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The context

I have been reading Gromov's Metric Structures..., and came upon result 1.14.(a), page 11, which states the following.

Let $K\subset\mathbb R^d$ be a compact subset, and $d_\ell$ its length metric.¹

If the distortion $\sup_{x\neq y}\frac{d_\ell(x,y)}{d_{\mathbb R^d}(x,y)}$ is less than $\frac\pi2$, then $K$ is simply connected.

The proof, if I understand correctly, goes roughly as follows [Edit: see below for more detail]:

  1. suppose there exists a compact $K$ satisfying the inequality but admitting a non-trivial loop;
  2. choose a non-trivial loop $\gamma$ of minimal length;
  3. show that the length metric on $\gamma$ is the same as that of $K$ (by minimality of the loop);
  4. argue that $\gamma$ must ‘make a turn’ at some point, contradicting the metric assumption.

My question

lies in Point 2 of this program. Of course, there is no reason for such a curve to exist (see e.g. the Hawaiian earring). [Edit: YCor's example, in the comments, shows in fact that even in a fixed free homotopy class, one may not find a length-minimising curve.]

  1. Am I mistaken in thinking that Point 2 is actually what is happening in the proof?

    [Edit: This point is somewhat answered by M. Dus in his answer, who notes that it might be suggested by the authors that the subset is supposed to be a submanifold, or a domain with smooth boundary.]

  2. If not, is there a simple way to get the argument back on track?

    [Edit: Following M. Dus's answer, the question might actually be: Is there a simple way to get the argument working for all compact sets?]

A bit more about the proof

I'll start by saying that a quick search in your favourite search engine will take you to a pdf version of the book, for anyone interested. However I thought I should make my question as self-contained as possible.

Point 1 of the proof is self-evident, and point 2 is given as is in the book (‘let $\alpha$ be a nontrivial homotopy class in which there exists a curve of minimal length among all homotopically nontrivial loops’). Say that we get a curve $\gamma$ of length $\ell$, parametrised by arc length by $c:\mathbb R/\ell\mathbb Z\to\mathbb R^d$.

Point 3 (after trimming) aims to show that the path length distance between $c(t)$ and $c(t+\ell/2)$ is the same, viewed as points in $K$ or $\gamma$. Otherwise, there must be a curve in $K$ of length $<\ell/2$ between these two points, creating two loops of length $<\ell$ whose product is $c$. One of them must be non trivial (because $c$ is), a contradiction (because $c$ is of minimal length among such curves).

Point 4, then, defines $\tilde c:t\mapsto c(t+\ell/2)-c(t)$, $r=|\tilde c|$ and $u=\tilde c/|\tilde c|$. We will show that the length of $u$ is less than $2\pi$, a contradiction, because $u$ takes values in the sphere, and $u(t+\ell/2)=-u(t)$.

Noting that $\frac{\mathrm du}{\mathrm dt}$ is orthogonal to $\tilde c$, we get $$ \left|\frac{\mathrm du}{\mathrm dt}\right|^2 = \left|\frac1{r(t)}\frac{\mathrm d\tilde c'}{\mathrm dt}\right|^2 - \left|\frac{-r'(t)}{r(t)^2}\tilde c(t)\right|^2 \leq \frac{4-r'(t)^2}{r(t)^2} \leq \frac4{r(t)^2}. $$ But according to Point 3, the path length distance in $K$ between $c(t)$ and $c(t+\ell/2)$ is precisely $\ell/2$, so we get $$ \frac{\ell/2}{r(t)}\leq\sup_{x\neq y}\frac{d_\ell(x,y)}{d_{\mathbb R^d}(x,y)}\leq\frac\pi2-\varepsilon<\frac\pi2. $$ This yields $|u'(t)|\leq(2\pi-\varepsilon)/\ell$ so the length of $u$ is at most $2\pi-\varepsilon<2\pi$, as announced.


¹ If $S\subset\mathbb R^d$, its associated length metric is $$d_\ell(x,y):=\inf\{\mathrm{length}(\gamma),\gamma\text{ a path from }x\text{ to }y\text{ with values in }S\}.$$ For instance, $d_\ell$ is infinite if $x$ and $y$ are not in the same connected component, or two different points in the Koch snowflake.

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    $\begingroup$ The fraction you wrote is clearly $\le 1$. You probably meant the inverse fraction. Also, $d_\ell$ might be infinite, but the non-distortion assumption indeed implies it is finite. $\endgroup$ – YCor Jan 31 at 18:33
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    $\begingroup$ You're right about the Hawaiian earring. It's tempting to instead consider a loop of minimal length within its free homotopy class. But there are compact spaces with examples of loops that are homotopic to loops of arbitrary small length, yet are not contractible... $\endgroup$ – YCor Jan 31 at 18:38
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    $\begingroup$ I am quite sure that you want to have a length minimal loop that is homotopic to the given one. Then the Hawaiian ring would not be a counterexample. $\endgroup$ – Piotr Hajlasz Jan 31 at 18:40
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    $\begingroup$ @Neal Yes. The 1-point union of two cones over the Hawaiian earring. It embeds into $\mathbf{R}^3$. $\endgroup$ – YCor Jan 31 at 18:49
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    $\begingroup$ @Neal it's precisely a counterexample to the claim that the wedge of two contractible spaces is contractible. To be more precise, let $H$ be the earring, $o$ its singular point, the cone over $H$ is $H\times [0,1]/(H\times\{0\})$, and its basepoint is $(o,1)$, and the gluing is over the given basepoint (if one were gluing over the crushed point $(\ast,0)$, the wedge would be contractible). $\endgroup$ – YCor Jan 31 at 19:00
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[Edit] I think the following argument, made possible by the help of Anton Petrunin in the comments (thank you very much!), does the trick. I left the original sketch below for historical/affective reasons.

Suppose that $K\subset\mathbb R^d$ is a compact satisfying the above non-distortion condition. Then the argument of Gromov described above shows that there is no curve (contractible or not) such that the path length distance between opposite points is no less than half the length of the loop. In other words, for all $\alpha>0$ small enough, any arc-length curve parametrisation $c:\mathbb Z/\ell\mathbb Z$ must admit a pair of points $(c(t),c(t+\ell/2))$ such that their path length distance is at most $(1-\alpha)\ell/2$. We now fix such an $\alpha>0$.

Cellulation of the unit disc

Let $c:\mathbb Z/\ell\mathbb Z$ be a loop in $K$ of finite length parametrised by arc length. Let $\mathbb D$ be the closed unit disc in $\mathbb C$, and define $\Omega_\varnothing=\mathring{\mathbb D}$. We identify $\partial\Omega_\varnothing$ with $\mathbb Z/2\pi\mathbb Z$ according to a parametrisation by arc length, so that there exists a parametrisation $\gamma_\varnothing:\partial\Omega_\varnothing\to\mathrm{im}c$ at constant speed; find it and fix it.

Because of the above argument, there exists two opposite points $a$ and $b$ on $\partial\Omega_\varnothing$ such that the path length distance between $\gamma_\varnothing(\partial)$ in $K$ is at most $(1-\alpha)\ell/2$. The straight line $E$ in $\mathbb D$ between $a$ and $b$ cuts $\Omega_\varnothing\setminus E$ into two open convex connected components $\Omega_0$ and $\Omega_1$. Identifying $\partial\Omega_i$ by arc length with some $\mathbb Z/L_i\mathbb Z$, there exists a curve parametrisation $\gamma_i:\partial\Omega_i\to K$ such that

  1. $\gamma_i$ agrees with $\gamma_\varnothing$ over the intersection of their domains,

  2. $\gamma_0$ and $\gamma_1$ agree and have constant speed over the intersection of their domains, and

  3. the total length of $\gamma_i$ is at most $(1-\alpha/2)\ell$.

Iterate this construction: find four disjoint open sets $\Omega_{00}$, ..., $\Omega_{11}$ with piecewise smooth boundaries such that $\overline{\Omega_{i0}}\cup\overline{\Omega_{i1}}=\overline{\Omega_i}$, and four loops $\gamma_{00}$, ..., $\gamma_{11}$ such that they are compatible with one another and with $\gamma_0$, $\gamma_1$, have constant speed over the newly created boundaries $\partial\Omega_{i0}\cap\partial\Omega_{i1}$, and most importantly have length at most $(1-\alpha/2)^2\ell$.

Almost-continuous maps

Construct, for each $n\geq0$, a map $f_n:\mathbb D\to K$ as follows. For each word $w$ on $\{0;1\}$ with length $|w|=n$, ${f_n}_{|\partial\Omega_w}=\gamma_w$. This is possible according to the compatibility conditions (1) and (2). Moreover, for $x$ not in any $\partial\Omega_w$, there exists a unique $w$ such that $x\in\Omega_w$; then $f_n$ sends $x$ to any point in $\gamma_w(\partial\Omega_w)$.

Two facts (pointed out by Anton Petrunin) will be relevent in the sequel. First, if $x\in\mathbb D$ and $n$ is fixed, we can define an open neighbourhood of $x$ whose points $y$ satisfy $|f_n(x)-f_n(y)|\leq(1-\alpha/2)^n\ell/2$. Indeed, if we set $$\mathcal U_x^n=\Big(\bigcup_{w:x\notin\overline{\Omega_w}}\overline{\Omega_w}\Big)^\complement, $$ then it is an open neighbourhood of $x$ and any of its points $y$ must be in some $\overline\Omega_w\ni x$. In particular, $f_n(x)$ and $f_n(y)$ both belong to $\gamma_w(\partial\Omega_w)$, and must lie at distance at most half of the length of $\gamma_w$, as awaited.

Second, $|f_n-f_{n+1}|$ is bounded by $(1-\alpha/2)^n\ell$. Indeed, if $x$ belongs to $\overline{\Omega_{wi}}$ with $w$ a word of length $n$, then for any point $y$ in $\partial\Omega_w\cap\partial\Omega_{wi}$ (the intersation in non-empty),

  1. $f_n(y)=f_{n+1}(y)$,

  2. $f_n(x)$ and $f_n(y)$ are at distance at most $(1-\alpha/2)^n\ell/2$, because they both lie on the image of $\gamma_w$, and

  3. $f_{n+1}(x)$ and $f_{n+1}(y)$ are at distance at most $(1-\alpha/2)^{n+1}\ell/2$, because they both lie on the image of $\gamma_{wi}$.

This imples that $f_n(x)$ and $f_{n+1}(x)$ are indeed at distance at most $(1-\alpha/2)^n\ell/2+(1-\alpha/2)^{n+1}\ell/2\leq(1-\alpha/2)^n\ell$.

Limit as the loops shrink

Let $f_n:\mathbb D\to K$ be a sequence of maps and $\varepsilon_n>0$ a sequence of positive numbers tending to zero. Suppose that $(f_n)_{n\geq0}$ is a Cauchy sequence with respect to the norm $|\cdot|_\infty$ and that for any $n\in\mathbb N$ and any $x\in\mathbb D$, there exists a neighbourhood $U$ of $x$ such that $|f(x)-f(y)|<\varepsilon_n$ for all $y\in U$. (These hypotheses are of course satisfied by the $f_n$ defined above.) Then $(f_n)_{n\geq0}$ converges to a continuous map $f:\mathbb D\to K$ with respect to the uniform norm.

Indeed, because it is Cauchy and $K$ is compact, $(f_n)_{n\geq0}$ must converge in the uniform norm to some $f:\mathbb D\to K$. Now for any $x\in\mathbb D$ and $\varepsilon>0$, there exists $n\geq0$ such that $\varepsilon_n<\varepsilon/3$ and $|f_n-f|_\infty<\varepsilon/3$. Let $U$ be a neighbourhood of $x$ such that $|f_n(x)-f_n(y)|<\varepsilon_n$ for all $y\in U$; then of course $|f(x)-f(y)|<2|f_n-f|+\varepsilon_n<\varepsilon$. This holds for any $x\in\mathbb D$ and $\varepsilon>0$, so $f$ is continuous.

In conclusion, we constructed a map $f:\mathbb D\to K$ whose restriction to $\partial\mathbb D$ is the initial curve $c$: the curve $c$ was contractible.


I'll give myself permission to give some ideas that may lead to a solution.

Note that in the proof given above, all that is needed to get to a contradiction is a loop $\gamma$ (of length $\ell$, parametrised by $c:\mathbb R/\ell\mathbb Z\to\mathbb R^d$ by arc length) such that for all $t$, the opposed points $c(t)$ and $c(t+\ell/2)$ are at path length distance at least $\ell/2$ in $K$. In fact, $(1-\delta)\ell/2$ is enough, provided $\delta>0$ is small enough (depending on how close the distortion actually is from $\frac\pi2$).

Suppose that is not the case, and let $\gamma$ be a loop of positive length $\ell$. We will try and show that $\gamma$ is in fact contractible. If it is, then perfect! If not, then there are two points $a$ and $b$ on $\gamma$ at distance $\ell/2$ in $\gamma$, but at path length distance $<(1-\alpha)\ell/2$ in $K$.

A loop with two opposite points at distance less than half the length

This is decomposes $\gamma$ as a product of $\gamma_0$ and $\gamma_1$, whose lengths are both $<(1-\alpha/2)\ell/2$.

A decomposition of a curve in two curves of less than the original length

If some of them are contractible, then great! Otherwise, do the same reasoning on $\gamma_i$, getting $\gamma_{i0}$ and $\gamma_{i1}$ of size $<(1-\alpha/2)^2\ell/2$. In the image below, $\gamma_0$ is contractible, whereas $\gamma_1$ isn't, and we decompose it in smaller loops.

Recursion on the loop decomposition

Here comes the shady point. Iterate this construction at infinity. Intuitively, you fill all the holes bounded by the curves $\gamma_*$, either because they are contractible, or because you created a web so dense that the compactness of $K$ makes it an actual surface. Congratulations! you proved that $\gamma$ was actually contractible in the first place.

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    $\begingroup$ For the "shady point" --- you do not need to use the contractable curves. After few iterations you get a map $f_n\colon\mathbb D\to K$, which is almost continuous --- meaning that for a small fixed $\varepsilon_n>0$ there is $\delta>0$ such that if $|x-y|_{\mathbb D}<\delta$, then $|f_n(x)-f_n(y)|_{K}<\varepsilon_n$. From the construction you also get that $|f_n-f_{n+1}|$ is bounded by a decreasing geometric progression, in particular it converges to a contunuous map $f_\infty\colon\mathbb D\to K$. $\endgroup$ – Anton Petrunin Feb 13 at 5:03
  • $\begingroup$ @AntonPetrunin Thank you for your comment! I'm not sure I understand how you would construct the $f_n$, though. It is not clear to me which curves in $\mathbb D$ would be suitably sent to the curves considered in $K$. $\endgroup$ – Pierre PC Feb 13 at 8:57
  • $\begingroup$ Any reasonable subdivision of $\mathbb{D}$ will do the job. $\endgroup$ – Anton Petrunin Feb 13 at 20:49
  • $\begingroup$ I think I get it, thank you very much! I added what I believe is close enough to a proof. $\endgroup$ – Pierre PC Feb 14 at 12:50
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This is an answer you might not accept.

Looking carefully at the book, I saw that the title of the chapter is "Riemannian manifolds with boundary and subsets of $\mathbb{R}^n$ with smooth boundary".

In a compact manifold, you can always choose such a length minimizing curve (the rough argument is that the unit tangent bundle is compact, you should maybe be more precise, but there is also a classical argument using Ascoli, see Remark 1.13.b just before the statement in question in Gromov's book).

Since the Hawaiian ring, embedded in $\mathbb{R^n}$ is not a sub-manifold, this is not a counter-example. However, it seems that $\mathrm{dil}(Id)\geq \pi/2$ in it, so the it is not a counter-example to the full statement. Maybe the right question is "Can we extend the statement to some (all ?) compact subsets of $\mathbb{R}^n$ ?". Since I'm absolutely not answering this new question, feel free not to accept my answer!

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  • $\begingroup$ This is arguably an answer to my question 1, and also a point in favour of the authors, so it is without a doubt a valuable one. :) I'll admit I'm more interested in my question 2, that I should now rewrite as ‘is there a simple way to make the proof work in the general case?’. I'll edit it. Thank you for your answer, and for the bounty by the way. $\endgroup$ – Pierre PC Feb 8 at 8:15

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