When does a pair of homotopic Lipschitz functions fail to admit a Lipschitz homotopy?

Question

Let $(M,d)$ and $(N,\rho)$ be metric spaces. A function $f: M \to N$ is Lipschitz if there exists some constant $\kappa \geq 0$ so that $\rho(f(x),f(y))$ is smaller than $\kappa d(x,y)$ for all points $x,y \in M$.

If $f$ and $g$ are two Lipschitz maps from $M$ to $N$, a Lipschitz homotopy between them is a Lipschitz map $H: M \times [0,1] \to N$ for which $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for each $x \in M$. Here we assume that the metric $\Delta$ on $M \times [0,1]$ is given by

$$\Delta\left((x,t),(y,s)\right) = \max\left\{d(x,y),|s-t|\right\}.$$

Here's the question which has been driving me nuts:

Is there an example of metric spaces $M, N$ and Lipschitz functions $f,g:M \to N$ which are homotopic, but not Lipschitz-homotopic?

If there is any justice in the world, the answer is an easy "yes". And in this case, I ask

Does the answer continue to be "yes" if $M$ and $N$ are assumed compact?

Answer 1

There is no "justice" in the world of arbitrary metric spaces:-) Let your first space $M$ be the ordinary circle with its ordinary metric, and the second space $N$ be homeomorphic to a cylinder with two border circles. Let the metric be usual on the border circles, but very rough (bumpy) on the middle part of the cylinder. You can certainly make it so rough that intermediate curves between the border circles will simply have infinite length... And of course you can make this example compact.

Answer 2

There are many situations where you can get Lipshitz homotopies, at least with restrictions on the topology of the spaces in question. See the paper "Quantitative algebraic topology and Lipschitz homotopy" by Ferry and Weinberger, available here.

They discuss a number of results and conjectures. One relevant one is the following, which is Theorem 1 in their paper. In it, simplicial complexes are given the path metric obtained by giving each simplex the standard Euclidean metric:

Theorem : Let $Y$ be a finite simplicial complex with finite homotopy groups in dimensions less than or equal to $d$. Then there exists $C(d)$ so that for all simplicial complexes $X$ of dimension less than or equal to $d$, if $f,g : X \rightarrow Y$ are homotopic $L$-Lipschitz maps with $L \geq 1$, then there is a $C(d)L$-Lipschitz homotopy $F$ from $f$ to $g$.

Conversely, if a Lipschitz homotopy always exists, then the homotopy groups of $Y$ are finite in dimensions less than or equal to $d$.

Answer 3

Let me give the answer in a clear and clean way.

Define $$ \forall_{\,x\ y\in\mathbb R}\ d(x\ y)\ :=\ |x-y|^\frac 12 $$

Thus $\ d(x\ x)=0\ $ and $\ d(x\ y) = d(y\ x)\ $ and

$$ (d(x\ y) + d(y\ z))^2 \ge |x-y|+|y-z| \ge |x-z| = d^2(x\ z) $$

thus $\ d(x\ y)+d(y\ z)\ge d(x\ z);\ $ function $\ d\ $ is a metric in $\mathbb R.\ $ And so is the induced metric in $\ \mathbb J:=[0;1].$

Thus let $\ M:=\{p\}\ $ be a 1-point metric space, and $\ N:=\mathbb J\ $ with metric $\ d\,|\,\mathbb J.$ Next, let the functions $\ f\ g: M\rightarrow N\ $ be given by $\ f(p):= 0\ $ and $\ g(p) := 1.\ $ These two functions are (topologically) homotopic.

Of course my spaces $\ M\ N\ $ are compact. But there is no Lipschytz homotopy as defined in the Question above.

INDEED:

let the Lipschytz constant $\ \kappa\ge 0\ $ be arbitrary. Then $\ \kappa\le n\ $ for a certain natural $\ n.\ $ Consider a homotopy $\ H:M\times[0;1]\rightarrow N\ $ between $\ f\ $ and $\ g.\ $ Thus $\ H(p,\ 0) = 0\ $ and $\ H(p,\ 1) = 1.$

Thus there exists an integer $\ k\ $ such that $\ 1\le k\le n^2\ $ and

$$ \left|H(p,\ \frac{k-1}{n^2}) - H(p,\ \frac k{n^2})\right|\ \ge\ \frac 1{n^2} $$

However

$$ d\left(H\left(p,\ \frac{k-1}{n^2}\right), \ H\left(p,\ \frac k{n^2}\right)\right)\ \ge\ \frac 1n \ \ge\ \kappa\cdot\left|\frac k{n^2}-\frac{k-1}{n^2}\right| $$

QED

Answer 4

I agree with Eremenko that there is no "justice" in the world of arbitrary metric spaces:-) However, some justice is provided by the following beautiful result of Lang and Schlichenmaier, Theorem 1.5 in [1]. From this result one can conclude some results about Lipschitz homotopy of maps.

Theorem. Suppose that $X$ and $Y$ are metric spaces such that the Nagata dimension of $X$ satisfies $\dim_N X\leq n$ and $Y$ is Lipschitz $(n-1)$-connected. Then there is a constant $C\geq 1$ such that for any closed set $Z\subset X$ and any $L$-Lipschitz map $f:Z\to Y$ there is a $CL$-Lipschitz extension $F:X\to Y$.

The definitions of the Nagata dimension and Lipschitz $k$-connectivity are provided below:

Definition. A metric space $Y$ is Lipschitz $n$-connected for some integer $n\geq 0$ if there is a constant $\gamma\geq 1$ such that for each $k\in \{0,1,\ldots,n\}$, every $L$-Lipschitz map $f:\mathbb{S}^k\to Y$ admits a $\gamma L$-Lipschitz extension $F:\mathbb{B}^{k+1}\to Y$.

Definition. The Nagata dimension $\dim_N X$ of a metric space $X$ is the least integer $n$ with the property that there is $C>0$ such that for any $s>0$, there is a covering $X=\bigcup_{i\in I} A_i$ such that

• $\operatorname{diam} A_i\leq Cs$ for all $i\in I$;
• Every ball $\mathbb{B}(x,s)$ intersects at most $n+1$ sets $A_i$.

If no such integer $n$ exists, then $\dim_N X=+\infty$.

The Nagata dimension can be regarded, in some sense, as a quantitative version of the topological dimension.

[1] U. Lang, T. Schlichenmaier, Nagata dimension, quasisymmetric embeddings, and Lipschitz extensions. Int. Math. Res. Not. 2005, no. 58, 3625-3655.

Answer 5

Two examples

(I am not able to select one of them over the other one, sorry):

EXAMPLE 1   Let   $M=N=[-1;1]$   have the same metric, which induces the ordinary topology in   $[-1;1]$,   and such that the length of   $[-1;1]$   is infinite. Let   $f$   be the identity map, and let   $g:=-f$.

EXAMPLE 2   The same, except that   $M:={1}$,   $f(1):=1$,   $g(1):=-1$.

In both cases the image of the length $1$ interval   $\{1\}\times[0;1]$   is   $N$   which has infinite length. This shows that there is no Lipschitz homotopy under the given circumstances.

A distance function in   $\mathbb R$

Every non-degenerated interval in   $\mathbb R$   has infinite length with respect to the distance   $d:\mathbb R^2\rightarrow\mathbb R$   given by:

$$\forall_{x\ y\in\mathbb R}\quad d(x\ y)\ :=\ \sqrt{|x-y|}$$

Distance   $d$   induces the Euclidean topology in   $\mathbb R$. This distance is of course invariant with respect to all translations (in this sense it is   $\mathbb R$-invariant;   and never mind my harmless typo   $\mathbb Z$   in a comment below); its isometry group is simply the same as the group of Euclidean isometries.