The commutator subgroups of free groups and surface groups
TThe commutator subgroups of free groups andsurface groups
Andrew Putman ∗ Abstract
A beautifully simple free generating set for the commutator subgroup of a freegroup was constructed by Tomaszewski. We give a new geometric proof of histheorem, and show how to give a similar free generating set for the commutatorsubgroup of a surface group. We also give a simple representation-theoreticdescription of the structure of the abelianizations of these commutator subgroupsand calculate their homology.
Let F n be the free group on { x , . . . , x n } . The commutator subgroup [ F n , F n ] is aninfinite-rank free group. The conjugation action of F n on [ F n , F n ] induces an action onits abelianization [ F n , F n ] ab that factors through F n / [ F n , F n ] = Z n , making [ F n , F n ] ab into a Z [ Z n ]-module. This paper addresses the following questions: • What kind of Z [ Z n ]-module is [ F n , F n ] ab ? Can we describe it in terms of simplerrepresentations and calculate its homology? • It is easy to write down free generating sets for [ F n , F n ]; indeed, this is oftengiven as an exercise when teaching covering spaces. However, doing this naivelygives a very complicated free generating set. Can it be made simpler?We also study the analogous questions for the fundamental groups of closed surfaces. Rank 2.
As motivation, we start with the case n = 2. Regarding F as thefundamental group of a wedge X of two circles, the subgroup [ F , F ] correspondsto the cover (cid:101) X shown in Figure 1.a. The abelianization [ F , F ] ab is H ( (cid:101) X ). This is ∗ Supported in part by NSF grant DMS-1811210 a r X i v : . [ m a t h . G T ] J a n x x (a) (b)(c) (d) (e) Figure 1: (a) The cover (cid:101) X → X corresponding to [ F , F ] . (b) A maximal tree in (cid:101) X .(c) A generator x x x x − x − ∈ [ F , F ] coming from the maximal tree. (d) The element [ x , x ] x x ∈ [ F , F ] . (e) The element [ x , x ] ∈ [ F , F ] . a free abelian group with basis the set of squares in Figure 1.a. The deck group Z permutes these squares simply transitively, so [ F , F ] ab is a rank-1 free Z [ Z ]-module:[ F , F ] ab ∼ = Z [ Z ] . There are several natural choices of free generating sets for the free group [ F , F ]:(i) Using the maximal tree shown in Figure 1.b, we obtain the free generating set (cid:8) x k x k x x − k x − k − | k , k ∈ Z and k (cid:54) = 0 (cid:9) illustrated in Figure 1.c.(ii) It is not hard to convert the free generating set (i) into the free generating set (cid:110) [ x , x ] x k x k | k , k ∈ Z (cid:111) . illustrated in Figure 1.d. Here our conventions are that [ y, z ] = y − z − yz andthat superscripts indicate conjugation: y z = z − yz . Unlike in (i), this does not arise from a maximal tree.(iii) A classical exercise in combinatorial group theory (see [MagKSo, p. 196, exer.24] or [Se2, Proposition I.1.4]) gives the free generating set (cid:8) [ x k , x k ] | k , k ∈ Z and k , k (cid:54) = 0 (cid:9) illustrated in Figure 1.e. Again, this does not arise from a maximal tree. Remark . It is hard to see that [ F , F ] ab ∼ = Z [ Z ] from the generating sets (i) and(iii) above, but this isomorphism can be easily deduced from the generating set (ii). We will prove a much more general result in Theorem A below. enerating sets in higher rank. How about [ F n , F n ] for n ≥
3? A maximal treeargument like (i) above leads to the very complicated free generating set (cid:110) ( x k · · · x k m m ) x (cid:96) ( x k · · · x k (cid:96) +1 (cid:96) · · · x k m m ) − | ≤ (cid:96) < m ≤ n , k , . . . , k m ∈ Z , k m (cid:54) = 0 (cid:111) . Tomaszewski generalized (ii) to give the following much simpler free generating set:
Theorem A (Tomaszewski [T]) . For n ≥ , the group [ F n , F n ] is freely generated by (cid:110) [ x i , x j ] x kii x ki +1 i +1 ··· x knn | ≤ i < j ≤ n and k i , . . . , k n ∈ Z (cid:111) . Theorem A plays an important role in the author’s work on the Torelli group [P1, P2].Tomaszewski’s proof of it was combinatorial and involved extensive calculations withcommutator identities. We will give a proof that is geometric and calculation-free.
Remark . It is unclear how to generalize the generating set (iii) above to n ≥ Generating sets for surface groups.
Consider the fundamental group of a closedoriented genus- g surface Σ g : π (Σ g ) = (cid:104) x , . . . , x g | [ x , x ] · · · [ x g − , x g ] = 1 (cid:105) . Just like [ F n , F n ], the commutator subgroup [ π (Σ g ) , π (Σ g )] is an infinite-rank freegroup . How can we find a free generating set for it?One idea is to try to use Theorem A. The group [ π (Σ g ) , π (Σ g )] can be obtained from[ F g , F g ] by quotienting out by the F g -normal closure of the surface relation[ x , x ] · · · [ x g − , x g ] . When you try to express the F g -conjugates of this in terms of the free basis (cid:110) [ x i , x j ] x kii x ki +1 i +1 ··· x k g g | ≤ i < j ≤ g and k i , . . . , k g ∈ Z (cid:111) for [ F g , F g ] given by Theorem A, it is natural to assume that the relations you imposemust each involve a generator of the form[ x , x ] x k ··· x k g g . Letting (cid:101)
Σ be the cover of Σ g corresponding to [ π (Σ g ) , π (Σ g )], this means that π ( (cid:101) Σ) is free. Infact, Johansson [J] proved that all non-compact connected surfaces have free fundamental groups. Aconceptual way to see this uses a theorem of Whitehead ([W]; see [P3] for an expository account)that says that any smooth connected noncompact n -manifold M deformation retracts to an ( n − (cid:101) Σ, this says that (cid:101)
Σ deformation retracts to a 1-dimensionalsimplicial complex, i.e. a graph. We conclude by observing that graphs have free fundamental groups. F g . This suggests that these [ F g , F g ]-generators can be eliminated,and the remaining ones should freely generate [ π (Σ g ) , π (Σ g )].Unfortunately, it seems quite hard to make this idea rigorous. Even expressing surfacerelations of the form ([ x , x ] · · · [ x g − , x g ]) x k ··· x k g g in terms of our generators seems quite messy. Nonetheless, we will give a differentgeometric argument to show that the above idea does lead to a free generating set: Theorem B.
For g ≥ , the group [ π (Σ g ) , π (Σ g )] is freely generated by (cid:110) [ x i , x j ] x kii x ki +1 i +1 ··· x k g g | ≤ i < j ≤ g , ( i, j ) (cid:54) = (1 , , and k i , k i +1 , . . . , k g ∈ Z (cid:111) . Remark . We do not known any other explicit free generating sets in the literaturefor [ π (Σ g ) , π (Σ g )]. Non-freeness and homology.
As we discussed above, [ F , F ] ab is a free Z [ Z ]-module. It is natural to wonder whether this holds in higher rank. The structure ofthe free generating set for [ F n , F n ] given in Theorem A suggests that it should not befree, and we prove that this indeed is the case.In fact, we prove even more. For a group G , if M is a free Z [ G ]-module thenH k ( G ; M ) = 0 for all k ≥
1. To prove that M is not free, it is thus enough to findsome nonvanishing homology group of degree at least 1. We prove the following: Theorem C.
For n ≥ and k ≥ , we have H k ( Z n ; [ F n , F n ] ab ) ∼ = ∧ k +2 Z n . Inparticular, [ F n , F n ] ab is not a free Z [ Z n ] -module for n ≥ .Remark . The case k = 0 says that H ( Z n ; [ F n , F n ] ab ) ∼ = ∧ Z n . Recall that the 0 th homology group is the coinvariants of the coefficient module, i.e. the largest quotientof the coefficient module on which the group acts trivially. Since the action of Z n on[ F n , F n ] ab is induced by the conjugation action of F n on [ F n , F n ], we haveH ( Z n ; [ F n , F n ] ab ) = [ F n , F n ][ F n , [ F n , F n ]] ∼ = ∧ Z n . This is a special case of a classical theorem of Magnus ([Mag2]; see [Se1] for anexpository account) that says that the graded quotients of the lower central series ofa free group are precisely the graded terms of the free Lie algebra.For surface groups, the abelianization [ π (Σ g ) , π (Σ g )] ab is a module over π (Σ g )[ π (Σ g ) , π (Σ g )] ∼ = H (Σ g ) ∼ = Z g . The analogue of Theorem C for these groups is as follows:4 heorem D.
For g ≥ and k ≥ , we have H k ( Z g ; [ π (Σ g ) , π (Σ g )] ab ) ∼ = (cid:40) ( ∧ Z g ) / Z if k = 0 , ∧ k +2 Z g if k ≥ . In particular, [ π (Σ g ) , π (Σ g )] ab is not a free Z [ Z g ] -module for g ≥ . Module structure.
If [ F n , F n ] ab is not a free Z [ Z n ]-module, what kind of moduleis it? We will prove that it has a filtration whose associated graded terms are not free,but are not too far from being free. In the following theorem, for 1 ≤ m ≤ n we regard Z m as a module over Z n via the projection Z n → Z m onto the final m coordinates. Theorem E.
For all n ≥ , the Z [ Z n ] -module [ F n , F n ] ab has a sequence M ⊂ M ⊂ · · · ⊂ M n − = [ F n , F n ] ab of submodules such that M k /M k +1 ∼ = (cid:0) Z (cid:2) Z n − k +1 (cid:3)(cid:1) ⊕ ( n − k ) for ≤ k ≤ n − . Question 1.5.
What is the nature of the extensions in the filtration in Theorem E?Do any of them split?
For surface groups, we relate the structures of [ π (Σ g ) , π (Σ g )] ab and [ F g , F g ] ab asfollows: Theorem F.
For all g ≥ , we have a short exact sequence −→ Z [ Z g ] −→ [ F g , F g ] ab −→ [ π (Σ g ) , π (Σ g )] ab −→ of Z [ Z g ] -modules.Remark . As one would expect, the Z [ Z g ]-submodule identified by Theorem F isgenerated by the image of the surface relation and its conjugates. Question 1.7.
Does the exact sequence in Theorem F split?
Outline.
The rest of this paper is divided into two sections: § § Conventions.
For a set S , we will let F ( S ) denote the free group on S . For S = { x , . . . , x n } , we thus have F ( S ) = F n . For a group G and x, y ∈ G , we define[ x, y ] = x − y − xy and x y = y − xy . This latter convention ensures that for x, y, z ∈ G we have ( x y ) z = x yz . 5 Commutator subgroup of free group
In this section, we prove Theorems A, C, and E.
We start with some preliminary lemmas. We emphasize in these lemmas that we allowfree groups to have infinite rank. Our first lemma will only be used in an example inthis section, but will be needed in a more serious way when we discuss the commutatorsubgroup of a surface group.
Lemma 2.1.
Let T be a set and let S ⊂ T . The normal closure of S in F ( T ) is thenfreely generated by { s w | s ∈ S and w ∈ F ( T \ S ) } . Proof.
This is a standard consequence of covering space theory applied to a wedge of | T | circles, whose fundamental group can be identified with F ( T ). Lemma 2.2.
Let F be a free group and let S ⊂ F be a generating set that projects toa basis for the free abelian group F ab . Then F is freely generated by S .Proof. To show that no nontrivial reduced word in S is trivial in F , it suffices to provethat all finite subsets S (cid:48) ⊂ S freely generate the subgroup F (cid:48) they generate. Thecomposition F (cid:48) −→ ( F (cid:48) ) ab −→ F ab takes S (cid:48) to a linearly independent subset of F ab . Since the image of S (cid:48) in ( F (cid:48) ) ab generates it, we deduce that ( F (cid:48) ) ab is a free abelian group of rank | S (cid:48) | , so F (cid:48) is a freegroup of rank | S (cid:48) | . Thus the map F ( S (cid:48) ) → F (cid:48) induced by the inclusion S (cid:48) (cid:44) → F (cid:48) isa surjective map between free groups of the same finite rank. Since finite-rank freegroups are Hopfian the map F ( S (cid:48) ) → F (cid:48) must be an isomorphism, so F (cid:48) is freelygenerated by S (cid:48) . Lemma 2.3.
Let F be a free group, let A < F be a subgroup, and let B (cid:67) F be anormal subgroup such that F = B (cid:111) A . Assume that B is is freely generated by a set S on which A acts freely by conjugation, and let S (cid:48) < S contain a single element fromeach A -orbit. We then have F = F ( S (cid:48) ) ∗ A . A group G is Hopfian if all surjective homomorphisms G → G are isomorphisms. More generally,Malcev ([Malc]; see [LSc, Theorem IV.4.10] for the short proof) proved that finitely generatedresidually finite groups are Hopfian. See [C] for a variety of proofs that free groups are residuallyfinite. My favorite is in [MaleP], which gives a stronger result and builds on a proof of Hempel [H].
6o help the reader understand this lemma, we give an example before the proof.
Example . Fix some 1 ≤ k < n , let A < F n be the subgroup generated by { x k +1 , . . . , x n } , and let B (cid:67) F n be the normal closure of { x , . . . , x k } . The group B isthe kernel of the evident retraction F n → A , so F n = B (cid:111) A . Lemma 2.1 says that B is freely generated by S = { x wi | ≤ i ≤ k and w ∈ A } . The conjugation action of A on B restricts to a free action on S , and S (cid:48) = { x , . . . , x k } ⊂ S contains a single element from each A -orbit. As is asserted in the lemma, we have F n = F ( S (cid:48) ) ∗ A . Proof of Lemma 2.3.
Let S (cid:48)(cid:48) ⊂ A be a free generating set for A . The set S (cid:48) ∪ S (cid:48)(cid:48) generates F , and we must prove that it freely generates it. Since F = B (cid:111) A , we have F ab = ( B ab ) A ⊕ A ab , where the subscript indicates that we are taking the A -coinvariants, i.e. the largestquotient on which A acts trivially. Since A acts freely on S and S (cid:48) contains a singleelement from each A -orbit, it follows that S (cid:48) ⊂ B projects to a free basis for the freeabelian group ( B ab ) A . Since S (cid:48)(cid:48) projects to a free basis for the free abelian group A ab , we conclude that S (cid:48) ∪ S (cid:48)(cid:48) projects to a free basis for the free abelian group F ab .Lemma 2.2 then implies that S (cid:48) ∪ S (cid:48)(cid:48) is a free generating set for F .Our final lemma contains the geometric heart of Theorem A. Lemma 2.5.
Fix some n ≥ . Let K (cid:67) F n be the normal closure of x and let B = K ∩ [ F n , F n ] . Set T = { x , . . . , x n } . Then B is freely generated by (cid:110) [ x , x j ] x k w | < j ≤ n and k ∈ Z and w ∈ F ( T ) (cid:111) . Proof.
Let X be the wedge of n circles and let p ∈ X be the wedge point. Identify π ( X, p ) with F n . Let ( (cid:101) X, (cid:101) p ) be the cover of ( X, p ) with π ( (cid:101) X, (cid:101) p ) = B . The group B is the kernel of the surjective homomorphism f : F n → F ( T ) × Z defined via theformula f ( x i ) = (cid:40) (1 ,
1) if i = 1 , ( x i ,
0) if 2 ≤ i ≤ n. Here the 1 in the first coordinate of (1 , ∈ F ( T ) × Z means the identity element in F ( T ). For 1 ≤ i ≤ n , let x i = f ( x i ). We can identify (cid:101) X with the Cayley graph of F ( T ) × Z with respect to the generating set { x , . . . , x n } . Using this identification,7 u,0)(u,1)(u,2)(u,-1)(u,-2) (v,0)(v,1)(v,2)(v,-1)(v,-2)L e (a) (w,0)(1,0) (wx j-1 ,0)x x j (b) (w,0)(1,0) (wx j-1 ,0)x x j (c) Figure 2: (a) The “ladder” L e corresponding to the edge e . (b) The space (cid:98) L e with the edge e pointing towards the basepoint, along with the corresponding maximal tree. (c) The space (cid:98) L e with the edge e pointing away from the basepoint, along with the corresponding maximaltree. the set of vertices of (cid:101) X is identified with F ( T ) × Z and the edges of (cid:101) X are orientedand labeled with elements of { x , . . . , x n } . The basepoint (cid:101) p of (cid:101) X is (1 , ∈ F ( T ) × Z .Let Y be the Cayley graph of F ( T ) with respect to T . Identify Y with the full subgraphof (cid:101) X whose vertices are F ( T ) × ⊂ F ( T ) × Z . The projection F ( T ) × Z → F ( T )induces a surjective map ρ : (cid:101) X → Y . For each edge e of Y , let L e = ρ − ( e ). If e goesfrom u ∈ F ( T ) to v ∈ F ( T ), then L e is the “ladder” depicted in Figure 2.a. Thegraph (cid:101) X is the union of the L e as e ranges over the edges of Y .We want to apply the Seifert–van Kampen theorem to express π ( (cid:101) X, (cid:101) p ) = B in termsof the fundamental groups of the L e . For this, we will have to enlarge the L e so thatthey include the basepoint (cid:101) p = (1 , e of Y , let (cid:98) L e be the union of L e and the shortest edge path in Y ⊂ (cid:101) X that starts at the basepoint (cid:101) p = (1 ,
0) and endswith e . The graph (cid:98) L e contains (cid:101) p and deformation retracts to L e .All finite intersections of more than one of the (cid:98) L e are contractible, so Seifert–vanKampen applies to show that B = π ( (cid:101) X, (cid:101) p ) = ∗ e π ( (cid:98) L e , (cid:101) p ) . Letting S = (cid:110) [ x , x j ] x k w | < j ≤ n and k ∈ Z and w ∈ F ( T ) (cid:111) be our purported free generating set for B , it is thus enough to find a subset S e ⊂ S for each edge e such that π ( (cid:98) L e , (cid:101) p ) is freely generated by S e and such that S = (cid:91) e S e . Fix an edge e of Y . Let x j be the label on e and let ( w, ∈ F ( T ) × Z be the endpointof e . What we claim is that the subgroup π ( (cid:98) L e , (cid:101) p ) of π ( (cid:101) X, (cid:101) p ) = B is freely generatedby S e = (cid:110) [ x , x j ] x k w − | k ∈ Z (cid:111) . x , x j ] x k w − = wx − k x − x − j x x j x k w. The proof of this claim divides into two cases: • The first is that e points towards the basepoint as in Figure 2.b. The purportedfree generating set is then the one associated to the maximal tree depicted inFigure 2.b. • The second is that e points away from the basepoint as in Figure 2.c. Thepurported free generating set is then the one associated to the maximal treedepicted in Figure 2.c.It is clear that S is the union of the S e , so the proof is complete.We now prove Theorem A. Proof of Theorem A.
Recall that we must prove that the group [ F n , F n ] is freelygenerated by (cid:110) [ x i , x j ] x kii x ki +1 i +1 ··· x knn | ≤ i < j ≤ n and k i , . . . , k n ∈ Z (cid:111) . The proof will be by induction on n . The base case n = 1 is trivial since [ F , F ] = 1and the purported free generating set is empty.Assume now that n > F n − , F n − ]. Set T = { x , . . . , x n } ⊂ F n and let K (cid:67) F n be the normal closure of x . The group K is the kernel of theevident retraction F n → F ( T ), so F n = K (cid:111) F ( T ). Define A = [ F ( T ) , F ( T )] and B = K ∩ [ F n , F n ]. Our semidirect product decomposition restricts to[ F n , F n ] = B (cid:111) A. By Lemma 2.5, the group B is freely generated by S = (cid:110) [ x , x j ] x k w | < j ≤ n and k ∈ Z and w ∈ F ( T ) (cid:111) . The conjugation action of F ( T ) on B preserves S and acts freely on it. Restrict thisaction to A = [ F ( T ) , F ( T )]. The group A still acts freely on S , and the set S (cid:48) = (cid:110) [ x , x j ] x k x k ··· x knn | < j ≤ n and k , . . . , k n ∈ Z (cid:111) A on S . Lemma2.3 then implies that [ F n , F n ] = F ( S (cid:48) ) ∗ A. By induction, A = [ F ( T ) , F ( T )] is freely generated by S (cid:48)(cid:48) = (cid:110) [ x i , x j ] x kii x ki +1 i +1 ··· x knn | ≤ i < j ≤ n and k i , . . . , k n ∈ Z (cid:111) . We conclude that [ F n , F n ] is freely generated by S (cid:48) ∪ S (cid:48)(cid:48) , as desired. We start with the following observation.
Lemma 2.6.
Let G be a group and let x, y, w, w (cid:48) ∈ G . Assume that w and w (cid:48) mapto the same element of G ab . Then [ x, y ] w and [ x, y ] w (cid:48) map to the same element of [ G, G ] ab .Proof. For z, z (cid:48) ∈ [ G, G ], write z ≡ z (cid:48) if z and z (cid:48) map to the same element of [ G, G ] ab .We can write w (cid:48) = wc with c ∈ [ G, G ], so[ x, y ] w (cid:48) = c − [ x, y ] w c ≡ c − c [ x, y ] w = [ x, y ] w . This allows us to make the following definition.
Definition 2.7.
Let G be a group. For x, y ∈ G and h ∈ G ab , write { x, y } h for theimage in [ G, G ] ab of [ x, y ] w , where w ∈ G is any element projecting to h .We now launch into the proof of Theorem E. Proof of Theorem E.
Recall that the theorem in question asserts that for all n ≥ Z [ Z n ]-module [ F n , F n ] ab has a sequence0 = M ⊂ M ⊂ · · · ⊂ M n − = [ F n , F n ] ab of submodules such that M k /M k +1 ∼ = (cid:0) Z (cid:2) Z n − k +1 (cid:3)(cid:1) ⊕ ( n − k ) for 1 ≤ k ≤ n −
1. The proof of this will be by induction on n . The base case n = 1 istrivial since [ F , F ] = 1 and the theorem simply asserts that [ F , F ] ab = 0.10ssume now that n > F n − . Let T = { x , . . . , x n } and let r : F n → F ( T ) be the evident retract whose kernel is the normal closure of x . The map r induces a surjective homomorphism p : [ F n , F n ] → [ F ( T ) , F ( T )]. Let B = ker( p ). Let p : [ F n , F n ] ab → [ F ( T ) , F ( T )] ab be the map induced by p and let B = ker( ρ ).The map r descends to the projection Z n → Z n − onto the last ( n −
1) coordinates.Use this projection to make [ F ( T ) , F ( T )] ab into a module over Z [ Z n ]. We then have ashort exact sequence0 −→ B −→ [ F n , F n ] ab p −→ [ F ( T ) , F ( T )] ab −→ Z [ Z n ]-modules. Since | T | = n −
1, our inductive hypothesis says that [ F ( T ) , F ( T )] ab has a filtration of the appropriate form. Using (2.1), to prove the theorem for [ F n , F n ] ab ,it is enough to prove that B ∼ = ( Z [ Z n ]) ⊕ ( n − as Z [ Z n ]-modules. Indeed, we can then take M = B and M k for 2 ≤ k ≤ n − F n , F n ] ab of the filtration for [ F ( T ) , F ( T )] ab .Lemma 2.5 says that the group B is freely generated by S = (cid:110) [ x , x j ] x k w | < j ≤ n and k ∈ Z and w ∈ F ( T ) (cid:111) . This implies that B is generated as an abelian group by S = (cid:8) { x , x j } h | ≤ j ≤ n and h ∈ Z n (cid:9) . Here we are using the notation introduced before the proof. Theorem A says that theset S (cid:48) = (cid:110) [ x , x j ] x k ··· x knn | ≤ j ≤ n and k , . . . , k n ∈ Z (cid:111) forms part of a free basis for [ F n , F n ]. This implies that S is a basis for the free abeliangroup B . Since the Z [ Z n ]-action on [ F n , F n ] ab is induced by the conjugation actionof F n on [ F n , F n ], this implies that B is freely generated as a Z [ Z n ]-module by the( n − {{ x , x j } | ≤ j ≤ n } . We conclude that B ∼ = ( Z [ Z n ]) ⊕ ( n − , as desired. We now prove Theorem C, which calculates the homology of Z n with coefficients in[ F n , F n ] ab . We could use Theorem E, but will instead give a shorter direct proof.11 roof of Theorem C. Fix some n ≥
2. Our goal is to prove thatH k ( Z n ; [ F n , F n ] ab ) ∼ = ∧ k +2 Z n for k ≥ . Consider the short exact sequence1 −→ [ F n , F n ] −→ F n −→ Z n −→ . The associated Hochschild–Serre spectral sequence is of the form E pq = H p ( Z n ; H q ([ F n , F n ])) ⇒ H p + q ( F n ) . Since [ F n , F n ] is a free group, all of these entries vanish except E p = H p ( Z n ) = ∧ p Z n and E p = H ( Z n ; [ F n , F n ] ab ) . The E -page of our spectral sequence is thus of the form H ( Z n ; [ F n , F n ] ab ) H ( Z n ; [ F n , F n ] ab ) H ( Z n ; [ F n , F n ] ab ) H ( Z n ; [ F n , F n ] ab ) · · · Z Z n ∧ Z n ∧ Z n · · · This has to converge to H k ( F n ) = Z if k = 0 , Z n if k = 1 , . We deduce that the differentials ∧ k +2 Z n = E k +2 , → E k, = H k ( Z n ; [ F n , F n ] ab )must be isomorphisms for all k ≥
0. The theorem follows.
In this section, we prove Theorems B, D, and F.
We begin by introducing some notation that will be used throughout this section: • Fix some g ≥
2, and let π = π (Σ g ) = (cid:104) x , . . . , x g | [ x , x ] · · · [ x g − , x g ] = 1 (cid:105) . S S -1,-1 S -1,0 S S S S S -1,1 Figure 3:
The Z -regular cover (cid:101) Σ of Σ g corresponding to the normal closure G of T = { x , . . . , x g } . Putting the basepoint in S , , the subgroup π ( S , ) of G = π ( (cid:101) Σ) is thesubgroup F generated by T . The flat region is X ⊂ R , and the orange lines are of the form ( n + 1 / × R and R × ( n + 1 / for n ∈ Z . • Let
F < π be the subgroup generated by T = { x , x , . . . , x g } . • Let G (cid:67) π be the normal closure of F .We now prove several preliminary lemmas. Lemma 3.1.
We have [ π, π ] < G .Proof. The group G is the kernel of the homomorphism f : π → (cid:104) x , x | [ x , x ] = 1 (cid:105) ∼ = Z defined via the formula f ( x i ) = (cid:40) x i if i = 1 , , ≤ i ≤ g. Since the codomain of f is abelian, the kernel of f contains [ π, π ].For the next lemma, for w ∈ π the notation F w means the w -conjugate of the subgroup F of G . Since G is a normal subgroup of G , we have F w < G . Lemma 3.2.
The subgroup
F < π is free on T = { x , . . . , x g } , and G = ∗ ( k ,k ) ∈ Z F x k x k . roof. Let ∗ ∈
Σ be the basepoint. Let ( (cid:101) Σ , (cid:101) ∗ ) be the based cover of (Σ g , ∗ ) with G = π ( (cid:101) Σ , (cid:101) ∗ ). We depict (cid:101) Σ in Figure 3. As is shown there, it can be decomposed as (cid:101)
Σ = X ∪ (cid:91) ( k ,k ) ∈ Z S k ,k , where X and S k ,k are as follows: • X = R \ (cid:0) ∪ ( k ,k ) ∈ Z U k ,k (cid:1) , where U k ,k is a small open ball around ( k , k ) ∈ R . • S k ,k is a genus ( g −
1) surface with 1 boundary component glued to X along ∂U k ,k .The deck group Z acts in the evident way with ( n, m ) ∈ Z taking S k ,k to S k + n,k + m .Choosing our identifications and basepoint correctly, we can ensure that (cid:101) ∗ ∈ S , and that π ( S , , (cid:101) ∗ ) < G is F . It is then clear from our picture that F is free on T = { x , . . . , x g } and that G = ∗ ( k ,k ) ∈ Z F x k x k , as desired.Our next lemma is a general one and does not make use of the notation we introducedat the beginning of this section. Lemma 3.3.
Let S be the set of formal symbols { y n,m | n, m ∈ Z } . Then the freegroup F ( S ) is freely generated by S (cid:48) = { y , } ∪ (cid:8) y − n +1 ,m y n,m | n, m ∈ Z (cid:9) ∪ (cid:8) y − ,m +1 y ,m | m ∈ Z (cid:9) . Proof.
Consider the oriented tree T from Figure 4. This tree lies in R and its verticesare Z . Define S (cid:48)(cid:48) = { y , } ∪ (cid:8) y n,m y − n (cid:48) ,m (cid:48) | T has an oriented edge from ( n (cid:48) , m (cid:48) ) to ( n, m ) (cid:9) . The set S (cid:48) can be obtained from S (cid:48)(cid:48) by inverting some elements of S (cid:48)(cid:48) , so it is enoughto prove that F ( S ) is freely generated by S (cid:48)(cid:48) .Every ( n, m ) ∈ Z except for (0 ,
0) is the endpoint of precisely one edge in T , so wecan define a homomorphism f : F ( S ) → F ( S ) via the formula f ( y n,m ) = (cid:40) y n,m y − n (cid:48) ,m (cid:48) if there is an edge of T from ( n (cid:48) , m (cid:48) ) to ( n, m ) ,y , if ( n, m ) = (0 , . This could also be deduced from the Freiheitsatz ([Mag1]; see [P4] for an expository account). Figure 4:
The oriented tree T ⊂ R used in the proof of Lemma 3.3. We have f ( S ) = S (cid:48)(cid:48) , so to prove that S (cid:48)(cid:48) freely generates F ( S ) it is enough to provethat f is an automorphism of F ( S ).We will do this by writing down an inverse g : F ( S ) → F ( S ) for f . For ( n, m ) ∈ Z ,there is a unique geodesic edge path in T from (0 ,
0) to ( n, m ). Let vertices traversedby this path be (0 ,
0) = ( a , b ) , ( a , b ) , . . . , ( a k , b k ) = ( n, m ) . We then define w n,m = y a k ,b k y a k − ,b k − · · · y a ,b . Having done this, we define the homomorphism g : F ( S ) → F ( S ) via the formula g ( y n,m ) = w n,m . It is clear that f ◦ g = g ◦ f = id, so g is the desired inverse for f .Our final lemma uses the decomposition from Lemma 3.2: Lemma 3.4.
Let φ : G = ∗ ( k ,k ) ∈ Z F x k x k → F be the homomorphism that takes the factor F x k x k to F via conjugation by x − k x − k .Then the kernel of φ is freely generated by (cid:110) [ x , x j ] x k x k w | ≤ j ≤ g , k , k ∈ Z , and w ∈ F (cid:111) ∪ (cid:110) [ x , x j ] x k w | ≤ j ≤ g , k ∈ Z , and w ∈ F (cid:111) . Proof.
Lemma 3.2 implies that G has the free generating set S = (cid:26) x x k x k j | ≤ j ≤ g and k , k ∈ Z (cid:27) . For each x j with 3 ≤ j ≤ g , we can apply Lemma 3.3 to the free group on the subset (cid:26) x x k x k j | k , k ∈ Z (cid:27) ⊂ S
15y identifying y n,m with x x n x m j . From this, we see that G is freely generated by S (cid:48) = { x j | ≤ j ≤ g } ∪ (cid:40)(cid:18) x x k x k j (cid:19) − x x k x k j | ≤ j ≤ g and k , k ∈ Z (cid:41) ∪ (cid:40)(cid:18) x x k j (cid:19) − x x k j | ≤ j ≤ g and k ∈ Z (cid:41) . Since (cid:18) x x k x k j (cid:19) − x x k x k j = x − k x − k − x − j x x j x k x k = [ x , x j ] x k x k , (cid:18) x x k j (cid:19) − x x k j = x − k − x − j x x j x k = [ x , x j ] x k , we can rewrite S (cid:48) as S (cid:48) = { x j | ≤ j ≤ g } ∪ (cid:110) [ x , x j ] x k x k | ≤ j ≤ g , k , k ∈ Z (cid:111) ∪ (cid:110) [ x , x j ] x k | ≤ j ≤ g , k ∈ Z (cid:111) . With respect to this free generating set for G , the homomorphism φ is the evidentretract onto the free group F on { x , . . . , x g } , so ker( φ ) is normally generated by thefollowing subset of S (cid:48) : (cid:110) [ x , x j ] x k x k | ≤ j ≤ g , k , k ∈ Z (cid:111) ∪ (cid:110) [ x , x j ] x k | ≤ j ≤ g , k ∈ Z (cid:111) . Lemma 2.1 shows that the normal closure of the above set is freely generated by (cid:110) [ x , x j ] x k x k w | ≤ j ≤ g , k , k ∈ Z , and w ∈ F (cid:111) ∪ (cid:110) [ x , x j ] x k w | ≤ j ≤ g , k ∈ Z , and w ∈ F (cid:111) , as desired.We can now prove Theorem B. Proof of Theorem B.
Recall that we must prove that the group [ π, π ] is freely generatedby (cid:110) [ x i , x j ] x kii x ki +1 i +1 ··· x k g g | ≤ i < j ≤ g , ( i, j ) (cid:54) = (1 , k i , . . . , k g ∈ Z (cid:111) . π, π ] < G . Let φ : G → F be the homomorphism from Lemma3.4. By construction, ker( φ ) (cid:67) [ π, π ] and φ ([ π, π ]) = [ F, F ]. We thus have a shortexact sequence 1 −→ ker( φ ) −→ [ π, π ] φ −→ [ F, F ] −→ . This exact sequence splits in the obvious way, so[ π, π ] = ker( φ ) (cid:111) [ F, F ] . Lemma 3.4 says that ker( φ ) is freely generated by S = (cid:110) [ x , x j ] x k x k w | ≤ j ≤ g and k , k ∈ Z and w ∈ F (cid:111) ∪ (cid:110) [ x , x j ] x k w | ≤ j ≤ g and k ∈ Z and w ∈ F (cid:111) . The group [
F, F ] acts freely on S by conjugation, and the set S (cid:48) = (cid:110) [ x , x j ] x k x k ··· x k g g | ≤ j ≤ g and k , . . . , k g ∈ Z (cid:111) ∪ (cid:110) [ x , x j ] x k x k ··· x k g g | ≤ j ≤ g and k , . . . , k g ∈ Z (cid:111) contains a single element from each [ F, F ]-orbit. Lemma 2.3 thus implies that[ π, π ] = F ( S (cid:48) ) ∗ [ F, F ] . By Theorem A, the group [
F, F ] is freely generated by S (cid:48)(cid:48) = (cid:110) [ x i , x j ] x kii x ki +1 i +1 ··· x k g g | ≤ i < j ≤ g and k i , . . . , k g ∈ Z (cid:111) . We conclude that [ π, π ] is freely generated by S (cid:48) ∪ S (cid:48)(cid:48) , as desired. We now prove Theorem F.
Proof of Theorem F.
Fix some g ≥
1. Recall that this theorem asserts that there is ashort exact sequence0 −→ Z [ Z g ] −→ [ F g , F g ] ab −→ [ π (Σ g ) , π (Σ g )] ab −→ Z [ Z g ]-modules.Set r = [ x , x ] · · · [ x g − , x g ] ∈ [ F g , F g ] , { r } ∈ [ F g , F g ] ab be the image of r . Let R be the Z [ Z g ]-span of { r } in[ F g , F g ] ab . We have a short exact sequence0 −→ R −→ [ F g , F g ] ab −→ [ π (Σ g ) , π (Σ g )] ab −→ Z [ Z g ]-modules, and to prove the theorem it is enough to prove that R ∼ = Z [ Z g ].Let Σ g be a compact oriented genus- g surface with 1 boundary component β . Fixa basepoint ∗ ∈ β , and identify F g with π (Σ g , ∗ ) in such a way that r is the looparound β . Let ( (cid:101) Σ , (cid:101) ∗ ) be the cover of (Σ g , ∗ ) with π ( (cid:101) Σ , (cid:101) ∗ ) = [ F g , F g ]. We then haveH ( (cid:101) Σ) = [ F g , F g ] ab .By construction, { r } ∈ H ( (cid:101) Σ) is the homology class of the component of ∂ (cid:101) Σ containing (cid:101) ∗ . The submodule R of H ( (cid:101) Σ) is the image of H ( ∂ (cid:101) Σ) in H ( (cid:101) Σ). The deck group Z g acts simply transitively on the set of components of ∂ (cid:101) Σ, so as a Z [ Z g ]-module wehave H ( ∂ (cid:101) Σ) ∼ = Z [ Z g ] . To prove the theorem, we must show that the map H ( ∂ (cid:101) Σ) → H ( (cid:101) Σ) is injective.This map fits into a long exact sequence of relative homology groups that containsthe segment H ( (cid:101) Σ , ∂ (cid:101) Σ) −→ H ( ∂ (cid:101) Σ) −→ H ( (cid:101) Σ) . We must therefore prove that H ( (cid:101) Σ , ∂ (cid:101) Σ) = 0. Collapse each component ∂ of ∂ (cid:101) Σ to apoint P ∂ to form a surface (cid:101) Σ (cid:48) . Let P = (cid:110) P ∂ | ∂ a component of (cid:101) Σ (cid:111) ⊂ (cid:101) Σ (cid:48) . We then have H ( (cid:101) Σ , ∂ (cid:101) Σ) ∼ = H ( (cid:101) Σ (cid:48) , P ) ∼ = H ( (cid:101) Σ (cid:48) ) = 0 , where the final = follows from the fact that (cid:101) Σ (cid:48) is a noncompact connected surface.The theorem follows. We close the paper by proving Theorem D.
Proof of Theorem D.
Fix some g ≥
1. We must prove thatH k ( Z g ; [ π (Σ g ) , π (Σ g )] ab ) ∼ = (cid:40) ( ∧ Z g ) / Z if k = 0 , ∧ k +2 Z g if k ≥ .
18y Theorem F, we have a short exact sequence0 −→ Z [ Z g ] −→ [ F g , F g ] ab −→ [ π (Σ g ) , π (Σ g )] ab −→ Z [ Z g ]-modules. Since Z [ Z g ] is a free Z [ Z g ]-module, we haveH k ( Z g ; Z [ Z g ]) = (cid:40) Z if k = 0 , k ≥ . Also, Theorem D says thatH k ( Z g ; [ F g , F g ] ab ) ∼ = (cid:40) ∧ Z g if k = 0 , ∧ k +2 Z g if k ≥ . The long exact sequence in Z g -homology associated to (3.1) thus immediately impliesthe result we want for k ≥
2. For k = 0 ,
1, this long exact sequence contains thesegment 0 → ∧ Z g → H ( Z g ; [ π (Σ g ) , π (Σ g )] ab ) → Z → ∧ Z g → H ( Z g ; [ π (Σ g ) , π (Σ g )] ab ) → . To prove the theorem, we must therefore prove that the map Z → ∧ Z g in this exactsequence is not the zero map. For 1 ≤ i ≤ g , let x i ∈ Z g be the image of x i ∈ F g in( F g ) ab = Z g . Tracing through all the maps involved, we see that the map Z → ∧ Z g takes the generator of Z to x ∧ x + · · · + x g − ∧ x g ∈ ∧ Z g . This is indeed nonzero, and the theorem follows.
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