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My question is why for an orthogonal action of G on $S^n$ the fixed point set is again sphere $S^r$

Orthogonal action of G means G has orthogonal representation that is there exist a homeomorphism of G into O(n).

Fixed point set of G on X is the set of those x in X such that g(x)=x for all g in G

For action of O(3) on $S^2$ the fixed point set is $S^0$ the two point set.but i don't know how to prove it generally . Thanks in advance

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Hint: The fixed point set of a single orthogonal transformation $T$ on a sphere centered at the origin is the intersection of the sphere with the $1$-eigenspace of $T$.

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    $\begingroup$ Slick - I like it! $\endgroup$ – Jason DeVito Nov 29 '16 at 18:47
  • $\begingroup$ Is it true that fixed point set will be always $S^0$ for orthogonal action on $S^n$?@Andrew D.Hwang $\endgroup$ – Shivani Sengupta Nov 29 '16 at 19:13
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    $\begingroup$ @prakriti: No. Consider the trivial action. Less trivially, consider a single reflection. $\endgroup$ – Jason DeVito Nov 29 '16 at 19:23
  • $\begingroup$ Your group $G$ could be the trivial group, in which case the fixed point set is the entire sphere, or the group of order $2$ generated by reflection in one axis, in which case the fixed point set is a great hypersphere, or.... :) $\endgroup$ – Andrew D. Hwang Nov 29 '16 at 19:31
  • $\begingroup$ Is it true that If I take full group O(n) then the fixed point set of this action on $S^n$ is $S^0$ ? $\endgroup$ – Shivani Sengupta Nov 29 '16 at 19:35

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