Math. Z.DOI 10.1007/s00209-013-1255-5 Mathematische Zeitschrift

On the spread of positively curved Alexandrov spaces

Takumi Yokota

Received: 2 October 2012 / Accepted: 29 October 2013© Springer-Verlag Berlin Heidelberg 2013

Abstract It was proved by F. Wilhelm that Gromov’s filling radius of closed positivelycurved manifolds with a uniform lower bound on sectional curvature attains the maximumwith the round sphere. Recently the author proved that this is also the case for closed finite-dimensional Alexandrov spaces with a positive lower curvature bound. These were provedas a corollary of a comparison theorem for the invariant called spread of those spaces. In thispaper, we extend the latter result to infinite-dimensional Alexandrov spaces.

Keywords Alexandrov space · Spread · Filling radius · Packing radius

Mathematics Subject Classification (2000) 53C23

1 Introduction

In the present paper, which is a continuation of the previous paper [17], we are concernedwith the geometry of possibly infinite-dimensional Alexandrov spaces with a positive lowercurvature bound. As the nature of infinite-dimensional Alexandrov spaces has not been fullyexplored, we are particularly interested in those spaces. Our goal is to establish a comparisontheorem for them with the round sphere as the model space, which extends the main resultof [17].

In order to state the main theorem, we prepare the following definition.

Partly supported by JSPS Postdoctoral Fellowships for Research Abroad.

T. YokotaResearch Institute for Mathematical Sciences, Kyoto University, Kyoto 606-8502, Japan

T. Yokota (B)Mathematisches Institut, University of Münster, Einsteinstr. 62, 48149 Münster, Germanye-mail: yokota@uni-muenster.de

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Definition 1 (Wilhelm [14]) For any metric space X = (X, d), we define its spread, denotedby Spread(X), as the infimum of λ > 0 for which there is a subset Y ⊂ X of Diam(Y ) ≤ λ

such that d(x, Y ) ≤ λ for any x ∈ X .

The spread of closed manifolds appears in the following lemma.

Lemma 2 (Katz [9]) The filling radius Fill Rad(V ) of any closed Riemannian manifold Vfulfills Fill Rad(V ) ≤ (1/2) Spread(V ).

For the definition and the fascinating theory of the filling radius of Riemannian manifolds,the reader is encouraged to consult Gromov’s seminal paper [5]. In the estimate in Lemma 2,the equality holds for S

n and RPn equipped with metrics of constant curvature [9].Using Lemma 2, Katz [9] obtained the exact value of the filling radius of the round sphere

Sn of constant curvature 1;

Fill Rad(Sn) = (1/2) Spread(Sn) = �n/2.

Here and throughout this paper, we use �n := arccos(− 1n+1 ). We know that �n is the

spherical distance between vertices of a regular (n + 1)-simplex inscribed in the unit sphereS

n ⊂ Rn+1 and the set Y := {p1, . . . , pn+2} ⊂ S

n of its vertices gives Spread(Sn) =Diam(Y ) = �n . We remember that �n is decreasing in n and �n → π/2 as n → ∞.

Using Lemma 2, Wilhelm [14] proved comparison and rigidity theorems which are statedas follows.

Theorem 3 (Wilhelm [14]) For any n-dimensional closed Riemannian manifold V of sec-tional curvature ≥ 1, either Spread(V ) < Spread(Sn) or V is isometric to the roundsphere S

n. In particular, for any such V , either Fill Rad(V ) < Fill Rad(Sn) or V is iso-metric to the round sphere S

n.

In [17], we extended Theorem 3 to finite-dimensional Alexandrov spaces of curvature≥ 1. By dimension, we mean the covering dimension or the Hausdorff dimension. They areknown to coincide for any Alexandrov space with a lower curvature bound.

Theorem 4 ([17]) For any n-dimensional Alexandrov space X of curvature ≥ 1, eitherSpread(X) < Spread(Sn) or X is isometric to the round sphere S

n. Moreover, for any n-dimensional Alexandrov space X of curvature ≥ 1 without boundary, either Fill Rad(X) <

Fill Rad(Sn) or X is isometric to the round sphere Sn.

Defining the filling radius of X requires a fundamental homology class of X . It was shownin the papers by Grove–Petersen [6] and Yamaguchi [15] that any finite-dimensional compactAlexandrov space without boundary admits a fundamental class. The second statement inTheorem 4 follows from the first one and Katz’s Lemma 2.

Here we state the main theorem of the present paper, which extends the first part ofTheorem 4 to infinite-dimensional Alexandrov spaces.

Theorem 5 Any infinite-dimensional Alexandrov space X of curvature ≥ 1 fulfillsSpread(X) ≤ π/2.

Remark 6 The estimate in Theorem 5 is sharp, because a unit sphere S∞, equipped with

the angle metric, of an infinite-dimensional Hilbert space has Spread(S∞) = π/2, e.g. [17,Remark 9]. At present, we do not have a classification of X in Theorem 5 with Spread(X) =π/2.

The estimate Spread(X) ≤ (2/3) Rad(X) for any length space X , cf. [9], yields that theradius Rad(X) := infx∈X supy∈X d(x, y) of any length space (X, d) with Spread(X) =π/2 is at least 3π/4. On the other hand, the Bonnet–Myers theorem says that Rad(X) ≤Diam(X) ≤ π for any Alexandrov space X of curvature ≥ 1.

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On the spread of Alexandrov spaces

Remark 7 As was remarked by Wilhelm [14], cf. [17, Corollary 8], Theorem 4 also yieldsthat Cont0 Rad(X) ≤ Cont0 Rad(Sn) = �n/2 for any n-dimensional Alexandtrov spaceX of curvature ≥ 1. The contractibility radius Contk Rad is defined for any metric space,see [5, Appendix 2]. However, an infinite-dimensional sphere S

∞ is contractible and henceContk Rad(S∞) = 0 for any k ≥ 0. This is why we do not expect a comparison theorem forCont0 Rad in infinite dimension.

Our proof of Theorem 5 is a modification of that of Theorem 4 given in [17], which is inturn an adaptation of Wilhelm’s argument in the proof of Theorem 3 given in [14]. In provingTheorem 5, we also re-prove Theorem 4 and the proof would become slightly simpler thanthat in [17]. For this reason, the present paper may have substantial overlap with [17]. Severaldifficulties in proving Theorem 5 arise from that an infinite-dimensional Alexandrov spaceof curvature ≥ 1, as well as the space of directions at its point, may be noncompact.

After we recall relevant facts, we extend some known results to the possibly infinite-dimensional setting in Sect. 2. Then, we present proofs of Theorems 4 and 5 in Sect. 3.

2 Preliminary

In this section, we recall the basic facts in Alexandrov geometry and prove two propositionsthat are required in our proof of the main theorem.

As in the previous paper [17], we call a metric space an Alexandrov space of curvature ≥ κ

for some real number κ ∈ R, or simply an Alexandrov space, if it is a complete length space,not necessarily a geodesic space, whose distance function enjoys the so-called quadruplecondition, e.g. [17, Definition 12]. We refer to [17] for the references on Alexandrov geometryas well. Here, instead of giving the precise definition of Alexandrov spaces, we collect someof their fundamental properties after setting some notations.

For a fixed number κ ∈ R, we denote by ˜�κ (x; y, z) ∈ [0, π] the comparison angle for a

triple {x, y, z} consisting of points of a metric space with y, z �= x . For example, if κ = 1,which is the case of our interest, it is defined by

cos ˜� 1(x; y, z) := cos(d(y, z)) − cos(d(x, y)) cos(d(x, z))

sin(d(x, y)) sin(d(x, z))

for any triple {x, y, z} in (X, d) whose perimeter peri(x, y, z) := d(x, y)+d(y, z)+d(z, x)

is less than 2π .For two points y, z of a metric space (X, d), we use the notation

(y, z) := {w ∈ X\ {y, z} : d(y, w) + d(w, z) = d(x, z)}.In the statement below, (Mκ , dκ ) denotes the model surface, i.e., the 2-dimensional simply-

connected complete Riemannian manifold of constant curvature κ ∈ R. When κ = 1, wealso use (S2, � ) instead of (M1, d1) later.

Proposition 8 Let (X, d) be an Alexandrov space of curvature ≥ κ and {x, y, z} be a triplein X, with peri(x, y, z) < 2π/

√κ if κ > 0. Then the following hold.

(1) (Triangle comparison) There is an isometric copy {x, y, z} ⊂ (M2κ , dκ ) of {x, y, z} ⊂

(X, d), and for any w ∈ (y, z) and the corresponding point w ∈ (y, z) with dκ (w, y) =d(w, y),

d(w, x) ≥ dκ (w, x).

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(2) (Angle monotonicity) For any point w ∈ (y, z),

˜�κ (y;w, x) ≥ ˜�

κ (y; z, x).

Due to the angle monotonicity stated above, the angle

� p(γ, η) := lims,t→0

˜�κ (p; γ (s), η(t))

is well-defined for any two geodesics γ, η : [0, δ) → X with p := γ (0) = η(0). Thespace of directions (p, � p) at a point p ∈ X is the metric completion of the set (′

p,� p) of

equivalence classes of unit speed geodesics emanating from p. We use the notations ⇑qp ⊂ ′

p

and ↑qp ∈ ⇑q

p , respectively, to denote the set of all equivalence classes of geodesics from pto q ∈ X and one of its elements. The equivalence class of a geodesic γ with γ (0) = p isdenoted by γ (0) ∈ p .

As our main interest is in infinite-dimensional Alexandrov spaces of curvature ≥ 1, wegive new examples of those spaces, which come from Wasserstein geometry. We refer to [13]and references therein for more information on it.

It is well-known that a metric space Y is an Alexandrov space of nonnegative curvature ifand only if the Wasserstein space P2(Y ) over Y is. We also recall the author’s joint work [13]with A. Takatsu in which we showed that a metric space Y is isometric to a Euclidean coneif and only if P2(Y ) is.

Now we give a construction of another Alexandrov space of curvature ≥ 1 from a givenone.

Example 9 Let X be an Alexandrov space of curvature ≥ 1. Then the Euclidean cone Y :=C(X) over X and the Wasserstein space P2(Y ) over Y are Alexandrov space of curvature ≥ 0.Moreover, the dimension of P2(Y ) is infinite and it is isometric to a Euclidean cone C(Q)

over some metric space Q. As a consequence, we obtain an infinite-dimensional Alexandrovspace Q of curvature ≥ 1.

In our proof of Theorem 5, we will exploit ultraproduct of metric spaces. Since its definitioncan be found in several papers in Alexandrov geometry and other areas of geometry, we donot repeat it here and we only try to fix the notation. For a thorough treatment, we refer toe.g. Mitsuishi [10] and references therein.

For a metric space (X, d), supposed to be bounded, i.e., Diam(X) < ∞, for simplicity,and a nonprincipal ultrafilter ω on the set N of natural numbers, we can define the ultraproduct(Xω, dω) of (X, d). Roughly speaking, (Xω, dω) is the set of equivalence classes of infinitesequences in XN equipped with the distance dω given by dω((xi ), (yi )) := limi→ω d(xi , yi )

for (xi ), (yi ) ∈ XN.We list relevant properties of ultraproducts.

– If (X, d) is proper, i.e, any closed bounded subset of X is compact, then (Xω, dω) isisometric to (X, d). In general, (X, d) is isometrically embedded in (Xω, dω). We identifya point x ∈ X with xω := (x, x, . . .) ∈ Xω.

– If (X, d) is a length spaces, then (Xω, dω) is a geodesic space.– If (X, d) is an Alexandrov space of curvature ≥ κ , then so is (Xω, dω).

In order to prove our main Theorem 5 for infinite-dimensional Alexandrov spaces, werequire the following two propositions, which partially generalize the results mainly knownin finite dimensions. We know that any finite-dimensional Alexandrov space is proper andits space of directions at any point is compact, but this is not the case in infinite dimensionin general.

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On the spread of Alexandrov spaces

In Propositions 10 and 11 below, we assume that (X, d) is isometrically embedded intoan ultraproduct (Xω, dω) for some ultrafilter ω on N. Hence, the space of direction (p, � p)

at p ∈ X is also isometrically embedded into (pω , � pω ), which shares some nice propertieswith the space of directions at a point of a proper Alexandrov space. This is the main reasonwhy we introduced the ultraproduct. In the sequel, we use the notation

[⇑qp]ω := ⇑qω

pω ⊂ pω

for p, q ∈ X .

Proposition 10 (First variation formula, cf. [1]) Suppose that x, y and p are points of anAlexandrov space (X, d) of curvature ≥ κ , with d(x, p) < π/

√κ if κ > 0, and that a

geodesic segment xy representing ↑yx ∈ x is given. We put � pxy := � xω (

[⇑px]ω

,↑yx ) =

inf{� xω (ξ,↑y

x ) : ξ ∈ [⇑px]ω}

. Then, for z ∈ xy, we have

d(z, p) = d(x, p) − d(x, z) cos � pxy + o(d(x, z))

as z → x.

In the literature, the above proposition is proved only for proper Alexandrov spaces,e.g. Burago–Burago–Ivanov [2, Corollary 4.5.7]. For the completeness of the argument, wegive a proof of this proposition; cf. Alexander–Petrunin–Kapovitch [1]. In the proof andhereafter, we use the notation

Jz :=⋂

k∈N

⋃

y∈X

{x ∈ X\{z} : ˜�

κ (x; y, z) > π − k−1}

for a point z ∈ X of an Alexandrov space X of curvature ≥ κ . It is easy to see that Jz is adense subset of X and there exists a unique ↑y

z ∈ ⇑yz for each y ∈ Jz , cf. Plaut [12].

Proof of Proposition 10 Inspired by the proof of [12, Proposition 49], we shall prove that

limz→x

˜�κ (x; p, z) = � pxy. (1)

We take a sequence {zi }i∈N ⊂ xy with zi → x as i → ∞ and a minimal unit speed geodesicγi in X from zi to pi ∈ Jzi for each i ∈ N with {pi }i∈N converging to p so fast that

limi→∞

˜�κ (x; pi , zi ) = lim

i→∞˜�κ (x; p, zi ).

Then, for each ε > 0, we can find δ > 0 and z(ε) ∈ xy such that

lim supi→∞

˜�κ (x; γi (δ), z(ε)) > � pxy − ε.

If this is not the case, we have a geodesic γ (t) := (γi (t)), t ∈ [0, d(x, p)] in Xω forwhich � xω (γ (0),↑y

x ) ≤ � pxy − ε, contradicting the definition of � pxy. By using the anglemonotonicity and Alexandrov’s lemma, e.g. [2], we also deduce

˜�κ (x; γi (δ), z(ε)) ≤ ˜�

κ (x; γi (δ), zi )

< ˜�κ (x; γi (δ), zi ) + ˜�

κ (x; , pi , γi (δ))

≤ ˜�κ (x; pi , zi ),

and hence taking i → ∞ and ε → 0 yields that limz→x ˜�κ (x; z, p) ≥ � pxy. The other

inequality implying Eq. (1) follows from the angle monotonicity. This completes the proofof Proposition 10. �

The following proposition is a partial extension of Perelman’s result to infinite dimension.

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Proposition 11 (cf. Perelman [11, Lemma 2.3]) Let (p, � p) be the space of directions at apoint p ∈ X of an Alexandrov space (X, d) of curvature ≥ κ . Suppose that Q ⊂ X\ {p} is afinite subset and a family B = {

B := [⇑qp]ω : q ∈ Q

}of subsets of (pω , � pω ) satisfies that

� pω (B, B ′) > λ > π/2 for each B �= B ′ ∈ B. We choose two elements B+, B− ∈ B. Then,for any δ > 0, there exists w ∈ ′

p ⊂ pω such that � pω (w, B+) > λ, � pω (w, B−) < π −λ,and � pω (w, B) ∈ (π/2, π/2 + δ) for any B ∈ B0 := B\ {B+, B−}.

In the proof of Proposition 11, we utilize Lemma 12 below. Before stating it, we makesome observation.

Let (X, d) be an Alexandrov space of curvature ≥ κ . We fix a sequence {εα} of positivenumbers with εα → 0 as α → ∞ and put dα(·, ·) := (εα)−1d(·, ·). Then, for any unit speedgeodesics γi , i ∈ {2, 3}, departing a point p ∈ X and for some t ∈ (0, 1), we can find asequence {xα} ⊂ Jp of points satisfying

limα→∞

dα(γ2(εα), xα)

t= lim

α→∞ dα(γ2(εα), γ3(ε

α)) = limα→∞

dα(xα, γ3(εα))

1 − t. (2)

It is easy to see in this situation, with ξi := γi (0) ∈ ′p being the initial vectors, that Eq. (2)

is equivalent to that the limit limα→∞ dα(xα, p) ∈ (0,∞) exists and

limα→∞

˜�κ (p; γ2(ε

α), xα)

s= � p(ξ2, ξ3) = lim

α→∞˜�κ (p; xα, γ3(ε

α))

1 − s

for some s ∈ (0, 1).

Lemma 12 Let (X, d) be an Alexandrov space of curvature ≥ κ and {εα} and dα(·, ·) be asabove. Suppose that γi , i ∈ {1, 2, 3}, are unit speed geodesics departing a point p ∈ X withinitial vectors ξi := γi (0) ∈ (′

p,� p) satisfying peri(ξ1, ξ2, ξ3) < 2π , and that {xα} ⊂ Jp

is a sequence of points satisfying Eq. (2) for some t ∈ (0, 1). Choose an isometric copy{ξ1, ξ2, ξ3

}⊂ (S2, � ) of {ξ1, ξ2, ξ3} ⊂ (p, � p) and the corresponding point ξ4 ∈ (ξ2, ξ3)

with � (ξ2, ξ4) = limα→∞ ˜�κ (p; γ2(ε

α), xα). Then, we have

lim infα→∞

� p(↑xα

p , ξ1) ≥ � (ξ4, ξ1).

As Lemma 12 is easy to verify, we omit its proof, cf. [16, Sublemma 14].

Proof of Proposition 11 In this proof, we use the notation � := � pω . First of all, we put

0+ := {v ∈ p : � (v, B) > π/2 for any B ∈ B0

},

and for sufficiently small ε = ε(λ, δ, � (B+, B−)) > 0, we take a point w ∈ 0+ ∩ ′p such

that

� (w, B+) > � (v, B+) − ε for any v ∈ 0+.

It is clear that 0+ contains B− and hence

� (w, B+) > � (B−, B+) − ε > λ.

We would like to prove that w satisfies the rest of the required properties. Suppose that� (w, B0) ≥ π/2 + δ for some B0 ∈ B0. We fix w0 ∈ B0 with � (w,w0) = � (w, B0) andgeodesics γ and γ0 starting from p with γ (0) = w and γ0(0) = w0. We also take sequences{xα} ⊂ Jp of points and εα → 0 of positive numbers such that

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On the spread of Alexandrov spaces

limα→∞

˜�κ (p; γ (εα), xα) + lim

α→∞˜�κ (p; xα, γ0(ε

α)) = � (w,w0),

and limα→∞ ˜�κ (p; xα, γ0(ε

α)) = (π + δ)/2.Put ξα := ↑xα

p ∈ ′p ⊂ pω . Here we claim that

lim infα→∞

� (ξα, B) = infv∈B

lim infα→∞

� (ξα, v) (3)

for any B := [⇑qp]ω ⊂ pω with some point q ∈ X\ {p}. It is clear that the left hand side

of (3) is less than or equal to the right hand side. If the inequality is strict, then there existsw0 ∈ B for which lim infα→∞ � (ξα,w0) is less than the right hand side of (3). This is acontradiction.

Now, we observe that

lim infα→∞

� (ξα, B) > π/2 for each B ∈ B0. (4)

This means that ξα ∈ 0+ for any large α � 1, because B is finite. To see (4) for B ∈B0\ {B0}, let v ∈ B and γv be the geodesic with γv(0) = v. Then, by Lemma 12 and˜� 1(w;w0, v) > λ,

lim infα→∞

� (ξα, v) ≥ lim infα→∞

˜�κ (p; xα, γv(ε

α))

> � (w, v) + ε > π/2 + ε,

and (4) follows from (3). To see (4) for B0, we let v ∈ B0 and γv be the geodesic withγv(0) = v. Then, by Lemma 12,

lim infα→∞

� (ξα, v) ≥ lim infα→∞

˜�κ (p; xα, γv(ε

α))

≥ � (w, v) − limα→∞

˜�κ (p; γw(εα), xα) ≥ (π + δ)/2,

and (4) follows from (3).Now let v ∈ B+ and γv be the geodesic with γv(0) = v. By Lemma 12 and ˜� 1(w;w0, v) >

λ, we have

lim infα→∞

� (ξα, v) ≥ lim infα→∞

˜�κ (p; xα, γv(ε

α))

> � (w, v) + 2ε ≥ � (w, B+) + 2ε.

By (3), this implies that

lim infα→∞

� (ξα, B+) > � (w, B+) + ε,

which contradicts to the choice of w. This implies that � (w, B) ∈ (π/2, π/2 + δ) for anyB ∈ B0.

If � (w, B−) ≥ π − λ, we note that

cos ˜� 1(w; B+, B−) < cos � (B+, B−) − cos λ < 0

with some abuse of notation, and we can derive a contradiction by the same argument. Thisfinishes the proof.

Remark 13 Perelman [11] proved Proposition 11 with δ = 0 and a general family B ofsubsets of for any compact Alexandrov space (, � ) of curvature ≥ 1. We can provethis for any not necessarily compact Alexandrov space. For future possible applications, wepresent a proof of this in the “Appendix”; see Proposition 22 below.

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In general, the space of directions at a point of an infinite-dimensional Alexandrov spaceis not an Alexandrov space. More precisely, we can prove that it is a metric space of curvature≥ 1, e.g. [16, Proposition 28], but it may fail to be a length space. This is why we requireProposition 11, not Proposition 22 below, in our proof of Theorem 5.

Proposition 11 plays a role in combination with the following definition. This is nothingbut a simplified version of the definition of regular points of admissible functions, which wasintroduced by Perelman and plays crucial role in the proof of his celebrated stability theoremfor finite-dimensional Alexandrov spaces. For details, we refer to Perelman’s paper [11] orKapovitch’s survey [8].

Definition 14 Let ε > 0 and f : U → Rk be a function defined on an open set U of

an Alexandrov space (X, d) of curvature ≥ κ , and suppose that each coordinate functionof f = ( f1, . . . , fk) is given by the distance function fi (·) := d(·, pi ) from a pointpi ∈ X . Following Perelman [11], we say that f is ε-regular (resp. regular) at x ∈ Uif ˜�

κ (x; pi , p j ) > π/2 + ε for any i �= j , and there is a point p0 ∈ X such that˜�κ (x; pi , p0) > π/2 + ε for any i (with some ε > 0).

We collect some facts on regular points, e.g. [11, Lemma 2.3 (2)]; cf. [8, Lemma 6.7].

Lemma 15 Suppose that f : U → Rk is regular at x ∈ U. Then

(1) k ≤ dim X;(2) The set of regular (resp. ε-regular) points of f is open;(3) f is 1-Lipschitz on U with respect to the norm |x |∞ := max1≤i≤k |xi | on R

k;(4) If X is proper, f is co-Lipschitz around x, i.e., for any small R > 0, f (B(x, R)) ⊃

I ( f (x), cR) for some small constant c > 0, where B(x, ·) and I ( f (x), ·) are openmetric balls with respect to the given metrics on X and R

k respectively.

In the literature, Lemma 15 is stated for finite-dimensional Alexandrov spaces. To provePart (4) when dim X = ∞, we would need Proposition 11.

3 Proof of the main theorem

In this section, we present proofs of Theorems 4 and 5. To do this, we state and proveLemma 18 below. Before stating this main lemma, we recall the packing radius of positivelycurved Alexandrov spaces.

Definition 16 Let (X, d) be a metric space and q ≥ 2 be an integer. We define its q-thpacking radius packq(X) by

packq(X) := 1

2sup

{min

1≤i< j≤qd(xi , x j ) : (xi )

qi=1 ∈ Xq

}. (5)

The sequence (xi ) ∈ Xq is called a q-th packer when it attains the supremum in (5).

We have the following comparison and rigidity results for packing radius of positivelycurved Alexandrov spaces.

Proposition 17 ([17], cf. Grove–Wilhelm [7]) Let (X, d) be an Alexandrov space of curva-ture ≥ 1 and q ≥ 2. Then

(1) packq(X) ≤ packq(Sn) = �q−2/2 for n ≥ q − 2.

(2) If packq(X) = packq(Sq−2) and there exists a q-th packer, then X is isometric to the

spherical join Sq−2 ∗ Y for some Alexandrov space Y of curvature ≥ 1.

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On the spread of Alexandrov spaces

Proposition 17 was proved for finite-dimensional spaces in [7] and for possibly infinite-dimensional spaces in [17]. The proof in [17] uses the Lang–Schroeder–Sturm inequality [16,Theorem 9] and the rigidity theorem [16, Theorem B].

Now we are ready to state our main lemma.

Lemma 18 (cf. [14, Main Lemma 8]) For any Alexandrov space X of curvature ≥ 1 andany natural number n ≤ dim X,

Spread(X) ≤ max{2 packn+2(X), π/2

} ≤ �n . (6)

In particular, any infinite-dimensional Alexandrov space X of curvature ≥ 1 fulfillsSpread(X) ≤ π/2.

The second inequality in (6) follows from Proposition 17.(1). Theorems 4 and 5 areimmediate consequence of Lemma 18 and Proposition 17.(2).

This lemma was proved implicitly in [14] for closed Riemannian manifolds of sectionalcurvature ≥ 1 and in [17] for finite-dimensional Alexandrov spaces of curvature ≥ 1. Ourproof of this lemma is a sight modification of the proof of [17, Lemma 25].

Proof of Lemma 18 First of all, we may assume that n ≥ 2, because the case dim X ≤ 1is trivial [17]. We also suppose that an Alexandrov space (X, d) of curvature ≥ 1 is isometri-cally embedded into an ultraproduct (Xω, dω) for some nonprincipal ultrafilter ω on N. Thisallows us to utilize Propositions 10 and 11.

We fix three numbers n ≤ dim X and λ′ > λ > max{2 packn+2(X), π/2

}with λ′ − λ

small and take a maximal subset Y := {p1, . . . , pk} of X such that d(pi , p j ) ∈ [λ, λ′] foreach i �= j . Note that λ > π/2 and we let λ > π/2 denote the angle at the vertices of theregular triangle of side length λ in S

2 as in [14]. We put Y := {p1, . . . , pk−1}.Next, we let I :=

{x ∈ X : d(x, pi ) ∈ [λ, λ′] for any pi ∈ Y

}and take a maximal subset

R := {r1, . . . , rs} of I containing r1 := pk ∈ I such that d(ri , r j ) > λ for any i �= j . Notethat |Y ∪ R| = k + s − 1 < n + 2.

We shall prove that Y is a λ-net in X , namely d(x, Y ) < λ for any x ∈ X . This implies thatSpread(X) ≤ λ′. The proof is divided into two steps. The first step is to prove the following

Lemma 19 (cf. [14, Lemma 10]) R contains a point other than r1 = pk, unless Y is a λ-netin X.

Proof We assume that Y is not a λ-net in X and hence the subset C := {x ∈ X : d(x, Y ) ≥ λ}is nonempty. Note that ˜� 1(x; pi , p j ) > λ for any x ∈ C and pi �= p j ∈ Y .

We fix small δ > 0 with δ � (λ′ − λ)/k and apply Proposition 11 for each x ∈ C and i0

with pi0 ∈ Y and d(x, pi0) > λ to find ξx ∈ ′x such that

(1) � xω (ξx ,[⇑pi0

x

]ω

) < π − λ;

(2) � xω (ξx ,[⇑pi

x]ω

) ∈ (π/2, π/2 + δ) for each pi ∈ Y\ {pi0

};

(3) � xω (ξx ,[⇑pk

x]ω

) > λ.

Now, we take a point z ∈ C and construct a sequence z0 := z, z1, . . . , zN of points ofC with N1 := 0 ≤ N1 ≤ · · · ≤ Nk =: N and L(N ) := ∑N

l=1 d(zl−1, zl) < C(λ, k),where C(λ, k) < ∞ is a constant depending only on λ and k, such that for each l ∈{

Ni0 + 1, . . . , Ni0+1}

with 1 ≤ i0 ≤ k − 1,

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T. Yokota

– d(zl , pi0) < d(zl−1, pi0);– |d(zl , pi ) − d(zNi0

, pi )| < ε · L(l)/L(N ) for each pi ∈ Y\ {pi0

};

– d(zl , pk) > d(zl−1, pk);– d(zNi , p j ) ∈ [λ, λ′] for each i, j with 1 ≤ j < i < k.

This is done as follows: for given zl , we take zl+1 ∈ X close to zl with ↑zl+1zl ∈ ′

zlbeing

equal to ξzl chosen above. Then, Proposition 10 yields the required properties. In the casethat this process does not terminate and yields an infinite sequence {zl}∞l=N ′ for some N ′, itis a Cauchy sequence in C since

∑∞l=N ′ d(zl+1, zl) < ∞, and hence it converges to some

point z′ ∈ C . Then, we can restart after putting zN ′+1 := z′.Now we know that there exists a point rs+1 := zN ∈ I and d(rs+1, pk) > λ, which means

that R contains at least two points. This completes the proof of Lemma 19. �We continue our proof of Main Lemma 18.

Lemma 20 (cf. [14, Lemma 9]) R consists of a single point pk .

Proof Suppose that there is a point r0 ∈ R ⊂ I satisfying d(r0, pk) > λ. Notice that|Y ∪ {r0} | = k + 1 < n + 2. Now, a map f : X → R

k−1 defined as f (·) :=(d(·, p1), . . . , d(·, pk−1)) is regular at r0 ∈ R.

Now we claim that for any small neighborhood Ur0 around r0, we can find a point p0 anda sequence

{pα

0

}in Ur0 such that

d(pα0 , p0) > ρ for some ρ > 0 and f (pα

0 ) → f (p0) in Rk−1

asα → ∞. Using Proposition 11, we know that for any points p and q in a small neighborhoodUr0 of r0, there exists a sequence

{pα

}with f ( pα) → f ( p) as α → ∞ in R

k−1 with d( pα, q)

bounded from above and below by a constant multiple of a norm of f ( p) − f (q) ∈ Rk−1. If

the claim above is not true, f is injective and pα → p as α → ∞ for any p ∈ Ur0 . Then wecan prove that f is bi-Lipschitz on Ur0 , cf. Lemma 15.(4). This is a contradiction becausethe Hausdorff dimension of f (Ur0) is at most k − 1 and dim X ≥ n > k − 1.

We put p′0 := pα

0 for some large α � 1. By exchanging p0 and p′0 if necessary, we may

assume that d(p0, pk) ≥ d(p′0, pk). We put Y := {p1, . . . , pk, p0} and fix ε > 0 so small

that p′0 is contained in the set

Y εreg :=

{x ∈ X : ˜� 1(x; pi , p j ) > π/2 + ε for any pi �= p j ∈ Y

}.

We notice that Y εreg contains the subset

{x ∈ f −1(I ( f (p0), ε

′)) : π/2 < d(x, pk) ≤ d(p0, pk) and d(x, p0) > ρ}

for some ε′ � ε, where

I ( f (p0), ε′) :=

{(vi ) ∈ R

k−1 : |vi − d(p0, pi )| < ε′ for any 1 ≤ i ≤ k − 1}

.

We fix δ > 0 with δ � ε′ and we apply Proposition 11 for each x ∈ Y εreg to find wx ∈ ′

xsuch that

(1) � xω (wx ,[⇑pk

x]ω

) < π − λ;

(2) � xω (wx ,[⇑pi

x]ω

) ∈ (π/2, π/2 + δ) for each pi ∈ Y ;(3) � xω (wx ,

[⇑p0x

]ω) > λ.

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On the spread of Alexandrov spaces

Now we construct a sequence q0 := p′0, q1, . . . , qN of points in Y ε

reg with L(N ) :=∑N

l=1 d(ql−1, ql) < C(λ), where C(λ) < ∞ is a constant depending only on λ, such thatd(qN , pk) ∈ [λ, λ′] and for each l ∈ {1, . . . , N },– d(ql , pk) < d(ql−1, pk);– |d(ql , pi ) − d(q0, pi )| < ε · L(l)/L(N ) for each pi ∈ Y ;– d(ql , p0) > d(ql−1, p0).

This is done as follows: If the point ql is given, we take a point ql+1 ∈ X close to ql with↑ql+1

ql ∈ ′ql

being equal to wql chosen above. Proposition 10 yields the required properties. Inthe case that this process does not terminate and yields an infinite sequence {ql}∞l=N ′ for someN ′, it is a Cauchy sequence in X and converges to some point q ′ ∈ X with d(q ′, pk) ≥ λ.Then we can restart after putting qN ′+1 := q ′.

The point pk+1 := qN lies in I and satisfies d(pk+1, pi ) ∈ [λ, λ′] for each pi ∈ Y . Thiscontradicts the maximality of Y and completes the proof of Lemma 20. �

Now, the combination of Lemmas 19 and 20 induces that Y is a λ-net of X with Diam(Y ) ≤λ′. Since λ′ is chosen arbitrarily, we complete the proof of Lemma 18. �

We conclude this section with a simple application of Theorem 5.

Proposition 21 Any sequence {Xi }i∈N of Alexandrov spaces of curvature ≥ 1 with n ≤dim Xi ≤ ∞ does not converge to a round sphere S

k of curvature 1 and dimension k < n inthe Gromov–Hausdorff topology.

This proposition is easy to prove because we know

Spread(Xi ) ≤ Spread(Sn) < Spread(Sk) for each i ∈ N

and that Spread(·) is continuous with respect to the Gromov–Hausdorff distance. This alsofollows from Colding’s argument in the proof of [3, Lemma 5.10] which relies on theBishop–Gromov inequality if supi∈N dim Xi is finite. However, this might be new whensupi∈N dim Xi = ∞.

Acknowledgments The author thanks Ayato Mitsuishi for discussions related to this work and the refereefor his/her comments with which he hopes the readability of this paper was improved. This work was doneduring the author’s stay in the University of Münster. He thanks for its hospitality and stimulating researchenvironment.

Appendix

As promised before, we present the proof of the following

Proposition 22 (cf. Perelman [11, Lemma 2.3]) Let (, � ) be an Alexandrov space of cur-vature ≥ 1. Suppose that a family B of subsets of satisfies that � (B, B ′) > λ > π/2 foreach B �= B ′ ∈ B. We choose two elements B+, B− ∈ B. Then there exists w ∈ such that� (w, B+) > λ, � (w, B−) < π − λ, and � (w, B) = π/2 for any B ∈ B0 := B\ {B+, B−}.

Our proof of Proposition 22 relies on the following lemma.

Lemma 23 (Ekeland principle, e.g. Ekeland [4]) For any continuous function f : X → R

on a complete metric space (X, d) with inf X f > −∞ and ε > 0, we can find a point x ∈ Xfor which the following folds:

f (y) ≥ f (x) − ε · d(y, x) for any y ∈ X.

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T. Yokota

Proof of Proposition 22 We put

0 := {v ∈ : � (v, B) ≥ π/2 for any B ∈ B0} ,

and for a small ε = ε(λ, � (B+, B−)) > 0, we apply Proposition 23 to find w ∈ 0 such that

� (v, B+) ≤ � (w, B+) + ε · � (v,w) for any v ∈ 0.

We verify that w satisfies the required properties. Since 0 is a closed subset of containingB− and ε is small enough, it is clear that

� (w, B+) ≥ � (B−, B+) − ε · � (B−, w) ≥ � (B−, B+) − πε > λ.

We suppose that � (w, B0) > π/2 for some B0 ∈ B0. Choose w0 ∈ Jw near B0 and w′ ∈(w,w0) with � (w′, w0) > π/2. Then, by the triangle comparison and ˜� 1(w;w0, v) > λ forany v ∈ B+, we deduce that w′ ∈ 0 and

� (w′, B+) > � (w, B+) + ε0 · � (w′, w)

with ε0 := − cos λ, which contradicts to the choice of w if ε < ε0. Thus implies that� (w, B) = π/2 for any B ∈ B0.

If � (w, B−) ≥ π − λ, then

cos ˜� 1(w; B+, B−) < cos � (B+, B−) − cos λ < 0

with some abuse of notation, and we can derive a contradiction by the same argument. Thisfinishes the proof. �

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