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 positively curved manifolds with a uniform lower bound on sectional curvature attains the maximum with the round sphere. Recently the author proved that this is also the case for closed finitedimensional Alexandrov spaces with a positive lower curvature bound. These were proved as a corollary of a comparison theorem for the invariant called spread of those spaces. In this paper, 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 concerned with the geometry of possibly infinite-dimensional Alexandrov spaces with a positive lower curvature bound. As the nature of infinite-dimensional Alexandrov spaces has not been fully explored, we are particularly interested in those spaces. Our goal is to establish a comparison theorem for them with the round sphere as the model space, which extends the main result of [17]. In order to state the main theorem, we prepare the following definition.

Partly supported by JSPS Postdoctoral Fellowships for Research Abroad. T. Yokota Research Institute for Mathematical Sciences, Kyoto University, Kyoto 606-8502, Japan T. Yokota (B) Mathematisches Institut, University of Münster, Einsteinstr. 62, 48149 Münster, Germany e-mail: [email protected]

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Definition 1 (Wilhelm [14]) For any metric space X = (X, d), we define its spread, denoted by 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 V fulfills 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 Sn and R P n 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. 1 ). We know that n is the Here and throughout this paper, we use n := arccos(− n+1 spherical distance between vertices of a regular (n + 1)-simplex inscribed in the unit sphere Sn ⊂ Rn+1 and the set Y := { p1 , . . . , pn+2 } ⊂ Sn 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 stated as follows.

Theorem 3 (Wilhelm [14]) For any n-dimensional closed Riemannian manifold V of sectional curvature ≥ 1, either Spread(V ) < Spread(Sn ) or V is isometric to the round sphere Sn . In particular, for any such V , either Fill Rad(V ) < Fill Rad(Sn ) or V is isometric to the round sphere Sn . 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 are known to coincide for any Alexandrov space with a lower curvature bound. Theorem 4 ([17]) For any n-dimensional Alexandrov space X of curvature ≥ 1, either Spread(X ) < Spread(Sn ) or X is isometric to the round sphere Sn . Moreover, for any ndimensional 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 shown in the papers by Grove–Petersen [6] and Yamaguchi [15] that any finite-dimensional compact Alexandrov space without boundary admits a fundamental class. The second statement in Theorem 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 of Theorem 4 to infinite-dimensional Alexandrov spaces. Theorem 5 Any infinite-dimensional Alexandrov space X of curvature ≥ 1 fulfills Spread(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 the radius Rad(X ) := inf x∈X sup y∈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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Remark 7 As was remarked by Wilhelm [14], cf. [17, Corollary 8], Theorem 4 also yields that Cont 0 Rad(X ) ≤ Cont0 Rad(Sn ) = n /2 for any n-dimensional Alexandtrov space X of curvature ≥ 1. The contractibility radius Cont k Rad is defined for any metric space, see [5, Appendix 2]. However, an infinite-dimensional sphere S∞ is contractible and hence Contk Rad(S∞ ) = 0 for any k ≥ 0. This is why we do not expect a comparison theorem for Cont0 Rad in infinite dimension. Our proof of Theorem 5 is a modification of that of Theorem 4 given in [17], which is in turn an adaptation of Wilhelm’s argument in the proof of Theorem 3 given in [14]. In proving Theorem 5, we also re-prove Theorem 4 and the proof would become slightly simpler than that in [17]. For this reason, the present paper may have substantial overlap with [17]. Several difficulties in proving Theorem 5 arise from that an infinite-dimensional Alexandrov space of 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 infinitedimensional 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 propositions that 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 quadruple condition, e.g. [17, Definition 12]. We refer to [17] for the references on Alexandrov geometry as well. Here, instead of giving the precise definition of Alexandrov spaces, we collect some of 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 simplyconnected complete Riemannian manifold of constant curvature κ ∈ R. When κ = 1, we also use (S2 ,  ) instead of (M1 , d1 ) later. {x, y, z} be a triple Proposition 8 Let (X, d) be an √Alexandrov space of curvature ≥ κ and in 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˜ } ⊂ (Mκ2 , 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)

:= lim  ˜ κ ( p; γ (s), η(t)) s,t→0

is well-defined for any two geodesics γ , η : [0, δ) → X with p := γ (0) = η(0). The space of directions ( p ,  p ) at a point p ∈ X is the metric completion of the set ( p ,  p ) of q equivalence classes of unit speed geodesics emanating from p. We use the notations ⇑ p ⊂ p q q and ↑ p ∈ ⇑ p , respectively, to denote the set of all equivalence classes of geodesics from p to q ∈ X and one of its elements. The equivalence class of a geodesic γ with γ (0) = p is denoted by γ˙ (0) ∈ p . As our main interest is in infinite-dimensional Alexandrov spaces of curvature ≥ 1, we give 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 if and 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 cone if and only if P2 (Y ) is. Now we give a construction of another Alexandrov space of curvature ≥ 1 from a given one. 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 Alexandrov space Q of curvature ≥ 1. In our proof of Theorem 5, we will exploit ultraproduct of metric spaces. Since its definition can be found in several papers in Alexandrov geometry and other areas of geometry, we do not repeat it here and we only try to fix the notation. For a thorough treatment, we refer to e.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 (X ω , d ω ) of (X, d). Roughly speaking, (X ω , d ω ) is the set of equivalence classes of infinite sequences in X N equipped with the distance d ω given by d ω ((xi ), (yi )) := limi→ω d(xi , yi ) for (xi ), (yi ) ∈ X N . We list relevant properties of ultraproducts. – If (X, d) is proper, i.e, any closed bounded subset of X is compact, then (X ω , d ω ) is isometric to (X, d). In general, (X, d) is isometrically embedded in (X ω , d ω ). We identify a point x ∈ X with x ω := (x, x, . . .) ∈ X ω . – If (X, d) is a length spaces, then (X ω , d ω ) is a geodesic space. – If (X, d) is an Alexandrov space of curvature ≥ κ, then so is (X ω , d ω ). In order to prove our main Theorem 5 for infinite-dimensional Alexandrov spaces, we require the following two propositions, which partially generalize the results mainly known in finite dimensions. We know that any finite-dimensional Alexandrov space is proper and its space of directions at any point is compact, but this is not the case in infinite dimension in general.

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In Propositions 10 and 11 below, we assume that (X, d) is isometrically embedded into an ultraproduct (X ω , d ω ) 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 properties with the space of directions at a point of a proper Alexandrov space. This is the main reason  q ω qω why we introduced the ultraproduct. In the sequel, we use the notation ⇑ p := ⇑ pω ⊂ pω for p, q ∈ X . Proposition 10 (First variation formula, cf. [1]) Suppose that x,√y and p are points of an Alexandrov space (X, d) of curvature ≥ κ, with d(x, p) < π/ κ if κ > 0, and that a y p ω y geodesic segment x y representing ↑x ∈ x is given. We put  px y :=  x ω ( ⇑x , ↑x ) =   p ω  y inf  x ω (ξ, ↑x ) : ξ ∈ ⇑x . Then, for z ∈ x y, we have d(z, p) = d(x, p) − d(x, z) cos  px y + 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, we give a proof of this proposition; cf. Alexander–Petrunin–Kapovitch [1]. In the proof and hereafter, we use the notation  Jz := x ∈ X \{z} :  ˜ κ (x; y, z) > π − k −1 k∈N y∈X

for a point z ∈ X of an Alexandrov space X of curvature ≥ κ. It is easy to see that Jz is a y y dense subset of X and there exists a unique ↑z ∈ ⇑z for each y ∈ Jz , cf. Plaut [12]. Proof of Proposition 10 Inspired by the proof of [12, Proposition 49], we shall prove that lim  ˜ κ (x; p, z) =  px y.

z→x

(1)

We take a sequence {z i }i∈N ⊂ x y with z i → x as i → ∞ and a minimal unit speed geodesic γi in X from z i to pi ∈ Jzi for each i ∈ N with { pi }i∈N converging to p so fast that lim  ˜ κ (x; pi , z i ) = lim  ˜ κ (x; p, z i ).

i→∞

i→∞

Then, for each ε > 0, we can find δ > 0 and z(ε) ∈ x y such that lim sup  ˜ κ (x; γi (δ), z(ε)) >  px y − ε. i→∞

If this is not the case, we have a geodesic γ (t) := (γi (t)), t ∈ [0, d(x, p)] in X ω for y which  x ω (γ˙ (0), ↑x ) ≤  px y − ε, contradicting the definition of  px y. By using the angle monotonicity and Alexandrov’s lemma, e.g. [2], we also deduce  ˜ κ (x; γi (δ), z(ε))

≤  ˜ κ (x; γi (δ), z i ) <  ˜ κ (x; γi (δ), z i ) +  ˜ κ (x; , pi , γi (δ)) ≤  ˜ κ (x; pi , z i ),

and hence taking i → ∞ and ε → 0 yields that lim z→x  ˜ κ (x; z, p) ≥  px y. The other inequality implying Eq. (1) follows from the angle monotonicity. This completes the proof of 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 a point p ∈ X of an Alexandrov space (X,  qd)  ≥ κ. Suppose that Q ⊂ X \ { p} is a ωof curvature finite subset and a family B = B := ⇑ p : q ∈ Q of subsets of ( pω ,  pω ) satisfies that  p ω (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 make some observation. Let (X, d) be an Alexandrov space of curvature ≥ κ. We fix a sequence {εα } of positive numbers with ε α → 0 as α → ∞ and put d α (·, ·) := (ε α )−1 d(·, ·). Then, for any unit speed geodesics γi , i ∈ {2, 3}, departing a point p ∈ X and for some t ∈ (0, 1), we can find a sequence {x α } ⊂ J p of points satisfying d α (γ2 (ε α ), x α ) d α (x α , γ3 (ε α )) = lim d α (γ2 (ε α ), γ3 (ε α )) = lim . α→∞ α→∞ α→∞ t 1−t lim

(2)

It is easy to see in this situation, with ξi := γ˙i (0) ∈ p being the initial vectors, that Eq. (2) is equivalent to that the limit limα→∞ d α (x α , p) ∈ (0, ∞) exists and lim

 ˜ κ ( p; γ2 (ε α ), x α )

α→∞

s

=

p (ξ2 , ξ3 )

= lim

 ˜ κ ( p; x α , γ3 (ε α ))

α→∞

1−s

for some s ∈ (0, 1). Lemma 12 Let (X, d) be an Alexandrov space of curvature ≥ κ and {εα } and d α (·, ·) be as above. Suppose that γi , i ∈ {1, 2, 3}, are unit speed geodesics departing a point p ∈ X with initial vectors ξi := γ˙i (0) ∈ ( p ,  p ) satisfying peri(ξ1 , ξ2 , ξ3 ) < 2π, and that {x α } ⊂ J p is of points satisfying Eq. (2) for some t ∈ (0, 1). Choose an isometric copy  a sequence ξ˜1 , ξ˜2 , ξ˜3 ⊂ (S2 ,  ) of {ξ1 , ξ2 , ξ3 } ⊂ ( p ,  p ) and the corresponding point ξ˜4 ∈ (ξ˜2 , ξ˜3 ) with  (ξ˜2 , ξ˜4 ) = limα→∞  ˜ κ ( p; γ2 (ε α ), x α ). Then, we have lim inf  α→∞

xα p (↑ p , ξ1 )

≥  (ξ˜4 , ξ˜1 ).

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 ) and geodesics γ and γ0 starting from p with γ˙ (0) = w and γ˙0 (0) = w0 . We also take sequences {x α } ⊂ J p of points and ε α → 0 of positive numbers such that 

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lim  ˜ κ ( p; γ (ε α ), x α ) + lim  ˜ κ ( p; x α , γ0 (ε α )) =  (w, w0 ),

α→∞

α→∞

and limα→∞  ˜ κ ( p; x α , γ0 (ε α )) = (π + δ)/2. α Put ξ α := ↑xp ∈ p ⊂ pω . Here we claim that lim inf  (ξ α , B) = inf lim inf  (ξ α , v) α→∞

v∈B α→∞

(3)

 q ω for any B := ⇑ p ⊂ 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 exists w0 ∈ B for which lim inf α→∞  (ξ α , w0 ) is less than the right hand side of (3). This is a contradiction. Now, we observe that lim inf  (ξ α , B) > π/2 α→∞

for each B ∈ B0 .

(4)

This means that ξ α ∈ 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 γ˙v (0) = v. Then, by Lemma 12 and  ˜ 1 (w; w0 , v) > λ, lim inf  (ξ α , v) ≥ lim inf  ˜ κ ( p; x α , γv (ε α )) α→∞

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 with γ˙v (0) = v. Then, by Lemma 12, lim inf  (ξ α , v) ≥ lim inf  ˜ κ ( p; x α , γv (ε α )) α→∞

α→∞

≥  (w, v) − lim  ˜ κ ( p; γw (ε α ), x α ) ≥ (π + δ)/2, α→∞

and (4) follows from (3). Now let v ∈ B+ and γv be the geodesic with γ˙v (0) = v. By Lemma 12 and  ˜ 1 (w; w0 , v) > λ, we have lim inf  (ξ α , v) ≥ lim inf  ˜ κ ( p; x α , γv (ε α )) α→∞

α→∞

>  (w, v) + 2ε ≥  (w, B+ ) + 2ε. By (3), this implies that lim inf  (ξ α , B+ ) >  (w, B+ ) + ε, α→∞

which contradicts to the choice of w. This implies that  (w, B) ∈ (π/2, π/2 + δ) for any B ∈ 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. This finishes the proof. Remark 13 Perelman [11] proved Proposition 11 with δ = 0 and a general family B of subsets of for any compact Alexandrov space ( ,  ) of curvature ≥ 1. We can prove this for any not necessarily compact Alexandrov space. For future possible applications, we present 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 space is 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 require Proposition 11, not Proposition 22 below, in our proof of Theorem 5. Proposition 11 plays a role in combination with the following definition. This is nothing but a simplified version of the definition of regular points of admissible functions, which was introduced by Perelman and plays crucial role in the proof of his celebrated stability theorem for finite-dimensional Alexandrov spaces. For details, we refer to Perelman’s paper [11] or Kapovitch’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 function of f = ( f 1 , . . . , f k ) is given by the distance function f i (·) := d(·, pi ) from a point pi ∈ X . Following Perelman [11], we say that f is ε-regular (resp. regular) at x ∈ U if  ˜ 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) (2) (3) (4)

k ≤ dim X ; The set of regular (resp. ε-regular) points of f is open; f is 1-Lipschitz on U with respect to the norm |x|∞ := max1≤i≤k |xi | on Rk ; If X is proper, f is co-Lipschitz around x, i.e., for any small R > 0, f (B(x, R)) ⊃ I ( f (x), c R) for some small constant c > 0, where B(x, ·) and I ( f (x), ·) are open metric balls with respect to the given metrics on X and Rk respectively.

In the literature, Lemma 15 is stated for finite-dimensional Alexandrov spaces. To prove Part (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 prove Lemma 18 below. Before stating this main lemma, we recall the packing radius of positively curved Alexandrov spaces. Definition 16 Let (X, d) be a metric space and q ≥ 2 be an integer. We define its q-th packing radius pack q (X ) by

 1 q pack q (X ) := sup min d(xi , x j ) : (xi )i=1 ∈ X q . (5) 1≤i< j≤q 2 The sequence (xi ) ∈ X q 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 positively curved Alexandrov spaces. Proposition 17 ([17], cf. Grove–Wilhelm [7]) Let (X, d) be an Alexandrov space of curvature ≥ 1 and q ≥ 2. Then (1) pack q (X ) ≤ pack q (Sn ) = q−2 /2 for n ≥ q − 2. (2) If pack q (X ) = pack q (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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Proposition 17 was proved for finite-dimensional spaces in [7] and for possibly infinitedimensional 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 and any natural number n ≤ dim X ,   Spread(X ) ≤ max 2 pack n+2 (X ), π/2 ≤ n .

(6)

In particular, any infinite-dimensional Alexandrov space X of curvature ≥ 1 fulfills Spread(X ) ≤ π/2. The second inequality in (6) follows from Proposition 17.(1). Theorems 4 and 5 are immediate consequence of Lemma 18 and Proposition 17.(2). This lemma was proved implicitly in [14] for closed Riemannian manifolds of sectional curvature ≥ 1 and in [17] for finite-dimensional Alexandrov spaces of curvature ≥ 1. Our proof 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 ≤ 1 is trivial [17]. We also suppose that an Alexandrov space (X, d) of curvature ≥ 1 is isometrically embedded into an ultraproduct (X ω , d ω ) for some nonprincipal ultrafilter ω on N. This allows us to utilize Propositions 10 and 11.   We fix three numbers n ≤ dim X and λ > λ > max 2 pack n+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 ] for each i  = j. Note that λ > π/2 and we let λ˜ > π/2 denote the angle at the vertices of the 2 ˇ regular triangle of side  length λ in S as in [14]. We put Y := { p 1 , . . . , pk−1 }.

Next, we let I := x ∈ X : d(x, pi ) ∈ [λ, λ ] for any pi ∈ Yˇ 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. Note that |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 that Spread(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-net in 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 i 0 with pi0 ∈ Yˇ and d(x, pi0 ) > λ to find ξx ∈ x such that p ω i (1)  x ω (ξx , ⇑x 0 ) < π − λ˜ ;  pi ω   (2)  x ω (ξx , ⇑x ) ∈ (π/2, π/2 + δ) for each pi ∈ Yˇ \ pi0 ;  p ω (3)  x ω (ξx , ⇑x k ) > λ˜ . Now, we take a point z ∈ C and construct a sequence z 0 :=  Nz, z 1 , . . . , z N of points of C with N1 := 0 ≤ N1 ≤ · · · ≤ Nk =: N and L(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 ≤ i 0 ≤ k − 1,

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– – – –

d(zl , pi0 ) < d(zl−1 , pi0 );   |d(zl , pi ) − d(z Ni0 , pi )| < ε · L(l)/L(N ) for each pi ∈ Yˇ \ pi0 ; d(zl , pk ) > d(zl−1 , pk ); d(z Ni , p j ) ∈ [λ, λ ] for each i, j with 1 ≤ j < i < k.

∈ z l being This is done as follows: for given zl , we take zl+1 ∈ X close to zl with ↑zl+1 l equal to ξzl chosen above. Then, Proposition 10 yields the required properties. In the case ∞

that this process does not terminate yields an infinite sequence {zl }l=N

for some N , it and ∞ is 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 z N +1 := z . Now we know that there exists a point rs+1 := z N ∈ I and d(rs+1 , pk ) > λ, which means that R contains at least two points. This completes the proof of Lemma 19. 

z

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 → Rk−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 and a sequence p0α in Ur0 such that d( p0α , p0 ) > ρ

for some

ρ>0

and f ( p0α ) → f ( p0 ) in Rk−1

as α → ∞. Using Proposition 11, weknow that for any points pˆ and qˆ in a small neighborhood Ur0 of r0 , there exists a sequence pˆ α with f ( pˆ α ) → f ( p) ˆ as α → ∞ in Rk−1 with d( pˆ α , q) ˆ bounded from above and below by a constant multiple of a norm of f ( p) ˆ − f (q) ˆ ∈ Rk−1 . If the claim above is not true, f is injective and pˆ α → pˆ as α → ∞ for any pˆ ∈ Ur0 . Then we can prove that f is bi-Lipschitz on Ur0 , cf. Lemma 15.(4). This is a contradiction because the Hausdorff dimension of f (Ur0 ) is at most k − 1 and dim X ≥ n > k − 1. We put p0 := p0α for some large α  1. By exchanging p0 and p0 if necessary, we may  := { p1 , . . . , pk , p0 } and fix ε > 0 so small assume that d( p0 , pk ) ≥ d( p0 , pk ). We put Y that p0 is contained in the set  rεeg := x ∈ X :  ˜ 1 (x; pi , p j ) > π/2 + ε for any pi  = p j ∈ Y  . Y rεeg contains the subset We notice that Y   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 ) ∈ Rk−1 : |vi − d( p0 , pi )| < ε for any 1 ≤ i ≤ k − 1 . rεeg to find wx ∈ x

We fix δ > 0 with δ  ε and we apply Proposition 11 for each x ∈ Y such that  p ω ˜ (1)  x ω (wx , ⇑x k ) < π − λ;  pi ω (2)  x ω (wx , ⇑x ) ∈ (π/2, π/2 + δ) for each pi ∈ Yˇ ;  p ω (3)  x ω (wx , ⇑x 0 ) > λ˜ .

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rεeg with L(N ) := Now we construct a sequence q0 := p0 , q1 , . . . , q N of points in Y N l=1 d(ql−1 , ql ) < C(λ), where C(λ) < ∞ is a constant depending only on λ, such that d(q N , 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 ∈ Yˇ ; – 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 q ↑ql+1 ∈ q l being equal to wql chosen above. Proposition 10 yields the required properties. In l ∞ the case that this process does not terminate and yields an infinite sequence {ql }l=N

for some



N , 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 q N +1 := q . The point pk+1 := q N lies in I and satisfies d( pk+1 , pi ) ∈ [λ, λ ] for each pi ∈ Y . This contradicts 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 {X i }i∈N of Alexandrov spaces of curvature ≥ 1 with n ≤ dim X i ≤ ∞ does not converge to a round sphere Sk of curvature 1 and dimension k < n in the Gromov–Hausdorff topology. This proposition is easy to prove because we know Spread(X i ) ≤ Spread(Sn ) < Spread(Sk ) for each i ∈ N and that Spread(·) is continuous with respect to the Gromov–Hausdorff distance. This also follows from Colding’s argument in the proof of [3, Lemma 5.10] which relies on the Bishop–Gromov inequality if supi∈N dim X i is finite. However, this might be new when supi∈N dim X i = ∞. Acknowledgments The author thanks Ayato Mitsuishi for discussions related to this work and the referee for his/her comments with which he hopes the readability of this paper was improved. This work was done during the author’s stay in the University of Münster. He thanks for its hospitality and stimulating research environment.

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 curvature ≥ 1. Suppose that a family B of subsets of satisfies that  (B, B ) > λ > π/2 for each 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 ∈ X for 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 containing B− 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 for any 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. This finishes the proof. 

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