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For some reason, today I want to understand better the group of diffeomorphisms of the 2-sphere, $S^2$.

After a few minutes I found this result by Smale in 1958.

The space $\Omega$ of all orientation-preserving $C^\infty$ diffeomorphisms of $S^2$ has a strong deformation retract [to] the rotation group $SO(3)$.

It sounds like any invertible, differentiable map $f:S^2 \to S^2$ can be "smoothed" to a 3D rotation of a certain angle $\theta$ around a certain axis $\vec{k}$.

Smale gives an explicit retraction from $f \in C^\infty(S^2)$ to the rotation

$$\begin{array}{rcc} f(x_0) &\mapsto& x_0 \\ df(e_1) &\mapsto& e_1 \\ df(e_2) &\mapsto& e_2 \end{array}$$

There seems to be an alternative proof in some topology notes. Jacob Lurie describes a fibration over the space of all conformal structures on the 2-sphere

\[ \mathrm{Diff}_{\mathrm{Conf}}(S^2) \to \mathrm{Diff}(S^2) \to \mathrm{Conf}(S^2) \]

and another diagram related to the existence of isothermal coordinates:

\[ \begin{array}{ccccc} SO(2) & \to & SO(3) & \to & S^2 \\ \downarrow & & \downarrow & & \downarrow \\ Aut(\mathbb{C}) & \to & \mathrm{Diff}_{\mathrm{Conf}}(S^2) & \to & S^2 \end{array} \]

These discussions of the diffeomorphism group 2-sphere were not very concrete, but after writing this I have answered much of my own questions...

I suppose can could get conformal structures like $e^{f(z)} dz\, d\overline{z}$.


What kind of information did we lose after these reductions?
What are some elements of $\mathrm{Diff}(S^2)$ you can build?

REMARK: One comment suggested I use vector fields to construct my diffeomorphism. If I have a smooth section of the tangent bundle $T(S^2)$ can it be "integrated" to a diffeomorphism of $S^2 \to S^2$ ?

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closed as unclear what you're asking by Yemon Choi, Ryan Budney, Mariano Suárez-Álvarez, Chris Godsil, Benoît Kloeckner Jun 30 '13 at 17:41

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    $\begingroup$ Regarding your question, it seems to me the clear answer is you lose essentially all information about the group structure. Smale's theorem is only about the homotopy-type of $Diff(S^2)$, it says little about $Diff(S^2)$ as a topological group. It depends on what you mean by "exotic" but diffeomorphisms with only one fixed point might qualify, as they are not conjugate to linear maps. $\endgroup$ – Ryan Budney Jun 30 '13 at 16:39
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    $\begingroup$ If you try to phrase the question in mathematical terms, it would be easier to answer, and most likely you will answer it yourself in the process. $\endgroup$ – Igor Belegradek Jun 30 '13 at 16:42
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    $\begingroup$ Ryan, a translation of $\mathbb R^2$ extends to a diffeomorphism of $S^2$ with one fixed point; it is hardly exotic. $\endgroup$ – Igor Belegradek Jun 30 '13 at 16:47
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    $\begingroup$ @Igor: not from the perspective of $SO(3)$. With no functional definition of exotic in this context, I suppose there's no real point of talking about things being exotic or not. $\endgroup$ – Ryan Budney Jun 30 '13 at 16:58
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    $\begingroup$ See survey "Diffeomorphism groups of compact manifolds" Journal of Mathematical Sciences 2007, (146) Issue 6, pp 6213-6312, by Smolentsev. link.springer.com/article/10.1007%2Fs10958-007-0471-0 $\endgroup$ – Igor Belegradek Jun 30 '13 at 17:13
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Some more exotic elements or even subgroups of exotic elements: The following groups extend to subgroups of $Diff(S^2)$ which fix the north pole via stereographic projection:

  • $Diff_{\mathcal B}(\mathbb R^2) = \lbrace Id + f: f\in \mathcal B(\mathbb R^2)^2, \det(df)\ge\epsilon>0\rbrace$ where $\mathcal B(\mathbb R^2)$ is the space of functions which are globally bounded, in each partial derivative separately (called $\mathcal D_{L^\infty}$ by L. Schwartz)

  • $Diff_{H^\infty}(\mathbb R^2) = \lbrace Id + f: f\in H^\infty(\mathbb R^2)^2, \det(I_2+df)>0\rbrace$ where $H^\infty(\mathbb R^2)$ is the intersection of all Sobolev spaces $H^k$ for $k\ge 0$.

  • $Diff_{\mathcal S}(\mathbb R^2) = \lbrace Id + f: f\in \mathcal S(\mathbb R^2)^2, \det(I_2+df)>0\rbrace$ where $\mathcal S$ is the Schwartz space of rapidly decreasing functions.

  • $Diff_{c}(\mathbb R^2) = \lbrace Id + f: f\in C^\infty_c(\mathbb R^2)^2, \det(\mathbb I_2+df)>0\rbrace$ where $C^\infty_c$ are the smooth functions with compact support.

Note that all these diffeomorphism groups are normal subgroups in each larger one, the largest being $Diff_{\mathcal B}$.

Edit:

The question changed when I wrote the above answer. The question now asks which element of $Diff(S^2)$ you get by integrating vector fields:

Answer: if you use time-dependent vector fields, you get each orientation preserving diffeomorphism. With autonomous vector fields you only get a connected subset which does not cover any neighborhood of the identity. The strongest result is due to Grabowski: There exists a smooth curve of diffeomorphisms in $Diff(S^2)$ starting at the identity, injective, such that the elements of the curve form free generators for a free group (with uncountably many generators) such that no element of this free group can be embedded into a flow. This free group is smoothly path-connected to the identity (slide the generators along the curve).

Second edit:

@Mariano Suárez-Alvarez: The paper is the following.

  • MR0974661 (90b:58031) Grabowski, Janusz(PL-WASW) Free subgroups of diffeomorphism groups. Fund. Math. 131 (1988), no. 2, 103–121.

The method consists of piecing together 1-dimensional diffeomorphisms which do not embed into a flow (published first by Nancy Kopell) in an intricate combinatorial way.

@John Mangual: Generators? On the circle trigonometric polynomial vector fields are dense in the Lie algebra --- probabaly this is what you mean if you talk about generators. On $S^2$ one can expand vector fields with respect to the eigenfunctions of the Laplacian (eigenvalues are multiple since the Laplacian has a large symmetry group $O(3)$). Finte sums are then dense.

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    $\begingroup$ What's the idea behind the proof that such a huge number of diffeos generate a free group? $\endgroup$ – Mariano Suárez-Álvarez Jun 30 '13 at 17:52
  • $\begingroup$ can you write a set of generators for $Diff(S^2)$ the way you can for $Diff(S^1)$ ? $\endgroup$ – john mangual Jul 2 '13 at 17:37

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