# Writing three dimensional lens space $L(p;a,b)$ in the form $L(p;1,b')$

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 $$L(b-d; c-a,c+a)=L(b-d;q)$$?

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 $$\mathbb{Z}_p$$.)
• 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. May 21 '20 at 12:09