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I have a question in mind for some time. That is, does there exist a smooth circle action on a closed manifold $M^{4n}$ ($n\geq 2$) with exactly three fixed points? Remarks:(1) For n=1, the examples are obvious (standard linear circle action on $CP^2$). (2) If a manifold admits a circle action with 3 fixed points, then the signature of this manifold is 1, so the dimension of this manifold is necessarily divisible by 4.

edited Nov 8 2011 at 11:43

Gerry Myerson
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asked Nov 7 2011 at 6:49

Ping
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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 $n\ge 2$. – Vitali Kapovitch Nov 7 2011 at 19:23

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). – Ping Nov 7 2011 at 23:57

Sorry "there are known examples in dimensions other than 4" should be "there are no known examples in dimensions other than 4" – Ping Nov 7 2011 at 23:58

In fact, I showed in mathjournals.org/mrl/2011-018-003/… 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). – Ping Nov 8 2011 at 0:04

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. – Vitali Kapovitch Nov 8 2011 at 4:40

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Ryan Budney · Answer 1 · 2011-11-08 07:52:15Z UTC

I think the answer to your question is yes, but perhaps you meant to ask a slightly different question than the precise one you pose.

For example, take the model of $\mathbb RP^{2n}$ where you view it as the unit ball in $\mathbb C^n$ modulo the antipodal map on the boundary.

So the action of $S^1$ on $\mathbb C^n$ factors to an action on $\mathbb RP^{2n}$, and it has precisely one fixed point.

Take the disjoint union of three of these $S^1$ spaces. By design, this has three fixed points.

Perhaps you wanted the manifold to be connected? So do an equivariant surgery, pairwise along a free orbit, drilling out $D^{2n-1} \times \{0,1\} \times S^1$ and gluing in an $S^{2n-2} \times [0,1] \times S^1$. Do this twice, and now you have a connected $S^1$-manifold with precisely three fixed points.

So my first guess is that you would prefer the manifold to be orientable?

Does there exist smooth circle action on manifolds M^{4n} with exactly three fixed points such that n\neq 1

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