DDIVISION IN GROUP RINGS OF SURFACE GROUPS
GRIGORI AVRAMIDI
Abstract.
We prove a division algorithm for group rings of high genus surface groups anduse it to show that some 2-complexes with surface fundamental groups are standard. We alsogive an application of division to cohomological dimension of 2-relator groups acting on H n . introduction The goal of this paper is to study 2-complexes X with a fixed fundamental group Γ up tohomotopy equivalence by means of a division algorithm over the group ring of Γ. These twothings are related through the second homotopy group, which is a Z Γ-module. Most of themathematical content of the paper consists of proving a division algorithm for group rings ofhigh genus surface groups. We find this interesting in its own right, even outside the contextof 2-complexes.
On division.
In the rational group ring of a free group there is a division algorithm analogousto polynomial long divison that was discovered by Moritz Cohn [5]. A division algorithm is aprocess that lets one divide one element x by another non-zero element y with a remainder r whose ‘size’ is smaller than that of y . In the group ring Q F n , the measure of ‘size’ we use is thediameter of the support of the group ring element (defined at the end of this section), whichwe denote by | · | . In symbols, a division algorithm asks for q, r ∈ Q F n such that x = qy + r and | r | < | y | or r = 0. Unlike in the case of polynomial long division, there cannot be a divisionalgorithm for nonabelian free groups that works for arbitrary x and y . In fact, for a genericpair of group ring elements, the diameter of the support of any linear combination will be atleast as large as that of either element, so there is no hope of obtaining a remainder of smallerdiameter. Therefore, in order to have hope there must be linear combinations of x and y ofsmall diameter. What Cohn discovered is that there is a division algorithm as long as x and y satisfy a non-trivial linear relation in the group ring . This condition means there are elements a, b ∈ Q F n , not both zero, such that ax + by = 0. In fact, a geometric picture of this relationis what dictates the process for actually running the algorithm (see section 2).In this paper, we show that the same division algorithm is true when Γ is the fundamentalgroup of a surface of sufficiently high genus. Theorem 1 (Division algorithm for surface groups) . Let Γ be the fundamental group of a closedsurface of genus ≥ e . Suppose x and y (cid:54) = 0 are elements in Q Γ satisfying a nontrivialrelation ax + by = 0 . Then there are q, r ∈ Q Γ such that x = qy + r and | r | < | y | or r = 0 . To take a concrete example, in the free group F on the letters g and h , any linear combination of g − h − ≥
1, so it is not possible to divide g − h − In the commutative case Q F = Q [ t, t − ] any pair of elements x and y satisfy the obvious relation xy − yx = 0,and the algorithm becomes the usual long division for Laurent polynomials. a r X i v : . [ m a t h . G T ] J a n GRIGORI AVRAMIDI
Our method is inspired by Hog-Angeloni’s geometric proof of Cohn’s division algorithm [10]and by Delzant’s proof that groups rings of hyperbolic groups with large infimum displacementhave no zero divisors [7].
Euclid’s algorithm for finding the greatest common divisor.
The process of applyingthe division algorithm repeatedly to a pair of elements, dividing at each stage the divisor fromthe previous stage by the remainder is called Euclid’s algorithm. Starting from the divisionalgorithm in the integers (or in the polynomial ring Q [ t ]) Euclid’s algorithm produces thegreatest common divisor of two integers (or polynomials). The same is true in our case. Corollary 2 (Euclid’s algorithm for surface groups) . Applying the division algorithm repeat-edly, first dividing x by y to obtain a remainder r , then dividing y by r to obtain a remainder r , and so on, eventually produces an element z := r k that divides the previous r k − with noremainder. The element z obtained in this way is a greatest common divisor of x and y . Algebraic application.
Rephrasing things a bit, Euclid’s algorithm implies that the (left)ideal ( x, y ) generated by the pair of elements x, y ∈ Q Γ is always free: If x and y do not satisfyany relation, then they are a free basis for the ideal, and if they do satisfy a relation then theideal is generated by their greatest common divisor z . But, by the theorem of Delzant alludedto earlier, z is not a zero-divisor, which is the same as saying that the ideal z generates is free.A similar argument shows any pair of vectors v, w ∈ Q Γ n generate a free Q Γ-module, and allthe arguments work over any field, in particular over the finite fields F p . Corollary 3.
For any field k , any submodule M of k Γ n generated by a pair of vectors is free. For topological applications, we need this sort of result over Z Γ. Using a ‘local-to-global’method of Bass ([1]) we assemble the Q and F p statements together to prove such a result underthe additional assumption that the quotient Z Γ n /M is torsion-free . This is good enough forus since the topologically meaningful modules associated to a 2-complex satisfy this condition. Corollary 4.
If a submodule M of Z Γ n is generated by a pair of vectors and Z Γ n /M istorsion-free, then M is free. Non-free examples.
To put the division algorithm and its corollaries into context, note thatcorollary 4 is false for the group Z : The ideal ( s − , t −
1) in Z [ Z ] = Z [ s, s − , t, t − ] is notfree since it has the obvious relation ( s − t −
1) = ( t − s −
1) and cannot be generated byone element. More generally, for any non-free group Γ generated by a pair of elements a and b , the ideal ( a − , b −
1) in Z Γ is not free. Remarkably, there is a 2-generator, 2-relator groupthat, by Thurston’s work (4.7 of [15]), arises as the fundamental group of a closed hyperbolic3-manifold obtained by Dehn filling on the figure-eight knot complement (see section 6). So,the division algorithm and its corollaries do not extend to fundamental groups of arbitraryhyperbolic manifolds. We say z is a divisor of x if x = az for some a ∈ Q Γ. It is a greatest common divisor of x and y if z is adivisor of x and y and for any other divisor z (cid:48) of x and y , z (cid:48) divides z . We say ‘a’ here instead of ‘the’ becausegreatest common divisors are only well-defined up to multiplication by a unit in Q Γ. An assumption is needed: the ideal (2 , t −
1) in Z [ Z ] = Z [ t, t − ] is not free even though all ideals in k [ Z ] are. If ( a − , b −
1) is free then Γ has cohomological dimension one, hence is free by Stallings’ theorem ([14]).
IVISION IN GROUP RINGS OF SURFACE GROUPS 3
Group theoretic application.
Our proof of the division algorithm does work word-for-wordfor any group that acts by isometries on hyperbolic space H n with large infimum displacement (this as a quantitative improvement on torsion-freeness). Free groups and high genus surfacegroups are low-dimensional groups that have such actions. As an application of division, weshow that any (cohomologically) higher dimensional group that has such an action requiresmore than two relations to present. In other words Corollary 5.
Suppose Γ is a finitely generated -relator group acting by isometries on H n withinfimum displacement ≥ . Then Γ has cohomological dimension ≤ . ( )It seems clear that the method should work for δ -hyperbolic groups of large infimum dis-placement and carrying out the details of this might make a good Master thesis. Topological application.
Let us now turn to the topological application mentioned at thebeginning of the introduction. An old theorem of Tietze [6] says that two 2-complexes with thesame fundamental group and Euler characteristic become homotopy equivalent after wedgingboth of them with the same suffiiently large number of 2-spheres. A basic question is to deter-mine whether wedging on these extra 2-spheres is really necessary. One of the first examples ofinequivalent 2-complexes with the same fundamental group and Euler characteristic involvesthe trefoil group T = (cid:10) a, b | a = b (cid:11) . Let Y be the presentation 2-complex corresponding tothis standard presentation. Dunwoody constructed another presentation 2-complex X for thetrefoil group whose second homotopy group π X is not free as a Z T -module ([8]). This complexhas two generators and two relations so it has the same Euler characteristic as Y ∨ S , but isnot homotopy equivalent to it ( π ( Y ∨ S ) is free since Y is aspherical). Dunwoody also showedthat the complexes X and Y ∨ S do become homotopy equivalent after wedging on another S , which on the level of π says that π X ⊕ Z T = Z T ⊕ Z T . So, π X is generated by twoelements and is stably free but not free. Corollary 4 implies that this algebraic phenomenondoes not happen for fundamental groups Γ of high genus surfaces.We can also ask whether a similar topological phenomenon to the one discovered by Dun-woody can happen for surface groups Γ = π Σ in place of the trefoil group T . If X is a2-complex with surface fundamental group and minimal Euler characteristic χ ( X ) = χ (Σ),then it is easy to see that X is homotopy equivalent to Σ. The first interesting case when theEuler characteristic is non-minimal is χ ( X ) = χ (Σ) + 1. The main point is to show π X is free.One way is to use a theorem of Louder ([12]) which implies (see section 7) that X becomesstandard after wedging on X ) − ( χ ( X ) − χ (Σ)) different 2-spheres. So, if X hastwo 2-cells then X ∨ S is homotopy equivalent to Σ ∨ S ∨ S . On π , this implies π X isstably free and generated by two elements. If the surface has high enough genus then Corollary4 implies that π X is free, and hence X is homotopy equivalent to Σ ∨ S . In summary Theorem 6.
Suppose X is a -complex with two -cells and surface fundamental group π X = π Σ . If the genus of the surface is ≥ e , then X is homotopy equivalent to Σ or Σ ∨ S . Torsion free 1-relator groups have the stronger property of having aspherical presentation complexes ([4]). The Klein bottle group K = (cid:10) a, b | a b = 1 (cid:11) also has such stably free but not free Z K -modules generatedby a pair of elements, but they have not yet been geometrically realized as π -modules of 2-complexes ([9]). Another way to show π X is free, which also works for the groups in Corollary 5, is given in section 6. GRIGORI AVRAMIDI On -complexes with more -cells. Let us finish this introduction with several remarksabout generalizations to 2-complexes with more than two 2-cells.For the torus group Z not every submodule of a free Z [ Z ]-module is free, but all the stably-free ones are (this is Serre’s conjecture proved by Quillen and Suslin, see [11]), and this is allone needs to show that any 2-complex with Z fundamental group is standard. For the freegroups F m , there is a generalization of Euclid’s algorithm (also due to Cohn) which shows thatany ideal in Q F m (on any finite number of generators) is free. It also works with coefficientsin F p instead of Q so Bass’s theorem implies any stably free Z F n -module is free. This impliesall finite 2-complexes with free fundamental group are standard. (See [10].)On the other hand, the fundamental group of an orientable genus g surface does have a non-free ideal on 2 g generators, namely its augmentation ideal. It is tempting to conjecture thatany ideal on fewer than 2 g generators is free. Let us only remark here that Cohn’s generalizedEuclid’s algorithm also has a surface version, which shows that any ideal on f ( g ) generatorsis free, where f ( g ) is a function such that f ( g ) → ∞ as g → ∞ . The details of this are moreinvolved than the 2-generator case and will (?) be the subject of a future paper. Combiningsuch an algorithm with Bass’s method and Louder’s result would imply that 2-complexes withΣ g fundamental group and ≤ f ( g ) 2-cells are standard. Plan of the paper.
We explain the division algorithm for free groups in Section 2. In Section3 we recall and derive properties of hyperbolic space that will be used in the proof of Theorem1 (the division algorithm for surface groups), which is given in Section 4. We then give a proofof Euclid’s algorithm for surface groups together with Corollaries 3 and 4 in Section 5. Thegroup theoretic application (Corollary 5) and one way to get Theorem 6 is proved in Section 6and the other way is given in Section 7.
Notation and terminology.
Before we start, let us fix some notation that will be usedthroughout the paper and describe how group ring elements can, to a large extent, be thoughtof geometrically.Throughout the paper Γ will denote either a free group or a surface group. The group Γacts by a covering space action on a space Y , which is a tree when Γ is free and the hyperbolicplane H when Γ is a surface group. Pick an orbit of Γ in Y and identify group elements withpoints of that orbit in Y . A group ring element x ∈ Q Γ is a finite formal linear combination x = (cid:80) x γ · γ . Support.
The support of x consists of all the group elements γ with non-zero coefficients x γ appearing in this sum, thought of as points in Y . We will denote the support of an element bythe corresponding capital letter. So, the support of x will be denoted X . Diameter.
The diameter of X is the maximal distance between a pair of points in X . It willbe denoted | x | (or | X | ), and we will also call it the diameter of x . Barycenter.
The barycenter of X is the center of the smallest ball containing X . It will bedenoted (cid:98) x (or (cid:98) X ), and we will simply call it the barycenter of x . Boundary points.
Let B (cid:98) x ( R ) be the smallest ball containing X . We will call points of X thatare a maximal distance R from the barycenter the boundary points of x . IVISION IN GROUP RINGS OF SURFACE GROUPS 5
Acknowledgements.
I would like to thank Tˆam Nguy˜ˆen Phan for suggesting writing section2, Ian Leary and Jean Pierre Mutanguha for pointing out some 3-dimensional 2-relator groups,and the Max Planck Institute for its hospitality and financial support.2.
The division algorithm for free groups
In this section, we will sketch the division algorithm for free groups.
The algorithm.
Suppose we have a pair of group ring elements x and y that are related bya nontrivial linear relation ax + by = 0. The main step in the division algorithm is to showthat if | x | ≥ | y | then we can subtract translates of y from x to obtain an element x = x − c y whose diameter is strictly smaller than that of x . Iterating this step will give division (we willsay a few more words about this iteration at the end of this subsection.)The choice of c is dictated by the relation ax + by = 0 as follows. Let o be the barycenterof the support of ax and R the radius of the smallest ball containing this support. There is an x -translate γx with a γ (cid:54) = 0 that contains a boundary point of ax . We can assume that γ = 1,so that x contains a boundary point. (If γ (cid:54) = 1, multiply the relation on the left with γ − andstart again.) Let us call the points of x that are on the boundary of ax the extremal points of x . Claim 1:
Any boundary point of ax = − by appears in a unique x -translate (that is, γx with a γ (cid:54) = 0) and also in a unique y -translate ( ρy with b ρ (cid:54) = 0.)Therefore, the extremal points of x can all be canceled by y -translates (weighted with ap-propriate coefficients) to obtain an element x = x − (cid:88) c γ γy whose support does not contain any of the extremal points from the support of x . The coefficients are c γ = − b γ /a if γy contains an extremal point of x , and c γ = 0 otherwise. GRIGORI AVRAMIDI
Claim 2: x has smaller diameter than x .If | x | < | y | then this finishes the division algorithm, since we can take x to be the remainder.If not, then we note that x and y are related by the non-trivial relation ax + ( b + ac ) y = 0and repeat the above argument. Each iteration decreases the diameter by at least one, so afterfinitely many steps we arrive at an element x n = x n − − c n y = x − ( c + · · · + c n ) y whosediameter is smaller than y . This is our remainder. Why it works.
The key behind everything is that we are on a tree.Denote the support of x by the corresponding capital letter X . We look at the set S = (cid:91) a γ (cid:54) =0 γX. It contains the support of ax but can be strictly bigger if ax has some cancellation. Let B o (cid:48) ( R (cid:48) )be the smallest ball containing S . We will show that any point in S ∩ S o (cid:48) ( R (cid:48) ) is in the supportof ax . For this, it is enough to show that any p ∈ S ∩ S o (cid:48) ( R (cid:48) ) lies in precisely one X -translate. Proof. If γX touches the boundary at p then, since we are on a tree, the barycenter γ (cid:98) x lies onthe geodesic from o (cid:48) to p and is precisely | X | / p . If there is another translate ρX containing p then γ (cid:98) x = ρ (cid:98) x and hence γ = ρ . So, the translates ρX and γX are the same. (cid:3) It follows from this that the points S ∩ S o (cid:48) ( R (cid:48) ) all appear in the support of ax . Therefore o = o (cid:48) , R = R (cid:48) , what we have called above the ‘boundary points of ax ’ are pIn other words,therecisely the set S ∩ S o ( R ), and every boundary point of ax appears in exactly one x -translate.All the same arguments apply to the expression by . This proves the first claim. Remark.
It also shows that the picture of ax on the previous page is accurate: The minimalball containing the support of ax entirely contains the supports of all the x -translates { γx } a γ (cid:54) =0 .To prove the second claim, one uses similar arguments to show (see figures on the next page)that all the points of x are ≤ | x | / (cid:98) x and are not extremal. Thus, | x | ≤ | x | . In the case of equality there is a diameter realizing segment in x whose midpointis (cid:98) x . But then, at least one of its endpoints is extremal, which is a contradiction. Where do relations come from?
We can work backwards, starting from an element z toproduce pairs of elements satisfying successively more complicated relations: ( z, → ( z, az ) → ( z + baz, az ) → ( z + baz, az + cz + cbaz ) → . . . . What the division algorithm implies is thatany pair s of theatisfying a relation is obtained by this process. IVISION IN GROUP RINGS OF SURFACE GROUPS 7 Tree-like properties of hyperbolic space
Our proof of the division algorithm for surface groups is based on the tree-like propertiesof hyperbolic space. In this section we recall these properties in a convenient form and derivesome specific consequences that will be used in the proof.3.1. δ -hyperbolicity. Everything can be easily obtained from the following basic property. • There is a universal constant δ so that if pq is a segment with midpoint m and o is anypoint in hyperbolic space, then one of the paths omp or omq cannot be shortened bymore than δ . In symbolsmax( d ( o, p ) , d ( o, q )) ≥ d ( o, m ) + 12 d ( p, q ) − δ. Remark.
In a tree we can take δ = 0 and in hyperbolic space we can take δ = 5.It is useful to note that one of the angles ∠ m ( o, p ) or ∠ m ( o, q ) is obtuse ( ≥ π/ • any geodesic segment connecting a sphere S o ( R ) to a larger concentric sphere S o ( R (cid:48) )and not intersecting the interior of B o ( R ) has length between | R (cid:48) − R | and | R (cid:48) − R | + δ . Proof.
Let m be a point on S o ( R ) and q a poin of thet on S o ( R (cid:48) ). The angle ∠ m ( o, q ) isobtuse, so d ( o, q ) ≥ d ( o, m ) + d ( m, q ) − δ . Plugging in d ( o, q ) = R (cid:48) and d ( o, m ) = R gives d ( m, q ) ≤ R (cid:48) − R + δ . The other inequality R (cid:48) − R ≤ d ( m, q ) is clear. (cid:3) GRIGORI AVRAMIDI
Midpoints and barycenters.
A consequence of δ -hyperbolicity is that if pq is a length L segment in an R -ball, then its midpoint m is within R − L/ δ of the center of the ball. Proof.
Let o be the center of the R -ball. Then R ≥ max( d ( o, p ) , d ( o, q )) ≥ d ( o, m )+ L/ − δ . (cid:3) Another consequence is that any set X of diameter D is contained in a ( D/ δ )-ball. Proof.
Let p, q be a pair of points realizing the diameter D and let m be their midpoint. If o isany point in X then D ≥ max( d ( o, p ) , d ( o, q )) ≥ d ( o, m ) + D/ − δ implies d ( o, m ) ≤ D/ δ .In other words, X is contained in the D/ δ ball centered at o . (cid:3) These two properties together imply that • the barycenter of a set is 2 δ -close to the midpoint of any segment realizing the diameter.So we can replace one with the other at the expense of a small error.Next, suppose that X is a set of diameter D , (cid:98) x is its barycenter and o is a point. Then forany diameter realizing segment pq of X with midpoint m we havemax( d ( o, p ) , d ( o, q )) ≥ d ( o, m ) + D − δ (1) ≥ d ( o, (cid:98) x ) + D − δ. (2)For any point p (cid:48) in X we have d ( o, p (cid:48) ) ≤ d ( o, (cid:98) x ) + d ( (cid:98) x, p (cid:48) ) ≤ d ( o, (cid:98) x ) + D/ δ and therefore(3) d ( o, (cid:98) x ) ≥ d ( o, p (cid:48) ) − D − δ, Putting these two inequalities together tells us how far the barycenter (cid:98) x is from a point o interms of the diameter of X and the radius of the smallest ball at o containing X . Lemma 7. If B o ( R ) is the smallest ball centered at o containing X , then R − D − δ ≤ d ( o, (cid:98) x ) ≤ R − D δ. Proof.
Plug d ( o, p (cid:48) ) = R into (3) and max( d ( o, p ) , d ( o, q )) ≤ R into (2). (cid:3) Shrinking the diameter of X . Another application of these two inequalities specifies par-ticular points of X to throw out in order to shrink its diameter. Lemma 8 (Extremal cancellation) . If B o ( R ) is the smallest ball centered at o containing X ,then the diameter of X ∩ B o ( R − δ ) is strictly less than the diameter of X .Proof. If the diameter of X ∩ B o ( R − δ ) is not smaller, one of its diameter realizing segments pq also realizes the diameter of X . Therefore, plugging (3) into (2) and using R = d ( o, p (cid:48) ) givesmax( d ( o, p ) , d ( o, q )) ≥ R − δ , so at least one of the points p or q is outside the R − δ ballcentered at o , which is a contradiction. (cid:3) IVISION IN GROUP RINGS OF SURFACE GROUPS 9
Fellow traveling.
The next treelike feature of hyperbolic space we need is fellow trav-eling . It says that for a pair of points p and q on the boundary of a ball centered at o , thesegments pq and po fellow travel until we reach the midpoint of pq , up to an error 4 δ . Toexpress it precisely, it is useful to parametrise geodesics. For a geodesic segment pq we denoteby pq ( t ) the point obtained by traveling from p to q for a time t along the geodesic. Lemma 9 (Fellow traveling property) . For a pair of points p, q ∈ S o ( R ) and t ≤ d ( p,q )2 we have d ( pq ( t ) , po ( t )) ≤ δ. Proof.
Let m be the midpoint of pq , L = d ( o, m ) and D = d ( m, p ). Let p (cid:48) = po ( D ) be the pointobtained by going for a time D from p to o and p (cid:48)(cid:48) = op ( L ) the point obtained by traveling fora time L from o to p . Finally, let m (cid:48) be the midpoint of the geodesic segment mp (cid:48) . Now, sincethe angle ∠ m ( o, p ) is right, it follows that R ≥ L + D − δ. It is also clear from the picture that d ( p, m (cid:48) ) ≥ d ( p (cid:48)(cid:48) , p ) = R − L, and plugging in the previous inequality gives d ( p, m (cid:48) ) ≥ D − δ. Since the angle ∠ m (cid:48) ( p, p (cid:48) ) is right, if follows that D ≥ d ( p, m (cid:48) ) + d ( m (cid:48) , p (cid:48) ) − δ. Therefore d ( m (cid:48) , p (cid:48) ) ≤ D + δ − d ( p, m (cid:48) ) ≤ δ. Since m (cid:48) is the midpoint of mp (cid:48) , it follows that d ( m, p (cid:48) ) ≤ δ . This proves the lemma for t = d ( p, q ) /
2. The lemma for smaller values of t follows from convexity. (cid:3) Large infimum displacement implies no zero divisors in the group ring.
Next, wegive a key application of fellow traveling. It was observed by Delzant in [7] and is the mainstep in his proof that group rings of some hyperbolic groups have no zero divisors. Lemma 10 (Delzant) . Suppose γ is an isometry of H n . If X and γX are contained in a ball B o ( R ) and their intersection contains a point p in the µ -neighborhood of the boundary of theball, then the midpoint m of the segment from p to γp is moved ≤ µ + 9 δ by γ − .Proof. Let L be the length of the segment from p to γp . Let q be the point obtained by goingfrom γp to p for a distance µ + δ . Then q ∈ B o ( R − µ ). Let m (cid:48) be the midpoint of the segmentfrom p to q . Then po fellow travels with pq for a time t = L − ( µ + δ )2 until it reaches m (cid:48) so, if wedenote by p (cid:48) = po ( t ) the point reached by traveling from p to o for a time t , then d ( m, p (cid:48) ) ≤ d ( m, m (cid:48) ) + d ( m (cid:48) , p (cid:48) ) ≤ µ + δ δ. The same argument applied to the segment from p to γ − p shows that its midpoint γ − m satisfies d ( γ − m, p (cid:48) ) ≤ µ + δ + 4 δ . Therefore d ( m, γ − m ) ≤ µ + 9 δ . (cid:3) In other words, if the infimum displacement of Γ acting on H n is > µ + 9 δ , then Γ-translatesof X that lie in a ball do not intersect in the µ -neighborhood of the boundary of that ball.Now we apply this to products in the group ring. The following corollary will be usedrepeatedly in the next section. It implies that any cancellation in a product ax happens awayfrom the boundary of ax , as long as the infimum displacement is sufficiently large. Corollary 11.
Suppose Γ has infimum displacement > µ + 9 δ . Let a and x be non-zerogroup ring elements. Then, the smallest ball containing ax also contains all the x -translates { γx } a γ (cid:54) =0 . Moreover, every point in the µ -neighborhood of the boundary of this ball is containedin at most one such x -translate.Proof. The support of the product ax is contained in the set S = (cid:91) a γ (cid:54) =0 γX. To be more precise, the µ = 0 case in the setting of δ -hypebolic groups. IVISION IN GROUP RINGS OF SURFACE GROUPS 11
Let B o ( R ) be the smallest ball containing S . Delzant’s lemma implies that on the µ -neighborhoodof the boundary of this ball the X translates { γX } a γ (cid:54) =0 do not intersect. Therefore, B o ( R ) isthe smallest ball containing the support of ax . The rest is clear. (cid:3) In particular, this says that once the infimum displacement is > δ the support of ax isnon-empty, so ax (cid:54) = 0. Corollary 12 (Delzant) . If Γ has infimum displacement > δ then Q Γ has no zero divisors. Approximating barycenters.
The following lemma is useful.
Lemma 13.
Suppose X is a set with diameter D and barycenter (cid:98) x contained in a ball B o ( R ) and q ∈ X is a point in the µ -neighborhood of the boundary of the ball. Let qo ( D/ be thepoint obtained by traveling for time D/ along the geodesic from q to o . Then d ( qo ( D/ , (cid:98) x ) ≤ δ + 32 µ. Proof.
We can assume that d ( q, o ) = R − µ . First, note that d ( (cid:98) x, o ) ≤ R − D + 3 δ implies thatthe distance from (cid:98) x to S o ( R − µ ) is at least s = D − δ − µ . Therefore, the segment q (cid:98) x can beextended by s before it reaches S o ( R − µ ), and hence the segment q (cid:98) x fellow travels with qo fora time t = d ( q, (cid:98) x )+ s . Also note that the distance from (cid:98) x to q is controlled by s ≤ d ( (cid:98) x, q ) ≤ D δ. Therefore | d ( q, (cid:98) x ) − t | = d ( q, (cid:98) x ) − s ≤ δ + µ s ≤ t ≤ D − δ − µ | t − D/ | ≤ δ + µ. Thus, the distance from (cid:98) x to qo ( D/
2) is bounded by d ( (cid:98) x, qo ( D/ ≤ | d ( (cid:98) x, q ) − t | + 4 δ + | t − D/ |≤ δ + 32 µ. (cid:3) An immediate consequence is the following.
Corollary 14. If X and Y are both sets in B o ( R ) containing a point q in the µ -neighborhoodof the boundary, then the barycenters of X and Y are || X |−| Y || -apart, up to an error δ + 3 µ . A more significant consequence for us is the following.
Corollary 15.
Suppose
X, γX, and Y are contained in a ball B o ( R ) , the intersection of Y and X contains q , and the intersection of Y and γX contains q (cid:48) , where q and q (cid:48) are in the µ -neighborhood of the boundary of the ball. If | X | ≥ | Y | then d ( (cid:98) x, γ (cid:98) x ) ≤ δ + 6 µ. Proof.
Look at the function f ( t ) = d ( qo ( t ) , q (cid:48) o ( t )). It is ≤ δ + 3 µ at t = | Y | / t goes from | Y | / | X | / | X | ≥ | Y | ).Finally, at t = | X | / d ( (cid:98) x, γ (cid:98) x ) by an error of at most 18 δ + 3 µ , again by Lemma13. This establishes the corollary. (cid:3) We will use these two corollaries in the proof of the division algorithm for surface groups inthe next section.4.
Proof of the division algorithm for surface groups
Setup.
Throughout the proof, we will keep track of how large the infimum displacementhas to be for the argument to work at that stage. The infimum displacement needed to makeeverything work is stated at the end, where we also translate it into a condition about thegenus of the surface. To start, we assume the infimum displacement is > δ so that there areno zero divisors.We are given a non-trivial relation ax + by = 0 where a, b, x and y (cid:54) = 0 are elements ofthe group ring Q Γ, and we want to show that there are q, r ∈ Q Γ such that x = qy + r and | r | < | y | or r = 0. If | x | < | y | then there is nothing to do, since we can take q = 0 , r = x .If | x | ≥ | y | , then it is enough to subtract a multiple b (cid:48) y of y from x for which the resultingelement x (cid:48) = x − b (cid:48) y has smaller diameter than x . Since the set of possible diameters is discreteand the elements x (cid:48) and y satisfy the non-trivial relation a ( x − b (cid:48) y ) + ( b + ab (cid:48) ) y = 0, iteratingthe process finitely many times will prove the division algorithm. If this relation were trivial then a = 0 and the original relation would imply that y is a zero-divisor. IVISION IN GROUP RINGS OF SURFACE GROUPS 13
Next, we will describe which points of x we will try to cancel out with translates of y inorder to reduce the diameter of the resulting group ring element x (cid:48) . This will be dictated bythe relation ax + by = 0. Let o be the barycenter of the support of ax and R the radius of ax .As long as the infimum displacement is > δ , by Corollary 11, all the x -translates { γx } a γ (cid:54) =0 are contained in the ball B o ( R ). Pick such an x -translate γx containing a boundary point of ax . After multiplying our relation on the left by ( a γ γ ) − , we can assume that this translate is x , i.e. that x contains a boundary point of ax and that a = 1.Let us call the points of x that are in the α -neighborhood of the boundary of ax the α -extremal points of x . We have shown in Lemma 8 that if we throw out the 5 δ -extremal pointsfrom the support of x , then the resulting set has strictly smaller diameter. So these are thepoints we will try cancel out. To that end, note that if the infimum displacement is > δ + 9 δ then, by Corollary 11, all these 5 δ -extremal points are not contained in any other x -translate γx with a γ (cid:54) = 0. The relation ax = − by implies that each one of them must be contained in a y -translate γy with b γ (cid:54) = 0, and Corollary 11 applied to by implies that there is a unique such y translate.Therefore, as long as the infimum displacement of Γ is sufficiently large ( > δ ), the 5 δ -extremal points of x can all be cancelled by y -translates (weighted with appropriate coeffi-cients) . Call the resulting element x (cid:48) = x − (cid:88) c γ γy. Our goal in the rest of the proof is to show the diameter of x (cid:48) is less than the diameter of x .4.2. Showing | x (cid:48) | < | x | . We will warm up by showing that if the diameter of x (cid:48) is greaterthan that of x , it cannot be much greater. We can estimate the distance from a y -point p of x (cid:48) to (cid:98) x using the estimate on the distance between barycenters given in Corollary 14: d ( (cid:98) x, p ) ≤ d ( (cid:98) x, γ (cid:98) y ) + d ( γ (cid:98) y, p ) ≤ (cid:18) | x | − | y | δ + 3 · δ (cid:19) + (cid:18) | y | δ (cid:19) = | x | δ. So, if | x (cid:48) | ≥ | x | it follows that a diameter realizing segment of x (cid:48) has midpoint 35 δ -close to (cid:98) x .Note for future use that this implies one of the endpoints of this segment is 37 δ -extremal.Let us look more at the extra y -points p of x (cid:48) that have been introduced by subtracting the y -translates { γy } c γ (cid:54) =0 from x . Fix a constant µ ≥ δ. First, we will show that if the infimumdisplacement is sufficiently large ( > δ + 6 µ ), then such a y -point p cannot be µ -extremal.If it was, then it would have to cancel with a unique x -translate ρx that is different from x . But then the barycenters (cid:98) x and ρ (cid:98) x would be too close! More precisely, we would have d ( (cid:98) x, ρ (cid:98) x ) ≤ δ + 6 µ by Corollary 15, which contradicts the infimum displacement assumption.Therefore, if | x (cid:48) | ≥ | x | then x (cid:48) has a 37 δ -extremal x -point. Call this point q (cid:48) . Remark.
In the case of free groups acting on trees, δ = 0 and above we can take µ = 0 sothat at this stage in the argument we have an element x (cid:48) with | x (cid:48) | ≤ | x | and if | x (cid:48) | = | x | then To be more precise, c γ = − b γ if γy contains a 5 δ -extremal point of x and c γ = 0 otherwise. If p canceled with x then it would not have appeared in x (cid:48) . x (cid:48) has an extremal point, which is not a y -point, hence must be an x point. But we assumedthat all the extremal x -points have been canceled out, so we arrive at a contradiction. In thesurface group case we have to work harder. The reason is because we have found a 37 δ -extremal x -point q (cid:48) in x (cid:48) , while only the 5 δ -extremal x -points have been canceled out.Now, for large enough infimum displacement ( > δ + 9 δ ) the point q (cid:48) appears in a unique y translate ρy that is different from all the y -translates { γy } c γ (cid:54) =0 that we subtracted from x toget x (cid:48) . By Corollary 14 we have d ( (cid:98) x, ρ (cid:98) y ) ≤ | x | − | y | δ + 3 · δ. The rest of the argument breaks up into two cases, depending on the size of | x | − | y | . First, we deal with the case is | x | − | y | ≤ µ . In this case, the barycenter of ρy and of allthe y -translates { γy } c y (cid:54) =0 are (cid:0) µ + 129 δ (cid:1) -close to (cid:98) x . Therefore, if the infimum displacement islarge enough ( > µ + 258 δ ) we must have ρ (cid:98) y = γ (cid:98) y and hence ρ = γ , which is a contradiction. Finally we deal with the case | x | − | y | ≥ µ . We will show that in this case the y -points of x (cid:48) are < | x | / − δ away from (cid:98) x . This will imply that the diameter of x (cid:48) is less than | x | and wewill be done.Let p be a y -point of x (cid:48) . Thus, there is a y -translate γy and a 5 δ -extremal point q of x sothat both p and q are in γy . Denote by L the length of the segment pq and m its midpoint.Since p is not µ -extremal, the segment qp can be extended by µ − δ before it reaches the5 δ -neighborhood of the boundary. Let m (cid:48) be the midpoint of this extended segment. It fellowtravels with the segment qo for a distance t = L + µ − δ . Let y = qo ( t ) be the point obtained bytraveling from q to o for a time t . Also, let x = qo ( | x | /
2) be the point obtained by travelingfrom q to o for a time | x | /
2. This is illustrated in the figure below.Note that | x | − t = | x |− L − ( µ − δ )2 is non-negative because of the case we are in, so we cancompute d ( (cid:98) x, p ) ≤ d ( (cid:98) x, x ) + d ( x , y ) + d ( y , m (cid:48) ) + d ( m (cid:48) , p ) ≤ (cid:18) δ + 32 · δ (cid:19) + (cid:12)(cid:12)(cid:12)(cid:12) | x | − t (cid:12)(cid:12)(cid:12)(cid:12) + 4 δ + L − ( µ − δ )2= | x | . δ − µ. IVISION IN GROUP RINGS OF SURFACE GROUPS 15
So, for this part of the argument to work, any µ > . δ will do.In summary, everything works for µ = 37 δ , and in that case the biggest displacementcondition we need is that the infimum displacement is µ + 258 δ = 295 δ . Since we are inhyperbolic space, we can take δ = 5. From infimum displacement to genus.
Buser showed in [2] that every surface of genus ≥ ≥ (cid:112) log( g ).For this metric and g ≥ e , we get infimum displacement ≥ > ·
5, which is goodenough. This finishes the proof.
Remark.
Buser and Sarnak show in [3] that there is a sequence of hyperbolic surfaces Σ g i with g i → ∞ for which one has a much better bound, namely infimum displacement ≥ log g i andthat every genus g surface has a hyperbolic metric with infimum displacement c log g where c is some small (unspecified) positive constant that doesn’t depend on the genus, but neither ofthese can be directly applied to get an explicit bound on how high the genus g has to be. Remark.
This last step is the only place in the proof where we use the fact that Γ is a surfacegroup. Everything else works word-for-word (with the same constants) for groups Γ actingisometrically on hyperbolic n -space H n with infimum displacement ≥ · A remark about fields.
Everything in the paper so far works with coefficients Q replacedby any field k , in particular by the finite fields F p . This will be used in the next section.5. Euclid’s algorithm and algebraic applications
Proof of Euclid’s algorithm.
We are given a pair of elements x, y ∈ Q Γ satisfying anon-trivial relation ax + by = 0. Dividing x by y we get q and r such that x = q y + r and | r | < | y | or r = 0. If r (cid:54) = 0 then the elements y and r satisfy the non-trivial relation ar + ( b + aq ) y = 0. So we can divide y by r to get q and r such that y = q r + r and | r | < | r | or r = 0, and so on. We iterate this process. Since at each step the diameter ofthe remainder decreases, the process stops after finitely many steps with an r k that divides r k − without remainder. All the pairs produced in this way generate the same ideal ( x, y ) =( y, r ) = ( r , r ) = · · · = ( r k − , r k ) = ( r k , z = r k is a greatest common divisor of x and y . Theelement z is a divisor of x and y since x, y ∈ ( z ). Suppose z (cid:48) is another divisor such that x = cz (cid:48) , y = c (cid:48) z (cid:48) . Since z ∈ ( x, y ) we can express it as Q Γ-linear combination z = a (cid:48) x + b (cid:48) y =( a (cid:48) c + b (cid:48) c (cid:48) ) z (cid:48) , so z (cid:48) is a divisor of z . Therefore, z is a greatest common divisor of x and y .5.2. Modules generated by pairs of vectors v, w in Q Γ n and F p Γ n . Delzant’s result that Q Γ has no zero-divisors implies that the submodule of Q Γ n generated by a single non-zerovector v = ( v , . . . , v n ) is free. Our division algorithm implies the analogous result for a pairof vectors. The proof is very similar to that of Euclid’s algorithm. Since we will need both the Q and F p versions in the next subsection, we state it for a general field k . Corollary 3.
Let k be a field. Any submodule M of k Γ n generated by a pair of non-zerovectors v, w is free. If there are no zero divisors, then the map Q Γ → Q Γ n , a (cid:55)→ av is an isomorphism onto its image, which isthe module generated by v . Proof.
We may assume that v (cid:54) = 0 and that | v | ≥ | w | . If for any relation av + bw = 0 both a and b are zero, then M is free of rank two. So, suppose there is such a relation with either a or b non-zero. We will show that this implies M is free of rank one.There are two cases to consider, depending on whether or not w is zero. Case 1: w = 0 . Looking at the first coordinate of the relation, we get av = 0 and since v (cid:54) = 0 we must have a = 0. Thus bw = 0. Since the relation was non-trivial, b (cid:54) = 0 so we musthave w = 0. But then M is generated by a single vector v , and hence it is free of rank one. Case 2: w (cid:54) = 0 . Then the relation av + bw = 0 implies both a and b are non-zero. We usethis relation to divide v by w and get v (cid:48) = v − qw satisfying | v (cid:48) | < | w | or v (cid:48) = 0. Then thevectors v (cid:48) , w still generate M and either v (cid:48) = 0 or the sum of diameters of their first entries | v (cid:48) | + | w | is strictly smaller than | v | + | w | . Moreover, av (cid:48) +( b − aq ) w = 0 is again a non-trivialrelation (with a (cid:54) = 0).At this point, we have arrived back at the situation of the two cases, with v (cid:48) in place of w .Moreover, if v (cid:48) (cid:54) = 0 then the sum of diameters of the first entries of generators | v (cid:48) | + | w | isstrictly smaller than | v | + | w | . Therefore, after iterating this process finitely many times itwill stop and we will arrive in the case 1 situation with M a free module generated by a singlevector. (cid:3) Bass’s ‘local-to-global’ method for Z Γ -modules. Obviously any zero divisor in Z Γis a zero-divisor in Q Γ, so submodules of Z Γ n generated by a single vector are free. Theanalogous statement for modules generated by a pair of vectors is not true. For example, theideal (2 , t −
1) in the group ring Z [ Z ] = Z [ t, t − ] is not free even though all ideals in k [ Z ]are. A general ‘local-to-global’ theorem of Bass ([1]) shows that this sort of thing doesn’thappen when the module splits off as a direct summand of Z Γ n (in other words, if the moduleis projective ). This is good enough for the proof of Theorem 6 given in section 7. The proofgiven below is Bass’s argument, specialized to our situation. Corollary 16.
If a submodule of Z Γ n generated by a pair of non-zero vectors v, w splits off asa direct summand, then it is free.Proof. If v and w do not satisfy any non-trivial relation, then ( v, w ) is a free Z Γ-module. Ifthey satisfy a non-trivial relation, then they generate the same Q Γ module as their greatestcommon divisor z ∈ Q Γ n . We rescale z (multiplying by a rational number if necessary) so that z ∈ ( v, w ) and z / ∈ ( kv, kw ) for any integer k >
1. Since z is a Q Γ-divisor of v and w , there isa positive integer m such that mv = az, mw = bz for some a, b ∈ Z Γ. Pick the smallest such m . In summary we have sandwiched the module generated by z in the following way:( mv, mw ) ⊂ ( z ) ⊂ ( v, w ) . Our goal is to show that m = 1. Suppose it is not, and let p be a prime dividing m . For any Z Γ-module M , denote the mod p reduction by M p := M/pM . Note that the composition ofinduced maps(4) ( mv, mw ) p → ( z ) p → ( v, w ) p is zero because p divides m . The key is to show the second map is injective. It is a ‘local-to-global’ theorem because it assembles Q and F p results to get a Z result. IVISION IN GROUP RINGS OF SURFACE GROUPS 17
Claim: The map i : ( z ) p → ( v, w ) p is injective. This is where we use the assumption that( v, w ) is a direct summand of Z Γ n . It implies that the inclusion ( v, w ) (cid:44) → Z Γ n induces aninclusion of mod p reductions ( v, w ) p (cid:44) → F p Γ n . By Corollary 3, ( v, w ) p is a free F p Γ-module.So, the image of i is a submodule of a free module and generated by one element, so it is freeby Corollary 3. Therefore, i is either injective or the zero map. The later happens precisely if z ∈ ( pv, pw ), but this is ruled out by our choice of z . So, the map i is injective.Since the composition (4) is zero, this implies the first map ( mv, mw ) p → ( z ) p is the zeromap, which is the same as saying ( mv, mw ) ⊂ ( pz ). But then z is a Z Γ-divisor of both mp v and mp w , which contradicts the minimality of m . So we are done. (cid:3) Proof of Corollary 4.
Note that all we used in the proof above is that the inclusion
M (cid:44) → Z Γ n induces an inclusion on mod p reductions. For this we don’t really need M to be adirect summand. Knowing that the quotient Q = Z Γ n /M is torsion-free as an abelian group isgood enough: Suppose v ∈ M and v = pw for some w ∈ Z Γ n . In the quotient Q we have v = 0and since Q is torsion-free also w = 0. But that means w ∈ M and therefore v ∈ pM . So,we’ve shown that M ∩ p Z Γ n = pM , which is the same as saying that M p → F p Γ n is injective.So, the proof of Corollary 16 also applies to such M . This proves Corollary 4.6. 2 -relator groups acting on hyperbolic space Until now, we have primarily focused on surface groups in this paper. They are two-dimensional groups acting on hyperbolic space, and requiring a single relation to present.Moreover, passing to finite index subgroups we get surface groups again but now of highergenus and (if we pick the subgroup correctly) with large infimum displacement. In higherdimensions n ≥ n, H n ( ). It well known that—in contrast to surfacegroups—these higher dimensional lattices require more than one relation. In fact, they requiremore than two. The reason is that these uniform lattices have a non-zero cohomology class indimension n ≥
3, while we will show next that 2-relator groups acting with sufficiently largedisplacement on H n are cohomologically 2-dimensional.Shifting attention for a moment to these 2-relator groups Γ and 2-complexes X presentingthem as such, the freeness of π X comes out in the wash. A new wrinkle is that we do notknow whether such 2-relator groups have aspherical presentation 2-complexes Y . For any thatdo (in particular, for the high genus surface groups) it is clear what a standard Y ∨ S ∨ · · · ∨ S ) and we get a versionof Theorem 6 from the introduction. In summary, we have Corollary 17.
Suppose X is a finite -complex with two -cells and fundamental group Γ . If Γ acts isometrically on H n with infimum displacement ≥ , then • the cohomological dimension of Γ is ≤ , • π X is free, and • if Γ has an aspherical presentation -complex Y then X is standard. For example, the group of invertible 4 × x + x + x − x and congruent to the identity matrix modulo N for large enough N acts on H = SO(3 , / SO(3) in this way. Homotopy equivalent to Y or Y ∨ S or Y ∨ S ∨ S . The third case happens only if Γ is a free. Proof.
Look at the chain complex on the universal cover: π ( X ) → C ( (cid:101) X ) → C ( (cid:101) X ) → C ( (cid:101) X ) → Z . The image of the second map is called the relation module R . It is generated by two elements,a submodule of a free module, and the quotient C /R is again a submodule of a free module.Therefore, by the remark at the end of the previous section, we can apply Bass’s method toconclude R is a free Z Γ-module. Since R is also the kernel of the third map, we get a freeresolution R → C → C → Z of length 2. This is the same as saying that the cohomologicaldimension of Γ is ≤
2, so we have proved the first bullet.Since R is free, C splits as a direct sum π ( X ) ⊕ R . So, π ( X ) is a stably free Z Γ-modulegenerated by two elements. Corollary 4 implies it is free. This proves the second bullet.Finally, suppose there is an aspherical presentation 2-complex Y . Start by building an arbi-trary π -isomorphism Y → X . Since π X is free, we can extend it to a homotopy equivalencefrom a standard complex Y or Y ∨ S or Y ∨ S ∨ S by mapping the 2-spheres to a basis for π X . In the third case π X = Z Γ , so the relation module R vanishes, so Γ has cohomologicaldimension one and hence, by Stallings’ theorem ([14]), is a free group. (cid:3) But, if Γ does not have an aspherical presentation 2-complex, then it is conceivable that thereis a pair of 2-complexes X and X (cid:48) that have the same π but are not homotopy equivalent. Flat and hyperbolic -dimensional -relator groups. Torsion-free 1-relator groups haveaspherical presentation 2-complexes ([4]), so they are at most 2-dimensional. This is no longertrue for 2-relator groups.The simplest 3-dimensional example of a 2-relator group was pointed out to me by IanLeary. It is the fundamental group of the mapping torus of the matrix (cid:18) − (cid:19) acting on T . Note that this is a closed, flat (cid:10) a, b | [ a, b ] = 1 , tat − = b, tbt − = a − (cid:11) . One caneliminate the generator b to get a 2-generator, 2-relator presentation.There are also hyperbolic 3-manifold examples that were explained to me by Jean PierreMutanguha. The mapping torus of the matrix (cid:18) (cid:19) acting on the punctured torus is ahyperbolic 3-manifold with a single cusp . Its presentation is (cid:10) a, b, t | tat − = a b, tbt − = ab (cid:11) and since the second relation says a = [ t, b ] one can elliminate a together with this relationto get a 1-relator presentation. One can close off the cusp by gluing in a solid torus, and forall but finitely many choices of gluing parameters (a pair of relatively prime numbers ( p, q ))one gets a closed hyperbolic 3-manifold (see 4.7 in [15]). On the level of fundamental groups,the gluing introduces a new relation of the form t p = [ a, b ] q . So, one ends up with a closedhyperbolic 3-manifold whose fundamental group has a 2-generator 2-relator presentation (cid:10) b, t | t [ t, b ] t − = [ t, b ] b, t p = [[ t, b ] , b ] q (cid:11) . The manifold is flat since it is obtained by gluing the ends of T × [0 ,
1] by an isometry. This manifold is homeomorphic to the figure-eight knot complement (see p. 177 of [16]).
IVISION IN GROUP RINGS OF SURFACE GROUPS 19 An improved Tietze’s theorem for surface fundamental groups
An old theorem of Tietze [6] says that two 2-complexes with the same fundamental groupbecome homotopy equivalent after wedging both of them with enough 2-spheres. This sectionis about improvements on this theorem when the fundamental group is that of a closed surfaceΣ. The main point is to interpret a Nielsen equivalence result of Louder in this light.
Minimal Euler characteristic.
First note that if X is a 2-complex with fundamental group π Σ and minimal Euler characteristic χ ( X ) = χ (Σ), then X is homotopy equivalent to Σ. Proof.
The complexes become homotopy equivalent after wedging both with the same largenumber of 2-spheres n . Since Σ is aspherical, on π this homotopy equivalence gives π S ⊕ Z Γ n ∼ = Z Γ n . So (see e.g. [13]) S is also aspherical, and hence homotopy equivalent to Σ. (cid:3) Nielsen equivalence for surface groups.
The orientable surfaces have presentations (cid:104) x , y , . . . , x g , y g | [ x , y ] · · · [ x g , y g ] = 1 (cid:105) while the nonorientable ones have presentations (cid:10) x , . . . , x r | x · · · x r = 1 (cid:11) . A standard generating set is one of these, possibly with some extra generators z , . . . , z k satis-fying the trivial relations z = 1 , . . . , z k = 1 thrown in at the end.Now, let X be a finite presentation 2-complex with n generators e , . . . , e n for the surfacegroup, and fix a π -isomorphism f : X → Σ. In [12], Louder showed • There is a free group automorphism ϕ : F n → F n so that f ◦ ϕ ( e ) , . . . , f ◦ ϕ ( e n ) is astandard generating set for π Σ. Interpretation as quantitative variant of Tietze’s theorem for surface groups.
Forconcreteness, suppose it is one representing a genus g orientable surface with k trivial generatorsat the end (the argument in the non-orientable case is similar). Form a new complex Y = X ∪ D ∪ D ∪ · · · ∪ D k by attaching k + 1 different 2-cells to X . The disk D is attached along the commutator[ ϕ ( e ) , ϕ ( e )] · · · [ ϕ ( e g − ) , ϕ ( e g )] and the other disks D i are attached along ϕ ( e g + i ). Byconstruction, these attaching maps are nullhomotopic in π X , so Y is homotopy equivalent X ∨ S ∨ · · · ∨ S . On the other hand, the map f extends to Y and its restriction to the union S = D ∪ · · · ∪ D k is a π -isomorphism. Since f : S → Σ is a π -isomorphism that extendsto the 2-cells of X , the attaching maps of the 2-cells of X are null-homotopic in S , and weconclude that Y is also homotopy equivalent to S ∨ · · · ∨ S ∨ S . Finally, since the 2-complex S has the minimal possible Euler characteristic χ ( S ) = χ (Σ) among 2-complexes with thisfundamental group, the map f : S → Σ is a homotopy equivalence. In summary, X becomesstandard after wedging on k + 1 different 2-spheres: X ∨ ( k + 1) S ∼ Σ ∨ ( X ) S . Second proof of Theorem 6.
The situation that our division algorithm can say somethingabout is when X has one vertex, two 2-cells and Euler characteristic χ ( X ) = χ (Σ) + 1. In thiscase, it is easy to see that k = 0 and the above homotopy equivalence becomes X ∨ S ∼ Σ ∨ S ∨ S . On π this says π X ⊕ Z Γ ∼ = Z Γ . Therefore, by Corollary 4, π X is free. From here we canproceed as in the proof of the third bullet of Corollary 17 to prove Theorem 6. References [1] H. Bass,
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