Timeline for Extension of a group action beyond the boundary
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 5, 2016 at 22:08 | comment | added | Maxim Braverman | @SebastianGoette I found a paper of Palais "Equivalence of nearby differentiable actions of a compact group" which shows that any two actions which are closed to each other are conjugated. Moreover, given a continuous family of actions, one can find a continuous family of conjugations. It is not clear to me from his proof whether one can find a smooth family of conjugations. However, I found a much more recent paper of Kankaanrinta, which proves all what I need, see my answer to this question. | |
| Mar 4, 2016 at 16:01 | comment | added | Sebastian Goette | @MaximBraverman Then I suggest that you put that discussion into your question. Maybe someone else has a solution. | |
| Mar 4, 2016 at 15:39 | comment | added | Maxim Braverman | @SebastianGoette I don't have an example. I would not be surprised if it is always the case. But I can not prove it. | |
| Mar 4, 2016 at 10:46 | comment | added | Sebastian Goette | @MaximBraverman I am trying to think of an example where $\phi_t$ is not conjugate to a constant action (this is the case I had in mind - respecting the product structure means mapping $\{y\}\times[0,1)$ to $\{g(y)\}\times[0,1)$ in my terminology). Do you have such an example where $G$ is compact? | |
| Mar 4, 2016 at 4:09 | comment | added | Maxim Braverman | @IgorRivin Actually, I don't understand why this construction gives a smooth action even if the boundary is totally geodesic. Suppose that a neighborhood of the the boundary is isometric to the product $N\times[0,1)$ and the group action is given by $\phi_t$ as in my comment to Sebastian, where each $\phi_t$ defines an isometry of $N$. Why this action extends to a smooth action beyond the boundary? | |
| Mar 3, 2016 at 22:04 | comment | added | Igor Rivin | @MaximBraverman You are right. Action by isometries is good if you can make the boundary totally geodesic, but this is, in general, not possible. | |
| Mar 3, 2016 at 21:53 | comment | added | Maxim Braverman | @SebastianGoette I think to take the double it is not enough to know that the action of $G$ preserves the product structure. Of course, if $G$ action is a product, then the action on the double is smooth. But in general you have a family $\phi_t:G\to Diff(N)$ of actions and the action on the collar $N\times[0,1)$ is given by $g\cdot (y,t)= (\phi_t(g)(y),t)$. When you extend it to the double, the action on the double of the collar will be $g\cdot (y,t)= (\phi_{|t|}(g)(y),t)$ for $t\in (-1,1)$. Why is it smooth at $t=0$? | |
| Mar 3, 2016 at 21:48 | comment | added | Igor Rivin | @MaximBraverman Sebastian tells you how to tweak a topological extension to a smooth extension. | |
| Mar 3, 2016 at 21:46 | comment | added | Maxim Braverman | @IgorRivin Topological extension is easy to construct. But I need a smooth extension. | |
| Mar 3, 2016 at 20:53 | comment | added | Igor Rivin | @SebastianGoette Good point! | |
| Mar 3, 2016 at 20:49 | comment | added | Sebastian Goette | You can always take a metric such that the boundary has a collar neighbourhood which looks like a product. You have to make sure first that the action of $G$ respects the product structure. If you average over $G$, the action will be by isometries. Then you can double. | |
| Mar 3, 2016 at 19:30 | history | answered | Igor Rivin | CC BY-SA 3.0 |