A topological proof that surface relators are test words

A TOPOLOGICAL PROOF THAT SURFACE RELATORS ARE TEST WORDS EDWARD C. TURNER Abstract. We use topological methods to show that all the surface relators [x1, x2]· · · [x2n−1, x2n] and x2 1 ···x2 min the free groups F2n and Fmare test words. 1. Introduction How much information does one need about an endomorphism (i.e., a self-homomorphism) of a finitely generated free group to be certain that it is an automorphism? The lazy group theorist could hardly hope for more than the existence of test words. Definition. The word w∈Fis a test word if any endomorphism fixing wis necessarily an automorphism. It is easy to check that if wis a test word, then an endomorphism ϕis an automorphism if and only if ϕ(w) = α(w) for some automorphism α—and this is an issue that is easy to decide in a free group by using the algorithms developed by J. H. C. Whitehead [11]. Nielsen proved [5] that the commutator of a pair of generators of a free group of rank two is a test word. This was improved by Zieschang [12], using standard techniques of combinatorial group theory, who showed (among other things) that the words cg= [x1, x2][x3, x4]· · · [x2g−1, x2g] and sh=x2 1x2 2· · · x2 hare test words. The words cgand share referred to as the surface relators since they are the relators that come from the standard representations of closed compact surfaces as CW-complexes with a single 2-cell. To the mathematician trained as a topologist, this fact cries out for a proof using the special properties of surfaces—this note provides such proof. I am indebted to Warren Dicks for pointing out to me that Dold [2] has a proof using two-dimensional cell complexes (quite different from the one presented here) and that he (Dicks) and Formanek have a rather short proof using Fox derivatives that appears in [1], p169. Algebraic proofs can be found in [7] and [10]. I am also indebted to the Centre de Recerca The author gratefully acknowledges the support of the Spanish Government through grant SAP2003-0253 and the hospitality of the Centre de Recerca Matem`atica in Bellaterra, Spain during the completion of this work. 1 2 EDWARD C. TURNER Matem`atica at the Unversitat Aut`onoma de Barcelona in Bellaterra Spain for their hospitality and support during the completion of this work. Theorem. All surface relators are test words. We assume that ϕis an endomorphism of the free group F2gor Fh, that w=cgor shis a surface relator and that Sis the corresponding compact surface—so if w=cg, then Sis the connected sum of gtori and if w=sh, then Sis the connected sum of hprojective planes. In either case, let B be an open ball in Swith closure B, boundary ∂B and center ∗so that π1(S\B) is a free group in which ∂B represents w. (The basepoint is no problem to deal with and will not be explicitly mentioned unless necessary— it will be assumed to be on ∂B or to be ∗as appropriate.) The surface groups are the groups (x1,··· , x2g|cg) and (x1,· · · , xh|sh). The proof of the Theorem proceeds in the following 4 steps, which are justified by the comments and the Lemmas. Outline of proof of the Theorem A) Represent ϕby a map f:S\B→S\Bso that f|∂B is the identity. Extend fto ˜ f:S→Sas the identity on B. B) Since e fis a degree 1 map of a surface, it is a homotopy equivalence. C) Let eg:S→Sbe a homotopy inverse for e fand e H:S×I→Sbe a homotopy so that e H|S×0= e f◦egand e H|S×1 is the identity. Choose egso that eg|Dis the identity and eg−1(D) = Dand choose e H so that e H|D×Iis projection on the first factor and that e H−1(D) = D×I. D) Define g:S\B→S\Bas eg|S\Band H: (S\B)×I→(S\B) as e H|(S\B)×I. Then ginduces a right inverse for ϕso ϕis onto. Therefore ϕis an automorphism. Steps Aand Dare applications of standard techniques as follows. The existence of fin step Afollows from the fact that ϕfixes wwhich is represented by ∂B. Since surface groups are Hopfian (see [4]), it suffices in Dto show that ϕis onto. Lemma 1. A degree 1 map of a surface is a homotopy equivalence. Lemma 2. If f: (S, ∗)→(S, ∗)induces an isomorphism on the fundamental group, then fhas an inverse gin the category of homotopy classes of pointed maps [(S, s0),(S, s0)]—in particular, there is a homotopy e H: S×I→Sso that e H|S×0= e f◦eg, e H|S×0is the identity and e H(∗, t) = ∗ for all t. Lemma 3. It is possible to alter e Has required in Step C. A TOPOLOGICAL PROOF THAT SURFACE RELATORS ARE TEST WORDS 3 Proof of Lemma 1: This follows as an application of Edmonds’ homotopy characterization of maps between surfaces [3] from which it follows that a degree 1 map is actually homotopic to a homeomorphism. We give an elementary proof. Let G=π1(S, ∗) and consider the map ψ:G→Ginduced by e fwith H=ψ(G). Since Gis Hopfian, to show that ψis an isomorphism, it suffices to show that H=G. Furthermore, since Sis an Eilenberg-MacLane space K(G, 1), fis a homotopy equivalence if and only if ψis an isomorphism. A degree 1 map e fof the surface Sinduces a map on first cohomology that preserves the cup product pairing, and since this pairing is non-singular, the induced map is an isomorphism. Dualizing, the induced map on first homology is an isomorphism. This completes the proof in the case that Sis the torus, whose fundamental group is abelian. If Sis the Klein bottle, then Gis the non-trivial semi-direct product of Zwith Zand the only endomorphisms that abelianize to automorphisms are automorphisms, completing the proof in this case as well. For the remainder of the proof, we assume that Sis a surface other than the torus or the Klein bottle; i.e., that the Euler characteristic of Sis non-zero. Recall that the rank of a group K—denoted rank(K)—is the minimum cardinality of a set of generators. Claim: If Hand Gare as above, then rank(H)= rank(G). Proof of claim: Since His a quotient of G, rank(H)≤rank(G). As a subgroup of a surface group, His the fundamental group of a (probably non-compact) surface. But a surface whose fundamental group is finitely generated is a compact surface with finitely many punctures (see [6]): thus His either a surface group or free. In either case, rank(H) = rank(Hab)— and of course rank(G) = rank(Gab). Now let j:H→Gbe the inclusion map and consider the sequence Gab ψab −→ Hab jab ∗ −→ Gab. Since the composition is an isomorphism, it follows that jab ∗is surjective and that rank(Gab)≤rank(Hab), completing the proof of the claim. Now we know that ψ:G→His a surjective map from a surface group Gto a group Hof the same rank which is either a surface group or a free group. It is not hard to see that the only way this can happen is that ψ is an isomorphism. For if Hwere free, then the images of the generators of Gwould be a basis contradicting the fact that the relator is mapped to the identity. If His a surface group, it must be isomorphic to G(by rank considerations), and it corresponds to a compact covering space of Sof the same Euler characteristic. Since Sis neither the torus nor the Klein bottle, 4 EDWARD C. TURNER this implies that the index of the covering is 1, completing the proof that Gand Hare isomorphic. ¤ Proof of Lemma 2: The map e f: (S, ∗)→(S, ∗) has a homotopy inverse eg: (S, s0)→(S, s0) and we need to know that egis an inverse in the category of homotopy classes of pointed maps. Applying Theorem 8.1.9 on page 427 of [9] to the map e f◦egwith the pointed spaces X=Y= (S, s0) and using the fact that e f◦eginduces the identity on π1(S, ∗) proves precisely that. ¤ Proof of Lemma 3: Step 1: We begin by making e Htransverse to the point ∗so that e H−1(∗) is ∗ × Iplus a number of circles. Step 2: Next, we combine all circles into one circle. If C1and C2are two circles in e H−1(∗) we alter e Hto splice C1and C2as follows. Join C1to C2 by an embedded path that meets e H−1(∗) only at its endpoints and enclose the path in a box R. Do this so that e His constant on vertical lines in the box, the parts of C1and C2in Rare vertical line segments and that cylinders about the pieces of C1and C2in Bare mapped to a small disc about ∗. See Figure 1a.      ¨ §¥ ¦¨ §¥ ¦ § ¦ § ¦ ppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppp C1C2 ± ° ± ° ± ° ± ° ± ° ± ° ± ° ± ° ± ° ± ° ± °± °± °± °± °± °± °± °± °± ° C ' $ ² ¯ ¾ » ¾ » ¾ » ¾ » ¾ » ¾ » ¾ » ¾ » ¾ » ¾ » ¾ »¾ »¾ »¾ »¾ »¾ »¾ »¾ »¾ »¾ » C & % ¹¸ ¨ §¥ ¦¨ §¥ ¦ § ¦ § ¦      Figure 1a Figure 1b The idea now is to redefine e Hin Rso that the preimages of ∗are line segments joining the same four points on the boundary of Rbut with the top two paired and the bottom two paired. If the interiors of the cylinders A TOPOLOGICAL PROOF THAT SURFACE RELATORS ARE TEST WORDS 5 around the parts of C1and C2are removed from the box, what is left is a solid two-torus T T with boundary T—in other words, we consider Ras being built up from its boundary by adding the two solid cylinders and then filling in with the solid two-torus. The map e Hon TT can be considered as a solution to the problem of extending the restriction of e Hto Tover T T as a map to S\ ∗. (We can in fact assume that all of Ris mapped to B, so that the obstructions encountered will be in π1(B\ ∗)∼ =Z.) Now view R in a different way as its boundary with two cylinders added and then a solid two-torus filled in as shown in Figure 1b. The parts of C1and C2shown in Figure 1a are replaced by the cores of the cylinders in Figure 1b. One of the cylinders joins the discs on the top and the other joins the discs on the bottom. Let T T 0and T0be the associated solid two-torus and its boundary. Now redefine e Hinside Rby first extending over these cylinders to have the same image as the original cylinders, namely a small neighborhood of ∗. Note that the discs transverse to the pieces of C1and C2can be rotated as many times as we wish. This leaves the problem of extending the resulting map on T0over T T 0(as a map to B\ ∗). The obstruction to this extension is the element of H2(T T 0, T 0, π1(B\ ∗)) = Z×Zwhose coordinates are the homotopy classes of the maps on the boundary circles of the two two-cells added in building TT0from T0. One of these circles runs along the top side of the upper handle and then across the top face and the other runs along the bottom side of the lower handle then across the bottom face. The rotations of the transverse discs can be chosen to make these homotopy classes vanish. Note that since S\ ∗ is aspherical, there are no other obstructions. The effect is to splice C1and C2together. If C1and C2are directed, this splicing can be done in a manor that preserves the directions. Repeating this process if necessary leads to a homotopy e Hfor which e H−1(∗) = ∗ × I∪C, Ca single circle. Step 3: Next we arrange that the circle Cdoes not link the arc ∗ × I. Consider first the following example of a homotopy G:D2×I→D2from the identity to itself so that G(s, t) = sfor all s∈S1and that G−1(∗) is ∗ × I∪C0where C0is a circle that links ∗ × Ionce. Start with the Hopf fibration map η:S3→S2, for which the preimage of any two distinct points is a pair of linked circles. By following this with a map of S2to itself that identifies two points, we get a map η0:S3→S2so that the preimage of a point is two linked circles. Now take a small box about a point on one of the circles and parameterize the map so that its restriction to the box is projection on one side—the desired homotopy is the restriction of this map to the closure of the complement of the box. The homotopy Gcan be extended over S×Ias the identity to produce e G:S×I→S. Now if Clinks ∗ × I k times, we can stack kcopies of e G on top of e Hand then use the splicing procedure of Step 1 to splice these 6 EDWARD C. TURNER circles onto Cbut in the opposite direction. The resulting circle does not link ∗ × I. Step 4: Next we remove Cfrom e H−1(∗). Since e His a homotopy equivalence and e H(C) is null homotopic, Cbounds an immersed disc in S×I. We may as well assume that Clies in B×I. If Dwere actually embedded, then since Cdoes not link ∗ × I, (and there are no other circles) we can pull D off of ∗ × Iso that Dlies in the complement of ∗ × I. Then the boundary of a small neighborhood of Dis a 2-sphere in S×Iwhose image lies in (S\ ∗)×I, which is aspherical. In this case, e Hcan be redefined inside the 2-sphere to miss ∗ × I, thus removing the circle Cfrom e H−1(∗) and completing the argument. The problem of finding an embedded disc bounded by Cis the problem of unknotting C. Any circle can be unknotted by a sequence of moves corresponding to the replacement of an overcrossing by an undercrossing or vice versa in a knot projection. We do this with Cas follows.              ±° ²¯ ²                                                                                                                                   ¨ §¥ ¦ § ¦ ppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppp C C C     ±° ²¯ ²                    qqqqqqqqqqqqqqqqqqqqqqqqqqqqqq                                                                                          C       Figure 2a Figure 2b ¨ §¥ ¦ § ¦ ¯¿ ¿¿¿¿¿¿¿¿¿¿¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ ¿ £ £ ¢ÀÀÀÀÀÀÀÀÀÀÀ À À À À À À À À À ° ¨ § ¤¤¤¤¤¤¤¤¤¤µ µ µ µ µ µ µ µ µ µ µµµµµµµµµµ " ² Assume that a path from one point on Cto another is enclosed in a box Rwith the potions of Cin Ras in Figure 2a with surrounding cylinders—so again Ris viewed as its boundary with two handles adjoined and filed in with a solid two-handled torus, but this time one cylinder goes form top to bottom and the other from front to back. Now view Rdifferently with the handles of Figure 2b adjoined and filled in with the corresponding solid two-torus T T 0with boundary T. Again we have freedom to rotate the discs transverse to the pieces of Cwhen defining the maps on the cylinders. A TOPOLOGICAL PROOF THAT SURFACE RELATORS ARE TEST WORDS 7 Again, we can choose the rotations to guarantee that the obstruction to the extension over T T 0vanish, completing the argument. In this case, one of the two boundary circles runs from front to back on the right side of the front-back tube and then along the back, right and front faces at constant height to close up. The other runs from top to bottom along the left side of the redirected tube and then along the bottom, left and top faces to close up. Once Cis unknotted, we find an embedded disc and remove Cas indicated above. Step 5: Finally, “blow up” ∗ × Ito B×Ito produce the desired e e H.¤ References [1] Dicks, W. and Dunwoody, M.J., it Groups acting on graphs, Cambridge atudies in advanced mathematics 17, Cambridge University Press, Cambridge, (1989). [2] Dold, A., it Nullhomologous words in free groups which are not nullhomologous in any proper subgroup, Arch. Math., 50 (1988), 564–569. [3] Edmonds, A.L., Deformation of maps to branched coverings in dimension two, Annals of Math. 110 (1979), 113–125. [4] Frederick, K. N., The Hopfian property for a class of fundamental groups, Comm. Pure Appl. Math. 16 (1963), 1–8. [5] Nielsen,J., Die Automorphismen der allgemeinen unendlichen Gruppe mit zwei Erzeugenden, Math. Ann. 78 (1918), 385–397. [6] Richards, I. On the classification of noncompact surfaces, Trans. AMS 106 (1963), 259–269. [7] Shpilrain, V., Recognizing automorphisms of the free groups, Arch. Math., 62 (1994), 385–392. 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