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Let $G$ be a finite group with the discrete topology. To set terminology:

  • a $G$-sphere is a sphere equipped with a continuous $G$-action

  • a $G$-representation sphere is a $G$-sphere obtained from an orthogonal $G$-representation $V$ by taking its unit sphere $S(V)\subset V$

From what I've heard, there exist $G$-spheres that are not homeomorphic (as $G$-spaces) to any $G$-representation sphere. However, I haven't been able to construct one, or find any reference for this; any help would be appreciated.

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  • $\begingroup$ Maybe working too hard, but: it would suffice to find a $G$-action that doesn't fix a smooth structure. I don't know if this is easy or hard. $\endgroup$
    – Qiaochu Yuan
    May 27, 2015 at 7:22
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    $\begingroup$ You can take the classical example of a Z/2Z-action on $S^4$ with non-manifold fixed set, e.g. Corollary 4.10 of math.rutgers.edu/~sferry/ps/geotop.pdf. $\endgroup$
    – skupers
    May 27, 2015 at 8:35
  • $\begingroup$ @QiaochuYuan, can a finite group not fix a smooth structure? $\endgroup$
    – Mariano Suárez-Álvarez
    May 27, 2015 at 12:10
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    $\begingroup$ @Zev: there are also examples of finite groups acting freely on spheres but the actions are not conjugate to linear actions. Do a Google search for "fake projective space" and you will find several papers. Generally people use this language in the manifold world rather than talking about $G$-spheres. $\endgroup$
    – Ryan Budney
    May 27, 2015 at 15:31

1 Answer 1

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There are involutions on some Brieskorn spheres which are true spheres whose fixed point sets are only Z/2-homology spheres. (But fixed sets of linear actions on spheres are spheres.)

Here it is easy to write down the Brieskorn sphere and the involution, it is more work to see that the Brieskorn sphere is topologically a sphere, and it is even more work if one is interested in the smooth structure.

Reference: Bredon, Exotic Actions on Spheres, Proceedings of the Conference on Transformation Groups 1968, pp 47-76. doi:10.1007/978-3-642-46141-5_2

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