aa r X i v : . [ m a t h . G R ] D ec Lipschitz 1-connectedness for some solvable Liegroups.
David Bruce CohenOctober 12, 2018
Abstract
A space X is said to be Lipschitz 1-connected if every Lipschitz loop γ : S Ñ X bounds a O p Lip p γ qq -Lipschitz disk f : D Ñ X . A Lip-schitz 1-connected space admits a quadratic isoperimetric inequality,but it is unknown whether the converse is true. Cornulier and Tesserashowed that certain solvable Lie groups have quadratic isoperimetricinequalities, and we extend their result to show that these groups areLipschitz 1-connected. Lipschitz 1-connectedness.
Consider a complete, simply connected Rie-mannian manifold X which is homogeneous in the sense that some Lie groupacts on X transitively by isometries. It is well known that in such a man-ifold, every loop admits a Lipschitz filling (see Proposition 3.5; for a muchstronger result see [9, Corollary 1.4]). A classical way to study the largescale geometry of X is by asking how hard it is to fill loops in X . Definition.
Let γ : S Ñ X be a loop. The filling span of γ denoted Span p γ q ,is defined to be inf t Lip p f q| f : D Ñ X ; f | S “ γ u . We say that X is Lipschitz 1-connected if there exists a constant C suchthat Span p γ q ď C Lip p γ q for all Lipschitz loops γ : S Ñ X .For instance, Euclidean n -dimensional space E n is Lipschitz 1-connectedbecause one may “cone off” loops in E n : given a loop γ : S Ñ X with γ p q “
0, we may take f p re iθ q “ rγ p e iθ q to obtain a 1-Lipschitz filling. Asimilar argument shows that all CAT(0) manifolds are Lipschitz 1-connected.1 olvable groups. In this paper, we will be interested in the case where X is some solvable real Lie group G equipped with a left invariant met-ric. To motivate this, note that Lipschitz 1-connectedness is a QI-invariant(Proposition 3.6), and that every Lie group is quasi isometric to a solvableLie group. We will further assume that G has the form U ¸ A , where thefollowing conditions hold. • A is abelian Lie group. • U is a closed subgroup of the group of real n ˆ n upper triangularmatrices with diagonal entries equal to 1 (for some n ). • A and U are contractible.We will now see some examples of groups which are not Lipschitz 1-connected. Groups of Sol type.
Fix real numbers t ą ą t ą G of matrices of the form »– t s x t s y fifl where s, x, y P R . Note that G decomposes as U ¸ A as above if wetake A to be the diagonal matrices of G and U to be the matrices withdiagonal entries equal to 1. It is known [5][Theorem 8.1.3] that there existsloops γ in G such that the minimal area of any filling of γ is on the orderof exp p Lip p γ qq . Hence, G is not Lipschitz 1-connected. Groups of this formare called groups of Sol type.More generally, Cornulier and Tessera have shown that if a group G “ U ¸ A as above surjects onto a group of Sol type, then it has loops of expo-nentially large area, and hence cannot be Lipschitz 1-connected [4, Theorem12.C.1]. If G surjects onto a group of Sol type we say that G has the SOLobstruction. Tame groups.
On the other hand, if the conjugation action of some a P A contracts U , then G “ U ¸ A will be Lipschitz 1-connected (Proposition5.2). For a to be a contraction means that there is some compact subset Ωof U such that for any compact subset K of U , some positive power of a conjugates K into Ω. If such a and Ω exist, G is said to be tame.2 he theorem of Cornulier and Tessera. It is clear that a space which isLipschitz 1-connected admits a quadratic isoperimetric inequality—i.e., anyloop of length ℓ must bound a disk of area O p ℓ q . Cornulier and Tessera havegiven a large class of solvable Lie groups admitting quadratic isoperimetricinequalities, and we shall extend their result to show that these groups areLipschitz 1-connected.Given an action of the abelian group A on a vector space V , let V Ă V be the subspace consisting of vectors v such that log } a n v } n Ñ n Ñ8 forall a P A . Their theorem [4, Theorem F] states that G “ U ¸ A satisfies aquadratic isoperimetric inequality if the following conditions hold. • p U {r U, U sq “
0. ( G is said to be standard solvable if this conditionholds.) • G does not surject onto a group of Sol type. • H p u q “
0, where u is the Lie algebra of U and H denotes secondLie algebra homology—see Definition 5.1. • Kill p u q “
0, where the killing module Kill p u q is the quotient of thesymmetric square u d u of u by the A -subrepresentation spanned ele-ments of the form r x, y s d z ´ y d r x, z s . Main Theorem.
Our primary objective in this paper is to establish thefollowing theorem, proved as Theorem 6.1, improving Cornulier and Tessera’sresult to show Lipschitz 1-connectedness.
Theorem 1.1.
Let G be a group of the form U ¸ A where U and A arecontractible real Lie groups, A is abelian, and U is a real unipotent group(i.e., a closed group of strictly upper triangular real matrices.)If p U {r U, U sq , H p u q and Kill p u q are all trivial and G does not surjectonto a group of Sol type, then G is Lipschitz -connected. Quadratic isoperimetric inequality versus Lipschitz 1-connectedness.
As noted above, if X is Lipschitz 1-connected, then it has a quadratic isoperi-metric inequality. It is not known whether the converse is true—that is, thereare no known examples where X has a quadratic isoperimetric inequalitybut is not Lipschitz 1-connected. Recently, Lytchak, Wenger, and Young[8]have shown some results about the existence of Holder fillings in spacesadmitting quadratic isoperimetric inequalities.3 .1 Organization. This paper is organized as follows. § § § § We wish to thank Robert Young and Yves Cornulier for productive conver-sations. This work has been supported by NSF award 1148609.
In this section, X will be a complete, simply connected Riemannian manifoldadmitting a transitive Lie group action by isometries. We will collect severalfacts about filling loops in X . Everything in this section is both trivial andwell known, but it seems easier to write the proofs than to find them in theliterature. The key facts we will prove are as follows. • Proposition 3.3 shows that X is Lipschitz 1-connected on a small scale.That is, there is some constant D such that Span p γ q “ O p Lip p γ qq whenLip p γ q ă D . • Corollary 3.4 shows that all loops in X have Lipschitz fillings. • Proposition 3.5 proves the existence a “filling span function”—meaningthat there is a function C p X, R q such that Span p γ q ă C p X, Lip p γ qq forall Lipschitz loops γ : S Ñ X . • Proposition 3.6 shows that Lipschitz 1-connectedness is a QI-invariantin this setting.
Let D be the unit disk, equipped with the Euclidean metric. Purely as amatter of convenience, we will think of D and the unit circle S as subsetsof C . We will need some convenient cellular decompositions of D .4igure 1: On the right, a decomposition of the unit disk into O p ǫ ´ q trianglesuniformly bilipschitz to D ǫ as in Proposition 3.1. On the right, a decom-position of D into a central disk together with O p ǫ ´ q “sectors” uniformlybilipschitz to D ǫ as in Proposition 3.2 Definition.
Let D ǫ be the Euclidean disk of radius ǫ . Proposition 3.1.
There exists a constant C such that for any ă ǫ ă there is a triangulation of D into at most Cǫ ´ triangles which are C -bilipschitz to D ǫ . See the left side of Figure 1 for a decomposition of this type. The proofof Proposition 3.1 is left to the reader.
Using templates.
Proposition 3.1 will help us fill loops in the followingway. Suppose that C is a class of loops in X such that Span p γ q “ O p Lip p γ qq for γ P C —typically, C will consist of loops with small Lipschitz constant.Now, given some other loop γ : S Ñ X , we may wish to find a O p Lip p γ qq -Lipschitz filling f : D Ñ X of γ . We can often use the following abstractstrategy to construct f . • For some sufficiently small ǫ , take a decomposition of D as in Propo-sition 3.1. For a 2-cell ∆ of this decomposition, let ψ ∆ : ∆ Ñ D ǫ bea C -bilipschitz map, where C is the universal constant guaranteed bythe proposition. • Take f to be γ on B D , and extend f over the 1-skeleton of the decom-position in such a way that Lip p f q “ O p Lip p γ qq and the restriction of f to the boundary of each 2-cell represents an element of C . That is,5e want f to be such that for every 2-cell ∆, the map γ ∆ : S Ñ X given by γ ∆ : e iθ ÞÑ f | B ∆ ˝ ψ ´ p ǫe iθ q satisfies γ ∆ P C . • For each 2-cell ∆, observe that Lip p γ ∆ q “ O p ǫ Lip p γ qq , so there existsa O p ǫ Lip p γ qq -Lipschitz filling f ∆ : D Ñ X of γ ∆ . • Extend f over ∆ with Lipschitz constant O p Lip p γ qq by taking, for each2-cell ∆, f | ∆ p z q “ f ∆ ˆ ǫ ψ ∆ p z q ˙ . The nontrivial part of this type of argument comes when we try to extend f over the 1-skeleton with the desired properties, so we will typically suppressthe other details. Proposition 3.2.
There exists a constant C such that, for each ă ǫ ă ,the unit disk D may be cellularly decomposed into an inner disk of radius ´ ǫ , surrounded by an annular region divided into 2-cells (we call thesesectors) with the following properties. • Each sector is bounded by two radial line segments, an arc of B D , andan arc of the boundary of the inner disk. • Sectors are C -bilipschitz to D ǫ . • The number of sectors is between Cǫ and Cǫ . See the right side of Figure 1 for a decomposition of this type. The proofof Proposition 3.2 is left to the reader. This proposition will typically usedto convert a filling of a loop ˜ γ to a filling f of a nearby loop γ , by taking f restricted to the inner disk to be a slightly rescaled filling of ˜ γ . Proposition 3.3.
Let X be a simply connected, complete homogeneous Rie-mannian manifold. There exist constants C p X q and D p X q such that thefollowing hold. • If x, y P X and d p x, y q ď D p X q , then there is a unique geodesic in X connecting x to y . Suppose γ : S Ñ X is such that ă Lip p γ q ă D p X q . Then Span p γ q ă C p X q Lip p γ q .Proof. It is clear that there is a uniform upper bound on sectional curvaturesof X , since X is homogeneous. Therefore, X is a CAT( κ ) space for some κ ě D p κ q depending only on κ such that points of X separated byless than D p κ q are connected by a unique geodesic, so taking D p X q ă D p κ q ensures unique geodesics.By the existence of normal coordinates in Riemannian manifolds [7,Proposition 8.2], there exists a bilipschitz map ψ : U Ñ V where U is aneighborhood in X and V a neighborhood in R dim p X q . Take D ą x P U with U Ą B D p x q and ψ p B D p X qq Ă V Ă V for some convex set V . Now let γ : S Ñ X be D -Lipschitz with γ p q “ x .Without loss of generality, ψ p x q “
0, and we may fill ψ ˝ γ by coning off–that is, we define a Lip p ψ q Lip p γ q -Lipschitz filling f : D Ñ V of ψ ˝ γ by letting f p re iθ q “ rψ p γ p e iθ qq . If we let f “ ψ ´ ˝ f , then f is a fill-ing of γ and Lip p f q ď Lip p ψ ´ q Lip p ψ q Lip p γ q . Taking D p X q ă D and C p X q ą Lip p ψ ´ q Lip p ψ q , the result follows. Corollary 3.4.
For all Lipschitz maps γ : S Ñ X , we have Span p γ q ă 8 .Proof. Let ˜ f : D Ñ X be a (continuous) filling of a Lipschitz γ : S Ñ X .We must find a Lipschitz filling f of γ . For ǫ ą
0, let τ ǫ be the triangulationof D given in Proposition 3.1, and for each 2-cell ∆ of τ ǫ , let ψ ∆ : ∆ Ñ D ǫ be the C -bilipschitz map guaranteed by the proposition. Let F p ǫ q be thelargest value of d p ˜ f p x q , ˜ f p y qq such that x, y are adjacent vertices of τ ǫ . If F p ǫ qÑ ǫ Ñ
0, then Proposition 3.3 lets us produce a Lipschitz filling f of γ as follows. Set f p x q “ ˜ f p x q for each vertex x of τ ǫ , where ǫ is small enoughthat CF p ǫ q ă D p X q where C is the constant given by Proposition 3.1 and D p X q is the constant given by Proposition 3.3. Set f to be a constant speedgeodesic on each edge of τ ǫ , so that the map S Ñ X given by e iθ ÞÑ f ´ ψ ´ ´ ǫe iθ ¯¯ is D p X q -Lipschitz. By Proposition 3.3, f now admits a Lipschitz extensionover 2-cells.On the other hand, if F p ǫ q does not go to 0 as ǫ Ñ
0, then for everynatural number n , we may find points x n , y n P D such that d p x n , y n q ă n but d p ˜ f p x n q , ˜ f p y n qq ą δ for some fixed δ ą
0. Because D ˆ D is compact,we know that p x n , y n q must subconverge to some point p x, x q in the diagonal7f D ˆ D . But p ˜ f p x n q , ˜ f p y n qq cannot subconverge to p ˜ f p x q , ˜ f p x qq because d p ˜ f p x n q , ˜ f p y n qq ą δ . This contradicts continuity of ˜ f . Proposition 3.5.
Let X be a simply connected, complete homogeneous Rie-mannian manifold. For any L ą , there exists C p X, L q ą such that if γ : S Ñ X is L -Lipschitz, then Span p γ q ď C p X, L q . Gromov [6, §
5] refers to C p X, L q as the filling span function of X . Proof.
Fix L (all constants from here on will be presumed to depend uncon-trollably on L ). Let C L denote the L -Lipschitz loops in X with some fixedbasepoint x , and equip C L with the uniform metric. By Arzela-Ascoli, C L iscompact. Certainly, Span is not continuous on C L , but we will show that itis bounded. Because C L is compact, for every ǫ ą C L may be covered bya finite number of ǫ balls. Hence, it suffices to find a constant C ą γ , γ P C L with d p γ, γ q ă C we haveSpan p γ q ă C max t , Span p γ qu . To do this, let D p X q be as in proposition 3.3 (assuming without loss ofgenerality that D p X q ă γ , γ P C L with d p γ , γ q ă D p X q . Let f : D Ñ X be a Lipschitz filling of γ . Let ǫ “ D p X q L and decompose D into an inner disk and sectors as in Proposition 3.2. We will producea filling f : D Ñ X of γ which is Lip p f q ´ ǫ -Lipschitz on the inner disk and O p q -Lipschitz on the annular region as desired.Define f | S to be equal to γ . Let R “ ´ ǫ be the radius of the inner diskand define f to be a rescaled copy of f on the inner disk—i.e., f p re iθ q “ f p rR e iθ q for r ď R . This implies that f is Lip p f q R -Lipschitz on the innerdisk. If x and y are the images under f of the endpoints of a radial segmentseparating two sectors, then d p x, y q ă D p X q because d p γ, γ q ă D p X q .Define f to be a minimal speed geodesic on each of the radial segmentsseparating two sectors, so that f is O p q -Lipschitz on the boundary of eachsector (considering L as fixed). Since sectors are uniformly bilipschitz to D ǫ , f admits a O p q -Lipschitz extension over each sector by Proposition 3.3.We see that f is O p q -Lipschitz on the annular region and Lip p f q ´ ǫ -Lipschitzon the inner disk, implying the desired bound for Span p γ q , taking any suf-ficiently large C . Proposition 3.6.
Let X and Y be simply connected, complete, homogeneousRiemannian manifolds, and suppose that X is quasi-isometric to Y . If Y isLipschitz 1-connected, then so is X .
8t seems likely that X and Y are bilipschitz if they are quasi-isometric. Proof.
Let ψ : X Ñ Y be a quasi isometry, and Ψ : Y Ñ X a quasi inverseto ψ . Let γ : S Ñ X be an L -Lipschitz loop, where L ą ǫ “ L . To obtain a O p L q -Lipschitz filling for γ , weproceed as follows.Using Proposition 3.1, subdivide D into O p L q triangles bilipschitz to D ǫ , so that adjacent vertices on the boundary are mapped to within O p q of each other by ψ ˝ γ . Let ˜ γ : S Ñ Y be a O p L q -Lipschitz loop in Y whichagrees with ψ ˝ γ on vertices of our triangulation. By assumption, Y isLipschitz 1-connected, so ˜ γ admits a O p L q -Lipschitz filling ˜ f : D Ñ Y .We wish to convert Ψ ˝ ˜ f into a filling f : D Ñ X of γ . Let f : D Ñ X be a O p L q -Lipschitz map which agrees with Ψ ˝ ˜ f on vertices ofour triangulation—such a map exists because we may fill edges with con-stant speed geodesics and then fill triangles in X by Proposition 3.5, asΨ ˝ ˜ f maps adjacent vertices to within O p q of each other. This gives us a O p L q -Lipschitz filling of f | S . Note that the distance from γ to f | S in theuniform metric is O p q because on f | S agrees with Ψ ˝ ψ ˝ γ on vertices.Now we take a new subdivision of D , using Proposition 3.2 to subdi-vide D into an inner disk surrounded by O p L q sectors which are uniformlybilipschitz to D ǫ . We build our filling f of γ as follows. On the inner disk,let f be given by a rescaled copy f . On radial segments, take f to be aconstant speed geodesic, and fill sectors using Proposition 3.5. We now specialize to the case where X is equal to some simply connected Liegroup G equipped with a left-invariant Riemannian metric. Our main goalin this section is to reduce questions about filling loops in G to questionsabout manipulating words in some compact generating set for G . Notation.
Any simply connected Lie group G admits a compact generat-ing set S . The set S ˚ consists of all words s s . . . s ℓ where s , . . . , s ℓ P S and ℓ P N , together with the empty word ε . The length of a word w P S ˚ will bedenoted ℓ p w q . Given w “ s . . . s ℓ P S ˚ , w ´ denotes the word s ´ ℓ . . . s ´ .If w P S ˚ represents the identity element 1 G of G , w is said to be a relation.If w, w P S ˚ represent the same element of G , we write w “ G w . The wordnorm with respect to S will be denoted by | ¨ | S . That is for g P G , we define | g | S to be inf t ℓ p w q : g “ G w P S ˚ u . 9 ssumptions. Any time we take a compact generating set S for G , weassume without loss of generality that S is symmetric—meaning s P S ô s ´ P S —and that 1 G P S , unless otherwise indicated. It suffices to fill over the unit square.
It will be convenient to considerloops as maps r , sÑ X rather than S Ñ X and fillings as maps from theunit square r , s ˆ r , sÑ X rather than D Ñ X . Definition.
Given β : r , sÑ X with β p q “ β p q , a filling of β over theunit square is a map f : r , s ˆ r , sÑ X with the following properties: • f agrees with β on the bottom edge of the unit square, meaning that f p t, q “ β p t q for all t P r , s . • f is constant on the set on the other three edges of the unit square,meaning that f p x, y q “ β p q for all p x, y q P pr , s ˆ q Y p ˆ r , sq Y p ˆ r , sq . We define Span p β q to be the infimum of Lip p f q as f ranges over Lipschitzfillings of β over the unit square.Similarly, given β, γ : r , sÑ X , a homotopy from β to γ is a map r , s ˆ r , sÑ X such that f p t, q “ γ p t q , f p t, q “ β p t q and the restrictions f | ˆr , s and f | ˆr , s are constant.We will now see that it makes no difference whether we consider loopsas maps from r , s or S . The key tool is the following lemma. Lemma 4.1.
There exists a constant C such that, given loops β : S Ñ X and γ : S Ñ X such that γ is a reparameterization of β , and given a filling f : D Ñ X of γ , we have Span p β q ď C max t Lip p f q , Lip p β qu . Proof.
An equivalent statement is proved in [10, Lemma 8.13].
Corollary 4.2.
There is a universal constant C ą with the followingproperty. Suppose that γ : S Ñ X is a Lipschitz loop and β : r , sÑ X aLipschitz path with β p t q “ γ p e πit q for all t P r , s . Then Span p β q C ă Span p γ q ă C Span p β q . roof. Let β : Bpr , s qÑ X be equal to β on the bottom edge and β p q on the other three edges, and fix a bilipschitz map ψ : D Ñr , s ˆ r , s .Observe that γ is a reparameterization of β ˝ ψ : S Ñ X . If f : D Ñ X is aLipschitz filling of γ , then by Lemma 4.1 there is some O p Lip p f qq -Lipschitzfilling ˜ f : D Ñ X of β ˝ ψ , so ˜ f ˝ ψ ´ is a O p Lip p f qq -Lipschitz filling of β . We see that Span p β q “ O p Span p γ qq . The reverse inequality is provedsimilarly. It suffices to fill words.
Let S be a compact generating set for G . Foreach s P S choose a Lipschitz curve γ s : r , sÑ G connecting 1 to s , suchthat the following properties hold. • There is some uniform bound on Lip p γ s q as s ranges over S . • γ G is constant. • γ s p ´ t q “ γ s ´ p t q for all s P S and t P r , s .Given a word w “ s s . . . s ℓ P S ˚ , let γ w : r , sÑ G denote the concate-nation of the paths γ s , s γ s , . . . , s . . . s ℓ ´ γ s ℓ , reparameterized so that the i -th of these paths is used on r i ´ ℓ , iℓ s . That is, for 0 ď t ď i “ , . . . ℓ ,we have γ w ˆ i ´ ` tℓ ˙ “ s s . . . s i ´ γ s i p t q . Proposition 4.3.
Suppose that there exists a constant C such that for every w P S ˚ with w “ G G , there exists a Cℓ p w q -Lipschitz filling of γ w over theunit square. Then G is Lipschitz 1-connected.Proof. Note that G “ Ť k “ S k , and each S k is compact. Hence, by the Bairecategory theorem, some power S k must contain an open neighborhood, so S k contains some open neighborhood of the identity, which in turn containsthe r -ball around the identity for some r ą
0. Take K to be a natural numberlarger than kr , so that S K contains the 1-ball around the identity in G .Given a loop γ : r , sÑ G , we produce a O p Lip p γ qq -Lipschitz filling f : r , s Ñ G of γ as follows. Let n “ r Lip p γ q s —by Propositions 3.3 and 3.5it suffices to only consider γ such that n is at least 2. Cellularly decomposethe unit square into the rectangle r , s ˆ “ n , ‰ together with the collectionof squares “ in , i ` n ‰ ˆ “ , n ‰ as i runs from 0 to n ´ f to be constant on all the vertical edges. Note that for each i ,the element γ ` in ˘ ´ γ ` i ` n ˘ lies in the 1-ball in G , so it may be represented11y some w i P S K . On the horizontal edge connecting ` in , n ˘ to ` i ` n , n ˘ ,let f be a copy of γ w i translated by γ ` in ˘ —that is for 0 ď t ď
1, take f ` i ` tn , n ˘ “ γ ` in ˘ γ w i p t q . Letting w “ w . . . w n , observe that f agrees with γ w the the bottom of the rectangle r , s ˆ “ n , ‰ , so f may be extended overthis rectangle withLip p f q “ O p Lip p γ w qq “ O p ℓ p w qq “ O p Lip p γ qq where the first bound comes from our hypothesis. Finally, f admits a O p Lip p γ qq -Lipschitz extension over each square by Proposition 3.5. Definition.
Suppose C is a collection of pairs p v, w q where v, w P S ˚ arewords in S . We say that v w for p v, w q P C if there exists some constant C depending only on C with the following property: given p v, w q P C , themap from Bpr , s ˆ r , sq to X which is equal to γ v on the bottom edge, γ w on the top edge, and constant on the sides admits a Cℓ p v q -Lipschitz filling.We now give some basic rules for manipulating fillings. Often, the set C will be inferred from context. Lemma 4.4.
The following rules hold in any simply connected Lie group G . • Suppose C Ă S ˚ ˆ S ˚ . Then v w for p v, w q P C if and only if vw ε for p v, w q P C and Lip p w q “ O p Lip p v qq for each pair p v, w q P C . • Fix n , and suppose given sets C i Ă S ˚ ˆ S ˚ for i “ to n . If, for each i “ to n , we have v w for p v, w q P C i , then v . . . v n w . . . w n for all p v , w q P C , . . . , p v n , w n q P C n . • Given C , C Ă S ˚ ˆ S ˚ , if v v for p v , v q P C and v v for p v , v q P C , then v v for pairs p v , v q such that there exists v P S ˚ with p v , v q P C and p v , v q P C . • ww ´ ε for w P S ˚ . • Let U Ă G be an bounded neighborhood of the identity in G . If D is thecollection of relations w “ w . . . w ℓ P S ˚ such every prefix w . . . w i represents an element of U , then w ε for w P D . • G is Lipschitz 1-connected if and only if and v ε for v P S ˚ . roof. The first item follows immediately from Lemma 4.1.To prove the second item, subdivide the unit square into rectanglesof the form “ i ´ n , in ‰ ˆ r , s for i “ , . . . , n . Suppose given p v , w q P C , . . . , p v n , w n q P C n . By hypothesis, there exists a homotopy f i : r , s ˆr , sÑ G from v i to w i with Lip p f i q “ O p Lip p v i qq . Define a map f : r , s ˆ r , s by putting a rescaled copy of f i in “ i ´ n , in ‰ ˆ r , s —that is, for0 ď t ď i “ , . . . , n , we take f ˆ i ´ ` tn , y ˙ “ f i p t, y q . Thus Lip p f q “ O p max t n Lip p f i q : i “ . . . n uq “ O p Lip p v qq because n isfixed. Since f restricted to the top of the unit square is a reparameteri-zation of v . . . v n , and f restricted to the bottom of the unit square is areparameterization of w . . . w n , we are done.The proof of the third item is similar: divide the unit square into tworectangles r , s ˆ “ , ‰ and r , s ˆ “ , ‰ and put a rescaled copy of thehomotopy from v to v into the bottom rectangle and a rescaled copy of thehomotopy from v to v into the top rectangle to get the desired homotopy v to v .To prove the fourth item, let w P S ˚ , and let f : r , s ˆ r , s be definedby f p t, s q “ γ w p t , min t t ´ s, ´ t ´ s uuq . The reader may check that f is a Lip p γ ww ´ q -Lipschitz filling of γ ww ´ .To prove the fifth item, let K be such that S K contains U . Given w P D ,let ℓ “ ℓ p w q so that we can write w “ s . . . s ℓ . We will find a O p ℓ q filling f of γ w as follows. Subdivide r , sˆr , s into the rectangle r , sˆ “ ℓ , ‰ togetherwith the squares of the form “ i ´ ℓ , iℓ ‰ ˆ “ , ℓ ‰ . Define f to be constant on r , s ˆ “ ℓ , ‰ . For each i , choose w i P S K such that w i “ G s . . . s i and take f to be γ w i on the vertical edge iℓ ˆ “ , ℓ ‰ —that is f ˆ iℓ , ´ tℓ ˙ “ γ w i p t q for 0 ď t ď i “ , . . . , n . We may extend f over each square withLipschitz constant O p ℓ q by Proposition 3.5.The sixth item is a consequence of Proposition 4.3. We now discuss normal forms ω and ω -triangles. Using a technique of Gro-mov, we shall see that G is Lipschitz 1-connected if ω -triangles are Lipschitz13-connected. Definition.
A normal form for a compactly generated group G equippedwith compact generating set S is a map ω : G Ñ S ˚ such that ℓ p ω p g qq “ O p| g | S q for g P G . If ω is a normal form, then an ω -triangle is a word in S ˚ of the form ω p g q ω p g q ω p g q where g g g “ G . Lemma 4.5.
Let ω : G Ñ S ˚ be a normal form. If ∆ ε for ω -triangles ∆ , then G is Lipschitz 1-connected.Proof. This is proved in [10, Proposition 8.14]. The proof is sketched inFigure 2.
Assumptions.
From here on, we specialize to the case where G “ U ¸ A ,where U and A are contractible Lie groups, with A abelian and U a closedgroup of strictly upper triangular real matrices. Standard solvable groups.
Observe that A acts on the abelianization U {r U, U s . Fix a norm on the vector space U {r U, U s . If there is no vector X such that lim n Ñ8 n log } a n ¨ X }Ñ a P A , we say that G is standardsolvable.In this section we will be interested in the structure and geometry ofstandard solvable groups. § G may be presented as the freeproduct of its standard tame subgroups modulo certain easily understoodamalgamation relations. 14 ω p γ p qq γ ω p γ p q ´ q γ ω p γ p qq γ ω p γ p q ´ γ p qq γ ω p γ p q ´ γ p qq γ ω p γ p q ´ q Figure 2: This figure indicates how to fill γ “ γ w for an arbitrary relation w P S ˚ given that one knows how to fill ω -triangles. The top edge andall the vertical edges are taken to be constant, and each horizontal edge isunderstood to be an appropriate translate of its label, so that each rectanglerepresents an ω -triangle, except the bottom row. The bottom edge is takento be γ w , and there is a row of squares along the bottom which may be filledby Proposition 3.5 Let G “ U ¸ A be standard solvable, and let u be the Lie algebra of U ,identified as usual with the tangent space of U at 1 U , and fix any norm } ¨ } on u . For a P A we denote the conjugation action of a on u by Ad p a q , so thatAd p a q X “ ddt ˇˇ t “ a ´ exp p tX q a . Observe that Hom p A, R q is a vector space. Definition. (See [4, § α : A Ñ R , define the α -weight space u α Ă u to consist of 0 together with all X P u such that for all a P A , lim n Ñ8 n log } Ad p a q n X } “ α p a q . Define the set of weights W to consist of all α P Hom p A, R q for which15im u α ą
0. By a conic subset, we mean the intersection of W with anopen, convex cone in Hom p A, R q that does not contain 0. Denote the setof all conic subsets by C . For C P C , let U C denote the closed connectedsubgroup of U whose Lie algebra is À α P C u α , and let G C “ U C ¸ A . Thesegroups G C are referred to as standard tame subgroups of G (see remarks in § C is finite, that u “ À α P W u α ,and that r u α , u β s Ă u α ` β for α, β P W [4, § H p u q andrecall the definition of Kill p u q so that we will be able to state Theorem 6.1,our main theorem. Definition.
Let d : Ź u Ñ Ź u and d : Ź u Ñ u be the maps of A -modules induced by taking d p x ^ y ^ z q “ r x, y s ^ z ` r y, z s ^ x ` r z, x s ^ yd p x ^ y q “ ´r x, y s . Define H p u q “ ker p d q{ image p d q . Define Kill p u q to be the quotient of the symmetric square u d u by thesubspace spanned by elements of the form r x, y s d z ´ y d r x, z s . H p u q and Kill p u q are A -representations. Observe that the natural A -action on Ź u descends to an A -action on H p u q because, by the Jacobi-likeidentity Ad p a qr X, Y s “ r Ad p a q X, Y s`r X, Ad p a q Y s , the subspaces image p d q and ker p d q are preserved by the action of A . Similarly, Kill p u q is also an A -representation. Recall that, for an A -representation V we define V toconsist of 0 together with vectors X such that lim n Ñ8 n log } a n ¨ X } “ a P A . We thus define the subspaces H p u q and Kill p u q . Definition.
Given a P A , a vacuum subset for a is a compact Ω Ă U suchthat for every compact K Ă U , there is some n ą p a q n K Ă Ω. Wesay that G is tame if there exists a P A with a vacuum subset. G is tame if and only if there is some a P A with α p a q ă α P W ,so G C is tame for C P C [4, Proposition 4.B.5]. We now wish to show thatif G is tame, then it is Lipschitz 1-connected. Our starting point is thefollowing. 16 roposition 5.1. Suppose G is tame, then there is some a P A and acompact generating set S Ă U for U such that Ad p a q S Ă S .Proof. By hypothesis, there is some b P A with a vacuum subset Ω. Let S be a compact generating set for U . As in the proof of Proposition 4.3,we see that some power of S contains an open ball around the identity,hence for some M ą
0, the set S M contains Ω. As Ω is a vacuum set for b , there exists L such that Ad p b q L S M Ă Ω. Taking a “ b L and S “ S M ,we have that S is a generating set because it contains S (because 1 P S by our standing assumption that generating sets contain the identity), andAd p a q S “ Ad p b q L S M Ă Ω Ă S as desired. Proposition 5.2. If G “ U ¸ A is tame, then G is Lipschitz 1-connected. Remark.
Probably, a tame group G is CAT(0) for some choice of metric,but we do not know how to prove this, so we give a combinatorial proofusing Lemma 4.5. Proof.
Fix a P A and a compact generating set S U Ă U such that Ad p a q S U Ă S U as in Proposition 5.1, let S A be a generating set for A with a P S A , andlet S “ S U Y S A , so that S is a generating set for G . Note that Ad p a qp s q “ s for s P S A , and observe that if ℓ ą r log j s ą
0, thenAd p a q ℓ ´ S jU ¯ Ă Ad p a q ℓ ´ ˆ S r j s U ˙ Ă Ad p a q ℓ ´ ˆ S r r j ss U ˙ Ă . . . Ă Ad p a q ℓ ´ r log j s p S U q Ă S U , because the function f : j ÞÑ r j s satisfies f r log j s j “ j . In other words, given u P U , there exist k ď r log | u | S U s and s P S U suchthat u “ G a k sa ´ k . It follows, letting φ A : G Ñ A denote projection, that wemay define a normal form ω : G Ñ S ˚ such that • for any g P G , the word ω p g q is given by ω p φ A p g qq ω p φ A p g q ´ g q , • for g P A , the word ω p g q P S ˚ A is a minimal length word representing g , • and for g P U zt u , ω p g q is of the form a k sa ´ k where s P S U and0 ď k “ O p log | g | S U q . 17o check that this is a normal form—i.e., that ℓ p ω p g qq “ O p| g | S q , note that | φ A p g q| S ď | g | S , and | φ A p g q ´ g | S U “ O p exp | g | S q , so that | φ A p g q ´ p g q| S “ O p log p O p exp | g | S qqq “ O p| g | S q because U is at most exponentially distorted in G [4, Proposition 6.B.2].It will suffice to show that ∆ ε for ω -triangles ∆ P S ˚ . An ω -triangle∆ has the form a ω p u q a ω p u q a ω p u q , where for i “ , ,
3, we have u i P U and a i P S ˚ A with the a i and u i satisfying p Ad p a a q u qp Ad p a q u q u “
1. To show that ∆ ε , it thussuffices to establish the following two facts. • ω p u q b bω p b ´ ub q for u P U and b P S ˚ A . • ω p u q ω p u q ω p u q ε for u , u , u P U such that u u u “ Conjugation.
To show that ω p u q b bω p b ´ ub q for u P U and b P S ˚ A ,note that ω p u q may be written as a k sa ´ k for some s P S U and k ě
0. BecauseAd p a q S Ă S , there is some C ě a P S A , we haveAd p a q C Ad p a q S Ă S . Let K “ Cℓ p b q ` k , so that Ad p a q K ´ k Ad p b q S Ă S for any word b P S ˚ A with ℓ p b q ď ℓ p b q , and let s , s P S be given by s “ Ad p a q K ´ k s and s “ b ´ s b . We homotope as follows, liberally usingLemma 4.4. ω p u q b “ a k sa ´ k b a K a k ´ K sa K ´ k a ´ K b a K s ba ´ K a K bb ´ s ba ´ K ba K s a ´ K bω p b ´ ub q . Filling ω -triangles in U . To show that ω p u q ω p u q ω p u q ε for u , u , u P U such that u u u “
1, write ω p u i q as a k i s i a ´ k i for i “ , ,
3, with k i ě s i P S U . Let K “ k ` k ` k and let s i “ Ad p a k i ´ K q s i P S U . Wehomotope as follows. ω p u q ω p u q ω p u q “ p a k s a ´ k qp a k s a ´ k qp a k s a ´ k q p a K a k ´ K s a K ´ k a ´ K qp a K a k ´ K s a K ´ k a ´ K qp a K a k ´ K s a K ´ k a ´ K q p a K s a ´ K qp a K s a ´ K qp a K s a ´ K q a K s s s a ´ K a K a ´ K ε. .3 Filling freely trivial words. We now return to the case where the standard solvable group G “ U ¸ A is not necessarily tame. Recall that the collection of conic subsets C isfinite. For C P C , let G C be the tame group U C ¸ A , and let S G C be acompact generating set for this group. Let H “ ˚ C P C G C , and let S H Ă H be the union of the S G C . There is a natural map from H to G . Lemma5.6 will show that if w P S ˚ H represents the identity in H , then its image in G admits a O p ℓ p w qq -Lipschitz filling. We will need the following auxiliaryresults first—the reader should probably skip directly to the proof of Lemma5.6 to understand the point of these propositions. • Proposition 5.3 shows that a word w P S ˚ H representing the identityin H may be reduced to the identity by repeated deletion of subwords r j P S ˚ G Cj such that r j represents the identity in G C j . • Proposition 5.5 describes an appropriately Lipschitz rectangular ho-motopy between words obtained by these deletions. • Proposition 5.4 describes a part of the homotopy given in Proposition5.5.
Proposition 5.3.
Given a word w P S ˚ H such that w “ H , there exists anatural number n ď ℓ p w q and, for j “ , . . . , n , words a j , r j , b j P S ˚ H withthe following properties. • w “ S ˚ H a r b . • a n , r n , b n “ ε . • For all j “ , . . . , n , there is some C j P C such that r j P S ˚ G Cj and r j “ G Cj . • a j b j “ S ˚ H a j ` r j ` b j ` for j “ , . . . , n ´ .Proof. Note that each element of S H lies in some G C . If w ‰ ε represents theidentity, then by the theory of free products, w has a (nonempty) subwordwhich is comprised of elements of some S G C and represents the identityin G C . Thus, we may write w “ S ˚ H a r b where r P S ˚ G C as desired.Applying this argument recursively, we obtain a j , r j , b j as desired. Proposition 5.4.
Given a natural number k and w P S ˚ H , there exists a O p k ` ℓ p w qq Lipschitz map f : r , s ˆ ” , kk ` ℓ p w q ı Ñ G with the followingproperties. Along the bottom, f is given by k w , meaning f p t, q “ γ k w p t q for ď t ď . • Along the top, f is given by w k , meaning f ´ t, kk ` ℓ p w q ¯ “ γ w k p t q for ď t ď . • f is constant on the sides, so f p , s q and f p , s q do not depend on s .Proof. Let γ : R Ñ G be given as follows. γ p t q “ G for t ď γ p t q “ γ k w p t q for 0 ă t ď
1, and γ p t q “ γ k w p q for t ą
1. Observe that Lip p γ q “ O p k ` ℓ p w qq Then we may take f p t, s q “ γ p t ` s q as our desired filling. Proposition 5.5.
Given a, b P S ˚ H , r P S ˚ G C a relation in G C for some C P C , and any natural number k , let ℓ “ ℓ p arb k q and h “ ℓ p r q ℓ . There existsa O p ℓ q -Lipschitz map f : r , s ˆ r , h s Ñ G with the following properties. • Along the bottom, f is given by arb k , meaning f p t, q “ γ arb k p t q for t P r , s . • Along the top, f is given by ab k ` ℓ p r q , meaning f p t, h q “ γ ab k ` ℓ p r q p t q for t P r , s . • f is constant on the sides, so f p , s q and f p , s q do not depend on s .Proof. (See the right hand side of Figure 3.) Subdivide the rectangle into r , sˆr , h { s and r , sˆr h { , s . Define f to be a ℓ p r q b k on r , sˆt h { u —i.e., f p t, h { q “ γ a ℓ p r q b k p t q . First, we extend f over the top rectangle r , s ˆ r h { , s . For p t, s q P „ , ℓ p a q ℓ ˆ r h { , s Y „ ´ kℓ , ˆ r h { , s we have that γ a r b k p t q “ γ ab ℓ p r q` k p t q , so we can just set f to be constantvertically—i.e., we define f p t, s q “ γ ab ℓ p r q` k p t q for these p t, s q . To extend f to p t, s q P „ ℓ p a q ℓ , ´ kℓ ˆ r h { , s ,
20e simply apply Proposition 5.4. Thus, we have given a O p ℓ q -Lipschitzextension of f over the top rectangle.Now we extend over the bottom rectangle r , s ˆ r , h { s . For p t, s q P „ , ℓ p a q ℓ ˆ r , h { s Y „ ℓ p ar q ℓ , ˆ r , h { s define f p t, s q “ γ arb k p t q . Finally, we apply Proposition 5.2 to extend f over ” ℓ p a q ℓ , ℓ p ar q ℓ ı ˆ r , h { s , since this is equivalent to filling r . Lemma 5.6.
Recalling the notation introduced at the start of § w ε (in G ) for all w P S ˚ H such that w “ H Proof. (See Figure 3.) Let w P S ˚ H be a relation in H and take a sequence ofwords a j , r j , b j , j “ , . . . , n as in Proposition 5.3. We will define a O p ℓ p w qq -Lipschitz filling f : r , s ˆ r , sÑ G of γ w as follows. Let ℓ k “ ř j ă k ℓ p r j q ,so that ℓ “ ℓ n “ ℓ p w q , and subdivide r , s ˆ r , s into rectangles r , s ˆ r ℓ j , ℓ j ` s for j “ , . . . , n ´
1. Set f p t, ℓ j q “ γ a j r j b j ℓj p t q , noting that ℓ p a j r j b j ℓ j q “ ℓ p w q .Proposition 5.5 now shows that f may be extended over each rectangle r , s ˆ r ℓ j , ℓ j ` s with Lipschitz constant O p ℓ p w qq . Distortion.
We now see that if G is standard solvable, then elements of U may be expressed much more efficiently in the generators of G than inthe generators of U . Proposition 5.7.
Suppose G “ U ¸ A is standard solvable, S is a compactgenerating set for G , and S U is a compact generating set for U .There exists C ą such that if u P U zt U u , then C log p ` | u | S U q ď | u | S ď C log p ` | u | S U q . Proof.
This follows, with some effort, from [4, Proposition 6.B.2].21 r b a r b ℓ a r b ℓ . . .r n ´ ℓ n ´ ℓ n a r b k r ba b k ` ℓ p r q Prop. 5.2Proposition 5.4
Figure 3: The figure on the left depicts our strategy for filling of the freelytrivial word w “ a r b , where the a j , r j , b j are as in Proposition 5.3. Thefigure on the right depicts the proof of Proposition 5.5 which allows us tofill in each rectangle in the left hand figure. In this subsection, we will define the multiamalgam ˆ G of a standard solvablegroup G “ U ¸ A (first introduced by Abels [1]), and quote a key theoremof Cornulier and Tessera, which states that certain conditions under which G – ˆ G —this means that G is put together from its standard tame subgroupsin a nice way, which will eventually let us build fillings in G from fillingsin standard tame subgroups. In order to state this theorem in the propergenerality, we must briefly discuss the theory of unipotent groups. Unipotent groups.
For a commutative R -algebra P , and a real unipotentgroup U (i.e., a closed group of upper triangular real matrices with diagonalentries equal to 1), the theory of algebraic groups allows us to define a group U p P q [2, § U Ă GL p n ; R q consists of all upper triangularmatrices with diagonal entries equal to 1, then U p P q consists of all uppertriangular n ˆ n matrices over P with diagonal entries equal to 1—suchmatrices are certainly invertible, having determinant equal to 1. Suppose P “ R Y , so that P consists of all functions f : Y Ñ R . Then there is anobvious bijection U p P q Ø U Y , and for y P Y and ˜ u P U p P q we may speakof ˜ u p y q P U . 22 efinition. (See [4, § G “ U ¸ A be real standard solvable. Themultiamalgam ˆ G of the standard tame subgroups G C is defined byˆ G “ ˚ C P C G C {xx R G yy where R G “ t i C p u q ´ i C p u q : u P G C X G C u and i C denotes the inclusion of G C in the direct product.Similarly, the multiamalgam ˆ U is defined byˆ U “ ˚ C P C U C {xx R U yy where R U “ t i C p u q ´ i C p u q : u P U C X U C u and i C denotes the inclusion of U C in the direct product.For any commutative R -algebra P , we define ˆ U p P q and ˆ G p P q similarly,where G C p P q is understood to be U C p P q ¸ A .Of course, ˆ U ¸ A – ˆ G . Recall that G admits the SOL obstruction if itsurjects onto a group of SOL type. Cornulier and Tessera give conditionsunder which ˆ U is isomorphic to U . Theorem 5.8.
Let G “ U ¸ A be a standard solvable real Lie group. If H p u q “ , Kill p u q “ , and G does not admit the SOL obstruction then ˆ U p P q – U p P q for all commutative R -algebras P .Proof. This follows from Corollary 9.D.4 of [4]. The 2-tameness hypothesesof the corollary is satisfied because of Proposition 4.C.3 of [4].
The rest of this paper is devoted to the proof of the following theorem.
Theorem 6.1.
Let G “ U ¸ A where U and A are contractible real Liegroups, A is abelian, and U is a real unipotent group (i.e., a closed group ofstrictly upper triangular real matrices.)If G is standard solvable and does not surject onto a group of Sol type,and H p u q and Kill p u q are trivial, then G is Lipschitz -connected.Proof. Lemma 6.2 will show that there exists a generating set S for G andnormal form ω : G Ñ S ˚ with certain properties.Lemma 6.6 will show that if ω has these properties, then ∆ ε for ω -triangles ∆. By Lemma 4.5, this will suffice to prove the theorem.23 .1 Defining ω . Standing assumptions.
Throughout the rest of this paper, we assumethat G satisfies the hypotheses of the theorem. That is, G “ U ¸ A is astandard solvable group such that H p u q “
0, Kill p u q “
0, and G doesnot surject onto a group of Sol type. Notation.
Let H “ ˚ C P C G C and H U r P s “ ˚ C P C U C p P q for any commu-tative R algebra P , and let i C : U C p P qÑ H U r P s denote inclusion. We willwrite H U for H U r R s . Let S A be a compact generating set for A . For C P C ,let S C be a compact generating set for U C . Let S U “ Ť C P C S C —by The-orem 5.8 this is a compact generating set for U . Let S “ S A Y S U , this isa compact generating set for G . Let S H be a generating set for H whichis equal to the union of compact generating sets for G C as C ranges over C , and let S H U Ă H U be the union of the S C —this is a generating set for H U . Given C P C and x P U C , let x P p S A Y S C q ˚ be a minimal length wordrepresenting x . Let φ A : G Ñ A be projection. Let the set theoretic map φ U : G Ñ U be defined by φ U p g q “ φ A p g q ´ g , so that g “ φ A p g q φ U p g q . Lemma 6.2.
Under our standing assumptions, there exists a finite sequence C . . . C k of conic subsets and a normal form ω : G Ñ S ˚ such that ω has thefollowing properties. • For any g P G , ω p g q “ ω p φ A p g qq ω p φ U p g qq . • For a P A , ω p a q P S ˚ A is a minimal length word representing a . • For u P U , ω p u q has the form x . . . x k , where x i P U C i .Proof. This follows from Proposition 6.B.2 of [4], but we will now give adifferent proof in order to introduce a trick that will be used later.
The Cornulier-Tessera trick.
For a set Y , define the commutative R -algebra P Y as the collection of all functions f : Y ˆr , R such that thereis some β P N with | f p y, t q| ă p ` t q β . Note that an element of U p P Y q maybe identified with a function from Y ˆ r , to U . Alternatively, one maythink of an element of U p P Y q as a family of functions r , U , indexed by Y with matrix coefficients uniformly bounded by some polynomial p ` t q β .Choose Y to have at least continuum cardinality, and let ˜ g P U p P Y q besuch that for every g P U , there is some y P Y and t “ O p| g | S U q such that˜ g p y, t q “ g . It is certainly possible to do this—for instance, one might take Y “ U and set ˜ g p y, t q to be 1 U for t ă | y | S U and y for t ě | y | S U . In claiming24hat ˜ g P U p P Y q , we have used the fact that the matrix coefficients of g P U are at most polynomial in | g | S U .Since, by Theorem 5.8, U p P Y q is generated by the union of the U C p P Y q ,we may write ˜ g “ ˜ x . . . ˜ x k where each ˜ x i is an element of some U C i p P Y q .Observe that there exists some α P N such that | ˜ x i p y, t q| S Ci ď p ` t q α forall i “ , . . . , k , y P Y , and t ě g be an element of U . By definition of ˜ g , there exist y P Y and t “ O p| g | S U q such that ˜ g p y, t q “ g . For i “ , . . . , k , let x i “ ˜ x i p y, t q . Wehave that g “ ˜ g p y, t q “ ˜ x p y, t q . . . ˜ x k p y, t q “ x . . . x k , and x i P U C i p P Y q with | x i | S Ci “ O p t α q “ O p| g | α S U q . Take ω p g q to be x . . . x k ,and note that | ω p g q| S “ O p log | g | S U q because | x i | S A Y S Ci “ O p log | x i | S Ci q “ O p log | g | S U q by 5.7.We have thus defined ω on elements of U , with the desired properties.Define ω on A by taking ω p a q to be the shortest word in S ˚ A representing a P A . Extend ω to all of g by setting ω p g q “ ω p φ A p g qq ω p φ U p g qq . We mustshow that ω is a normal form—i.e., that, for g P G , ℓ p ω p g qq “ O p| g | S q .We have ℓ p ω p φ A p g qqq “ | φ A p g q| S “ O p| g | S q , so it suffices to show that ℓ p ω p φ U p g qqq “ O p| g | S q . If w P S ˚ is a minimal length word representingsome g P G , note that ω p φ A p g qq ´ w represents φ U p g q . By 5.7, there is someconstant C ą g such that | ω p φ A p g qq ´ w | S ě C log | φ U p g q| S U . Thus, since | ω p φ U p g qq| S “ O p log | φ U p g q| S U q , we have that ω p φ U p g qq “ O p| φ A p g q| S ` | w | S q “ O p| g | S q as desired. ω -triangles. We wish to show that we may fill ω triangles, where ω is a normal formproduced by Lemma 6.2. Proposition 6.5 will allow us to homotope ω -triangles into relations of the form x i . . . x K where each x i is a word in S A Y S C i efficiently representing an element x i of U C i , where C , . . . , C K issome fixed sequence of conic subsets. In order to fill such relations, recallfrom Theorem 5.8 that under our standing assumptions, U p P q – ˆ U p P q forany commutative R -algebra P . Consequently, the kernel of H U r P sÑ U p P q is normally generated by elements of the form i C p u q ´ i C p u q r u P U C p P q X U C p P q . s
25n order to fill x . . . x K in Corollary 6.4, we will need to factor x . . . x K in the free product H U as a product of a bounded number of elements ofthe form g ´ i C p u q ´ i C p u q g , where each g P H U is a product of a boundednumber of elements living in some factors U C , with | g | S U and | u | S U controlledby some polynomial of ř Kj “ | x j | S Cj . Lemma 6.3. (See [4, Lemma 7.B.1]). Suppose that our standing assump-tions are satisfied. Given a sequence of conical subsets C , . . . C K , thereexist natural numbers N, µ, β such that for any sequence x i P U C i with x x . . . x K “ U U there is an equality of the form x . . . x K “ H U p g r g ´ q . . . p g N r N g ´ N q satisfying • g j “ H U g j . . . g jµ where g jk P U C jk for some C jk P C . • Each r j is of the form i C j p u j q i C j p u j q ´ for some conical subsets C j , C j and some u j P U C j X U C j . • | g jk | S Cjk “ O p ℓ β q , | u j | S C j “ O p ℓ β q , and | u j | S C j “ O p ℓ β q where ℓ “ ` ř Ki “ | x i | . Proof.
This is a special case of [4, Lemma 7.B.1], but we will reprise mostof the details here. We will use the same Cornulier and Tessera trick weused to prove Lemma 6.2. Take P Y as in the proof of Lemma 6.2. Recallthat ˜ x P U C p P Y q may be thought of as a function from Y ˆ r , to U C .Let Y be a set with at least continuum cardinality, so that there exists p ˜ x , . . . , ˜ x K q P U C p P Y q ˆ . . . ˆ U C K p P Y q which has the following strongsurjectivity property: for any p x . . . x K q P U C p P Y q ˆ . . . ˆ U C K p P Y q with x . . . x K “ U y P Y and t “ O p| x | S C ` . . . ` | x K | S CK q with˜ x i p y, t q “ x i .By Theorem 5.8, we know that there is some equality of the form˜ x . . . ˜ x K “ H U r P Y s p ˜ g ´ ˜ r ˜ g q . . . p ˜ g ´ N ˜ r N ˜ g N q where ˜ g P ˚ U C p P Y q and each ˜ r j has the form i C j p ˜ u j q ´ i C j p ˜ u j q for someconical subsets C j , C j and ˜ u j P U C j p P Y q X U C j p P Y q . Since the U C p P Y q generate H U r P Y s , there must be some µ such that for all j “ , . . . , N ,˜ g j “ ˜ g j . . . ˜ g jµ , g jk lives in U C jk for some C jk P C . Note that by definition of P Y , there is some β such that all | ˜ g jk p y, t q| S U and | ˜ u j p y, t q| S C j are O p t β q .Given, x , . . . , x K P U C ˆ . . . ˆ U C K , let ℓ “ ř Ki “ | x i | S Ci and choose y P Y and t “ O p ℓ q such that ˜ x i p y, t q “ x i for i “ , . . . , K . For j “ , . . . , N and k “ , . . . , µ , let g j “ ˜ g j p y, t q , g jk “ ˜ g jk p y, t q , u j “ ˜ u j p y, t q ,and r j “ i C p u j q ´ i C p u j q . It follows that x . . . x K “ H U p g r g ´ q . . . p g N r N g ´ N q , and the g j and r j satisfy the desired conditions. Corollary 6.4.
Suppose that our standing assumptions are satisfied. Givena sequence of conical subsets C , . . . C K , we have, in G , that x . . . x K ε. for any sequence x i P U C i with x x . . . x K “ U U .Proof. By Lemma 6.3, we have (for β, N, µ independent of the x j ), x . . . x K “ H U p g r g ´ q . . . p g N r N g ´ N q where g j “ H U g j . . . g jµ for g jk P U C jk , each r j is of the form i C p u j q ´ i C p u j q for some conical subsets C j , C j and some u j P U C j X U C j , and | g jk | S Cjk , | u j | S C j “ O p ℓ β q where ℓ “ ` ř Ki “ | x i | . For each j “ , . . . , N , let g j P S ˚ be g j . . . g jµ . Let r j P S ˚ be equalto p u j q ´ u j , where u j P S A Y S C j is a minimal length word representing u j in G C j and u j P S A Y S C j is a minimal length word representing u j in G C j —this implies that r j represents r j in H .First we show that r j ε . Note that C j X C j is itself a conic subset,so there is some u j P p S A Y S C j X C j q ˚ which represents u j in G C j X C j with ℓ p u j q “ O p ℓ p r j qq . Thus we have (by Proposition 5.2), r j “ p u j q ´ u j u j ´ u j ε. Next, observe that ℓ p g j q “ µ ÿ k “ ℓ p g jk q “ µ ÿ k “ O p log | g jk | S Cjk q“ O ˜ β log ˜ ` K ÿ i “ | x i | ¸¸ “ O p ℓ p x . . . x K qq ,
27y Proposition 5.7, and ℓ p r i q “ O p ℓ p x . . . x K qq similarly.Because x . . . x K “ H p g r g ´ q . . . p g N r N g N ´ q , and ℓ pp g r g ´ q . . . p g N r N g N ´ qq “ O p ℓ p x . . . x K qq , we have by Lemma 4.4 and Lemma 5.6 that x . . . x K p g r g ´ q . . . p g N r N g N ´ q p g g ´ q . . . p g N g N ´ q ε because r j ε as noted above.We need one more proposition before we can fill ω -triangles. Given a, x P G , let a x denote axa ´ . Proposition 6.5.
Fix a sequence of conical subsets C , . . . , C K P C . Wehave that ω p a q x . . . x K a x . . . a x ω p a q for all a P A and x i P U C i .Proof. ω p a q x . . . x K p ω p a q x ω p a q ´ qp ω p a q x ω p a q ´ q . . . p ω p a q x K ω p a q ´ q ω p a q a x . . . a x K ω p a q , by Proposition 5.2.We now conclude the proof of our main theorem by showing that we mayfill ω -triangles. Lemma 6.6.
Under our standing assumptions, if g , g , g P G with g g g “ G G , we have ω p g q ω p g q ω p g q ε. Proof.
Recall that ω p g q has the form ω p a q x . . . x k where u i P S A Y S C i forsome fixed sequence C , . . . C k of conical subsets. Let g , g , g P G with g g g “ G , and let a i “ φ A p g i q for i “ , ,
3. Let ω p φ U p g qq “ x . . . x k , ω p φ U p g qq “ x . . . x k and ω p φ U p g qq “ x . . . x k . Expanding and applyingProposition 6.5 repeatedly, we slide the a -words to the right to see that ω p g q ω p g q ω p g q “ ω p a q x . . . x k ω p a q x . . . x k ω p a q x . . . x k a x . . . a x ka a x . . . a a x ka a a x . . . a a a x k ω p a q ω p a q ω p a q a x . . . a x ka a x . . . a a x ka a a x . . . a a a x k . This resulting word admits a Lipschitz filling by Corollary 6.4.28 eferences [1] Herbert Abels. Finite presentability of S-arithmetic groups.
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Pacific Journal of Mathemat-ics , 270(2):393–421, 2014.[10] Robert Young. The dehn function of SL p n, Z q . Annals of Mathematics ,177(3):969–1027, 2013.David Bruce CohenDepartment of MathematicsUniversity of Chicago5734 S. University Avenue,Room 208CChicago, Illinois 60637E-mail: [email protected]@gmail.com