Timeline for Does there exist smooth circle action on manifolds M^{4n} with exactly three fixed points such that n\neq 1
Current License: CC BY-SA 3.0
16 events
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| Jul 5, 2017 at 15:47 | answer | added | Nick L | timeline score: 1 | |
| Nov 9, 2011 at 0:15 | comment | added | Ping | Vitali, I understand your meaning. But in fact in my proof we essentially use a result of Kawakubo-Uchida, which localizes the computation of the cobordism class of the global mfd to the fixed-point components and need the action to be semi-free. | |
| Nov 8, 2011 at 15:27 | comment | added | Vitali Kapovitch | @Ping Li If you require an action to be semi-free then you don't allow any points with finite isotropies. what I'm saying is that in your theorem you don't need to require the whole action to be semifree. you only need to assume that you have finitely many fixed points and the action is free in small punctured balls around these points. this allows for some points with finite isotropies so long as the fixed points are not limits of such points. | |
| Nov 8, 2011 at 11:46 | comment | added | Gerry Myerson | If you really want to have "cirlce" instead of "circle", and "Remmark" instead of "Remark", then you can undo my editing again - but I have to wonder why you would do such a thing. | |
| Nov 8, 2011 at 11:43 | history | rollback | Gerry Myerson |
Rollback to Revision 2
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| Nov 8, 2011 at 7:52 | answer | added | Ryan Budney | timeline score: 4 | |
| Nov 8, 2011 at 6:19 | history | rollback | Ping |
Rollback to Revision 1
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| Nov 8, 2011 at 5:23 | comment | added | Ping | Thanks. Sorry, "It can also be shown that if a manifold admits a semi-free circle action with isolated fixed points, then the number of this action must be even" should be "the number of the isolated fixed points of this action must be even". So you in fact provides a simple proof of this fact. BTW, I don't understand what you mean in "BTW...". If points are with finite isotropies, they of cousre are not fixed points. | |
| Nov 8, 2011 at 4:42 | comment | added | Vitali Kapovitch | BTW, I think you prove something slightly stronger than stated. you only need that the action is semi-free near the fixed points and points with finite isotropies are allowed so long as they don't touch the fixed points. | |
| Nov 8, 2011 at 4:40 | comment | added | Vitali Kapovitch | Thanks, this is a nice theorem. I had something much easier in mind - if the action of $S^1$ is semi-free and the number of fixed points is odd then you can get rid of them in pairs by cutting out little disks around them and gluing together the boundary $S^{4n−1}$'s preserving the $S^1$ action (this is possible since the action is semi-free). In the end you'll end up with a semifree $S^1$ action on some manifold with a single isolated fixed point. the manifold you get might be non-orientable but that doesn't matter when you look at the semi-free $Z_2$ action. | |
| Nov 8, 2011 at 0:04 | comment | added | Ping | In fact, I showed in mathjournals.org/mrl/2011-018-003/2011-018-003-005.pdf that if a manifold admits a semi-free circle action with isolated fixed points, then the manifolds bound, i.e., the Pontrjagin numbers and Stiefiel-Whitney numbers of the manifold vanish (Theorem 1.6). | |
| Nov 7, 2011 at 23:58 | comment | added | Ping | Sorry "there are known examples in dimensions other than 4" should be "there are no known examples in dimensions other than 4" | |
| Nov 7, 2011 at 23:57 | comment | added | Ping | Thanks very much for your remark. In fact, any circle or Z_P action on a manifold cannot have exactly one fixed point, which is quite well-known. If they have two fixed points, then the representations on the tangent spaces are isomorphic and this case have examples in each dimension (rotation on S^{2n}). So the next case should be with three fixed points. As far as I know, there are known examples in dimensions other than 4. It can also be shown that if a manifold admits a semi-free circle action with isolated fixed points, then the number of this action must be even (Bott residue formula). | |
| Nov 7, 2011 at 19:23 | comment | added | Vitali Kapovitch | Is this known for $Z_p$ actions with prime $p$? I can show using rather elementary tools that a $Z_p$ action can not have exactly one fixed point (I'm sure this must be very well-known) but I don't know about 3. Another obvious observation is that an $S^1$-action like this can not be semi-free for $n\ge 2$. | |
| Nov 7, 2011 at 11:21 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
fixed some typos, improved formatting
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| Nov 7, 2011 at 6:49 | history | asked | Ping | CC BY-SA 3.0 |