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In the following paper

" On actions of $SO(3)$ on lens spaces II" by S. Kim and J. Pak the following result (Lemma 1.1) has been used

Any effective action of $SO(3)$ on $L_{2n+1}(m)$, $m$ odd can be lifted equivariantly to an effective action of $SO(3)$ on $S^{2n+1}$. (Here $S^{2n+1}$ is sphere of dimension $2n+1$ and $L_{2n+1}(m)$ denotes Lens space.)

My query is if this result is true for free action? Can a free action of SO(3) on $L_{2n+1}(m)$, $m$ odd can be lifted equivariantly to a free action of $SO(3)$ on $S^{2n+1}$.

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Yes. A free action is by definition effective, so by their result, the action lifts equivariantly. Unless I misunderstand what `equivariant lift' means, then a lift of a free action is automatically free. For if $g\in SO(3)$ fixes $x \in S^{2n+1}$, then $g$ fixes the image of $x$ in $L_{2n+1}(m)$.

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  • $\begingroup$ I'm playing around with embedding a homogeneous space G/H into the space G. Picking G=SO3, H=Zn, yields a family of lens spaces parameterized by n. Projecting the left and right regular representation operators of G, which form SO3 x SO3, onto SO3/Zn seems to yield a representation of SO3 x O2, with the SO3 being the induced representation of SO3 on the lens space. Anyone know what to think of the O2 part? (Picking G=SO3, H=SO2, and doing the same thing yields the action of SO3 x Z2 on the two sphere, which is clearly O3, the group of proper and improper rotations.) $\endgroup$
    – Victor V Albert
    Commented Sep 15, 2019 at 1:16

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