On the spread of positively curved Alexandrov spaces

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 finite-dimensional 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.

Appendix

As promised before, we present the proof of the following

Proposition 22

(cf. Perelman [11, Lemma 2.3]) Let $(\Sigma , \angle )$ be an Alexandrov space of curvature $\ge 1$. Suppose that a family $\mathcal {B}$ of subsets of $\Sigma$ satisfies that $\angle (B, B^\prime ) > \lambda > {\pi / 2}$ for each $B \ne B^\prime \in \mathcal {B}$. We choose two elements $B_+, B_- \in \mathcal {B}$. Then there exists $w \in \Sigma$ such that $\angle (w, B_+) > \lambda$, $\angle (w, B_-) < \pi - \lambda$, and $\angle (w, B) = {\pi / 2}$ for any $B \in \mathcal {B}_0 := \mathcal {B}{\setminus } \left\{ B_+, B_- \right\}$.

Our proof of Proposition 22 relies on the following lemma.

Lemma 23

(Ekeland principle, e.g. Ekeland [4]) For any continuous function $f: X \rightarrow \mathbb {R}$ on a complete metric space $(X, d)$ with $\inf _{X} f > -\infty$ and $\varepsilon >0$, we can find a point $x \in X$ for which the following folds:

$$\begin{aligned} f(y) \ge f(x) -\varepsilon \cdot d(y, x)\quad \text { for any } y \in X. \end{aligned}$$

Proof of Proposition 22  We put

$$\begin{aligned} \Sigma ^0 := \left\{ v \in \Sigma : \angle (v, B) \ge {\pi / 2}\text { for any } B \in \mathcal {B}_0 \right\} , \end{aligned}$$

and for a small $\varepsilon = \varepsilon (\lambda , \angle (B_+, B_-))>0$, we apply Proposition 23 to find $w \in \Sigma ^0$ such that

$$\begin{aligned} \angle (v, B_+) \le \angle (w, B_+) +\varepsilon \cdot \angle (v, w)\quad \text { for any } v \in \Sigma ^0. \end{aligned}$$

We verify that $w$ satisfies the required properties. Since $\Sigma ^0$ is a closed subset of $\Sigma$ containing $B_-$ and $\varepsilon$ is small enough, it is clear that

$$\begin{aligned} \angle (w, B_+) \ge \angle (B_-, B_+) - \varepsilon \cdot \angle (B_-, w) \ge \angle (B_-, B_+) - \pi \varepsilon > \lambda . \end{aligned}$$

We suppose that $\angle (w, B_0) > {\pi / 2}$ for some $B_0 \in \mathcal {B}_0$. Choose $w_{0} \in J_w$ near $B_0$ and $w^\prime \in (w, w_0)$ with $\angle (w^\prime , w_0) > {\pi / 2}$. Then, by the triangle comparison and $\tilde{\angle }_1 (w; w_0, v) > \lambda$ for any $v \in B_+$, we deduce that $w^\prime \in \Sigma ^0$ and

$$\begin{aligned} \angle (w^\prime , B_+) > \angle (w, B_+) + \varepsilon _0 \cdot \angle (w^\prime , w) \end{aligned}$$

with $\varepsilon _0 := -\cos \lambda$, which contradicts to the choice of $w$ if $\varepsilon < \varepsilon _0$. Thus implies that $\angle (w, B) = {\pi / 2}$ for any $B \in \mathcal {B}_0$.

If $\angle (w, B_-) \ge \pi - \lambda$, then

$$\begin{aligned} \cos \tilde{\angle }_1 (w; B_+, B_-) < \cos \angle (B_+, B_-) -\cos \lambda <0 \end{aligned}$$

with some abuse of notation, and we can derive a contradiction by the same argument. This finishes the proof. $\square$