When a homeomorphism can be upgraded into an isometry?

Question

Let $X$ be a metrizable topological space. Let $f:X \rightarrow X $ be a homeomorphism. When can we find some metric $d$ which induces the original topology on $X$ making $f$ an isometry?

Partial Results:

$\text{(1)}$ If $f$ is of finite order the answer is yes. (Take any metric and sum on translates of it by $f$).

$\text{(2)}$ We found some obstructions (based on suggestion by PVAL, communicated by Mike:)

Roughly, a necessary condition for $f$ to be a potential isometry, is that it won't be contractive or expansive (for any realizing metric) on subspaces of $X$. The main idea is that if $f$ is behaving very badly w.r.t one realizing metric, then it will behave badly for any other realizing metric.

An accurate formulation is found below. (see also my detailed answer).

This rules our for example $f(x)=x^3,f(x)=2x,f(x)=\frac{1}{2}x $ on $X=\mathbb{R}$ .

While the above obstructions seems quite useful in eliminating candidates, I would still like to find more necessary\sufficient conditions for this property to hold.

(In particular, I feel that the although the obstructions found are ingenious, there is something cumbersome in their formulation, since verifying them would require us to go over all possible metrics for $X$).

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Proposition (precise statement of the obstructions found):

The following two conditions imply $f$ cannot be an isometry of any realizing metric for the topology of $X$.

$\text{(1)}$ There exists a realizing metric $d$ on $X$ such that $f$ restricts to a contracting (or expanding) map on a (non-trivial) compact subspace of $X$.

$\text{(2)}$ There exists a realizing metric $d$ on $X$ such that $f$ restricts to a contracting (or expanding) map on a subspace of finite diameter of $X$ which contains a fixed point of $f$ .

Remark: In Condition2 we do not assume the relevant subspace is compact.

Answer 1

This answer is essentially due to PVAL, who gave me a similar if not identical idea recently to a similar problem.

Let $f: \Bbb R \to \Bbb R$ be $x \mapsto x/2$. For any metric (that induces the standard topology on $\Bbb R$), some iterate $f^n$ will send $[0,1]$ to an interval of arbitrarily small diameter; in particular, we can make the diameter smaller than $\text{diam }[0,1]$. If $f$ were an isometry, $f^n$ would be an isometry; but it changes the diameter of a set, so it's clearly not.

This same argument should work given any $f: X \to X$ that restricts to a contracting map on a compact subspace.

Answer 2

I am writing a detailed solution based on the suggestion of Mike Miller:

Proposition (proofs below):

The following two conditions imply $f$ cannot be an isometry of any realizing metric for the topology of $X$.

$\text{(1)}$ There exists a realizing metric $d$ on $X$ such that $f$ restricts to a contracting (or expanding) map on a (non-trivial) compact subspace of $X$.

$\text{(2)}$ There exists a realizing metric $d$ on $X$ such that $f$ restricts to a contracting (or expanding) map on a subspace of finite diameter of $X$ which contains a fixed point of $f$ .

Remark: In Condition2 we do not assume the relevant subspace is compact.

Proof:(Condition1) First we note that since an inverse of a bijective isometry is also an isometry it is enough to prove the contractive case.

Let $d$ be a metric on $X$, $K\subseteq X$ a compact subspace such that $f:(K,d) \rightarrow (K,d)$ is contractive with constant $c<1$. (We assume $K$ contains more than one point so its diameter w.r.t any metric is positive). Assume $d'$ is some (compatible) metric on $X$ making $f$ an isometry.

Since in every metric space, compact sets are bounded, it follows that $L = \text{diam}(K,d) < \infty$.

Note that $\text{diam}(f^n(K),d) \le c^n \cdot L \stackrel{\mathrm{n\rightarrow \infty}}{\longrightarrow} 0$. (Here we explicitly use the fact $f(K)\subseteq K$ so we can use the retraction bound again and again. If we only assumed $f:K \rightarrow X$ is contractive, then after one iteration of $f$ we coudn't be sure f is still contracting for the points we landed upon).

Let $0 < \epsilon < \text{diam}(K,d')$, by compactness of $K$, there exists a $\delta > 0$ such that $\forall x,y \in K, d(x,y)< \delta \Rightarrow d'(x,y)< \epsilon $. (Here we used also the fact that $d,d'$ both generate the same topology).

Now for $n$ large enough, $\text{diam}(f^n(K),d) < \delta$, so $\forall x,y \in K , d(f^n(x),f^n(y)) < \delta \Rightarrow d'(f^n(x),f^n(y)) < \epsilon \Rightarrow $

$ \text{diam}(f^n(K),d') \le \epsilon < \text{diam}(K,d')$.

But $f:(K,d') \rightarrow (K,d')$ is an isometry, and so $f^n$. But this is a contradiction to the fact that an isometry must preserve diameters of sets.

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Proof:(condition2)

(We continue with same notations from proof 1 above).

Let $K \subseteq X, o \in K$ a fixed point of $f$ such that $f(K) \subseteq K$ and $f:(K,d) \rightarrow (K,d)$ is contractive with constant $c<1$. We assume $ \text{diam}(K,d) < \infty$

Since a metric is always continuous w.r.t the topology it induces, it follows that the function $h(x) = d'(x,o)$ is continuous on $K$ w.r.t the topology on $X$ induced by $d'$. Since $d$ also induces this topology it follows that $h:(K,d) \rightarrow \mathbb{R}$ is continuous.

Let $0 < \epsilon < \text{diam}(K,d')$, by continuity of $h$ at $o$, there exists $\delta > 0$ such that $d(x,0)< \delta \Rightarrow |h(x)-h(0)|=d'(x,o)< \frac{1}{2} \epsilon $.

Now for $n$ large enough, $\text{diam}(f^n(K),d) < \delta$ (here we use the assumption $ \text{diam}(K,d) < \infty$ and contraction), so $\forall x \in K , d(f^n(x),f^n(o))=d(f^n(x),o) < \delta \Rightarrow d'(f^n(x),o) < \frac{1}{2} \epsilon $

$ \Rightarrow \text{diam}(f^n(K),d') \le \epsilon < \text{diam}(K,d')$