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Let $G$ be the group of orientation-preserving homeomorphisms (or, if you prefer, diffeomorphisms) of the real line. Does there exist a natural way to associate, to each function $f \in G$, a homomorphism $\phi_f \colon \mathbb{R} \to G$ such that $\phi_f(1) = f$?

Motivation: Many of us, in high school or before, have wondered whether for a given function $f$, it is possible to find a function $f^{1/2}$ such that $f^{1/2} \circ f^{1/2} = f$. I have never seen a satisfactory answer. This is intended as a more "grown-up" version: Is there some natural definition of $f^r$ ($= \phi_f(r)$), for $r \in \mathbb{R}$, such that $f^1 = f$ and $f^r \circ f^{-s} = f^{r-s}$?

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  • $\begingroup$ I am not sure if I got you correctly, but is $\phi_f(t, x) = (1-t) \cdot x + t \cdot f(x)$ what you are seeking? $\endgroup$
    – Bo Peng
    Commented Jul 8, 2010 at 14:59
  • $\begingroup$ Bo, in general, $2 f(x) - x$ is not equal to $f \circ f$. $\endgroup$ Commented Jul 8, 2010 at 15:32
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    $\begingroup$ Related: mathoverflow.net/questions/17614/solving-ffxgx $\endgroup$ Commented Jul 8, 2010 at 15:53
  • $\begingroup$ The question has been answered when the question is asked without the word "natural": yes for self-homeomorphisms and no for $C^1$-diffeomorphisms. But with the word natural (for self-homeomorphisms)? for instance, one can ask whether there is a continuous map $\Phi:\mathrm{Homeo}^+(\mathbf{R})\times\mathrm{R}\to \mathrm{Homeo}^+(\mathbf{R})$ such that $\Phi(f,-)$ is a homomorphism $\mathrm{R}\to \mathrm{Homeo}^+(\mathbf{R})$ for every $f$, and $\Phi(f,1)=f$ for all $f$. If not, one can ask about Borel. $\endgroup$
    – YCor
    Commented Feb 20, 2020 at 0:41
  • $\begingroup$ Of course in "natural" we can make further requirements, e.g., $\Phi(gfg^{-1},t)=g\Phi(f,t)g^{-1}$ for all $f,g,t$. $\endgroup$
    – YCor
    Commented Feb 20, 2020 at 1:06

2 Answers 2

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This has some relation with this question, but it is obviously different.

In general, a homeomorphism of $\mathbb{R}$ which preserves the orientation may have or not fixed points.

If it has no fixed points, then it is conjugated to a translation and thus, one can easily construct such $\phi_f$.

The other case is not much more difficult, since one can consider the set of fixed points of $f$ (that is, such that $f(x)=x$) which is closed and then do the trick in the complement of that set and leave fixed the set of fixed points for every $t$ (when defining $\phi_f(t)$). Notice that in either case, there is in general no unique way to do this.

It is also interesting that the diffeomorphism case is quite different, in particular, one can easyly construct a diffeomorphism $f:\mathbb{R}\to \mathbb{R}$ such that there is no diffeomorphism $g$ such that $g\circ g =f$. This can be seen in the paper provided by Helge (in fact it has to do with distortion and the fact that if you take one contracting point, there are restrictions to construct, for example a square root, see Section 1 of this paper).

ADDED RELATED REFERENCE: In this paper, Palis gives a not so difficult proof that $C^1$-generic diffeomorphisms (which belong to a $G_\delta$-dense subset of $Diff^1(M)$) of a compact manifold, the diffeomorphisms are not the time one map of a flow.

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  • $\begingroup$ "There is in general no unique way to do this." Is it thus unlikely that we could put such things together into a nice map $\mathbb{R} \times G \to G$? $\endgroup$ Commented Jul 8, 2010 at 15:36
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    $\begingroup$ It depends on what is nice: Assuming no fixed points, one considers the interval $[0,f(0))$. Choosing an arbitrary map $h:[0,f(0)) \to [0,1)$ one can conjugate $f$ to $x\mapsto x+1$. This is done as follows, consider $x\in \mathbb{R}$ and define $H(x)= h(f^{-n}(x)) + n$ where $n$ is the unique integer such that $f^{-n}(x)\in [0,f(0))$. Now, the map is given by: $f_t= H^{-1}\circ T_t \circ H$ (or something similar) with $T_t(x) = x + t$. $\endgroup$
    – rpotrie
    Commented Jul 8, 2010 at 15:44
  • $\begingroup$ Do you know for $C^\infty$-diffeomorphisms? analytic? $\endgroup$
    – YCor
    Commented Feb 20, 2020 at 0:38
  • $\begingroup$ I do not know precise conditions. But the following reference may be relevant: arxiv.org/pdf/0811.1173.pdf $\endgroup$
    – rpotrie
    Commented Feb 22, 2020 at 22:12
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I commented earlier, without finishing reading google's reply:

http://www.math.northwestern.edu/~wilkinso/papers/announce04152008.pdf

shows the answer is no if you replace $\mathbb{R}$ by a compact manifold. Key is googling for "Is every diffeomorphism a time one map?". I'm not sure what the answer for $\mathbb{R}$ is, and if it has been looked at.

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  • $\begingroup$ The link is broken. $\endgroup$
    – YCor
    Commented Feb 20, 2020 at 0:37

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