circle action on sphere - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:08:10Z http://mathoverflow.net/feeds/question/18569 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18569/circle-action-on-sphere circle action on sphere student 2010-03-18T11:43:04Z 2010-03-18T12:29:34Z <p>surely $S^1$ can act on $S^n$ as a rotation.I want to know if there is some other way that a circle can act on sphere.</p> http://mathoverflow.net/questions/18569/circle-action-on-sphere/18571#18571 Answer by Mariano Suárez-Alvarez for circle action on sphere Mariano Suárez-Alvarez 2010-03-18T11:53:22Z 2010-03-18T12:06:50Z <p>One example: The sphere $S^{n+m+1}$ is the <a href="http://en.wikipedia.org/wiki/Join_%28topology%29" rel="nofollow">join</a> $S^n*S^m$of $S^n$ and $S^m$, so having $S^1$ act on each of the factors you get an action on $S^{n+m+1}$.</p> http://mathoverflow.net/questions/18569/circle-action-on-sphere/18575#18575 Answer by Sebastian for circle action on sphere Sebastian 2010-03-18T12:16:12Z 2010-03-18T12:18:00Z <p>A nice example is the Hopf fibration: $S^1 \subset C$ acts on $S^3\subset \mathbb C^2$ by scalar multiplication. The quotient is $\mathbb CP^1.$ Of course there are various generalizations.</p> http://mathoverflow.net/questions/18569/circle-action-on-sphere/18577#18577 Answer by Igor Belegradek for circle action on sphere Igor Belegradek 2010-03-18T12:26:12Z 2010-03-18T12:26:12Z <p>Classification of circle actions on (standard and on exotic) spheres is a classical activity in transformation groups, see e.g. the article of Schultz in the collection "Group actions on manifolds", proceedings from 1983 conference, or earlier account in Bredon's book "Introduction to compact transformation groups". </p> <p>One standard example is this. Take a smooth compact contractible $(n+1)$-manifold $C$ and consider $C\times D^k$ where circle acts trivially on $C$ and linearly on the $k$-disk $D^k$, $k>0$. Now if $n+k>4$, the boundary of $C\times D^k$ is diffeomorphic to the standard sphere (after the corners of $C\times D^k$ are rounded). But the fixed point set of the action is the original homology sphere that bounds $C$. I think it is clear that the fixed point sets of linear actions are standard spheres.</p> <p>Incidentally, in dimensions $k>4$ any homology $k$ sphere bounds a contractible manifold after possibly changing its smooth structure; also any homology $4$-sphere bounds a smooth contractible manifold, while any homology $3$-sphere bounds a topological contractible manifold, but not necessarily smooth ones. References to the above results can by found e.g. in my <a href="http://arxiv.org/abs/math/0302221" rel="nofollow">paper pp.8-9</a>.</p> http://mathoverflow.net/questions/18569/circle-action-on-sphere/18578#18578 Answer by HenrikRüping for circle action on sphere HenrikRüping 2010-03-18T12:29:34Z 2010-03-18T12:29:34Z <p>I do not know, in which category your actions live. In the following answer I want to consider isometric actions.</p> <p>Another classification problem might be:</p> <p>Classify all periodic one parameter subgroups of $SO(n)$ up to conjugacy, which is the same problem as the classification of all such actions up to isometry.</p> <p>There is a normal form for orthogonal matrices over the reals, which says, that every such matrix is conjugated to a block matrix consisting of $(2,2)$ rotation matrices and $(1,1)$ diagonal entries, which are $\pm 1$. </p> <p>My claim would be that every isometric action decomposes (after congugation) uniquely as a induced action on each of the components of the decomposition</p> <p>$S^n = S^1 * S^1 * \ldots * S^1 * S^0 * \ldots * S^0$ ($*$ should denote the join here). The isometric actions of $S^1$ on $S^1$ can be classified by the integers. The only possibilities are $(t,x)\mapsto mt+x$ for $m\in\mathbb{Z}$.</p> <p>I think the upper normal forms for matrices should imply this. But I am not sure.</p>