What are sufficient condition on a given diffeomorphism of the sphere (say, given explicitly with formulas) that can ensure that it is conjugate to a rotation, in the group of diffeomorphism of the sphere?

Also, if I take a given rotation of some fixed angle, is there a neigbourhood of it which consists only of elements conjugate to a rotation?

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    $\begingroup$ For the second question, take the identity map. There are clearly maps which are arbitrarily close to it which are not rotations. I am not sure what to make of the 1st question. In low dimensions, a diffeomorphism of finite order is always conjugate to a rotation, but it fails in higher dimensions. $\endgroup$
    – Misha
    Commented Sep 7, 2013 at 14:25
  • $\begingroup$ Thx for the comment. I was thinking about an angle which is irrational in fact, but I guess that the same problem arises. $\endgroup$ Commented Sep 7, 2013 at 17:18

3 Answers 3


I doubt there can be a simple answer to the first question. Even the analogous question regarding $S^1$ has no simple answer. According to work of Denjoy, given a homeomorphism $f : S^1 \to S^1$ with irrational rotation number, if $f$ is $C^2$ then it is conjugate to a rotation, but there are $C^1$ examples that are not conjugate to any rotation. Furthermore, according to work of Arnold (same link as above), one has little control over the smoothness of the conjugating map. Even in the nice case cited, the "rotation number" is not so easy to compute from a formula; it is more of an asymptotic invariant that describes the behavior of higher and higher iterates of $f$.

Things will only get worse in $S^2$.

EDIT: Here's an example of how it gets worse. On the circle, a diffeomorphism with rational rotation number is conjugate to a rotation.

Let's throw the book at $f : S^2 \to S^2$. Suppose $f$ has exactly two periodic points $p,q$, restricting to a homeomorphism of the annulus $A = S^2 - \lbrace p,q \rbrace$. Suppose furthermore that the restriction $f | A$ has an invariant area form. Suppose furthermore that $f | A$ has a constant rational rotation number on each point (defined on an annulus similar to the definition on a circle). So far these conditions are all necessary for $f$ to be conjugate to a finite order rotation. Can we add simple additional sufficient conditions? Well, it doesn't seem so; for instance, $f$ could be the composition of a finite order rotation by a map which twists a little tiny disc.

  • $\begingroup$ Thanks for the answer. I was also thinking that the situation could be bad, but what I would like is to have a "nice" situation of a map that I really think to be conjugate to a rotation, and I would like some results which allows to prove the statement, without finding the conjugating map. $\endgroup$ Commented Sep 7, 2013 at 17:15
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    $\begingroup$ Regarding the issue of the conjugating map, what Arnold proved is that there exist analytic maps of the circle which are conjugate in the homeomorphism group of $S^2$ but not in the diffeomorphism group. One would expect this to be just as bad in $S^2$. $\endgroup$
    – Lee Mosher
    Commented Sep 7, 2013 at 20:15
  • $\begingroup$ Perhaps the condition "for all $n$, the fixed point set of $f^n$ is topologically a sphere with even codimension" would be helpful? $\endgroup$
    – Will Sawin
    Commented Sep 9, 2013 at 1:17
  • $\begingroup$ Or "connected components of closures of orbits of $f$ are tori". $\endgroup$
    – Will Sawin
    Commented Sep 9, 2013 at 1:42

An irational rotation $f$ is arbitrarily distorted in $\text{Diff}^{\infty}S^2$, this was proved by Calegari and Freedman http://arxiv.org/pdf/math/0509701.pdf and in a more general setting by Militon, see http://arxiv.org/abs/1005.1765. This means that for any diverging sequence $g(n)$(diverging as slow as one wants), there is a subsequence $f^{n_i}$ and a finite set $S$ in $Diff(S^2)$ generating a group $G = <S>$ such that $\{f^{n_i}\} \in G$ for all $i$ and that $l_S(f^{n_i}) \leq g(n_i)$. ($l_S(g)$ is the word length of $g$ in the generating set $S$). To make it a little bit more clear, let's consider the following example: Let's take the translation $a: x \to x+1$ in $Diff(R)$. If we consider the subgroup $a:x \to x+1$, and $b : x \to 2x$ in $Diff(\mathbb{R})$, observe that $b^{n}ab^{-n} = a^{2^n}$, this implies that $a$ is at least exponentially distorted in $Diff(\mathbb{R})$, because we can write $a^{2^n}$ as a product of $O(n)$ elements in $a$ and $b$.

One condition that ensures that a given $f: S^2 \to S^2$ is conjugate to a rotation is that the sequence of derivatives of powers of $f$, $(Df^k)_{k \in \mathbb{z}}$, is a bounded sequence. The idea is that in that case, one can take an arbitrary riemannian metric $g$, take and averaging sequence $g_n = \frac{1}{n}\sum_{i=1}^n ({f}^n)^*g$, and observe that the bounds in derivatives imply that $g_n$ has a subsequence converging to a metric $g_{av}$ invariant by $f$. If $g_{av}$ is smooth, by the uniformization theorem, $g$ is conformally equivalent to the standard $S^2$, where $f$ is acting by rotations.

Using these two ideas, one can get a group theoretical way of characterizing any $f: S^2 \to S^2$ that is conjugate to a rotation: $f$ is conjugate to a rotation if and only if for any subsequence $(f^{n_i})_i$ of powers of $f$ there exist a subsequence $m_k$ of $n_i$ and a finite set $S$ of diffeomorphisms generating a group $G = <S>$ such that $\{f^{m_k}\} \subset G$ and $l_S(f^{m_k}) \leq k$. For more details, see the introduction of my paper http://arxiv.org/abs/1307.4447.

As rpotrie mentioned, it might be of interest to you some diffeomorphisms called "pseudorotations", that can be defined as diffeomorphisms of $S^2$ having exactly two periodic points. Birkhoff conjectured that any real analytic measure preserving pseudorotations is an actual rotation. There are many counterexamples in the $C^{\infty}$ category, in the case that the rotation angle is not diophantine, all of them as far as I know constructed by the same method. See http://www.math.psu.edu/katok_a/pub/Herman-survey-horo-corrected.pdf for a survey on this kind of constructions. There is a very positive result known as Herman's last geometric theorem in the case that $\alpha$ is diophantine. See http://www.math.jussieu.fr/~bassam/herman_ann_ens_second_revision.pdf. Using this method one can also construct diffeomoprhisms of $S^2$ that are arbitrarily distorted but not conjugate to rotations.


I guess by: $f: S^2 \to S^2$ is conjugate to a rotation in the group of diffeomorphisms you mean that there exists a rotation $R_\alpha$ and a diffeomorphism $\varphi: S^2 \to S^2$ such that $f = \varphi^{-1} \circ R_\alpha \circ \varphi$.

In that case, the second question admits a negative answer. Given such an $f$, you can first make an arbitrary small perturbation (compose with $\varphi^{-1} \circ h \circ \varphi$ where $h_\varepsilon$ is the very small diffeomorphism diffeomorphism of the sphere which is the identity in neighborhoods of the fixed points of the north and south poles and which rotates the equator by $\varepsilon$, this can be made as small as desired with small $\varepsilon$) to make the diffeomorphism to have some circles around the fixed point to have the same rotation number and change it to the image by $\varphi^{-1}$ of the equator. If you allow $\varphi$ to be a homeomophism, the answer is still negative (for example, compose the fixed point with a small homothety) I even in the volume preserving category (but this is slightly more delicate, I can amplify if you need it, but I am not sure I am answering the question you ask, so...).

The first question seems more difficult to respond. I can figure out no condition in finitely many iterates which might respond to that. On the other hand, it seems quite easy to figure out some criteria not to be conjugated that can be checked quite easily (look at the derivative at the fixed points, look for at least 3 periodic points of different period...) which of course are not enough to say it is conjugate if they fail...


If one looks at the neighborhood of diffeomorphisms being ``irrational pseudo-rotations'' (see these lecture notes) I think that the answer will depend on both the smoothness and some properties on the irrational numbers (being diophantine vs liouville). A positive answer for your second question could be in the lines of: Every diffeomorphism $C^r$-close (with large $r$) to a diophantine rotation which is still an irrational pseudo-rotation with the same rotation number (in particular, one must demand it to be volume preserving) is smoothly conjugated (with some loss in regularity) to the rotation. Some versions of this must have been proved by M. Herman, but I do not remember specific references, look at the lecture notes I linked which may have plenty of references. Hope it helps.

  • $\begingroup$ Thanks for the answer. If we only look at the neighbourhood in the subgroup of maps fixing the north pole and south pole, and which behave locally like rotations at these points, is the answer still false for my second question? (of course taking an angle irrationnal). $\endgroup$ Commented Sep 7, 2013 at 17:17
  • $\begingroup$ I added some remarks which I hope respond your question. $\endgroup$
    – rpotrie
    Commented Sep 7, 2013 at 17:44
  • $\begingroup$ Thanks a lot, I will have a look at the notes you mention. $\endgroup$ Commented Sep 7, 2013 at 17:48
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    $\begingroup$ A simple example that also addresses the first comment: take an irrational rotation and perturb it slightly (e.g. by composition with the flow map of a horizontal vector field vanishing outside a small set; the time parameter allows you to adjust things) to give it an additional periodic point on the equator. $\endgroup$ Commented Sep 7, 2013 at 21:05

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