is not a Polish group

Abstract

The group $\text{PL}_{+}(I)$ of increasing piecewise-linear self-homeomorphisms of the interval $I=[0,1]$ may not be assigned a topology in such a way that it becomes a Polish group. The same statement holds for the groups $\text{Homeo}_{+}^{\text{Lip}}(I)$ of bi-Lipschitz homeomorphisms of $I$, and $\text{Diff}_{+}^{1+\unicode[STIX]{x1D716}}(I)$ of diffeomorphisms of $I$ whose derivatives are Hölder continuous with exponent $\unicode[STIX]{x1D716}$, as well as the corresponding groups acting on the real line and on the circle.

Polishability of some groups of interval and circle diffeomorphisms

“…Our first main result may be viewed as an extension of Solecki's theorem for Homeo AC + (I) to higher degrees of smoothness: 2 The group Diff 1+Lip + (M ) is not mentioned explicitly in [3], but the arguments given for Diff 1+ǫ + (M ) hold just as well for ǫ = 1, which implies the result described. 3 A subgroup G of Homeo + (M ) is called locally moving if whenever U ⊆ M is open, there is a non-identity f ∈ G whose support is contained in U . A summary of the Polishability results of [17], [3], and the present work is diagrammed in Figure 1.…”

“…The results of Solecki, Kallman and the author, and those of the present work seem to suggest that absolute continuity is the "right" intermediate smoothness condition to impose as opposed to Lipschitz/Hölder conditions or bounded variation, at least from the viewpoint of topological group theory. We once again emphasize this contrast with our secondary result, which builds upon the work of[3]:Theorem 1.2 (see Theorem 4.7). Let M = I or M = S 1 .…”

“…The argument is based in part on the following key lemma which we present without proof here. The lemma is stated explicitly and proved for the case M = I as Lemma 2.3 in [3]; for the case where M = S 1 , it may be easily extrapolated from the proof of Theorem 3.3.2 in [5]. Proof.…”

“…3 A subgroup G of Homeo + (M ) is called locally moving if whenever U ⊆ M is open, there is a non-identity f ∈ G whose support is contained in U . A summary of the Polishability results of [17], [3], and the present work is diagrammed in Figure 1.…”

$\operatorname{PL}(M)$ admits no Polish group topology

“…In [3], Cohen and Kallman showed that PL(I) and PL(S 1 ) have no Polish group structure. However, their proof uses 1-dimensionality in an essential way.…”