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SOME REMARKS ON SUBGROUPS OF HYPERBOLIC GROUPS
S. M. Gersten
Abstract. If G is a hyperbolic group, where G = H
, H is finitely presented,and is an automorphism of H, then H satisfies a polynomial isoperimetric in
equality. Necessary and sufficient conditions of homological character are given for
a finitely presented subgroup H of a hyperbolic group to be hyperbolic (resp. aquasiconvex subgroup). If Y is a connected subcomplex of the finite connected 2
complex X, where X/Y is of strictly negative curvature (in the sense of the weight
test), then 1(Y ) is hyperbolic iff 1(X) is hyperbolic, and in this situation the pair(1(X), 1(Y )) is relatively hyperbolic in the sense of Farb. A relation between these
results and the Whitehead asphericity question is discussed.
1. Introduction. There is an extensive literature on hyperbolic groups, beginning with Gromovs fundamental paper [Gr1] and its exegeses [GH] [CDP] [Berk][Bow]. However, remarkably little is known about subgroups of general hyperbolicgroups; for example, it is unknown how distorted the word metrics of finitely generated subgroups can be and how distorted the areas of finitely presented subgroups[Ge1] can be.1
Here is a brief sketch of what is known about such subgroups. A finitely generated subgroup of a hyperbolic group G has a solvable word problem (since Gitself has a solvable word problem) and a subgroup of G has finite rational cohomological dimension (since G itself has finite rational cohomological dimension).From the action of G on the Gromov boundary one knows that solvable (and moregenerally amenable) subgroups of G are virtually cyclic, and if a subgroup is notvirtually solvable then it contains a nonabelian free subgroup [Gr]. FurthermoreRips showed there can exist finitely generated subgroups of hyperbolic groups whichare not finitely presentable [Ri] (i.e. in general hyperbolic groups are not coherent).N. Brady gave one example of a hyperbolic group G = H
where H is finitely
presented but not of type FP3 [Bra1]; since hyperbolic groups are of type FP, His not hyperbolic. In this example, G is of cohomological dimension 3; this contrastswith the result of [Ge3] that a finitely presented subgroup of a hyperbolic group
cS. M. Gersten 1999, all rights reserved1991 Mathematics Subject Classification. 20F05, 20F32, 57M07.
Key words and phrases. finitely presented group, word metric, distortion, isoperimetric function, hyperbolic group.
Partially supported by NSF grant DMS98001581An open test question is whether a finitely generated subgroup of a hyperbolic group can be
distorted more than exponentially in its word metric. Is even exponential distortion possible in
area for a finitely presented subgroup?
1

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of cohomological dimension 2 is hyperbolic. Finally, there is a universal bound onthe order of a finite subgroup of a hyperbolic group G that depends only on thenumber of generators and the Rips constant for that set of generators [Bra2].
This paper is concerned with several results related by a common theme ofisoperimetric inequalities concerning subgroups of hyperbolic groups.
Theorem A. If G = H
is hyperbolic, where H is finitely presented and isan automorphism of H, then H satisfies a polynomial isoperimetric inequality.
In particular, in Bradys example of the preceding paragraph, H satisfies a polynomial isoperimetric inequality. This answers a question raised in [Bra1], and improves the result of [Ge1], where I showed that H had an exponential isoperimetricfunction.
In order to state the next result we need some notation. Let H be a finitelypresented subgroup of a hyperbolic group G, and let X be a complex of typeK(G, 1) with a finite 3skeleton such thatX contains a finite connected subcomplexY of dimension 2 with fundamental group H.2 Let X be the universal cover ofX , and let Y be a connected component of the pullback of Y to X; Y is then acopy of the universal cover of Y . Let Z = X/Y , so Z is a cell complex acyclic indimensions at most 2, as follows from the homology exact sequence.
Theorem B2. In the situation above H is hyperbolic iff the filling norm onB2(Z,
) induced from the boundary map d3 : C3(Z,
) B2(Z,
) is equivalent to
the `1norm induced from the inclusion B2(Z,) = Z2(Z,
) C2(Z,
).
Suppose now that H is a finitely generated subgroup of the hyperbolic group G.We choose a finite set T of generators for G containing a subset S of generators forH and choose a finite presentation P for G with T as generators. Let X be the2complex canonically associated to P and let X be the universal cover of X . LetY be copy of the Cayley graph of H for generators S contained in X; precisely, Yis a connected component of the subgraph of X generated by all edges with labelin S. We choose the base point for X in Y and let Z = X/Y .
Theorem B1. In the situation above H is a quasiconvex subgroup of G iff thefilling norm on B1(Z,
) induced from the boundary map d2 : C2(Z,
) B1(Z,
)
is equivalent to the `1norm induced from the inclusion B1(Z,) = Z1(Z,
)
C1(Z,).
Next let Y be a nonempty connected subcomplex of the finite connected 2complex X , and let Z = X /Y . Let H = 1(Y
) and let G = 1(X). Then G
may be regarded as obtained from H be adding generators and relators modeledon the cells of Z .3
2Such a complex is easily constructed by starting with a finite presentation for H, adding
finitely many generators and relations to get a finite presentation for G, and then adding cellsin dimension 3 and higher to kill off the higher homotopy groups. Since G is of type F, only
finitely many 3 cells are needed to kill 2.3We reserve the symbol Z for X/Y , where X and Y are analogous to the complexes in Theo
rem B2.

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Theorem C. If Z is DR in the setting above, then any isoperimetric functionfor G is also an isoperimetric function for H, and the map H G induced byinclusion Y X and choice of base point in Y is injective.
A 2complex Z is called DR (i.e. diagrammatically reducible) if there are noreduced (in the sense of [LS]) spherical diagrams (i.e. combinatorial maps of S2)in Z . We shall review the notion DR and its consequences in 4. A referencefor isoperimetric functions is [Ge4]. The injectivity statement about fundamentalgroups in Theorem C is a consequence of the reciprocity law of [Ge2], but thestatement about isoperimetric functions is new. Another formulation of Theorem Cis that there is no area distortion, in the sense of [Ge1], for the injection H G ifZ is DR.
Corollary 1. If Z is DR and G is hyperbolic, then H is also hyperbolic.
Combining Theorem C with a result of [Bri] we obtain the following consequence.
Corollary 2. In Theorem C, if Z is strictly negatively curved in the sense of theweight test [Ge2], then the pair (G,H) is relatively hyperbolic in the sense of Farb[Fa]. Furthermore, H is hyperbolic iff G is hyperbolic.
We shall review the weight test and its connection with curvature in 4.In fact both Theorems C and its corollaries admit considerable generalization.
For convenience assume that both X and Y have only one vertex, which servesas the base point for their fundamental groups. Then associated to the corners of2cells of Z are elements of H, and a spherical diagram in Z is called admissible ifeach of the products of elements of H in the circuits in the links of each its verticesis the identity. Then we have
Theorem C. Under the assumptions above for the cofibration Y X Z ,
(1) if all admissible spherical diagrams in Z are reducible, then any isoperimetric function for G is an isoperimetric function for H, and the inducedmap H G is injective;
(2) if, in addition, there are positive weights on the corners of the 2cells of Z
with the property that admissible nontrivial reduced circuits in the link ofits vertex are very large (i.e. the sum of the weights is strictly larger than2) and such that the sum of the weights for each 2cell is no more than thecorresponding Euclidean sum, then H is hyperbolic iff G is also hyperbolic,and the pair (G,H) is relatively hyperbolic in the sense of Farb.
This result is illustrated in 5 by examples arising from the presentation t ttt1 whose associated 2complex is the dunce hat. Our method gives a systematicway of producing examples of relative hyperbolicity from elementary combinatorialconsiderations. We also make some observations about relations between our resultsand the Whitehead asphericity question.
In the following all 2complexes are combinatorial, in the sense that the attachingmaps of 2cells are given by maps of circles finitely subdivided into intervals, on

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each open interval of which the attaching map is a homeomorphism onto an open1cell. This restriction is needed since we shall be making geometrical constructionsinvolving weight test.
2. Hyperbolic mapping tori and proof of Theorem A.In this section we let be the Cayley graph of the hyperbolic group G for
finite generating set S, and assume is hyperbolic as a metric space for the pathmetric d, where each edge has length 1. We take this to mean the the Rips condition,that all geodesic triangles in are thin, in the sense that the neighborhood oftwo sides of a triangle contains the third side. We take the identity element 1 ofG = V () as base point and let w be an edgecircuit of length n, so w(0) = w(n) =1. Choose a geodesic combing of w from 1, so let i be a geodesic segment startingat 1 and ending at w(i), 0 i n. We extend i to all positive real numbersby setting i(t) = w(i) for t d(1, w(i)). Then by [Ge3] Lemma 5.1 one hasd(i(t), i+1(t)) 2(+1) for all t 0 and all i. This enables us to construct a vanKampen diagram f : D X for w, where X is the Cayley 2complex associated tothe finite presentation for G with generators SS1 and relators all labels of edgecircuits in of length at most 4 + 6, by joining i(j) to i+1(j) by an edgepathof length at most 2( + 1) and filling in quadrilaterals in X (cf. [Ge3] 5.2).
Lemma 2.1. There is a constant A > 0 so that for all edge circuits w in , andfor all vertices v of the van Kampen diagram f : D X just constructed for w onehas d(f(v), w) A+ log2(n+ 1).
Proof. From the construction of D, the vertex v is at distance at most 2(+1) fromone of the geodesics i, so it suffices to prove the Lemma for vertices on i. If v issuch a vertex, then it is at distance at most from a point v1 on one of two geodesics[w(0), w(i/2)] or [w(i/2), w(i)]. The point v1 is at distance at most from a point v2on one of four geodesics [w(ki/4), w((k + 1)i/4], 0 k 3. Continue this processp times where 2p1 i + 1 < 2p, obtaining points v = v0, v1, v2, . . . , vp, whered(vi, vi+1) . The final point vp is at distance at most 1 from some vertex w(j),and it follows that d(v, w(j)) p+1 (log2(i+1)+1)+1 (log2(n+1)+1)+1.This completes the proof.
We recall from [Gr2] p. 81 that the filling radius of an edgeloop S in is theleast number R such that S is homologous to zero in the Rneighborhood of S inX. The filling radius function f(n) is the maximum filling radius of all edgeloopsin of length at most n.
Corollary 2.2. The filling radius function of is bounded by A+ log2(n+1).
Now suppose that G = H
, where is an automorphism of the finitelypresented group H. By the theorem Rapaport [Ra] Theorem 2, it follows that is Ptame for some finite presentation P = x1, x2, . . . , xp  R1, R2, . . . , Rq of H.This means that there is an automorphism of the free group F on the generators{x1, x2, . . . , xp} of P inducing the automorphism on the quotient group G. We
can take Q = x1, x2, . . . , xp, t  xit = (xi), xi
t1 = 1(xi), Rj ; 1 i p, 1

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j q as a presentation for G. We shall call the relators Ri red relators and the
relators xit = (xi) and xi
t1 = 1(xi) blue relators. Faces of a van Kampendiagram are colored according to the color of the relator they correspond to.
Our strategy in calculating an isoperimetric function for H is to take an edgecircuit w in the generators xi for H and fill it with the special van Kampen diagram D above, which was constructed making use of the hyperbolicity of G. Thenwe modify D to make it taut, in a sense to be described shortly, by introducingblue relators but not increasing the number of red relators. Finally we eliminate allthe blue relators by surgery, as in [Ge1], using the tautness to control the numberof red relators.
Let X be the Cayley 2complex of Q, so X is the universal cover of the 2complexassociated to the presentation Q. Then there is a mapping X
, where
is the
tree of the HNN extension G = H . On group elements g = (h, ts), h H,
s
, the map is given by g s.
Let w be an edgecircuit of length n in the Cayley graph = X (1) of Q whichonly involves generators xi of H, and let f : D X be the van Kampen diagramconstructed above for w. Note that the boundary D of D maps to 0 in
. We
call the depth of a vertex v of D the absolute value of the image of f(v) under themapping X
; the depth of D is the maximum depth of its vertices.
Lemma 2.3. The depth of D is bounded above by A+ log2(n+ 1).
Proof. Let v be a vertex of D of maximal depth. By Lemma 2.1 there is a vertexw(i) such that d(f(v), w(i)) A + log2(n + 1). But the depth of v is boundedabove by d(f(v), w(i)), since the depth changes by at most one by moving over anedge of and all vertices of w are at depth 0. It follows that the depth of D isbounded by A+ log2(n+ 1).
Now the edges labelled t occur in D in disjoint annular corridors.4 We call twocorridors adjacent if they can be joined in D by an edgepath involving no tedges.If two corridors are adjacent, then they are either coherently oriented or oppositelyoriented, the latter case corresponding to a backtrack in
when is extended in
both directions to include t edges in each of the corridors it joins.
Lemma 2.4. It is possible by changing D to decrease the number of oppositelyoriented adjacent pairs of corridors without increasing the number of red faces and
without increasing the depth and without changing the boundary of the diagram D.
Proof.
Step 1: Choose an arc in D joining the initial points of two tedges in theoppositely oriented adjacent corridors. Make a cut along (all this takes placein the 2sphere, since, we recall, van Kampen diagrams are planar) creating twocopies of joined at their end points, and sew an arc labelled either t1()t1
or t1()t at these endpoints, depending on the orientations of the tedges in the
4These have also been called bands and trings by various authors.

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corridors, to make them clash at their ends. Then complete by adding blue facesto get a van Kampen diagram.
Step 2: Do Dehn surgeries5 at the two ends of . The effect is to replace theoriginal two annular corridors and added blue faces by a single annular corridor.
The number of blue faces increases in Step 1, but no red faces are added. InStep 2, which is variously called a Dehn surgery or diamond move, there is nochange in the number of faces, but only their arrangement. The net result is todecrease the number of such corridors by one without changing the boundary northe number of red faces. The depth of a vertex can remain the same or decreasebecause the minimal number of corridors that one must pass through to reach theboundary can only decrease in the process. This completes the proof of the Lemma.
Definition 2.5. We call a van Kampen diagram for w taut if there are no pairs ofoppositely oriented adjacent corridors.
If we repeat the process of the preceding lemma we obtain in a finite number ofsteps a taut diagram; this is because the number of oppositely oriented adjacentcorridors decreases each time. We summarize this in the next result.
Lemma 2.6. For every edgecircuit w in the generators of H there is a taut vanKampen diagram D for w such that
(1) depth(D)depth(D), and(2) the number of red faces of D is no more than the number of red faces of D.
Remark. It is instructive to understand what the taut van Kampen diagram Dlooks like. One can think of it as consisting of two copies of the towers of Hanoiparlour game, one extending in the positive direction in the tree
, the other in
the negative direction, where the rings are corridors. Then the reason for the termdepth becomes clear. It is just the larger of the number of nested corridors inthe two towers of Hanoi, so it measures the maximal amount of nesting of corridorsthat occurs in D.
Lemma 2.7. There is a constant C > 0 so that Area(D) Cn2, where n is thelength of w.
Proof. This follows since D was constructed by a geodesic combing of w, andgeodesics from the base point satisfy the synchronously combable condition in ahyperbolic group (cf. [Ge3] Theorem 9.7).
Remark. Although G is hyperbolic, and hence satisfies the linear isoperimetric inequality for fillings of edgecircuits, the diagramD was constructed by combing, andhence need not be a minimal van Kampen diagram for w. The special propertiesof D that are crucial for our argument arise from Lemma 2.1.
5A Dehn surgery in a diagram can take place when two edges with the same label meet in a
vertex with opposite orientations. One then makes cuts along the edges and reassembles them
with different identifications to create a new diagram (possibly cutting off one or more sphericalcomponents) with the same boundary label.

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Corollary 2.8. The number of red faces of the taut diagram D is bounded byCn2.
Now let N be the kernel of the canonical epimorphism F H, where we recallthat F is the free group on the generators of the presentation P for H.
Lemma 2.9. There is a constant M > 1 so that for all relations R N one hasmax(AreaP((R)),AreaP(
1(R))) MAreaP(R)
Proof. Take M = max(AreaP((Ri)),AreaP(1(Ri)), 1 i q), where Ri are
the defining relators of P, and check that this works.
Proof of Theorem A. We start with the van Kampen diagram D for w and replaceit with the taut diagram D. Denote by d the depth of D, so d A+ log2(n+ 1).Take an innermost corridor in D and let R be the inner boundary label of theannulus , so R N . Then the outer boundary label of is 1(R) := R, wherethe sign 1 depends on the orientation of the tedges. It follows from Lemma 2.9that AreaP (R
) MAreaP (R). So we surger D by removing and its interior
and replace it by a van Kampen diagram for R in P. The result is a taut vanKampen diagram for w with one fewer corridor. The number of red faces is at mostmultiplied by M in the process.
If we repeat this process 2d times (corresponding to the two towers of Hanoi, eachof depth at most d), the result is a van Kampen diagram E for w in P (so there areno blue faces), and the number of (red) faces of E is at most M 2d times the numberof red faces of D. It follows that Area(E) M 2dCn2 = M2A+2 log2(n+1)Cn2. SinceM log2(n+1) = (n + 1)log2 M (2n)log2 M for n > 0, we obtain Area(E) C1n
k
for some constant C1 > 0 and constant k = 2 + 2 log2M 2. It follows thatAreaP(w) C1n
k, and H satisfies a polynomial isoperimetric inequality. Thiscompletes the proof of Theorem A.
3. Filling norms and proofs of Theorems B2 and B1.We begin with a review of filling norms. Let X be a CWcomplex and equip
the cellular chain group Cn+1(X,) with the `1norm for a basis of oriented n+ 1
cells; since we only use real chains, this group will be abbreviated Cn+1(X). Theboundary map induces a surjection dn+1 : Cn+1(X) Bn(X), and we define thefilling norm fill = inf{c1  c Cn+1(X), dn+1(c) = }. In general this is only apseudonorm unless one imposes additional restrictions on X.
Since Bn Cn, there is a second norm on Bn given by the restriction to Bnof the `1norm on Cn. In general the two norms are unrelated, but if there is auniform bound M on dn+1e
(n+1)1, where e(n+1) ranges over the (n + 1)cells of
X, then dn+1 is a bounded linear map of normed linear spaces, and in this case,1 M fill for all Bn(X). If this is the case, then it follows that, since the`1norm is a norm (and not just a pseudonorm), the filling norm is actually a norm.
There are two important special cases when the bound M exists, namely, whenX is a simplicial complex and when X admits a group action by cellular homeomorphisms so that the orbit complex has a finite (n + 1)skeleton. Even when Mexists the two norms are not in general (bilipschitz) equivalent.

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We need two preliminary results for the proofs of Theorems B2 and B1.
Proposition 3.1 [Mi]. Let G be a hyperbolic group and let X be a complex of typeK(G, 1) with finite 3skeleton.6 Then the filling norm on B2(X) is equivalent tothe `1norm, where X is the universal cover of X
.
Mineyev establishes this result by exhibiting a thincombing of the Cayley
graph of G which is analogous to a combing by edge paths in X (1) but consistsrather of real 1chains with boundary the difference of two vertices. The supportsof these chains are required to lie in a uniform Hausdorff neighborhood of a geodesicjoining the same two vertices. The thinness condition means that two combingchains beginning at the base point and ending at vertices a unit distance apart inthe word metric determine, along with the edge joining these vertices, a real 1cyclewhich can be filled by a 2chain of uniformly bounded `1norm. For details of theconstruction consult [Mi].
Next let the finitely generated group H be a subgroup of the finitely generatedgroup G. Let T be a finite set of generators for G containing as a subset S, a finiteset of generators for H. We say that H is undistorted in G (or, equivalently, theinclusion H < G is a quasiisometric imbedding) if the restriction to H of the wordmetric on G is equivalent to the word metric on H. This notion is independentof generating sets. Also, it is known that a finitely generated subgroup H of ahyperbolic group G is undistorted iff H is a quasiconvex subgroup of G. Note alsothe equality dG(g, g
) = g gfill, where dG is the word metric on G and wherethe filling norm on the right is for the Cayley graph of G with respect to the givengenerators.
We denote B0(G) the real 0boundaries of the Cayley graph of G equipped withthe filling norm. As an vector space it is just the set of linear combinations of groupelements with real coefficients such that the sum of those coefficients is equal tozero; in other words, it is additively the augmentation ideal of the real group ring
G. The inclusion H < G induces a bounded mapping B0(H) B0(G) of normed
linear spaces.
Proposition 3.2. The following are equivalent for H < G as above.
(1) H is undistorted in G.(2) The inclusion B0(H) B0(G) is undistorted as a map of normed linear
spaces for their respective filling norms.
Proof. We remind the reader that the inclusion B0(H) B0(G) is said to beundistorted if the restriction of the filling norm on B0(G) is equivalent to the fillingnorm on B0(H) [Ge5]. One direction (2)(1) is immediate from the formula for theword metric given above. The converse result (1)(2) is nontrivial and is provedin [Ge5] Corollary 3.8.
6By the theorem of Rips [GH] one can find a K(G, 1) with finite nskeleton for all n for ahyperbolic group G.

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Proof of Theorem B2. We assume H is a finitely presented subgroup of the hy
perbolic group G, X is a K(G, 1) complex with finite 3skeleton such that X (2)
contains the presentation complex Y of a finite presentation for H. We let X bethe universal cover of X and let Y be a connected component of the pullback toX of Y ; Z = X/Y .
Assume first that the filling norm on B2(Z) is equivalent to the `1norm. Wewant to prove that H is hyperbolic. By the result of [Ge3], it is equivalent to provethat the filling norm on B1(Y ) = Z1(Y ) is equivalent to the `1norm. So let w be areal 1cycle in Y . Since G is hyperbolic, there is a linearly bounded filling c C2(X)for w, so d2(c) = w, c1 Kw1, where K > 0 is a constant independent of w.Upon passage to the quotient, c determines a 2cycle c Z2(Z) = B2(Z), wherethe last equality follows from the fact that Z is acyclic in degree 2. Since the fillingnorm on B2(Z) is equivalent to the `1norm, there is a 3chain x C3(X) whichdetermines upon passage to the quotient x C3(Z) so that (1) d3(x) = c and (2)x1 = x1 K1c1 KK1w1, where K1 > 0 is a constant.
We claim now that c d3x is a linear filling of w in Y . Observe first thatc d3x = 0 so c d3x C2(Y ). Also d2(c d3x) = d2c = w, so c d3x is a fillingfor w in Y . Finally, c d3x1 c1 + d3x1 Kw1 + M x1, where M is thebound on the map d3, which exists since X
(3) admits a cocompact group action.Since x1 KK1w1, we get that c d3x1 (K+MKK1)w1, establishing ourclaim. It follows from [Ge3] that H is hyperbolic.
For the converse, assume H is hyperbolic. We want to show that the fillingnorm on B2(Z) is equivalent to the `1norm. So let B2(Z). We can lift tob C2(X) so that b = and b1 = 1. Then d2b B1(Y ) (since d2b is a cyclein X which lies in C1(Y )), so by hyperbolicity of H there exists y C2(Y ) withd2y = d2b and y1 Kd2b1 KM b1 = KM 1, where M is the bound onthe norm of the map d2. Thus b y Z2(X), so by Proposition 3.1, b y = d3x,where x C3(X) and x1 Kb y1, where K is a constant. Thus = d3x, andx1 x1 Kb y1 K(b1 + y1) K(b1 + KM 1) = (K + KM)1,and it follows that the filling norm on B2(Z is equivalent to the `1norm. Thiscompletes the proof of Theorem B2.
Proof of Theorem B1. Let H be a finitely generated subgroup of the hyperbolicgroup G. We choose a finite set T of generators for G containing a subset S ofgenerators for H and choose a finite presentation P for G with T as generators.Let X be the 2complex canonically associated to P and let X be the universalcover of X . Let Y be copy of the Cayley graph of H for generators S containedin X, so Y is a connected component of the subgraph of X generated by all edgeswith label in S. We choose the base point for X in Y and let Z = X/Y .
Assume first that the filling norm onB1(Z) is equivalent to the `1norm. We mustprove that the inclusion Y X does not distort word metrics. So let v0, v1 Y
(0)
and let c be a geodesic edgepath in X joining them, so c1 = dX(v0, v1). Let c bethe reduction of c modulo C(Y ), so c Z1(Z) = B1(Z). Then there is a linearfilling of c, so C2(Z), d2 = c, and 1 Kc1 Kc1 = KdX(v0, v1),where K > 0 is a constant. We can lift to b C2(X) so that b = and

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b1 = 1. Consider c d2b C1(X). We have c d2b = 0, so c d2b C1(Y ).Also d1(cd2b) = d1c = v1v0, and cd2b1 c1+M b1 = c1+M 1 c1+MKc1 (1+MK)dX(v0, v1), where M is the norm of the map d2. Since the fillingnorm of v1 v0 in Y is dY (v0, v1), it follows that dY (v0, v1) (1+MK)dX(v0, v1),and hence the inclusion Y X does not distort word metrics.
Conversely, assume that H is undistorted in G, and let B1(Z). We can lift to b C1(X) so that b = and 1 = b1. Then d1b = d1 = 0, so d1b C0(Y ) with augmentation equal to zero. Hence d1b B0(Y ). But it follows fromProposition 3.2 that B0(Y ) is undistorted in B0(X). Thus there exists c C1(Y )so that c1 Kb1, where K > 0 is a constant, and so that d1c = d1b. It followsthat b c Z1(X), and b c1 b1 + c1 (1 + K)b1 = (1 + K)1. SinceG is hyperbolic, there exists x C2(X) with d2x = b c and x1 K1b c1 K1(1 + K)1, where K1 > 0 is a constant. Then x C2(Z), d2x = b = , andx1 K1(1+K)1, which establishes the linear filling for B1(Z). This completesthe proof of the theorem.
Remark. As was the case with the results of [Ge3] and [Ge5], Theorems B1 and B2reinforce our contention that the theory of hyperbolic groups is a subset of homological algebra. It is still a moot question whether geometric group theory itself isa branch of homological algebra.
4. Weight test, relative hyperbolicity, and Theorems C and C.
We begin by recalling the notion of diagrammatic reducibility (DR) and itsconsequences [Ge2]. The (combinatorial) 2complex Z is said to be DR if everyspherical diagram f : S2 X (so f is a combinatorial map from a subdivision ofS2 to Z) contains a pair of faces F, F in the domain with an edge e in commonso that their extended boundary labels (that is, the images of edges and cornersunder f of the extended boundary circuits, namely, edges and corners in order, ofthese faces) read beginning with e are equal. One says more intuitively that F andF are mapped by reflection across the edge e.
Convention. In the remainder of this article, we shall always assume without furthermention that when we take a connected subcomplex it is nonempty.
An important result is
Theorem 4.1 [Ge2]. Suppose that Y is a subcomplex of the 2complex X so thatZ = X/Y is DR. Then for each choice of base point v in Y (0) the homomorphism1(Y, v) 1(X, v) induced by inclusion Y X is injective.
Remark. A connected DR 2complex is aspherical, as is shown in [Ge2]. An exampleof an aspherical 2complex which is not DR is that associated to the presentationt  ttt1, which is known familiarly as the dunce hat.
I studied the notion DR in [Ge2] in connection with the Kervaire problem ofsolving equations with coefficients in a group in an overgroup, and I gave thereseveral different criteria for proving a 2complex is DR. The one that has proved

11
most useful in this and in many other contexts I could not envisage at that time isthe socalled weight test , which we now recall.7
Theorem 4.2 [Ge2]. Suppose we are given a function w from the corners of faces Fof a 2complex Z to the reals so that
(1) if F is an ngon, then the sum of the weights of the corners of F is at most(n 2), and
(2) for every nontrivial reduced circuit in the link of every vertex of Z thesum ( ) of the weights of the edges occurring in is at least equal to 2.
Then Z is DR.
Remark. In (2) above, an edge of the link of a vertex corresponds to a corner ofa face of X, so we can assign weights to the edges of the link of vertices. Thecurvature assigned to the circuit is ( ) := 2 ( ), so condition (2) says thateach nontrivial reduced circuit in the link of a vertex of Z has nonpositive curvature.
Definition 4.3. We say the weight w on the corners of 2cells of the 2complex Zis of strictly negative curvature if
(1) all weights of corners are positive,(2) there exists > 0 so that for every nontrivial reduced circuit in the link
of a vertex on Z one has ( ) , and(3) for each 2cell F of Z, if F is an n(F )gon, then n(F ) 3 and the sum of
the weights of the corners of F is at most (n(F ) 2).
For a finite complex Z it is enough to say that ( ) < 0 for all nontrivial reducedcircuits in (2), for the existence of follows as a consequence.
Remark. We could have defined a weight to be of nonpositive curvature if (1) and(3) hold, but (2) is replaced by ( ) 0 for all . This does not seem to be auseful notion, since the BaumslagSolitar group G = x, y  yxy1 = x2 has aweight of nonpositive curvature, assigning all corners the weight /2 except thecorner corresponding to the subword x2 of the relator, which gets assigned . Thisweight cannot correspond to any CAT(0) structure, since G has an exponentialDehn function, and not quadratic.
However a weight of strictly negative curvature does have useful consequences,as is shown by the next two results.
Proposition 4.4. Suppose Z is a 2complex (not necessarily finite) possessing aweight of strictly negative curvature, so there exists > 0 so that ( ) for allnontrivial reduced circuits in the links of vertices of Z. Then for all reduced discdiagrams f : D Z one has Area(D) (1 + 2/)`(D).
Proof. This is proved in [Ge4] Proposition 6.3. Here `(D) is the number of edgeson the boundary circle.
7I omitted the factor of in the original statement of the weight test in [Ge2]. Although the
plays no role, it is nevertheless important to include it to connect the weight test with the Gauss
Bonnet theorem and nonpositive curvature. The connection was first made in print explicitly in[GS].

12
Corollary 4.5. Suppose Z is a 1connected 2complex possessing a weight ofstrictly negative curvature. If there is a uniform bound on n(F ) over the 2cells Fof Z, then Z(1) is a hyperbolic metric space with the path metric, where all edgeshave length 1.
Proof. It follows from the proposition that Z satisfies the linear isoperimetric inequality for disc fillings of edgecircuits. Then the argument of [Berk] Theorem 2.5shows that Z(1) has thin geodesic triangles for some 0. We remark herethat although the argument of [Berk] is written in the context of finitely presentedgroups, what is actually used is that there is a bound on the length of relators,which corresponds to the bound on n(F ) in the hypothesis of the corollary. Thesimple connectivity of Z is used to guarantee the existence of (reduced) disc diagram fillings of geodesic triangles, to which the linear isoperimetric inequality ofthe proposition applies.
An application of these ideas is given by the next result.
Theorem 4.6. Let Y be a finite connected subcomplex of the finite connected 2complex X so that Z = X/Y has a weight of strictly negative curvature. Then forbase point v in Y one has
(1) the induced map H = 1(Y, v) 1(X, v) = G is injective, and(2) if H is hyperbolic, then G is also hyperbolic.
Proof. The first conclusion follows from the weight test for DR. The second conclusion follows as special case of a result of [Bri], since 1(Z) is hyperbolic fromCorollary 4.5.
Notation. We shall keep the following notation for the remainder of this section.We let Y be a connected subcomplex of the finite connected 2complex X and letZ = X /Y . X denotes the universal cover of X and Y is the pullback of Y to
X. Y denotes a connected component of Y and Z = X/Y . We also need Z, which
is obtained from X by collapsing each connected component of Y in X to its ownpoint. The complex Z is coarsely equivalent to Farbs complex [Fa]; precisely, Farb
adjoins cones on each connected component of Y so that each edge from each conepoint has length 1/2. This latter complex has a hyperbolic 1skeleton iff the graph
Z(1) is hyperbolic with the path metric where all edges of Z(1) have length 1, asFarb remarks, so we shall deal exclusively with Z in the sequel.
Suppose in addition that the induced homomorphism H := 1(Y, v)
1(X, v) := G is injective. In this situation Farb calls the pair (G,H) relatively
hyperbolic provided Z(1) is hyperbolic as a path metric space, where all edges havelength 1.
We can now prove Theorem C of the introduction.
Proof of Theorem C. We must show that if Z is DR then any isoperimetric functionforG is also an isoperimetric function forH and the induced homomorphismH Gis injective. But the injectivity of the map H G follows from Theorem 4.1. We

13
shall show the statement about isoperimetric functions by proving that for any discfilling in X of a simple edgecircuit in Y , there exists a disc filling in Y with nolarger area.
Let u be a simple edgecircuit in Y (1) and let f : D X be a disc filling ofu in X. Now compose f with the projection X Z to get the map D Z .Since f D takes values in Y , it follows that the last map factors to give a mapS2 Z . If we factor out additional cells in the domain of f which are mappedto Y , we obtain components which are spherical diagrams in Z . Since Z is DR,it follows that if the mapping S2 Z is not constant, then there exists a pair offaces F , F in the domain of one of these diagrams with an edge e in common whichare mapped by reflection across e into Z . These faces are images of faces F, F
respectively in D with the edge e in common, so in particular e is in the interiorof D . It follows from the properties of the quotient mapping X X Z thatthe faces F and F are mapped by reflection across e to the same face of X. Thuswe can perform a reduction on the diagram D by removing the cells F, F and eand sewing the boundary created. This yield a new disc diagram filling u with atleast two fewer faces than D (and a number of spherical diagrams created by thesewing, which we throw away).
We can continue this process as long as there are faces of D mapped nondegenerately into Z , each time reducing the area. In a finite number of steps, we arrive ata disc filling D of u in X which maps to a point in Z . It follows that D maps intoY , a disjoint union of copies of Y . Since u maps into Y , by connectivity it followsthat D maps to Y . Thus AreaY (u) Area(D
) Area(D), which establishes thatAreaY (u) AreaX(u), and the proof of Theorem C is complete.
Remark. The injectivity of the map H G can be proved by the same argumentas that giving the lack of area distortion above, without recalling the reciprocitylaw of [Ge2]. One takes an edgecircuit in Y , fills it in X, and then one uses thefact that Z is DR to remove in pairs faces not in Y .
Corollary 1 to Theorem C follows immediately, since hyperbolic groups are characterized as finitely presented groups satisfying the linear isoperimetric inequality[Gr1].
Proof of Corollary 2 to Theorem C. We assume here that Z is strictly negativelycurved. Theorems 4.6 and Theorem C tell us thatG is hyperbolic iffH is hyperbolic.So it remains to verify that the pair (G,H) is relatively hyperbolic. Thus we must
prove that Z(1) is a hyperbolic metric space.
But Z is a branched covering space of Z branched over Z (0)
.8 It follows thatthe weight on Z lifts to a weight on Z. Conditions (1) and (3) in Definition 4.3 forthe pullback weight are immediate. As for condition (2), the link of each vertex of
Z is a covering space of the link of the corresponding vertex of Z . Thus a nontrivialreduced circuit in a link of a vertex of Z maps onto a nontrivial reduced circuit
8This means that the map Z \ Z(0) Z \ Z(0) induced from X X by passage to thequotient is a covering map.

14
in the link of a vertex of Z . In particular, the sum of the weights of the edges of
is at least as large as the sum of the weights of their images, and hence at least aslarge as 2 + , where is the number occurring in Definition 4.3 for the weight ofstrictly negative curvature on Z . Thus the curvature ( ) of satisfies ( ) ,and condition (2) is satisfied for the pullback weight.
Since n(F ) is bounded where F ranges over the 2cells of Z by the correspond
ing bound for the finite complex Z , it follows from Corollary 4.5 that Z(1) is ahyperbolic metric space, and the proof of the corollary is complete.
Open question. If Z is only assumed aspherical, are the conclusions of Theorem Cstill valid? That is, is H G injective (this is a variant of the Kervaire question ofsolving equations over groups) and can the map H G distort areas in the senseof [Ge1] if Z is aspherical? We have some more to say about this situation in thediscussion following Proposition 5.6 below.
The explicit examples I know of a nontrivial kernel for H G and for areadistortion arise from HNN extensions. In an HNN extension with one stable letter t,Z is the 2complex associated to a presentation t  tt1, tt1, . . . , tt1, where therelator tt1 is repeated a number n of times. This is never aspherical if n 1.
We now assume in addition that X has only one vertex. Consider the effect ofthe quotient map p : X Z of a 2cell whose attaching map w in not entirely
in Y . Here we consider w as a cyclic word in X (1)
. Let u be a maximal subword of
w with values in Y (1)
. Then u is collapsed to a point under p, and corresponds toan oriented corner of the quotient 2cell on Z .9 We assign to that oriented cornerthe element [u] 1(Y
) = H represented by the circuit u in Y and call this thelabel assigned to that corner. In this way we obtain a map from oriented cornersof 2cells of Z to elements of H. Similarly, the branched covering Z Z inducesa mapping of corners of 2cells of Z to corners of 2cells of Z , and this map can becomposed with the preceding one to map the corners of 2cells of Z to H. If is acircuit in the link of a vertex of Z, then it is assigned an element of H by takingthe product in order of the labels of its edges, bearing in mind that edges of thelink correspond to corners of 2cells.
Lemma 4.7. For every circuit in the link of a vertex of Z, the group element inH assigned to it by the procedure above is trivial.
Proof. Let = 12 . . . n, where i is a corner of a 2cell of Z. Let i correspondto the group element hi H by the procedure above. Then
corresponds toh1h2 . . . hn H. But
is the image of a circuit in Y under the quotient mapp : X Z and each connected component of Y is simply connected (its a copy ofthe Cayley 2complex of H). It follows that h1h2 . . . hn = 1.
9Precisely, a corner of a 2cell is an edge of the link of a vertex, so an oriented corner correspondsto an oriented edge of the link.

15
Definition 4.8. Let f : S Z be a spherical diagram in Z and fix an orientationon S = S2. Then each vertex v of S has an oriented circle link, so determines anelement of H by assigning to it the product of the labels in H assigned to the imagecorners in Z in order in one circuit. This determines a conjugacy class fv in Hfor each such vertex v. We say that the diagram f is admissible if each conjugacyclass fv is that of the identity element.
Similarly a circuit in the link of the vertex of Z is called admissible if the productof labels associated to it is trivial.
From covering space theory and the preceding lemma we deduce
Proposition 4.9. A spherical diagram in Z lifts to one in Z iff it is admissible.
Proof. That a spherical diagram which lifts to Z is admissible follows fromLemma 4.7.
So suppose conversely that f : S Z is an admissible spherical diagram. Thenusing the fact that the vertex labels fv are trivial, we can fill them in with vanKampen diagrams in Y to form a map g of the 2sphere into X which yieldsunder the quotient the given map f . From covering space theory g lifts a sphericaldiagram G in X, and G yields under the quotient map X Z a lift of f to Z.This completes the proof.
Corollary 4.10. The complex Z is DR iff every admissible spherical diagram inZ is reducible.
We can now proceed to the proof of Theorem C of the introduction.
Proof of Theorem C.
(1) We assume that all admissible spherical diagrams in Z are reducible. We carryout the argument for Theorem C, and note that the spherical diagrams there all takevalues in Z, and are hence reducible, by Corollary 4.10. The result is that for everyedgecircuit u in Y we have AreaY (u) AreaX(u), and hence any isoperimetricfunction for G is an isoperimetric function for H.
As for the injectivity of the homomorphism H G, this follows from the reciprocity law of [Ge2] for spherical diagrams in Z, which follows from the fact all
spherical diagrams in Z are reducible by Corollary 4.10.
(2) Assume now that there are positive weights on the corners of the 2cells of Z
with the property that all admissible circuits in the link of the (unique) vertex arevery large (i.e. the sum of the weights is strictly larger than 2). and such that thesum of the weights of the corners of a 2cell F is at most (n(F ) 2). We claimthere is an > 0 so that all admissible circuits in the link of the vertex of Z havecurvature at most . Since Z is a finite complex, for each K > 2 there are onlyfinitely many circuits in the link whose total weight is < K. If such a circuit isadmissible, then its weight is > 2 and < K. Thus there are only finitely manyadmissible circuits 1,
2, . . . ,
m with sum of weights 2 < (
i) < K. Then we
take = min1im((i) 2) > 0. It follows that for all admissible circuits
inthe link one has ( ) 2 + , and hence ( ) .

16
Next we lift the weight on Z to a weight on Z via the branched covering mapZ Z . Since all circuits in the link of vertices of Z map to admissible circuits inthe link of Z , it follows that all reduced nontrivial circuits in the link of Z havecurvature < . It follows from Corollary 4.5 that Z(1) is a hyperbolic metric space,and hence the pair (G,H) is relatively hyperbolic in the sense of Farb.
If G is hyperbolic, then by part (1) of the proof, H is also hyperbolic. Conversely,if H is hyperbolic, then one goes through the arguments of [Bri] to see that onlyadmissible circuits occur at interior vertices in van Kampen diagrams in Z usedthere. This is because such diagrams arise by passage to the quotient from vanKampen diagrams in X . So it follows that G is hyperbolic. This completes theproof of Theorem C.
5. Applications and examples.Theorem C is a useful technique for producing examples of relatively hyperbolic
pairs of groups and for constructing new hyperbolic groups from old ones. Just afew examples will be given here to illustrate the technique.
Let Z be the 2complex associated to the presentation t  ttt1, so Z is thedunce hat. Let H be a group, let , , H and consider ttt1 H twhere t is an infinite cycle. Then the quotient group G = H t/N , where Nis the normal closure of ttt1, is isomorphic to H t/N , where N is thenormal closure of at2bt1, where a = and b = 1 , as one sees by substitutingt for t. So we can always take = 1 without loss of generality. Thus we takeG = H t/N where N is the normal closure of at2bt1, for elements a, b chosenin H, in the sequel.
The injectivity of the map H G induced by inclusion H < H t andprojection to G is a special case of the Kervaire problem of solving nonsingularequations over groups. In this situation injectivity always holds by a result ofHowies [Ho1]. We are interested in the questions of relative hyperbolicity of thepair (G,H) and whether hyperbolicity of H implies that of G and conversely. Wehave
Proposition 5.1. Let H be a finitely presented group and let G = H t/N , wheret is an infinite cycle and where N is the normal closure of the single element{at2bt1}, where a, b are two given elements in H.
(1) Assume that each of a, b is of order at least 6 and ap 6= bq for p + q < 4.If G is hyperbolic, then H is also hyperbolic.
(2) Assume that each of a, b is of order at least 7 and ap 6= bq for p + q < 5.Then the pair (G,H) is relatively hyperbolic, and G is hyperbolic iff H ishyperbolic.
Proof. Choose a finite presentation P for H and let Q be the presentation for Gwith one additional generator t and one additional defining relator at2bt1, wherea, b are words in the generators of P representing the elements a, b respectively.Let Y , X be the presentation complexes of P,Q respectively, so Z = X /Y .
The link of the vertex of Z has two vertices A,B with looplike edges eA, eB atA,B, respectively, and an edge e joining A to B. The labelling of the edges by

17
elements of H from 4 is given by eA 7 a, eB 7 b, e 7 1. Let us assign weight
/3 to each of the three corners of Z . Then one sees from the weight test that (1)if there are no nontrivial admissible circuits of length 5 in the link, every admissible spherical diagram in Z will be reducible, and (2) if there are no nontrivialadmissible circuits of length 6 in the link, then all such admissible circuits inthe link are very large. The hypotheses (1) and (2) in the proposition guaranteerespectively these situations, and the proposition follows from Theorem C.
Examples:
5.2. Let H = x be an infinite cycle and let a = xm, b = xn. Let Gm,n =H t/Nm,n, where Nm,n is the normal closure of x
mt2xnt1. Thus the groupGm,n is parametrized by integral points in the plane. Then the set of points (m,n)where the pair (Gm,n, H) is relatively hyperbolic (and Gm,n is hyperbolic) containsthe complement of a finite number of a finite number of lines through the origin inthe (m,n)plane, an open set in the Zariski topology.
In (5.3) and (5.4) we work out two of these groups, G1,2 and G2,3.
Remark. Something similar must hold for equations modeled on a finite complex Z
for which n(F ) 3 for all faces F (assuming one knew what the Zariski topologymeant for groups). That is, I expect that given such a Z , the pair (G,H) shouldbe generically relatively hyperbolic as the coefficients of the equations vary inH. What is not clear to me is how to handle the general case when Z has digons;these are important since they occur in HNN extensions. This is related to theopen question raised in 4 after the proof of the corollaries of Theorem C.
5.3. If H is the infinite cycle generated by x and a = x, b = x2, then G isthe 1relator group x, t  xt2 = tx2. Here a2 = b1, so p + q = 3 in theproposition. We claim that G is not hyperbolic. To see this, we subdivide thedefining relator by introducing two subdivision edges y, z and no new vertices toget G = x, t, y, z  xy = t, yx = tz, zx = t. Then eliminate t to get G = x, y, z 
yx1
= yz, zx1
= y. It follows that G is the mapping torus of the automorphismof the free group F (y, z) given by y 7 yz, z 7 y. However the mapping torus ofan automorphism of a free group of rank 2 is never hyperbolic, since the conjugacyclass of the commutator [y, z] is invariant up to inversion.
5.4. Take H to be the infinite cycle generated by x, a = x2, and b = x3. Herep = 3 and q = 2, so it follows from the proposition that G is hyperbolic and thepair (G,H) is relatively hyperbolic. We can identify the 1relator group G = x, t x2t2 = tx3 in this case. Introduce three subdivision edges in the relator, y, z, w,to get G = x, t, y, z, w  xy = t, yx = xz, zx = tw, wx = t. Now eliminate t to
get G = x, y, z, w  wx1
= y, yx1
= z, zx1
= yw. Thus we see that G is themapping torus of the automorphism of the free group F (y, z, w) of rank 3 given byw 7 y, y 7 z, z 7 yw. This is a PV automorphism and the results of [GSt][BF]and [Brk] show that its mapping torus is hyperbolic, thereby giving another proofthat G is hyperbolic.

18
5.5. Another instructive example, which was discussed in Farbs paper [Fa] from thepoint of view of geometry, is given by H = x, an infinite cycle, and G = a, b, x x = [a, b], a free group of rank 2. Take Y and X to be the 2complexes associatedto these presentations. Then Z is the 2complex associated to the presentationa, b  [a, b], a torus, but no nontrivial reduced circuits in the link of its vertexare admissible; from this and from Theorem C it follows that the pair (G,H) isrelatively hyperbolic. To relate this example to geometry, let X be the universalcover of X and let Y be the complete preimage of Y in X. Then Farbs complex Zis obtained by collapsing each connected component of Y to its own vertex. The2complex Z considered as a point set is the tesselation of the hyperbolic plane
2
generated by reflections of an ideal quadrilateral together with points on the idealboundary (Z however is a CW complex, which is a different topology); pictures of
X and Z appear side by side in [Fa] 3.1 Figure 1. The cusps in the first of thosefigures correspond to the connected components of Y , so are copies of
subdivided
at integral points with edges labelled x.10 In this case Z(1) is a hyperbolic graph,corresponding to the relative hyperbolicity of the pair (G,H).
We note the following consequence of the methods of 4. Let Y be a subcomplexof the 1vertex 2complex X , and let Z = X /Y . Let X be the universal cover of
X and let Y be the complete preimage of Y in X under the covering map. Let Ybe a connected component of Y containing the base point of X.
Proposition 5.6. If Z is DR, then 2(X) is generated as a 1(X
) module by2(Y
).
Proof. Given a spherical diagram in X , we lift it to X and use the DR property,as in the proof of Theorem C, to remove in pairs faces in X Y . In the process,spherical diagrams with values in Y split off. But this means that 2(X
) is gen
erated by spherical diagrams in Y . Taking into account the base point in Y and1(X
) action, this means that 2(X) is generated as a 1(X
) by 2(Y).
Corollary 5.7. If Y is aspherical and Z is DR, then X is aspherical.
It is natural to ask whether the conclusions of Proposition 5.6 and its corollaryremain valid if Z is only assumed to be aspherical. More precisely, we are motivatedto raise the
5.8. Question: Let Y be a subcomplex of the 1vertex 2complex X , and letZ = X /Y . If Z is aspherical,
(1) when is 1(Y) 1(X
) injective, and(2) when is 2(X
) generated as a 1(X) module by 2(Y
)?
We shall see in Example 5.11 below that 5.8 (2) is not satisfied in general; thereis no known example where 5.8 (1) fails. So the real question is to find appropriate
10The fact that there are no essential circuits in the links of vertices of Z is reflected in thefact the tesselation of 2 is ideal, with no interior vertices.

19
general conditions that guarantee that 5.8 (2) is satisfied. If the reader will bearwith me for a short while, it will become clear why this question is of interest.
Remarks.
(1) One can show by a covering space argument that, under the assumption thatZ is aspherical, the conclusion 5.8 (1) is implied by Howies conjecture[Ho2], that the induced homomorphism 1(Y
) 1(X) is injective if
H2(Z,
) = 0.
(2) Assuming 5.8 (1), 5.8 (2) is equivalent to H2(Z,
) = H2(X, Y ;) = 0; here
Z is Farbs complex, discussed in 4, obtained by collapsing each connectedcomponent of Y in X to its own point. The proof is a consequence of theHurewicz theorem and homology exact sequences.
(3) If Y is a graph and Z is aspherical, then 5.8 (1) is true, since Howiesconjecture is true for free groups.11
(4) Note that 5.8 (2) implies an affirmative answer to the follow open question.
5.9. Open question: Is X aspherical if both Y and Z are aspherical?
If true, 5.9 implies an affirmative answer to Whiteheads asphericity question[Wh], whether a subcomplex of an aspherical 2complex is aspherical, as the nextresult shows.
Proposition 5.10. Question 5.9 implies the Whitehead conjecture. In fact, an af
firmative answer to the special case of 5.9 when Y is a graph implies the Whiteheadconjecture.
Proof. Easy arguments show that it suffices to prove that a subcomplex W of an
aspherical 1vertex 2complex Z obtained by removing a family {e(2)i  i I} of
2cells e(2)i from Z
(so W (1) = Z (1)
) is aspherical. Choose one corner in each
2cell e(2)i and imbed a small circle Si in that corner with one point at the vertex of
Z . Then remove the interiors of all circles Si to obtain the CWcomplex X. In the
cell structure of X , the 2cell e(2)i is replaced by the exterior of Si in e
(2)i and one
additional 1cell. Let Y = iISi, so the bouquet of circles Y is a subcomplex of
X. Note that X /Y = Z . We are assuming that Z is aspherical, and Y , a graph,is aspherical, so 5.9 implies that X is aspherical. But X deformation retracts ontothe original subcomplex W of Z , so it follows that W is aspherical. This completesthe proof.
Now we shall give the counterexample to question 5.8 (2) in the full generalitywith which it is stated.
5.11. An example where Z is aspherical but 5.8 (2) fails, so 2(X) is not generated
as a 1(X) module by 2(Y
). By Remark 2 following 5.8, it suffices to prove that
H2(Z) 6= 0 provided we know that 1(Y) 1(X
) is injective.
11More generally, Howies conjecture is known to be true for the classes of locally residuallyfinite groups and locally indicable groups [Ho2].

20
We start with Z the 2complex of the presentation t  ttt1, so Z is the duncehat. Let H be the coefficient group of the equation at2bt1 = 1, so a, b H; weshall determine H in such a way that a spherical diagram D, to be constructed, isadmissible.
Let D be my favorite diagram in Z , so D is the unique twofaced reducedspherical diagram in Z . A picture of D appears in [Ge6] p. 172. It has onesource and one sink vertex and the remaining vertex is a saddle. We construct anew spherical diagram D by taking the branched cover of degree 2 of D branchedat the source and sink vertices. It has four vertices, a source, a sink, and twosaddles. The vertex labels are a2 for the source, b2 for the sink, and ab1 foreach of the saddles. Thus D is admissible iff a2 = b2 = ab1 = 1 in H. Sowe take H to be defined by these relations, whence H is cyclic of order 2. ThenG = 1(X
) is defined by adjoining one free generator t and subjecting it to therelation at2bt1b = a1t2at1 = 1.
Lemma 5.12. G = S3.
Proof. We have ata1 = t2, so conjugating by a gives t = t4, or t3 = 1. Then onesees that a 7 (12), b 7 (123) gives an isomorphism of G with S3.
Next we calculate the chain associated to D in C2(Z). To do the calculation,one removes a small disc neighborhood of each of the vertices, to get a punctureddisc diagram D1. The labels on the edges around the punctures are a, b, and 1, sofor example what was the source of D is replaced by a circle subdivided twice witheach of the two edges labelled a. One of the vertices of D1 is taken as a base point,
and paths are chosen in D(1)1 to the base points of each of the four 2cells. Since all
2cells of Z are equivalent under the action of G (the action of G on 2cells is free
since Z is a branched cover of Z , and Z has only one 2cell, so there is one Gorbitof 2cells), one obtains the 2chain associated to D as an element z of
G times one
of the 2cells. I calculate z to be the sum of four distinct group elements, two withplus sign and two with minus sign.12 It follows that the chain associated to D isnonzero, and since it is a cycle, it follows that H2(Z) 6= 0. Since 1(Y
) 1(X)
is injective in this example, it follows that 2(X) is not generated as a 1(X
)module by 2(Y
).
There remains to consider what additional hypotheses are needed for question 5.8 (2) to have an affirmative answer. Question 5.9 gives one conjecturaladditional hypothesis, that Y be aspherical. Another comes from the work ofKlyachko on the Kervaire conjecture for torsionfree groups [Kl].
5.13. Conjecture: If Z is the 2complex associated to the presentation P = t t1t2 . . . tr , where each i = 1 and
i i = 1 (we call such P a Kervaire
12I calculate z to be (23) + (132) 1 (13) for my choice of base point, in terms of the
representation in S3. Another choice of base point will give gz, where g G. The fact that the
signs have this form, two plus and two minus, is a reflection of the fact that Z is contractible,and hence Cockcroft, and serves as a check on the calculation.

21
complex because of its role in the Kervaire conjecture, that you cannot kill anontrivial group by adjoining one new free generator and one additional relator[MKS] p. 403) and if 1(Y
) = H is a torsionfree group, then 2(X) is generated as
a G = 1(X) module by 2(Y
). In this situation, conclusion 5.8 (1) is Klyachkostheorem [Kl].
In fact this conjecture would hold if every reduced spherical diagram in P hadeither a monochromatic source or a monochromatic sink, where one associates toeach corner of the relator of P a distinct color. This is precisely what Klyachkosmethods establish in the case of the relator (t1t)nt, n 1, by [FR] Theorem 4.1.I suspect this also holds in the case of other Kervaire complexes, but it is notimmediately evident from the arguments in the literature [Kl][FR], which involvechanging the combinatorial structure of the 2complex (cf. [FR] Lemma 4.2).
References
[Berk] H. Short et. al , Notes on word hyperbolic groups, Group Theory from a GeometricalViewpoint (E. Ghys, A. Haefliger, and A. Verjovsky, eds.), World Scientific, pp. 363.
[BF] M. Bestvina and M. Feighn, A combination theorem for negatively curved groups, J.
Differential Geom. 35 (1992), 85101.
[Bow] B. Bowditch, Notes on Gromovs hyperbolicity criterion for pathmetric spaces, Group
Theory from a Geometrical Viewpoint (E. Ghys, A. Haefliger, and A. Verjovsky, eds.),World Scientific, pp. 64167.
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Mathematics Department, University of Utah, Salt Lake City, UT 84112, USA
Email address: [email protected]