Timeline for What is the normalizer of the circle in the diffeomorphism group of the 2-sphere?
Current License: CC BY-SA 3.0
7 events
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| Aug 16, 2011 at 2:29 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
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| Aug 16, 2011 at 1:42 | comment | added | Igor Belegradek | @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. | |
| Aug 16, 2011 at 1:12 | comment | added | Tom Goodwillie | And as to your second, I just said "nice enough at the endpoints", but I did not discuss what condition this really is on $f$. | |
| Aug 16, 2011 at 0:33 | comment | added | Tom Goodwillie | Igor, as to your second comment, I am only asserting that $h$ has this form if the diffeomorphism commutes with every element of $SO(2)$. | |
| Aug 15, 2011 at 21:42 | comment | added | Igor Belegradek | 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? | |
| Aug 15, 2011 at 21:32 | comment | added | Igor Belegradek | 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 $S^2$. | |
| Aug 15, 2011 at 20:50 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |