Nonnegative Ricci curvature, metric cones, and virtual abelianness

Let $M$ be an open $n$-manifold with nonnegative Ricci curvature. We prove that if its escape rate is not $1/2$ and its Riemannian universal cover is conic at infinity, that is, every asymptotic cone $(Y,y)$ of the universal cover is a metric cone with vertex $y$, then $\pi_1(M)$ contains an abelian subgroup of finite index. If in addition the universal cover has Euclidean volume growth of constant at least $L$, we can further bound the index by a constant $C(n,L)$.

Geometry & Topology msp Volume 28 (2024) Nonnegative Ricci curvature, metric cones and virtual abelianness 1 Introduction We study the virtual abelianness/nilpotency of fundamental groups of open (complete and noncompact) manifolds with Ric 0. According to the work of Kapovitch and Wilking [8], these fundamental groups always have nilpotent subgroups with index at most C.n/; also see Gromov [7] and Milnor [9].In general, these fundamental groups may not contain any abelian subgroups with finite index, because Wei [16] has constructed examples with torsion-free nilpotent fundamental groups.This is different from manifolds with sec 0, whose fundamental groups are always virtually abelian; see Cheeger and Gromoll [4].
A question raised from here is, for an open manifold M with Ric 0, under what conditions is 1 .M / virtually abelian?To answer this question, one naturally looks for indications from the geometry of nonnegative sectional curvature.Ideally, if the manifold M fulfills some geometric conditions modeled on nonnegative sectional curvature, even in a much weaker form, then 1 .M / may turn out to be virtually abelian.In other words, when 1 .M / is not virtually abelian, some aspects of M should be drastically different from the geometry of nonnegative sectional curvature.
We have explored this direction in [12; 13], from the viewpoint of escape rate.Recall that each element in 1 .M; p/ can be represented by a geodesic loop at p, denoted by c , with the minimal length in its homotopy class.If M has sec 0, then all these representing loops must stay in a bounded ball; however, this property in general does not hold for nonnegative Ricci curvature.The escape rate measures how fast these loops escape from bounded balls: E.M; p/ WD lim sup j j!1 d H .p; c / j j ; where j j is the length of c and d H is the Hausdorff distance.As the main result of [13], if M satisfies Ric 0 and E.M; p/ Ä .n/,then 1 .M / is virtually abelian.We also mention that from the definition, the escape rate always takes values between 0 and 1 2 .To the author's best knowledge, all known examples of open manifolds with Ric 0 have escape rate strictly less than 1  2 .In this paper, we study how virtual abelianness/nilpotency is related to conic asymptotic geometry.Recall that an asymptotic cone of M is the pointed Gromov-Hausdorff limit of a sequence .r 1 i M; p/ GH !.Y; y/; where r i ! 1.We say that M is conic at infinity if any asymptotic cone .Y; y/ of M is a metric cone with vertex y.We do not assume the asymptotic cone to be unique in this definition.If the manifold has sec 0, then its asymptotic cone .Y; y/ is unique as a metric cone with vertex y.If the manifold has Ric 0 and Euclidean volume growth, then it is conic at infinity; see Cheeger and Colding [1].
We state the main result of this paper.
(1) If its Riemannian universal cover is conic at infinity, then 1 .M / is virtually abelian.
(2) If its Riemannian universal cover has Euclidean volume growth of constant at least L, then 1 .M / has an abelian subgroup of index at most C.n; L/, a constant only depending on n and L.
The contrapositive of Theorem 1.1 (1) shows that the nilpotency of 1 .M / leads to asymptotic geometry of the universal cover that is very different from the one with nonnegative sectional curvature; also see Conjecture 1.3.
Before proceeding further, we make some comments about the conditions in Theorem 1.1.
We emphasize that in Theorem 1.1 (1), the conic at infinity condition is imposed on the Riemannian universal cover of M, not M itself.In fact, Wei's example [16] has the half-line .OE0; 1/; 0/ as the unique asymptotic cone of M. Therefore, in general, virtual abelianness does not hold when M is conic at infinity.
Regarding the condition E.M; p/ ¤ 1 2 in Theorem 1.1, it is unclear to the author whether it can be dropped.On the one hand, at present we do not know any examples with Ric 0 and E.M; p/ D 1 2 .On the other hand, we are unable to show E.M; p/ ¤ 1 2 even when 1 .M / D Z.
Question 1.2 Let .M; p/ be an open n-manifold with Ric 0 and a finitely generated fundamental group.Is it true that E.M; p/ < 1 2 ?
The converse of Question 1.2 is known to be true: if E.M; p/ < 1 2 , then 1 .M / is finitely generated; see Sormani [15, Lemma 5].We believe that answering Question 1.2 will lead to a better understanding of the Milnor conjecture [9]: the fundamental group of any open n-manifold with Ric 0 is finitely generated.

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We compare Theorem 1.1 with previous results.In [11; 10] we have shown that if the universal cover is conic at infinity and satisfies certain stability conditions (for example, the universal cover has a unique asymptotic cone), then 1 .M / is finitely generated and virtually abelian; in fact, these manifolds have zero escape rate [12,Corollary 4.7].In contrast, here we do not assume any additional stability conditions in Theorem 1.1 (1), and the escape rate may not be equal or close to 0 in general.In Theorem 1.1(2), if L is sufficiently close to 1, then the universal cover fulfills the stability condition in [10] and thus E.M; p/ D 0. Theorem 1.1(2) also confirms [10, Conjecture 0.2] on the condition E.M; p/ ¤ 1 2 .
We briefly state our approach to proving Theorem 1.1 and give some indications as to why 1 .M / cannot be the discrete Heisenberg 3-group H 3 .Z/.Let be a generator of the center of H 3 .Z/.Then b can be expressed as a word in terms of two elements ˛and ˇoutside the center; moreover, this word has word length comparable to b 1=2 as b ! 1.This expression provides an upper bound on the length growth of : for all b large.If one could find a lower bound violating the above upper bound, then it would end in a desired contradiction.To prove a lower bound, we study the equivariant asymptotic cones of .z M ; h i/: .r where r i ! 1, z M is the universal cover of M, and h i is the group generated by .The limit .Y; y; H / may depend on the sequence r i , so we shall study all equivariant asymptotic cones.One key intermediate step is to show that the asymptotic orbit Hy is always homeomorphic to R (Proposition 4.1).This actually requires some understanding of equivariant asymptotic cones of .z M ; 1 .M; p// beforehand.Therefore, we shall first show that for any equivariant asymptotic cone .Y; y; G/ of .z M ; 1 .M; p//, the asymptotic orbit Gy is homeomorphic to R k (Proposition 3.6), by applying a critical rescaling argument and the metric cone structure.After knowing the orbit Hy is homeomorphic to R, we can further deduce some uniform controls on Hy by using the metric cone structure; see Lemmas 4.8 and 4.9.These estimates lead to an almost linear growth estimate of : for any s 2 .0;1/, we have for all b large (Theorem 5.3).The index bound in Theorem 1.1(2) follows from the above almost linear growth estimate and the universal bounds in [8; 10].We point out that this almost linear growth estimate and its proof are quite different from the case of small escape rate [12; 13], where a stronger almost translation estimate holds and the understanding of Hy is not required; see Remark for details.
Motivated by [14] and the work in this paper, we propose the conjecture below.
Conjecture 1.3 Let .M; p/ be an open n-manifold with Ric 0 and E.M; p/ ¤ 1 2 .Suppose that 1 .M / contains a torsion-free nilpotent subgroup of nilpotency length l.Then there exist an asymptotic cone .Y; y/ of the universal cover and a closed R-subgroup of Isom.Y / such that the orbit Ry is homeomorphic to R but has Hausdorff dimension at least l.
The first examples of asymptotic cones that admit isometric R-orbits with Hausdorff dimension strictly larger than 1 were discovered in Pan and Wei [14].It was suspected in [14,Remark 1.7] that this feature of extra Hausdorff dimension might be related to the nilpotency of fundamental groups.Conjecture 1.3 is a formal description of the question first raised in [14,Remark 1.7].
We organize the paper as follows.We start with some preliminaries in Section 2; this includes some results of nilpotent isometric actions on metric cones and the escape rate.In Section 3, assuming that M satisfies the conditions in Theorem 1.1(1) and 1 .M / is nilpotent, we study the equivariant asymptotic geometry of .z M ; 1 .M; p//.Then in Section 4, we further study the asymptotic orbits coming from h i-action, where 2 1 .M; p/ has infinite order.The properties of these asymptotic orbits lead to the almost linear growth estimate and virtual abelianness in Section 5.
Acknowledgements The author was supported by a Fields Postdoctoral Fellowship from the Fields Institute during the preparation of this work.The author is grateful to Vitali Kapovitch and Guofang Wei for many helpful conversations.

Equivariant asymptotic geometry and metric cones
Let M be an open n-manifold with Ric 0. Recall that for any sequence r i ! 1, after passing to a subsequence if necessary, the corresponding blowdown sequence converges in the pointed Gromov-Hausdorff topology: .r 1 i M; p/ GH !.X; x/: We call the limit space .X; x/ an asymptotic cone of M. In general, the limit X may not be unique and may not be a metric cone; see examples in [2].
Recall that an open n-manifold M is said to have Euclidean volume growth of constant where p 2 M and B n R .0/ is an R-ball in Euclidean space R n .By Bishop-Gromov volume comparison, the above limit always exists and is no greater than 1.
Theorem 2.1 [1] Let M be an open n-manifold with Ric 0. If M has Euclidean volume growth , then M is conic at infinity.
For the purpose of understanding fundamental groups, we study the asymptotic geometry of the Riemannian universal cover z M with the isometric -action, where D 1 .M; p/.Let r i ! 1 be a sequence.We can pass to a subsequence and consider the pointed equivariant Gromov-Hausdorff convergence [6] .r where G is a closed subgroup of the isometry group of Y.It follows from the work of Colding and Naber [5] that G is always a Lie group.We call the above limit space .Y; y; G/ an equivariant asymptotic cone of .z M ; /.
As a matter of fact, if M has nonnegative sectional curvature, then the equivariant asymptotic cone of .z M ; / is unique as .C.Z/; z; G/, where C.Z/ is a metric cone with vertex z; moreover, the orbit Gz is a Euclidean factor of C.Z/.In fact, by the splitting theorem [3] we write z M as a metric product R k W, where W does not contain any line.The isometry group of z M also splits as Isom.R k / Isom.W /; we denote by q 1 and q 2 the natural projections from Isom.z M / to Isom.R k / and Isom.W /, respectively.By setting the basepoint p in a soul of M [4], all minimal representing loops of D 1 .M; p/ are contained in the soul.Then one can show that the set fq 2 .w/ j 2 g is bounded in W, where z p D .0;w/; see [12, Proof of Proposition A.1]. Thus the limit orbit Gz can be viewed as the one in the asymptotic cone of .R k ; 0; q 1 .//,which is a Euclidean subspace in R k .(Also see [11,Theorem 1.4  M ; / as the Gromov-Hausdorff limit of any sequence f.Y j ; y j ; G j /g Â .z M ; /.Also, note that if .Y; y; G/ 2 .z M ; /, then its scaling .sY;y; G/ is also an equivariant asymptotic cone for any s > 0. Therefore, the Gromov-Hausdorff limit of any sequence .sj Y j ; y j ; G j / in .z M ; /, where s j > 0, is an equivariant asymptotic cone as well.The connectedness of .z M ; / will be used implicitly in Section 3.
For an asymptotic cone that is a metric cone, by Cheeger and Colding's splitting theorem [1], any line in the space must split off isometrically.
Let .M; p/ be an open manifold with the assumptions in Theorem 1.1(1) and a nilpotent fundamental group .Lemma 2.4 implies that G is always virtually abelian for any .Y; y; G/ 2 .z M ; /.One may wonder whether the virtual abelianness of all asymptotic limit groups of indicates that itself should be virtually abelian as well.However, it is possible that is a torsion-free nilpotent nonabelian group, while all asymptotic limit groups of are abelian; see the appendix for the example.In other words, the nilpotency length of may not be well-preserved in the asymptotic limits.

Escape rate
The notion of escape rate was introduced in [12] to study the structure of fundamental groups.It measures where the minimal representing geodesic loops of 1 .M; p/ are positioned in M. We assign two natural quantities to any loop c based at p 2 M : its length and its size.Here, size means the smallest radius R such that c is contained in the closed ball x B R .p/, or equivalently, the Hausdorff distance between the loop c and the basepoint p.For each element 2 1 .M; p/, we choose a representing geodesic loop of at p, denoted by c , such that c has the minimal length in its homotopy class; if there are multiple choices of c , we choose the one with the smallest size.We write j j WD d. z p; z p/ D length.c/ for convenience.The escape rate of .M; p/ is defined as As a convention, if 1 .M / is a finite group, then we set E.M; p/ D 0.
In [15], Sormani proved that if 1 .M / is not finitely generated, then there is a consequence of elements Because F is a finite cover, F 1 .p/consists of finitely many points.Let D > 0 be the diameter of

Asymptotic orbits of nilpotent group actions
In this section, we always assume that .M; p/ is an open n-manifold with Ric 0 and E.M; p/ ¤ 1 2 .Due to Lemma 2.5, we will also assume that 1 .M; p/ is an infinite nilpotent group, denoted by N.
The goal of this section is to study the properties of asymptotic equivariant cones of .z M ; N / when z M is conic at infinity.In particular, we will show that there is an integer k such that for any .Y; y; G/ 2 .z M ; N /, the orbit Gy must be homeomorphic to R k ; see Proposition 3.6.
For any point gy 2 Gy that is not y, there is a minimal geodesic from y to gy and an orbit point g 0 y 2 Gy such that d.m; g 0 y/ < 1 2 d.y; gy/; where m is the midpoint of .
The above is a limit minimal geodesic from y to gy.Moreover, is contained in B R .Gy/; in particular, let m be the midpoint of , then d.m; g 0 y/ Ä R for some g 0 2 G. Thus  By our choice of Ꮿ 1 as the closest component to Ꮿ 0 , Ꮿ 0 is also the closest component to Ꮿ 1 .The first inequality above implies g 0 y 2 Ꮿ 0 , while the second one implies g 0 y 2 Ꮿ 1 , a contradiction.
Starting from Lemma 3.3 below, we will assume that the universal cover z M is conic at infinity for the rest of this section.
Proof Let G 0 be the identity component subgroup of G.Because Gy is connected by Proposition 3.2, we have Gy D G 0 y.In other words, for any g 2 G, we can write gy D hy for some h 2 G 0 .Then any g 2 G can be written as the product of an element in G 0 and an element in the isotropy subgroup at y; namely, g D h .h 1 g/, where h 2 G 0 and h 1 g fixes y.
Let g 1 ; g 2 2 G.We write where h 1 ; h 2 2 G 0 and ˛1 and ˛2 belong to the isotropy subgroup at y.Because N is nilpotent, G must be nilpotent as well.According to Lemma 2.4, G 0 is central in G. Thus Geometry & Topology, Volume 28 (2024) so that each E i is .G/-invariant, where i D 1; : : : ; m.We write v 2 D P m iD1 v i , where v i 2 E i .By the hypothesis that .A; v 2 / does not have fixed points, there exists j 2 f1; : : : ; mg such that Aj E j D idj E j and v j ¤ 0. From the maximality of E in its definition, we can find some element .B; w/ 2 G such that Bv j ¤ v j .We write where 2).This kind of argument is also used in [11; 10; 12; 13], in different contexts, to prove certain uniform properties among all equivariant asymptotic cones.This method requires an equivariant Gromov-Hausdorff distance gap between certain spaces, which we establish below.Lemma 3.7 Given any integer n 2, there is a constant ı.n/ > 0 such that the following holds.
Let .C.Z j /; z j / 2 ᏹ.n; 0/ be a metric cone with vertex z j and let G j be a closed nilpotent subgroup of Isom.C.Z j //, where j D 1; 2. Suppose that 2 G 2 such that each e 0 j is ı.n/-close to e j , where j D 1; : : : ; k 1 .Let L 0 be the subgroup generated by fe 0 1 ; : : : ; e 0 k 1 g.Though elements in L 0 may not be commutative, by Lemma 3.3, the orbit L 0 z 2 can be identified as By the inequality k 2 > k 1 in the second condition, there exists an element g 0 2 G 2 such that d.g 0 z 2 ; z 2 / D d.g 0 z 2 ; L 0 z 2 / 2 .7;8/: By the triangle inequality, we have According to [12, Lemma 4.10], G 1 acts as translations on G 1 z 1 .Recall that each e j 2 G 1 has displacement 1=n at z 1 , thus each l j Ä 10n.Together with the choice of ı.n/, we see that This is a contradiction to d.g 0 z 2 ; L 0 z 2 / > 7, and thus d GH .C.Z 1 /; z 1 ; G 1 /; .C.Z 2 /; z 2 ; G 2 / ı.n/: Now we use Lemma 3.7 and a critical rescaling argument to prove Proposition 3.6.
Proof of Proposition 3.6 We argue by contradiction.
Claim 1 Suppose that the statement is not true.Then there exist spaces .Y 1 ; y 1 ; G 1 / and .Y 2 ; y 2 ; G 2 / in .z M ; N / such that for j D 1; 2, the orbit G j y j is a Euclidean factor of dimension k j , with k 1 > k 2 .
In fact, if the statement of Proposition 3.6 fails, then either there exists a space .W; w; H / 2 .z M ; N / of type .k;d/ with d > 0, or for j D 1; 2 there exist .W j ; w j ; H j / 2 .z M ; N / of type .kj ; 0/, with k 1 > k 2 .For the first case above, we consider the blowup and blowdown limits of .W; w; H /: .lW;w; H / GH !.W; w; H 1 /; .l 1 W; w; H / GH !.W; w; H 2 / where l !C1.Because d > 0, it is clear that the orbit H 1 w is a Euclidean factor with dimension strictly larger than k.By Lemma 3.5, the orbit H 2 w is a k-dimensional Euclidean factor.Then .W; w; H 1 / and .W; w; H 2 / are the desired spaces in Claim 1.For the second case, for j D 1; 2 the blowup limit of .W j ; w j ; H j / clearly satisfies the requirements; alternatively, one can also use the blowdown limits of .W j ; w j ; H j / and then apply Lemma 3.5.This proves Claim 1.
For j D 1; 2 let .Y j ; y j ; G j / 2 .z M ; N / be as described in Claim 1.Let r i ; s i ! 1 be such that .r By passing to a suitable subsequence of r i or s i , we can assume that t i WD r i =s i ! 1.We put Recall that in .Y 2 ; y 2 ; G 2 /, the orbit G 2 y is a k 2 -dimensional Euclidean factor with k 2 < k 1 , thus t i 2 L i for all i large; in particular, L i is nonempty.We choose l i 2 L i with inf L i Ä l i Ä inf L i C1 as a sequence of critical scales.
Suppose that l i subconverges to a number l 1 < C1.Then Next, after passing to a convergent subsequence, we consider the rescaling limit Let .k 0; d 0 / be the type of .Y 0 ; y 0 ; G 0 /.It has the following two possibilities.
Case 1 (k 0 k 1 ) Recall that each .li M i ; q i ; N i / satisfies d GH ..l i M i ; q i ; N i /; .W i ; w i ; H i // Ä 1 10 ı.n/ for some .W i ; w i ; H i / 2 .z M ; N / whose orbit H i w i is a Euclidean factor of dimension < k 1 .Since .li M i ; q i ; N i / converges to .Y 0 ; y 0 ; G 0 /, we have for all i large, where G 0 y 0 is of type .k 0; d 0 / with k 0 k 1 and H i w i is a Euclidean factor of dimension < k 1 .This contradicts Lemma 3.7.Thus Case 1 cannot happen.
Case 2 (k 0 < k 1 ) We consider the blowdown limit of .Y 0 ; y 0 ; G 0 /: .j 1 Y 0 ; y 0 ; G 0 / GH !.Y 0 ; y 0 ; H 0 /; where j ! 1.By Lemma 3.5, the orbit H 0 y 0 is a Euclidean factor of dimension k 0 .Let J 2 N be large such that d GH ..J 1 Y 0 ; y 0 ; G 0 /; .Y 0 ; y 0 ; H 0 // Ä 1 100 ı.n/:Note that .J .Y 0 ; y 0 ; H 0 // Ä 1 10 ı.n/ for all i large.Because l i ! 1 and H 0 y 0 is a Euclidean factor of dimension < k 1 , we conclude that J 1 l i 2 L i for all i large.However, this contradicts our choice of l i as inf With all possibilities of .Y 0 ; y 0 ; G 0 / being ruled out, we reach the desired contradiction and thus complete the proof of statement.
As a direct consequence of Proposition 3.6, in any .Y; y; G/ 2 .z M ; N /, any compact subgroup of G must fix the basepoint y.This implies the lemmas below.Proof We first prove that if h m y D y for some integer m 2, then hy D y.In fact, let H Â G be the closure of the subgroup generated by h.Because h m y D y, the orbit Hy consists of at most m 1 points; in particular, the orbit Hy is closed and bounded.Thus H is a compact subgroup of G.Because .Y; y; G/ is of type .k;0/ by Proposition 3.6, H must fix y.Thus hy D y.

Asymptotic orbits of Z-actions
Throughout this section, we always assume that an open manifold M satisfies the assumptions in Theorem 1.1(1) and has an infinite nilpotent fundamental group N. We fix an element 2 N with infinite order.We will study the equivariant asymptotic cones of .z M ; h i/.Our first goal of this section is to prove the result below.Proposition 4.1 Any space .Y; y; H / 2 .z M ; h i/ must be of type .1;0/.Consequently, the orbit Hy is connected and homeomorphic to R.
We remark that the group H could be strictly larger than R, because H may have a nontrivial isotropy subgroup at y. Also, recall that H is a closed subgroup of Isom.Y /, thus the orbit Hy is embedded in Y, that is, the subspace topology of Hy matches with the quotient topology from H =K, where K is the isotropy subgroup of H at y.
Here is the rough idea to prove Proposition 4.1: suppose that .Y; y; H / is not of type .1;0/.Then we shall find a space .Y 0 ; y 0 ; G 0 / 2 .z M ; N / violating Proposition 3.6.
We need some preparations first.
Definition 4.2 Let G be a group.We say a subset S of G is symmetric if S satisfies the following conditions: (1) id 2 S.
Definition 4.3 Let .X i ; x i ; G i / be a pointed equivariant Gromov-Hausdorff convergent sequence with limit .Y; y; H /. Recall that this means there is a sequence of triples of i -approximation maps .fi ; ' i ; i /: with the properties described in [6,Definition 3.3]).For each i , let S i be a closed symmetric subset of G i .We write ' i .S i / for the closure of ' i .S i / in H.We say that the sequence S i Gromov-Hausdorff converges to a limit closed symmetric subset S Â H, denoted by if S is the limit of ' i .S i / with respect to the topology on the set of all closed subsets of H induced by the compact-open topology.Equivalently, the closed symmetric subset S Â H satisfies: (1) For any h 2 S, there is a sequence of isometries g i 2 S i converging to h.
(2) Any convergent sequence of isometries g i 2 S i has the limit h in S.
It follows directly from the proof of [6, Proposition 3.6] that we have the precompactness result below.Because the orbit Gy is connected, we can assume g 2 G 0 .Let exp be the exponential map from the Lie algebra of G 0 to the Lie group G 0 ; note that exp is surjective because G 0 is abelian.Then g D exp.v/ for some v in the Lie algebra.We define the following subsets of Gy: P .g/yD fexp.tv/y j t 2 OE 1; 1g and R.g/y D fexp.tv/y j t 2 Rg: Lemma 4.6 In Definition 4.5, the sets P .g/y,and thus R.g/y, are uniquely determined by the orbit point gy.
Proof We first show that the set P .g/y is independent of the choice of v in Definition 4.5.Suppose that where v; w are elements in the Lie algebra of G 0 .By Lemma 3.8, we have In other words, we have shown that exp.tv/yD exp.tw/y holds for all t 2 Q.Because P .g/y is the closure of the set fexp.tv/y j t 2 OE 1; 1 \ Qg; we conclude that P .g/y is independent of the choice of v.
Next, we show that P .g/yonly depends on the orbit point gy, but not the choice of g 2 G 0 .Suppose that h 2 G 0 is such that gy D hy.Let v and w be vectors in the Lie algebra of G 0 such that exp.v/D g and exp.w/D h: Following a similar argument to that in the first paragraph of the proof and applying Lemmas 3.3 and 3.8, one can clearly verify the result.
Lemma 4.7 Let .Y; y; G/ 2 .z M ; N / and let S be a closed symmetric subset of G. Suppose that the set Sy satisfies the following properties: (1) Sy is closed under multiplication; that is, if g 1 ; g 2 2 S, then g 1 g 2 y 2 Sy.
Then Sy D fyg.
Proof Let H be the closure of the subgroup generated by S. The first assumption implies that Hy D Sy.Because Hy D Sy is bounded, we conclude that H must be a compact subgroup of G. Since .Y; y; G/ 2 .z M ; N / is of type .k;0/ by Proposition 3.6, H fixes y.In other words, we have Sy D Hy D fyg.
We are in a position to prove Proposition 4.1.
Proof of Proposition 4.1 Let r i ! 1 be a sequence.We consider the convergence We shall show that .Y; y; H / is of type .1;0/.
For each i , let l i D minfl 2 Z C j d. l z p; z p/ r i g and let S .li / D fid; ˙1; : : : ; ˙li g be a sequence of symmetric subsets of h i.By the triangle inequality, we have Passing to a subsequence, we obtain convergence where A is a closed symmetric subset of H and g 2 A with d.gy; y/ D 1.

Claim 1
The set Ay contains P .g/y.
where b c means the floor function.Thus the sequence bl i =bc subconverges to some limit ˛2 A. Because passing to a subsequence if necessary, we can assume that b bl i =bc D l i C b 0 for some b 0 D 0; : : : ; b and all i .For this subsequence, we have .r where g 0 2 A fixes y; moreover, g g 0 D ˛b.Thus ˛satisfies ˛by D g g 0 y D gy: where exp.v/D g.By construction, the limit symmetric subset A contains the set fid; ˛˙1 ; : : : ; ˛˙b g.Claim 2 Suppose that m i = l i !C 2 OE1; 1/ for a subsequence.We write We result in a contradiction to d.hy; R.g/y/ 2. This proves Claim 2.
For each i , let Suppose the contrary, that is, The hypothesis implies h j y j GH !y 0 with respect to the above convergence.We consider the closed symmetric subset S j D ft h j j t 2 OE0; 1g of H j and let S 0 H 0 be its limit symmetric subset, that is, .R 1 j Y j ; y j ; S j / GH !.Y 0 ; y 0 ; S 0 /: We claim that the set S 0 y 0 is closed under multiplication; the proof is similar to Claim 4 in the proof of Proposition 4.1.In fact, for any ˇ1; ˇ2 2 S 0 , we have t j ;1 ; t j ;2 2 OE 1; 1 such that .R 1 j Y j ; y j ; t j ;1 h j ; t j ;2 h j / GH !.Y 0 ; y 0 ; ˇ1; ˇ2/: If not, we write where o j 2 OE 1; 1.The sequence o j h j 2 S j subconverges to a limit ˇ0 2 S 0 .Then .o j h j / .˙hj /y j D ˇ0y 0 2 S 0 y 0 : Since the set S 0 y 0 is closed under multiplication and is contained in x B 1 .y 0/, by Lemma 4.7, we obtain S 0 y 0 D y 0 .On the other hand, by the construction of S j and R j , S 0 y 0 must have a point at distance 1 from y 0 ; a contradiction.Lemma 4.9 Given s; 2 .0;1/, there exists a constant L 0 .z M ; ; s; / such that for any .Y; y; H / 2 .z M ; h i/ and any h 2 H 0 with d.hy; y/ D 1, there exists an integer 2 Ä L Ä L 0 with Proof We argue by contradiction.Suppose that for each integer L j D j , there are .Y j ; y j ; H j / 2 .z M ; h i/ and h j 2 H j such that d.h j y j ; y j / D 1 and for all 2 Ä L Ä L j .After passing to a subsequence, we consider the convergence .Y j ; y j ; H j ; h j / GH !.Y 0 ; y 0 ; H 0 ; h 0 /: Claim For any integer L 2, we have In fact, due to Lemma 4.8, there is a constant C 1 such that for any integer L 2. Thus after passing to a subsequence, we can assume that .1=L/hj converges to some limit isometry ˇ2 H 0 as j ! 1.Note that h j y j D h 0 y 0 : Applying Lemma 3.8, we see that ˇy0 D ..1=L/h 0 /y 0 and the claim follows.
The above claim and the hypothesis together imply that for any integer L 2, as L ! 1.This shows that in .Y 0 ; y 0 ; H 0 /, the path P .h 0/y 0 from y 0 to h 0 y 0 has infinite length, which cannot be true since P .h 0/y 0 comes from an R-orbit of some isometric actions embedded in a Euclidean factor R k .

Almost linear growth and virtual abelianness
We prove the almost linear growth estimate (Theorem 5.3) and Theorem 1.1 in this section.
In Lemmas 5.1 and 5.2 below, we always assume that the manifold M satisfies the assumptions in Theorem 1.1(1) and the fundamental group is an infinite nilpotent group N. We fix as an element of infinite order in N. The purpose of Lemma 5.2 is to transfer Lemma 4.9, as an estimate in the asymptotic limits, to an estimate on z M at large scale.where d e means the ceiling function.
The above statement also holds if one replaces the power db i =Le by bb i =Lc.We use the ceiling function in Lemma 5.1 for later applications.
Proof of Lemma 5.1 The statement of Lemma 5.1 is to some extent similar to the claim in the proof of Lemma 4.9, but at the moment we don't have an estimate similar to Lemma 4.8 on the sequence.So, we need to first derive a similar estimate on the distance.
Claim There is a number C such that This claim assures that db i =Le z p subconverges to some limit point in Y. Suppose the contrary, that is, where d e means the ceiling function.
Proof Let L 0 D L 0 z M ; ; s; 1 2 , the constant in Lemma 4.9.We argue by contradiction to prove the statement.Suppose that there is a sequence b i ! 1 such that for all L D 2; : : : ; L 0 .Let r i D j b i j ! 1.We consider .rfor all L 2 f2; : : : ; L 0 g.On the other hand, by the choice L 0 D L 0 z M ; ; s; 1 2 and Lemma 4.9, we have for some L 2 f2; : : : ; L 0 g, a contradiction.
We are ready to prove the almost linear growth estimate.
Theorem 5.3 Let M be an open n-manifold with the assumptions in Theorem 1.1 (1).Suppose that its fundamental group is an infinite nilpotent group, denoted by N. Let 2 N be an element of infinite order.Given any s 2 .0;1/, there are positive constants C 0 D C 0 .z M ; ; s/ and P 0 D P 0 .z M ; ; s/ such that j b j C 0 b 1 s holds for all integers b > P 0 .
Proof Let P 0 be a large constant such that j b j R 0 .z M ; ; s/ for all b P 0 , where R 0 .z M ; ; s/ is the corresponding constant in Lemma 5.2.
Let b > P 0 .By Lemma 5.2, we have j b j L 1 s 1 j db=L 1 e j for some integer 2 Ä L 1 Ä L 0 , where L 0 D L 0 .z M ; ; s/ is the constant in Lemma 5.2.If db=L 1 e < P 0 , then we stop right here.If not, we can apply Lemma 5.2 again to find some integer 2 Ä L 2 Ä L 0 such that j b j L 1 s 1 j db=L 1 e j .L 1 L 2 / 1 j ddb=L 1 e=L 2 e j: Repeating this process, we eventually derive where C is a positive constant.Let N r D .N; g r / be the quotient Riemannian manifold .z N ; z g r /= .
Next, we construct an open Riemannian manifold .M; g/ as a warped product M D OE0; 1/ f S p N r ; g D dr 2 C f .r/ 2 ds 2 p C g r ; where .S p ; ds 2 p / is the standard p-dimensional sphere and f .r/ D r .1 C r 2 / 1=4 : Following the calculation in [16], one can verify that .M; g/ has positive Ricci curvature when p is sufficiently large (depending on ˛and ˇ).
Let p 2 M at r D 0. We explain that for ˇ> 1, the above constructed open manifold .M; p/ satisfies the required condition (2).Let hold for all l large.When ˇ> 1, ˇOE l 1 ; l 2 ˇis much shorter than j l 1 j and j l 2 j as l ! 1. Below, we fix a ˇ> 1.Let r i ! 1 be a sequence and consider an equivariant asymptotic cone .rAs a side note, we mention that by a similar argument in [14], one can check that the orbit Hy has Hausdorff dimension 1 C ˇC 4˛ 2. This supports Conjecture 1.

Lemma 3 . 8
Let .Y; y; G/ 2 .z M ; N / and let h 1 ; h 2 2 G.If h m 1 y D h m 2 y for some integer m 2, then h 1 y D h 2 y.

Ã 1 s r 0 ; where ˙ db=L 1 e=L 2 db=L 1 with OEX 1 ;
=L k < P 0 and r 0 D min m2Z C j m j > 0. Noting that e=L 2 =L k < P 0 ; we result inj b j Â b P 0 Ã 1 s r 0 D C 0 b 1 s ;where C 0 D r 0 =.P 1 s 0 /.Let z N be the simply connected 3-dimensional Heisenberg group and let be the discrete Heisenberg 3-group; that is, X 2 D X 3 as the only nontrivial Lie bracket.Given ˛> 0 and ˇ 1, we assign a family of norms k k r , where r 2 OE0; 1/ is the parameter, on this Lie algebra bykX 1 k r D kX 2 k r D .1 C r 2 / ˛and kX 3 k r D .1 C r 2 / ˇ=2 2˛:Geometry & Topology, Volume 28 (2024)The family of norms k k r uniquely determines a family of left-invariant Riemannian metrics z g r on z N, and z g r satisfies an almost nonnegative Ricci curvature bound Ric.z g r / C.1 C r 2 / ˇ;

3 .
Issue 3 (pages 1005-1499) 2024 1005 Homological mirror symmetry for hypertoric varieties, I: Conic equivariant sheaves MICHAEL MCBREEN and BEN WEBSTER 1065 Moduli spaces of Ricci positive metrics in dimension five MCFEELY JACKSON GOODMAN 1099 Riemannian manifolds with entire Grauert tube are rationally elliptic XIAOYANG CHEN 1113 On certain quantifications of Gromov's nonsqueezing theorem KEVIN SACKEL, ANTOINE SONG, UMUT VAROLGUNES and JONATHAN J ZHU 1153 Zariski dense surface groups in SL.2k C 1; Z/ D DARREN LONG and MORWEN B THISTLETHWAITE 1167 Scalar and mean curvature comparison via the Dirac operator SIMONE CECCHINI and RUDOLF ZEIDLER 1213 Symplectic capacities, unperturbed curves and convex toric domains DUSA MCDUFF and KYLER SIEGEL 1287 Quadric bundles and hyperbolic equivalence ALEXANDER KUZNETSOV 1341 Categorical wall-crossing formula for Donaldson-Thomas theory on the resolved conifold YUKINOBU TODA 1409 Nonnegative Ricci curvature, metric cones and virtual abelianness JIAYIN PAN 1437 The homology of the Temperley-Lieb algebras RACHAEL BOYD and RICHARD HEPWORTH and Remark 5.12].)See [11, Proposition 2.1] for a proof.The compactness allows us to obtain new spaces in .z Also note that since z is the unique vertex of C.Z/, any isometry of C.Z/ must fix the vertex z.Consequently, for any element g 2 Isom.Y /, the orbit point gy must be contained in the Euclidean factor R k fzg.Next, we go through some basic results about nilpotent isometric actions on metric cones.Recall that a group N is called nilpotent if its lower central series terminates at the trivial identity subgroup, that is,N D C 0 .N / F C 1 .N / F F C l .N / D feg;where the subgroup C j C1 .N / D OEC j .N /; N is inductively defined.The smallest integer l such that C l .N / D feg is called the nilpotency length of N. Let G be a nilpotent subgroup of Isom.R k /.Then two elements .A; v/ and .B; w/ in G commute if and only if A and B commute.Let .Y; y/ 2 ᏹ.n; 0/ be a metric cone with vertex y.Let G be a closed nilpotent subgroup of the isometry group of Y. Then the center of G has finite index in G; in particular, the identity component subgroup G 0 must be central in G.
Proof Write Y D R k C.Z/, where C.Z/ does not contain lines, and consider the group homomorphism W Isom.Y / !O.k/ Isom.C.Z//; .A; v; ˛/ 7 !.A; ˛/: Note that .G/, the closure of .G/, is a compact nilpotent Lie group.It follows from a standard result of group theory that the identity component of .G/, denoted by K, is central and of finite index in .G/; see, for example, [10, Lemma 5.7].Now we consider the subgroup H of G defined by ¤ 1 2 .In order to prove Theorem 1.1, it suffices to show that N is virtually abelian and further bound the index when z M has Euclidean volume growth.
y p; / D: For a sequence of elements i 2 1 .y M ; y p/ and their corresponding minimal representing geodesic loops i such that lim i!1 d H . y p; i / length.i / D E. y M ; y p/; we have d H .p; F. i // length.F.i // d H . y p; i / length.i / !E. y M ; y p/: This shows that E.M; p/ E. y M ; y p/.With Lemma 2.5, we can assume that 1 .M; p/ is nilpotent without loss of generality when proving Theorem 1.1.In fact, because E.M; p/ ¤ 1 2 , 1 .M / is finitely generated.By [9; 7], 1 .M / has a nilpotent subgroup N of finite index; moreover, according to [8], we can assume that the index of N is bounded by some constant C.n/.Let y M D z M =N be an intermediate cover of M and let y p 2 y M be a lift of p. Lemma 2.5 assures that E. y M ; y p/ Lemma 3.1 states that for some minimal geodesic from y to gy, its midpoint is closer to Gy than the endpoints of .Next, we show that this property implies the connectedness of Gy.In fact, suppose that z 0 2 Ꮿ 0 and z 1 2 Ꮿ 1 are such that d.z 0 ; z 1 / < d.y; Ꮿ 1 /.We can write z 0 D g 0 y and z 1 D g 1 y, where g 0 2 G 0 and g 1 2 G G 0 .Thus d.y; Ꮿ 1 / > d.g 0 y; g 1 y/ D d.y; g 1 0 g 1 y/:It follows from the choice of Ꮿ 1 that g 1 0 g 1 y 2 Ꮿ 0 D G 0 y.This leads to z 1 D g 1 y 2 G 0 y D Ꮿ 0 ; .y; gy/: Proposition 3.2 The orbit Gy is connected for all .Y; y; G/ 2 .z M ; N /.Proof We argue by contradiction.Suppose that Gy is not connected.Let Ꮿ 0 be the connected component of Gy containing y.Note that Ꮿ 0 D G 0 y, where G 0 is the identity component subgroup of G.Because G is a Lie group and Gy is not connected, there is a different component Ꮿ 1 of Gy such that d.Ꮿ 0 ; Ꮿ 1 / D d.y; Ꮿ 1 / D min gy2Gy Ꮿ 0 d.y; gy/ > 0: Let gy 2 Ꮿ 1 be such that d.y; gy/ D d.y; Ꮿ 1 /.Claim d.y; Ꮿ 1 / D d.Ꮿ 0 ; Ꮿ 1 /.By Lemma 3.1, there is a minimal geodesic from y to gy and a point g 0 y 2 Gy such that d.m; g 0 y/ < 1 2 d.y; gy/; where m is the midpoint of .Then Definition 3.4 Let .Y; y; G/ be a space, where G is a closed nilpotent Lie subgroup of Isom.Y /.Let T be a maximal torus of G 0 .Let k 2 N and d 2 OE0; 1/.We say that .Y; y; G/ is of type .k;d/ if the orbit Gy is connected and dim G dim T D k and diam.T y/ D d: Lemma 3.5 Let C.Z/ be a metric cone with a vertex z.Let G be a closed nilpotent subgroup of Isom.C.Z//.Suppose that .C.Z/; z; G/ is of type .k;d/.For a sequence r i ! 1, we consider the corresponding blowdown sequence GH! .C.Z/; z; G 0 /:Then the orbit G 0 z is a k-dimensional Euclidean factor in C.Z/.Proof Because the orbit Gz is connected and contained in a Euclidean factor of C.Z/, it suffices to prove the statement when C.Z/ is a Euclidean space R l and G is a connected Lie group.By Lemma 2.4, G is abelian.We set z as the origin 0 of R l , then we can write any element of G in the form of.A; v/ 2 SO.l/ Ë R l .Let W Isom.R l / !SO.l/; .A; v/ 7 !A;be the natural projection.Because .G/ is abelian, we can decompose R k into an orthogonal direct sum E C E ?, where E is the maximal subspace such that Aj E D idj E for all A 2 .G/.Note that by the above construction, any translation in E and any g 2 G must commute.Let .A; v/ 2 G.We write v D v 1 C v 2 , where v 1 2 E and v 2 2 E ? .Let ı 2 Isom.R l / be the translation by v 1 .We claim that ıg D .A; v 2 / must have a fixed point.The proof of this claim is by linear algebra.We argue by contradiction.Because .G/ is an abelian subgroup of SO.l/, we can further decompose E ?into an orthogonal direct sum of subspaces with dimension at most 2, w 1 2 E, w 2 2 E ? and w i 2 E i .As .A; v/ commutes with .B; w/, by direct calculation we have By the claim, any element g 2 G can be written as a product g D ı˛D ˛ı, where ı is a translation in E and ˛has a fixed point; moreover, this expression is unique.This enables us to define a group homomorphismF W G ! E; g D ı˛7 !ı:Noting that .R l ; z; G/is of type .k;d/, we can write G D H T , where H is a closed subgroup isomorphic to R k and T is a torus subgroup.We remark that the choice of H is not unique in general.It is clear that F.h/ ¤ 0 for all nontrivial elements h 2 H ; otherwise, hhi would be contained in a compact subgroup of Isom.R l /, which is not true.Therefore, F.H / consists of all translations in a k-dimensional subspace in V 1 .After blowing down By the proof of [12, Lemma 3.1], the finite generation of N implies that the orbit Gy is always noncompact.In other words, letting .k;d/ be the type of .Y; y; G/, we always have k 1. Proposition 3.6 Let .M; p/ be an open n-manifold with the assumptions in Theorem 1.1(1).Suppose that the fundamental group N is an infinite nilpotent group.Then there is an integer k such that all .
.B I /v i D .A I /w i for all i D 1; : : : ; m:Taking i D j , we derive that 0 ¤ .B I /v j D .A I /w j D 0;a contradiction.We have verified the claim.Geometry & Topology, Volume 28 (2024) GH !.R l ; z; G 0 /;it is clear that the limit orbit G 0 z is formed exactly by translations in F.H /.In particular,G 0 z is a k-dimensional Euclidean subspace of R l .Let .Y; y; G/ 2 .zM ; N /.Y; y; G/ 2 .zM ; N / are of type .k;0/.The proof of Proposition 3.6 is by contradiction and a critical rescaling argument, which implicitly uses the connectedness of .z M ; N / (Proposition 2. 2) the orbit G 2 z 2 is connected and is of type .k 2 ; d 2 /, where k 2 > k 1 .
Then d GH ..C.Z 1 /; z 1 ; G 1 /; .C.Z 2 /; z 2 ; G 2 // ı.n/: Proof We set ı.n/ D 1=.100n 2 /.Suppose that d GH ..C.Z 1 /; z 1 ; G 1 /; .C.Z 2 /; z 2 ; G 2 // < ı.n/: Let e 1 ; : : : ; e k 1 2 G 1 be such that their orbit points fe 1 z 1 ; : : : ; e k 1 z 1 g form an orthogonal basis of G 1 z 1 ' R k 1 and d.e j z 1 ; z 1 / D 1=n for all j D 1; : : : ; k 1 .Letting L be the subgroup generated by Geometry & Topology, Volume 28 (2024) fe 1 ; : : : ; e k 1 g, then Lz 1 is 1-dense in G 1 z 1 .Let e 0 1 ; : : : ; e 0 k 1 Let ı.n/ be the constant in Lemma 3.7.For each i , we define a set of scales L i byL i D ˚l 2 OE1; t i j d GH ..lM i ; q i ; N i / GH !.Y 1 ; y 1 ; G 1 / and .tiM i ; q i ; N i / GH !.Y 2 ; y 2 ; G 2 /: i ; q i ; N i /;.W; w; H // Ä 1 10 ı.n/; where .W; w; H / 2 .z M ; N / has the orbit H w as a Euclidean factor with dimension < k 1 « : 11 h 2 / m y:It follows from the previous paragraph that h 1 y D h 2 y.Let H be a closed R-subgroup of N and let ˇ2 G be such that ˇy is outside of Hy.Then d.ˇmy; Hy/ is unbounded as m !1.Proof We argue by contradiction.Suppose that there is a number C > 0 such that d.ˇmy; Hy/ Ä C for all m 2 Z.By the connectedness of Gy, we can assume that ˇ2 G 0 without loss of generality.Because H is central in G by Lemma 2.4, we can consider the quotient of .Let K Â G=H be the closure of the subgroup generated x ˇ.Then K is a compact subgroup in the identity component of G=H withOn the other hand, because G 0 is abelian and .Y; y; G/ is of type .k;0/, we can write G 0 D R k T , where T is a torus group fixing y.After taking the quotient by the R-subgroup H, any compact subgroup in G 0 =H must fix x y; a contradiction.
Y; y; G/ by the H -action, denoted by .Y =H; x y; G=H /.Let x ˇ2 G=H be the quotient of ˇ.By hypothesis, we have d.x ˇx y; x y/ > 0 and d. x ˇm x y; x y/ Ä C for all m 2 Z.
Proposition 4.4 Let .X i ; x i ; G i / be a pointed equivariant Gromov-Hausdorff convergent sequence with limit .Y; y; H /. For each i , let S i be a closed symmetric subset of G i .Then passing to a subsequence, we have the convergence .X i ; x i ; S i / GH !.Y; y; S/ for some limit closed symmetric subset S of H. Definition 4.5 Let .Y; y; G/ 2 .z M ; N / and let gy 2 Gy fyg.
Because b is an arbitrary positive integer and Ay is closed, we conclude that Ay contains P .g/y.
By Claim 1, the limit orbit Hy must contain R.g/y.To this end, we argue by contradiction to prove Proposition 4.1.Suppose that .Y; y; H / is not of type .1;0/, then there exists an element ˇ2 H such that ˇy 6 2 R.g/y.Because .Y; y; G/ is of type .k;0/, by Lemma 3.9 we can choose an element as a power of ˇ, denoted by h, such that d.hy; R.g/y/ 2. Let m i ! 1 be such that .r 1 i z M ; z p; m i / GH !.Y; y; h/: Because d.hy; y/ 2, it is clear that m i > l i by our choice of l i .
Recall that Ay contains P .g/ybyClaim 1. Together with Claim 2, that m i = l i ! 1, we see that By must contain R.g/y, which is unbounded.On the other hand, by the choice of d i , By should be contained in x B 1 .y/;acontradiction.This proves Claim 3.GH!.Y 0 ; y 0 ; G 0 ; H 0 ; h 0 ; B 0 /:Due to the choice of d i , it is clear thatd H .B 0 y 0 ; y 0 / D 1:Also, it follows from Claim 3 that h 0 y 0 D y 0 .The set B 0 y 0 is closed under multiplication; that is, if ˇ1; ˇ2 2 B 0 , then ˇ1ˇ2y 0 2 B 0 y 0 .In fact, let bi;1 ; b i;2 2 OE m i ; m i be two sequences of integers such that If b i;1 C b i;2 2 OE m i ; m i , then ˇ1ˇ2 2 B 0and the claim holds trivially.If not, we can writeb i;1 C b i;2 D ˙mi C o i ;where o i 2 OE m i ; m i .Let ˇ0 2 B 0 be the limit of o i after passing to a convergent subsequence.Lastly, we apply Lemma 4.7 to B 0 and conclude that B 0 y 0 D y 0 .We end in a contradiction to d H .B 0 y 0 ; y 0 / D 1.This contradiction completes the proof of Proposition 4.1.Let z 2 Hy be an orbit point.Because Hy is connected, we can write z D hy for some h 2 H 0 .Let v in the Lie algebra of H 0 be such that exp.v/D h.For convenience, in the rest of the paper, we will denote the orbit point exp.tv/y by .th/y, where t 2 R. By the proof of Lemma 4.6, this point .th/y is independent of the choice of v and h.Also, with Proposition 4.1, we have Hy D R.h/y.There exists a constant C 1 D C 1 .z M ; / such that the following holds.For any .Y; y; H / 2 .z M ; h i/ and any h 2 H 0 with d.hy; y/ ¤ 0, we have d..t h/y; y/ Ä C 1 d.hy; y/ for all t 2 OE0; 1: Proof Without loss of generality, we assume that d.hy; y/ D 1 by scaling .Y; y; H /. We argue by contradiction to prove the lemma.Suppose that we have a sequence of spaces .Y j ; y j ; H j / 2 .z M ; h i/ and h j 2 H j with d.h j y j ; y j / D 1, but R j WD max OE1; 1/.Let S .m i / D fid; ˙1; : : : ; ˙mi g: GH !.Y 0 ; y 0 ; ˇ1; ˇ2/: Lemma 5.1 Let b i ! 1 be a sequence of positive integers and let r i D d. b i z GH! .Y; y; H; h/:Then for any integer L 2 Z C , we have If follows from the hypothesis that r 1 i R i !1.We consider an asymptotic cone from the sequence R i , By Lemma 4.7 and the same argument as Claim 4 in the proof of Proposition 4.1, we see that By 0 is closed under multiplication and thus By 0 D y 0 .On the other hand, by the construction S .bi/ and R i , By 0 should contain a point with distance 1 to y 0 .This contradiction verifies the claim.For convenience, below we write m i D db i =Le.With the claim, we can pass to a subsequence such that Applying Lemma 3.8, we conclude that ..1=L/h/y D ˛y, that is, m i z p GH !.Y 0 ; y 0 ; h 0 ; B/; where S .bi/Dfid; ˙1; : : : ; ˙bi g and d.h 0 y 0 ; y 0 / D 0. GH !.Y; y; ˛/; where ˛2 H. Since Lm i L Ä b i Ä Lm ifor each i , we can pass to a subsequence such that b i D Lm i K, where K is some integer between 0 and L.Thus b i D Lm i K GH! ˛L for some ˇ2 H with ˇy D y. Recall that h 2 H is the limit of b i .It follows that ˛LˇD h and For each integer L 2, by Lemma 5.1, db i =Le z p GH! .Y; y; H; h/;where h 2 H satisfies d.hy; y/ D 1.
By construction, it is clear that H is a closed R-subgroup of G. Let l i ! 1 be a sequence of integers ; g 2 /; where g 1 ; g 2 2 G satisfy d.g 1 y; y/ D d.g 2 y; y/ D 1:It follows from the length estimates that OEg 1 ; g 2 y D y.Note that OEg 1 ; g 2 is also the limit of .l 2 OEg 1 ; g 2 2 H.Because H is a closed R-subgroup of G, we see that OEg 1 ; g 2 D id; in other words, G is abelian.