I have a question in mind for some time. That is, does there exist a smooth circle action on a closed manifold M^{4n} $M^{4n}$ (n\geq 2$n\geq 2$) with exactly three fixed points? Remmarks:(1)For Remarks:(1) For n=1, the examples are obvious (standard linear circle action on CP^2). $CP^2$). (2)If 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.
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Does there exist smooth cirlce circle action on manifolds M^{4n} with exactly three fixed points such that n\neq 1 |
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Does there exist smooth circle cirlce action on manifolds M^{4n} with exactly three fixed points such that n\neq 1I have a question in mind for some time. That is, does there exist a smooth circle action on a closed manifold $M^{4n}$ M^{4n} ($n\geq 2$n\geq 2) with exactly three fixed points? Remarks:(1) For Remmarks:(1)For n=1, the examples are obvious (standard linear circle action on $CP^2$). CP^2). (2) If 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. |
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Does there exist smooth cirlce circle action on manifolds M^{4n} with exactly three fixed points such that n\neq 1I have a question in mind for some time. That is, does there exist a smooth circle action on a closed manifold M^{4n} $M^{4n}$ (n\geq 2$n\geq 2$) with exactly three fixed points? Remmarks:(1)For Remarks:(1) For n=1, the examples are obvious (standard linear circle action on CP^2). $CP^2$). (2)If 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. |
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