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I'm totally new to equivariant topology. I would like to ask what is the decomposition of orbit type when we treat $O(3)/SO(3)$ as a $O(2)^-$ complex. Here $O(2)^-$ is the twisted subgroup of $O(3)$, where $O(2)^- \cap SO(3) =SO(2)$.

My understanding is $O(3)/SO(3)\cong Z/2= \{+I,-I\}$, where $+I$ denotes the $3\times 3$ matrices with determinant 1 and $-I$ the $3\times 3$ matrices with determinant -1. So here are only two points. Since $SO(2)$ fixes both $+I$ and $-I$ and they are on the same orbit, that's why the decomposition is just $(0, SO(2))$. I would like to know if it is correct or not. Thanks!

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  • $\begingroup$ I'm not sure what you mean by $(0,SO(2))$. Is that zero meant to be an O or similar? $\endgroup$ Commented May 19 at 22:51
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    $\begingroup$ Are you aware that $O(3)$ is just a direct product, $SO(3) \times \mathbb{Z}/2$? $\endgroup$ Commented May 19 at 22:55

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