Let $S^3/\Gamma_i\,(i=1,2)$ be a $3$-dimensional spherical space form, where $\Gamma_i \subset SO(4)$ is a finite subgroup acting freely on $S^3$. If $S^3/\Gamma_1$ is homotopy equivalent to $S^3/\Gamma_2$, can we find a $g \in SO(4)$ such that $$ g\Gamma_1 g^{-1}=\Gamma_2? $$
2026-03-25 06:33:58.1774420438
The fundamental groups of 3-dimensional spherical space forms
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This is not true in general, in fact it can fail for the lens spaces which are the quotients $$L(p;q) = S^3 / \Gamma(p;q) $$ where $\Gamma(p;q)$ is the cyclic subgroup of order $p$ in $SO(4)$ that is generated by the element described, in complex coordinates, by the formula $$(z_1,z_2) \mapsto (e^{2 \pi i/p} \cdot z_1, e^{2 \pi i q/p} \cdot z_2) $$ The link given explains how to determine when $L(p;q_1)$ and $L(p;q_2)$ are homotopy equivalent and when they are homeomorphic, and these do not match up. And if they are not homeomorphic then there cannot be any $g \in SO(4)$ for which $g \Gamma(p;q_1) g^{-1} = \Gamma(p;q_2)$, because the map of $SO(4)$ determined by $g$ would descend to a homeomorphism between $L(p;q_1)$ and $L(p;q_2)$.