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It is known that $L(p,q)\cong L(p,q')$ if and only if $q\equiv \pm (q')^{\pm 1}\mod p$ where

$$ L(p,q)\cong{\mathbb{S}^3}/\mathord{\sim} $$ is the quotient space generated by the $\mathbb{Z}_p$-action $$ \rho(z_1,z_2)=(\zeta z_1,\zeta^q z_2) $$

where $\zeta$ is a pth root of unity and $p$ and $q$ are coprimes.

Right to left implication can be proven using Reidemeister torsion (https://www3.nd.edu/~lnicolae/Torsion.pdf page 104). However, left to right proof is supposed to be verified easily by the reader. Does someone know how to construct the homeomorphism between $L(p,q)$ and $L(p,q')$ assuming the congruence on $q$ and $q'$?

EXTRA QUESTION: Does the Poincaré conjecture imply that every closed compact orientable 3-manifold with $\pi_1(X)\cong \mathbb{Z}_p$ is a lens space?

Aner
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    I think your last question is true. The universal cover is a closed, connected, simply connected 3-manifold, and hence a sphere, so $X$ is a quotient of $S^3$ by some covering space action by $\mathbb Z/p$. – Andres Mejia Jan 19 '21 at 22:09
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    An answer to your first question is here (although I admit to not reading it carefully.) – Andres Mejia Jan 19 '21 at 22:10
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    @AndresMejia how could you be sure that there are no more $\mathbb{Z}_p$-actions on $S^3$ other than the ones that define lens spaces? – Aner Jan 20 '21 at 01:10
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    great question. I don’t think I have a good answer off the top of my head. I hope somebody with a better background in low dimensional topology has something to say. – Andres Mejia Jan 20 '21 at 01:20
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    For your extra question: PC is not enough but it is a special case of Thurston's Geometrization Conjecture (called Spherical Space Form Conjecture), proven by Perelman. – Moishe Kohan Jan 20 '21 at 09:55
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    @AndresMejia Uniqueness of the $\mathbb{Z}_p$-action on $S^3$ follows from linear algebra. If $M$ is a matrix in $O(4)$ such that $M^p = I$, then the minimal polynomial of $M$ divides $x^p - 1$. You can then use some primary decomposition theorem to show that $M$ is always conjugate to some combination of classical $2 \times 2$ rotation matrices plus reflections. (This should, in fact, be true for higher-dimensional spheres as well.) – Jeroen van der Meer Jan 20 '21 at 12:39

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For two roots of unity $\lambda,\mu$ define $\rho_{\lambda,\mu}: (z,w)\mapsto (\lambda z, \mu w)$. Then $\rho_{\lambda,\mu}$ and $\rho_{\lambda,\bar\mu}$ are conjugate via the orthogonal map $$ (z,w)\mapsto (z, \bar{w}). $$ This shows that $L(p,q)$ is homeomorphic to $L(p,-q)$.

Next, if $q$ is coprime to the orders of $\lambda,\mu$ then the subgroup generated by $\rho_{\lambda,\mu}$ is the same as the one generated by $(\rho_{\lambda,\mu})^q$.

Lastly, assuming that $p$ is coprime to $q$, $q'$, $qq'\equiv 1$ mod $p$, and
$$ \lambda=\exp(i 2\pi/p), \mu= \exp(i 2\pi q/p), \mu'= \exp(i 2\pi q'/p), $$ then $$ \lambda^{q'}= \exp(i 2\pi q'/p), $$ $$ \mu^{q'}= \exp(i 2\pi /p). $$ In other words, the subgroup generated by $\rho_{\lambda,\mu}$ is the same as generated by $\rho_{\mu',\lambda}$. However, $\rho_{\mu',\lambda}$ is conjugate in $U(2)$ to $\rho_{\lambda,\mu'}$ via $$ (z,w)\mapsto (w,z). $$ This proves that $L(p,q)$ is homeomorphic to $L(p,q')$.

Lastly, for your extra question, PC is not enough, you need SSFC. The later was proven by Perelman in the same paper(s) where he proves the Poincare Conjecture.

Moishe Kohan
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