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Consider the group $\mathrm{Diff}(\mathbb R^n)$ of smooth diffeomorphisms. It has two interesting subgroups:

  • the orthogonal group $O(n)$,
  • the group of "diffeomorphisms applied along each axis" $\mathrm{Diff}(\mathbb R)^n$, so that $f(x_1, \dotsc, x_n) = \big(f_1(x_1), \dotsc, f_n(x_n)\big)$, where each $f_i \in \mathrm{Diff}(\mathbb R)$.

My question is whether the group generated by these two subgroups is the whole $\mathrm{Diff}(\mathbb R^n)$ or not. (I suspect it is not for $n\ge 2$, so any results characterizing this subgroup would be very welcome. I did a literature search and I have not found the answer to this problem, but I am not an expert, so any references on this problem or similar would be great).


  1. In dimension $n=1$ this is trivially true.
  2. Sole $\mathrm{Diff}(\mathbb R)^n \neq \mathrm{Diff}(\mathbb R^n)$ for $n\ge 2$, as the elements of the first group can only have diagonal jacobians.
  3. The linearised version of the problem is true: $GL(n)$ is the same as the group generated by $O(n)$ and invertible diagonal matrices $(\mathbb R^\times)^n$, which is a corollary of singular value decomposition.
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Here's a proof that $\mathrm{Diff}(\mathbf{R})^2\cup\mathrm{O}(2)$ doesn't generate $\mathrm{Diff}(\mathbf{R}^2)$. Identify $\mathbf{R}^2$ with $\mathbf{C}$.

Let $f:\mathbf{R}_{\ge 0}\to\mathbf{C}$ be a path going to infinity. We can write $f(t)=|f(t)|\exp(i\theta(t))$, so that $\theta$ is eventually continuous (because $f$ is eventually nonzero). For two choices such lifts, say with $\theta_1,\theta_2$, $\theta_1-\theta_2$ is eventually constant. Hence the condition that $\theta$ is eventually bounded is choice-free. Informally, it means that $f$ doesn't wind infinitely around the origin. Call it "non-winding".

Then both $\mathrm{Diff}(\mathbf{R})^2$ and $\mathrm{O}(2)$ preserve this property (i.e., if $u$ is such a diffeomorphism and $f$ is non-winding, then $u\circ f$ is non-winding).

But it is easy to construct $u$ without this property. For instance, take $z\mapsto z e^{i |z|^2}$, which is even real-analytic.

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    $\begingroup$ What if one adds $z\mapsto ze^{i\theta(|z|)}$ for all smooth $\theta:\mathbf{R}_{\ge0}\to\mathbf{R}/\mathbf{Z}$? Will they generate together? $\endgroup$ Aug 17 at 20:45
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    $\begingroup$ @მამუკაჯიბლაძე this need not be smooth at zero, but assuming you only take those maps that are smooth, I don't know (and would rather believe that this is not enough to generate). $\endgroup$
    – YCor
    Aug 17 at 22:02
  • $\begingroup$ @YCor I think the following example cannot lie in subgroup generated with addition of those parametric rotations. Take an embedding of regular rooted tree into the plane such that i-th level is evenly spaced on circle of radius N. Our diffeo would expand nbhd of a tree that shrinks rapidly to infinity such that it now shrinks much slower. Idea is that both diagonal elements and parametric rotations are uniformly continuous on part of circle of each radius (if we treat each circle as having radius 1) that has measure not tending to 0 as radius goes to infinity. $\endgroup$
    – Denis T
    Aug 19 at 17:37
  • $\begingroup$ ...maybe this approach is better explained in another way. One may try to measure minimal "complexity" of a compact (convex) covering with a property that diffeo is $(C, D)$-quasi-Lipschitz on each compact of a cover (so, we fix a "scale" on which diffeo is "locally tame" and try to estimate its global tameness). For example, one may take as a complexity (weighted) sum of $I$, maximal incidence number, and $W^{-1}$, where $W$ is infimum of diameters of largest inscribed balls (thought as functions on some "tame" in sense as above filtration of the plane). $\endgroup$
    – Denis T
    Aug 19 at 19:31

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