group) presentation P = 〈G, x | r〉 in which r is a set of words which is cyclically reduced belongs to G. All the definitions concerning relative presentations can be seen in [9]. In this paper, we will discuss the equation s1(t) = atbtctdtetft−1gthtit−1 = 1 in detail. It is well known that the group equation s1(t) = atbtctdtetft−1gthtit−1 is solvable if the natural homomorphism τ : G→ P(G) is injective. The sufficient condition for the injectivity of natural map from G to P = 〈G, x | r〉 is that the relative presentation is orientable and aspherical [9]. The notion of asphericity is discussed in detail in [9]. In our case s1(t) = atbtctdtetft−1gthtit−1, therefore r is singleton set. As stated in [9], if r is singleton then P is always orientable, therefore asphericity of P establishes that s1(t) = 1 has solution. In order to establish the validity of Levin’s conjecture, it is only left to prove that the presentation P is aspherical. So, in this paper we apply two tests for showing asphericity of P: Weight test and curvature distribution method. All the necessary definitions concerning weight test can be found in [9]. The weight test states that if the star graph Γ of P admits an aspherical weight function θ, then P is aspherical [9]. All the definitions related to pictures can be found in [14]. The curvature distribution asserts that if K is a reduced picture over P then by Euler (or Gauss-Bonnet) formula, the sum of the curvature of all regions of K is 4π, that is, K contains regions of positive curvature [14]. Then, if for every region ∆ of K of positive curvature c(∆), there is a neighbouring region ∆̂, uniquely associated with ∆, like c(∆̂) + c(∆) ≤ 0, then the sum of the curvature of all regions of K is non-positive, which implies that P is aspherical [6, 8]. Consider a torsion free group G. By applying the transformation u = tb on s1(t) = atbtctdtetft−1gthtit−1 = 1 it can be assumed that b = 1. Recall that P = 〈G, t | s1(t)〉 in which s1(t) = atbtctdtetft−1gthtit−1 (a, f, g, i ∈ G \ {1}, b = 1, c, d, e, h ∈ G). Moreover, G is not cyclic and G = 〈a, b, c, d, e, f, g, h, i〉 given in [13]. Suppose that K is a reduced spherical diagram over P. Up to cyclic permutation and inversion, the regions of K are given by ∆ as shown in Figure 1(i). The star graph Γ of P is shown in Figure 1(ii). Figure 1: Region ∆ of K and star graph Γ of P Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 917 Looking at closed paths in star graph Γ, using the fact that G is torsion free and working modulo cyclic permutation and inversion, the possible labels of vertices of degree 2 for a region ∆ of K are S = {ag, ag−1, fi, fi−1, hb−1, hc−1, hd−1, he−1, eb−1, ec−1, ed−1, dc−1, db−1, cb−1}. We can work modulo equivalence, that is, modulo t↔ t−1, cyclic permutation, inversion, and a↔ f−1, g ↔ i−1, b↔ e−1, c↔ d−1, h↔ h−1. We will proceed according to the number N of labels in S that are admissible [9] and clas- sify the cases correspondingly [8]. The following remark substantially reduce the number of cases to be considered. Remark 1. The following observations holds trivially. (i) If all the admissible cycle has length greater than 2 in region ∆ then c(∆) ≤ c(3, 3, 3, 3, 3, 3, 3, 3, 3) = −π. (ii) If ag and ag−1 are admissible then g2 = 1, a contradiction. (iii) If fi and fi−1 are admissible then i2 = 1, a contradiction. (iv) At most two of ag, ag−1, fi, fi−1 are admissible. (v) If any two of hb−1, hc−1, cb−1 are admissible then so is the third. (vi) If any two of hb−1, hd−1, db−1 are admissible then so is the third. (vii) If any two of hb−1, he−1, eb−1 are admissible then so is the third. (viii) If any two of hc−1, hd−1, dc−1 are admissible then so is the third. (ix) If any two of hc−1, he−1, ec−1 are admissible then so is the third. (x) If any two of hd−1, he−1, ed−1 are admissible then so is the third. (xi) If any two of eb−1, ec−1, cb−1 are admissible then so is the third. (xii) If any two of eb−1, ed−1, db−1 are admissible then so is the third. (xiii) If any two of ec−1, ed−1, dc−1 are admissible then so is the third. (xiv) If any two of dc−1, db−1, cb−1 are admissible then so is the third. The above remark reduces the number of cases to 245 for the group equation s1(t) = 1. From these 245 cases, 41 cases are solved in [4] so there remains 204 cases that needs to be solved. Among these 204 remaining cases, 159 cases are solved by weight test, 43 cases are solved by curvature distribution method and 2 cases are still open. Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 918 3. Main Results: A further 159 cases can be straightforwardly solved using the weight test. For example, consider the case, a = g−1, f = i−1 and h = b. In this case, the relator is s(t) = at2ctdtetft−1a−1t2f−1t−1. We put x = t−1a−1t to obtain v1 = x−1tctdtetfxtf−1 and v2 = x−1t−1a−1t. The presentation P has star graph Γ which is shown in Figure 2 in which µ1 = c, µ2 = d, µ3 = e, µ4 = 1, µ5 = f , µ6 = f−1, µ7 = 1; and ω1 = a−1, ω2 = 1, ω3 = 1. Figure 2: Star graph Γ We define a weight function θ such that θ(µ4) = θ(µ6) = θ(ω1) = θ(ω2) = 0 and θ(µ1) = θ(µ2) = θ(µ3) = θ(µ5) = θ(µ7) = θ(ω3) = 1. Then Σ(1−θ(µi)) = Σ(1−θ(ωj)) = 2 indicates that the first condition of weight test is fulfilled. Moreover, every cycle in Γ of weight smaller than 2 has label am, where m ∈ Z \ {0} and a ∈ G \ {1}, which implies a is torsion element in G, a contradiction, so the second condition of weight test is fulfilled. Furthermore, since θ assigns non-negative weights to each edge, so the third condition of weight test is obviously fulfilled. A further 24 cases are solved in Lemma 1 by an immediate application of curvature distribution method [10]. In what follows, the vertex labels correspond to the closed paths in the star graph Γ. From now onward, the label and the degree of a vertex v of region ∆ will be denoted by l∆(v) and d∆(v) respectively. Furthermore, l∆ ∈ {ww1, . . . , wwk} will be indicated by l∆(v) = {ww1, . . . , wwk}. Lemma 1. The presentation P = 〈G, t | s1(t)〉 is aspherical if any one of the following holds: (i) a = g; (ii) h = b; (iii) h = c; Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 919 (iv) e = b; (v) e = c; (vi) e = d; (vii) d = c; (viii) e = b, d = c; (ix) a = g, h = c, d = b; (x) a = g, h = e, d = b; (xi) a = g, h = d, e = b; (xii) a = g, h = d, c = b; (xiii) h = b, h = c, c = b; (xiv) h = b, h = d, d = b; (xv) h = b, h = e, e = b; (xvi) h = c, h = d, d = c; (xvii) e = b, e = c, c = b; (xviii) e = c, e = d, d = c; (xix) a = g, h = c, h = d, d = c; (xx) a = g, h = e, h = d, e = d; (xxi) a = g, e = b, d = b, e = d; (xxii) h = b, h = c, c = b, h = d, d = b, d = c; (xxiii) h = b, h = d, d = b, h = e, e = b, e = d; (xxiv) e = b, e = c, c = b, e = d, d = b, d = c. Proof. Here, ∆ has at most three vertices of degree 2, so has non-positive curvature for all of these cases. Consider the case, • e = b, d = c. Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 920 In this case ∆ is given in Figure 3. Figure 3: Region ∆ Since d(vb) = d(vc) = 2 or d(vc) = d(vd) = 2 or d(vd) = d(ve) = 2 can not occur together so c(∆) ≤ 0. A further 19 cases are solved in Lemma 2 by the application of curvature distribution method [10]. Lemma 2. The presentation P = 〈G, t | s1(t)〉 is aspherical if any one of the following holds: (i) a = g, h = c, e = b; (ii) a = g, h = c, e = d; (iii) a = g, h = e, c = b; (iv) a = g, h = e, d = c; (v) a = g, h = d, e = c; (vi) a = g, e = b, d = c; (vii) a = g, e = c, d = b; (viii) a = g, e = d, c = b; (ix) a = g, e = b, e = c, c = b; (x) a = g, e = c, e = d, d = c; (xi) a = g, h = e, h = c, e = c; (xii) a = g, d = b, c = b, d = c; (xiii) a = g, h = c, h = d, d = c, e = b; (xiv) a = g, h = e, h = d, e = d, c = b; (xv) a = g, h = e, h = c, e = c, d = b; (xvi) a = g, h = d, e = b, e = c, c = b; Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 921 (xvii) a = g, h = c, e = b, d = b, e = d; (xviii) a = g, h = e, h = d, e = d, h = c, e = c, d = c; (xix) e = b, e = c, c = b, e = d, d = b, d = c, h = b, h = e, h = c, h = d. Proof. 1. In this case ∆ is given in Figure 4(i). Figure 4: Region ∆ The subcases which are to be examined are given below: (a) d∆(va) = d∆(vc) = d∆(ve) = d∆(vg) = 2; (b) d∆(va) = d∆(vc) = d∆(ve) = d∆(vh) = 2. (a) Here d∆(va) = d∆(vc) = d∆(ve) = d∆(vg) = 2 which implies l∆(va) = ag−1, l∆(vc) = ch−1, l(ve) = eb−1, and l∆(vg) = ga−1, as given in Figure 4(ii). Notice that l∆(vc) = ch−1 and l∆(ve) = eb−1 implies that l∆(vd) = i−1da−1w in which w ∈ {b−1, c−1, e−1, h−1} which implies d∆(vd) > 3. Similarly notice that l∆(vg) = ga−1 and l∆(ve) = eb−1 implies that l∆(vf ) = fi−1c−1w in which w ∈ {b, d, e, h} which implies d∆(vf ) > 3. Since d∆(vd) > 3 and d∆(vf ) > 3 so c(∆) ≤ 0. (b) Here d∆(va) = d∆(vc) = d∆(ve) = d∆(vh) = 2 which implies l∆(va) = ag−1, l∆(vc) = ch−1, l(ve) = eb−1, and l∆(vh) = hc−1, as given in Figure 4(iii). Notice that l∆(vc) = ch−1 and l∆(ve) = eb−1 implies that l∆(vd) = i−1da−1w in which w ∈ {b−1, c−1, e−1, h−1} which implies d∆(vd) > 3. Similarly notice that l∆(va) = ag−1 and l∆(vh) = hc−1 implies that l∆(vi) = if−1d−1w in which w ∈ {b, c, e, h} which implies d∆(vi) > 3. Since d∆(vd) > 3 and d∆(vi) > 3 so c(∆) ≤ 0. 2. In this case ∆ is given in Figure 5(i). Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 922 Figure 5: Regions ∆ and ∆̂ The subcases which are to be examined are given below: (a) d∆(va) = d∆(vc) = d∆(ve) = d∆(vg) = 2; (b) d∆(va) = d∆(vc) = d∆(ve) = d∆(vh) = 2. (a) Here d∆(va) = d∆(vc) = d∆(ve) = d∆(vg) = 2 which implies l∆(va) = ag−1, l∆(vc) = ch−1, l∆(ve) = ed−1, l∆(vg) = ga−1, as shown in Figure 5(ii). Notice that l∆(vg) = ga−1 and l∆(ve) = ed−1 implies that l∆(vf ) = fi−1e−1w in which w ∈ {b, c, d, h} which implies d∆(vf ) > 3. Add c(∆) ≤ π 6 to c(∆̂) is given by Figure 5(iii). Notice that d ∆̂ (va−1) = d ∆̂ (ve−1) = 2. Similarly notice that either d ∆̂ (vg−1) = 2 or d ∆̂ (vh−1) = 2 otherwise contradiction occur and all other vertices have degree atleast 3. Therefore c(∆̂) ≤ c(2, 2, 2, 3, 3, 3, 3, 3, 4) = −π 6 . (b) Here d∆(va) = d∆(vc) = d∆(ve) = d∆(vh) = 2 which implies l∆(va) = ag−1, l∆(vc) = ch−1, l∆(ve) = ed−1, l∆(vh) = hc−1, as shown in Figure 5(iv). Notice that l∆(va) = ag−1 and l∆(vh) = hc−1 implies that l∆(vi) = if−1d−1w in which w ∈ {b, c, e, h} which implies d∆(vi) > 3. Add c(∆) ≤ π 6 to c(∆̂) is given by Figure 5(v). Notice that d ∆̂ (vc−1) = d ∆̂ (ve−1) = 2. Similarly notice that either d ∆̂ (vg−1) = 2 or d ∆̂ (vh−1) = 2 otherwise contradiction occur. Observe that either d ∆̂ (vb−1) = 3 or d ∆̂ (va−1) = 2 since d ∆̂ (vb−1) = 3 already present so d ∆̂ (va−1) > 2 otherwise contradiction occur and all other vertices have degree atleast 3. Therefore c(∆̂) ≤ c(2, 2, 2, 3, 3, 3, 3, 3, 4) = −π 3 . 3. In this case ∆ is given in Figure 6(i). Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 923 Figure 6: Regions ∆ and ∆̂ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. 4. In this case ∆ is given in Figure 7(i). Figure 7: Regions ∆ and ∆̂ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. 5. In this case ∆ is given in Figure 8(i). Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 924 Figure 8: Region ∆ The subcases which are to be examined are given below: (a) d∆(va) = d∆(vc) = d∆(ve) = d∆(vg) = 2; (b) d∆(va) = d∆(vc) = d∆(ve) = d∆(vh) = 2. (a) Here d∆(va) = d∆(vc) = d∆(ve) = d∆(vg) = 2 which implies l∆(va) = ag−1, l∆(vc) = ce−1, l(ve) = ec−1, and l∆(vg) = ga−1, as shown in Figure 8(ii). Notice that l∆(va) = ag−1 and l∆(vc) = ce−1 implies that l∆(vb) = bd−1h−1w in which w ∈ {b, c, d, e} which implies d∆(vb) > 3. Similarly notice that l∆(vg) = ga−1 and l∆(ve) = ec−1 implies that l∆(vf ) = fi−1d−1w in which w ∈ {b, c, e, h} which implies d∆(vf ) > 3. Since d∆(vb) > 3 and d∆(vf ) > 3 so c(∆) ≤ 0. (b) Here d∆(va) = d∆(vc) = d∆(ve) = d∆(vh) = 2 which implies l∆(va) = ag−1, l∆(vc) = ce−1, l(ve) = ec−1, and l∆(vh) = hd−1, as shown in Figure 8(iii). No- tice that l∆(va) = ag−1 and l∆(vc) = ce−1 implies that l∆(vb) = bd−1h−1w in which w ∈ {b, c, d, e} which implies d∆(vb) > 3. Similarly notice that l∆(va) = ag−1 and l∆(vh) = hd−1 implies that l∆(vi) = if−1e−1w in which w ∈ {b, c, d, h} which implies d∆(vi) > 3. Since d∆(vb) > 3 and d∆(vi) > 3 so c(∆) ≤ 0. 6. In this case ∆ is given in Figure 9(i). Figure 9: Region ∆ Here d∆(va) = d∆(vc) = d∆(ve) = d∆(vg) = 2 which implies l∆(va) = ag−1, l∆(vc) = cd−1, l(ve) = eb−1, and l∆(vg) = ga−1, as shown in Figure 9(ii). Notice that l∆(va) = Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 925 ag−1 and l∆(vc) = cd−1 implies that l∆(vb) = bc−1h−1w in which w ∈ {b, c, d, e} which implies d∆(vb) > 3. Similarly notice that l∆(vg) = ga−1 and l∆(ve) = eb−1 implies that l∆(vf ) = fi−1c−1w in which w ∈ {b, d, e, h} which implies d∆(vf ) > 3. Since d∆(vb) > 3 and d∆(vf ) > 3 so c(∆) ≤ 0. 7. In this case ∆ is given in Figure 10(i). Figure 10: Regions ∆ and ∆̂ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. 8. In this case ∆ is given in Figure 11(i). Figure 11: Region ∆ Here d∆(va) = d∆(vc) = d∆(ve) = d∆(vg) = 2 which implies l∆(va) = ag−1, l∆(vc) = cb−1, l(ve) = ed−1, and l∆(vg) = ga−1, as shown in Figure 11(ii). Notice that l∆(ve) = ed−1 and l∆(vc) = cb−1 implies that l∆(vd) = dc−1c−1w in which w ∈ {b, d, e, h} which implies d∆(vd) > 3. Similarly notice that l∆(vg) = ga−1 and l∆(ve) = ed−1 implies that l∆(vf ) = fi−1e−1w in which w ∈ {b, c, d, h} which implies d∆(vf ) > 3. Since d∆(vd) > 3 and d∆(vf ) > 3 so c(∆) ≤ 0. 9. In this case ∆ is given in Figure 12(i). Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 926 Figure 12: Regions ∆ and ∆̂ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. 10. In this case ∆ is given in Figure 13(i). Figure 13: Regions ∆ and ∆̂ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 927 11. In this case ∆ is given in Figure 14(i). Figure 14: Regions ∆ and ∆̂ Figure 15: Regions ∆ and ∆̂ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. 12. In this case ∆ is given in Figure 16(i). Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 928 Figure 16: Region ∆ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. 13. In this case c(∆) ≤ 0 for all of its subcases as shown in Figure 17. Figure 17: Region ∆ 14. In this case ∆ as given in Figure 18(i). Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 929 Figure 18: Region ∆ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. 15. In this case ∆ is given in Figure 19(i). Figure 19: Regions ∆ and ∆̂ Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 930 Figure 20: Regions ∆ and ∆̂ Figure 21: Regions ∆ and ∆̂ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 931 16. In this case c(∆) ≤ 0 for all of its subcases as shown in Figure 22(i). Figure 22: Region ∆ 17. In this case ∆ is given in Figure 23(i). Figure 23: Region ∆ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. 18. In this case ∆ is given in Figure 24(i). Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 932 Figure 24: Region ∆ Figure 25: Region ∆ Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 933 Figure 26: Region ∆ Figure 27: Region ∆ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 934 19. In this case ∆ is given in Figure 28(i). Figure 28: Region ∆ Figure 29: Region ∆ Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 935 Figure 30: Region ∆ Figure 31: Region ∆ Muhammad Saeed Akram, Maira Amjid, Sohail Iqbal / Eur. J. Pure Appl. Math, 13 (4) (2020), 914-938 936 Figure 32: Region ∆ Figure 33: Region ∆ REFERENCES 937 Figure 34: Region ∆ By adding c(∆) to c(∆̂) we get c(∆) ≤ 0. There remains only 2 cases given in Theorem that still needs to be solved. In fact, weight test and curvature distribution method can not be applied to these cases. The weight test can not be applied to these cases as it is not possible to find a weight function θ that satisfies all the three conditions of weight test simultaneously, whereas curvature test can not be applied to these cases as it is impossible to find any neighbouring region ∆̂ in the neighbourhood of region ∆ that cancels the curvatures of region ∆. Remark 2. We remark that the list of exceptional cases given in Theorem in section 1 is open for the equation s1(t). Since weight test and curvature distribution can not be applied to these cases to prove Levin conjecture, therefore, some new methods needs to be developed to establish the validity of the remaining cases of Levin conjecture for these group equations of length 9. References [1] Muhammad Saeed Akram and Maira Amjid. Solving a group equation of length nine. arXiv preprint arXiv:2008.09508, 2020. 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