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According to Wikipedia: https://en.wikipedia.org/wiki/Lens_space

In general, $n$-dimensional lens spaces are written as $L(p;q_1,..,q_n)$ for integers $q_1$,...,$q_n$ relatively prime to $p$. In dimension three, these lens spaces are the quotient by the $\mathbb{Z}_p$ action on $S^3 \subset \mathbb{C}^2$ generated by the homeomorphism $(z_1,z_2) \mapsto (e^{2\pi i q_1/p}z_1,e^{2\pi i q_2/p}z_2)$.

In dimension three, lens spaces are standardly written in the form $L(p;q)$ for a single integer $q$ relatively prime to $p$ and $L(p;q)=L_(p;1,q)$ using the more general $n$-dimensional notation.

I have an action of $\mathbb{Z}_{b-d} \subset S^1$ on $S^3$ given by $\zeta \cdot (z_1,z_2)= (\zeta^{c-a}z_1, \zeta^{c+a}z_2)$ which is free given the $\gcd$ conditions $\gcd(a,b,c,d)=gcd(a^2-c^2,b^2-d^2)=1$.

This lens space is then $L(b-d;c-a,c+a)$.

My question is, how can I figure out the integer $q$ for which $L(b-d; c-a,c+a)=L(b-d;q)$?

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Since $q_1$ is coprime to $p$, it has an inverse $q_1'\in\mathbb{Z}_p$. Precompose your action with the homeomorphism $(z_1, z_2)\mapsto (e^{2\pi i q_1' / p}z_1, e^{2\pi i q_1' / p}z_2)$ to show that the space $L(p;q_1,q_2)$ is identical to the space $L(p; 1, q_2q_1')$. (The intuition is that it doesn't matter which primitive $p^{th}$ root of unity you choose to generate the action of $\mathbb{Z}_p$.)

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    $\begingroup$ I believe in your intuition, you want to use primitive $p$th roots. Of course, if $p$ is prime, all the non-trivial roots are primitive. $\endgroup$ May 21 '20 at 12:09
  • $\begingroup$ @JasonDeVito yes that is correct, thank you for the catch $\endgroup$
    – Neal
    May 21 '20 at 13:28

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