What is the normalizer of the circle in the diffeomorphism group of the 2-sphere?

What is the normalizer of $SO(2)$ in $\mathrm{Diff}(S^2)$?

Remarks:

1. We let $SO(2)$ act on $S^2$ via the rotation about the $z$-axis. It is immediate that each element of the normalizer must map any parallel (i.e. a fiber of the projection of $S^2$ onto the $z$-axis) to a parallel.

2. Of course, the normalizer contains $O(2)$, as well as the following elements that push along the meridians''. Let $(\phi, r)\in [0,2\pi]\times [0, \pi]$ be coordinates on $S^2$ with parallels given by $r=\mathrm{const}$. Define a self-map of $S^2$ by $H(\phi, r)=(\phi, r+h(r))$ where $h$ is some smooth function subject to the boundary conditions ensuring that $H$ is a diffeomorphism (e.g. $h$ vanishes near $0$ or $\pi$). Then $H$ commutes with the $SO(2)$-action, that is given by translation in $\phi$.

-
$Aut(SO(2))=\mathbb{Z}/2$ (here, I'm taking Aut of the topological group structure), so an index 2 subgroup of the normalizer is the centralizer. The centralizer has Lie algebra given by vector fields on $S^2$ commuting with a non-zero vector field which is an infinitesimal rotation, and in local coordinates may be written as $\partial/\partial \phi$. So the Lie algebra is vector fields of the form $f(r)\partial/\partial \phi + g(r)\partial/\partial r$ (i.e., they are independent of $\phi$). – Ian Agol Aug 15 '11 at 23:44
@Agol: thank you, this clears things up somewhat. – Igor Belegradek Aug 16 '11 at 1:43

Let's call your coordinates $\phi$ and $\theta$, as is more usual. Thus $(x,y,z)=(sin \phi\ cos\theta, sin\phi\ sin\theta, cos\phi)$.

Yes, in the centralizer of $SO(2)$ there is the group that leaves $\theta$ unchanged, $(\phi, \theta)\mapsto (f(\phi ),\theta)$ where $f$ is a diffeomorphism from $[0,\pi]$ to itself that is nice enough at the endpoints. Also there is the group that leaves $\phi$ unchanged, $(\phi, \theta)\mapsto (\phi ,\theta +g(\phi))$ where $g$ is a smooth map from $[0,\pi]$ to $\mathbb R/2\pi \mathbb Z$. The centralizer is the semidirect product of these, and the normalizer is bigger by a factor of two.

EDIT Let's consider the related and slightly easier problem of maps $F$ from the plane to itself that commute with all of $SO(2)$, in other words maps $F:\mathbb C\to \mathbb C$ such that $F(re^{i\theta} )=F(r)e^{i\theta}$. Such a function is determined by its restriction to $\mathbb R$. That restriction must be odd, $F(-x)=-F(x)$, by considering $\theta=\pi$. Thus if $F$ is smooth then the restriction can be written as $x\mapsto xG(x^2)$ for a smooth $G:[0,+\infty)\to\mathbb C$. Conversely, given any such smooth $G$ we can write $F(z)=zG(|z|^2)$ and get a smooth map $\mathbb C\to \mathbb C$, the unique such map commuting with rotations and having $x\mapsto xG(x^2)$ as its restriction to $\mathbb R$.

If $F$ is a diffeomorphism then $G(0)$ is different from $0$ and also $G(u)$ is different from $0$ when $u>0$. Thus $G$ can be written $G(u)=a(u)e^{ib(u)}$ where $a>0$ and $b$ are smooth real functions of $u\ge 0$. The only further constraint on $a$ or $b$ is that $x\mapsto xa(x^2)$ must be a diffeomorphism from the positive reals to itself.

-
Fix an element of the normalizer and see what it does to the set of parallels, which in your notations defines a map $(\phi,\theta)\to (f(\phi), \theta))$. I think, you say that this $f$ must be a self-diffeomorphism of $[0,\pi]$. Why? It is clear that $f$ is a homeomorphism, but have no idea why it is smooth; a propri a non-smooth homeomorphism $f$ could be "covered" by a self-diffeomorphism of $S^2$. – Igor Belegradek Aug 15 '11 at 21:32
Also consider a self-diffeomorphism of $S^2$ that preserves every parallel as a set. Then it is given by $(\phi, theta)\to (\phi, h(\theta,\phi))$ for some function $h$. You say that $h$ is $\theta+g(\phi)$. Why? – Igor Belegradek Aug 15 '11 at 21:42
Igor, as to your second comment, I am only asserting that $h$ has this form if the diffeomorphism commutes with every element of $SO(2)$. – Tom Goodwillie Aug 16 '11 at 0:33
And as to your second, I just said "nice enough at the endpoints", but I did not discuss what condition this really is on $f$. – Tom Goodwillie Aug 16 '11 at 1:12
@Tom: thank you, I am beginning to understand what you said about $h$ even though I still cannot make your sketch into a proof. As for $f$, the difficulty is not limited to the endpoints, as far as I can see. – Igor Belegradek Aug 16 '11 at 1:42