3

Let $p,q \in \mathbb{Z}$, with $p > 0$ and $gcd(p,q) = 1$. The Lens space $L(p,q)$ is defined by $\mathbb{S}^3/\mathbb{Z}_p$ where the action is given by $n \cdot (z_1,z_2) = (e^{2\pi i n /p}z_1,e^{2\pi i n q/p}z_2)$. Here we are identifying $\mathbb{S}^3$ with the unit sphere of $\mathbb{C} \times \mathbb{C}$. I wish to show that if $q + q^\prime\equiv 0 \;(mod\; p)$ or $qq^\prime \equiv 1 \;(mod \; p)$ then $L(p,q)$ is diffeomorphic to $L(p,q^\prime)$.

I tried defining a map from $\mathbb{S}^3$ to $\mathbb{S}^3$ that could induce such diffeomorphisms but I wasn't able to find it. Any sugestion?

  • Any easy proof of this is not known to me. There is a nice paper by Bonahon, namely, Mapping class group of lens space. Have a look. The idea is very beautiful though. They essentially proved that genus one surface of Heegard splitting of lens spaces are isotopic...and then they tried to do the rest of the computation. – Anubhav Mukherjee May 23 '18 at 21:45

1 Answers1

1

You can do this by thinking carefully about the orbits of points under the various group actions. I'll look at the case where $q'q\equiv 1 \bmod p$ first.

Under the action of $\mathbb{Z}/p\mathbb{Z}$, given by $(z_1,z_2)\mapsto (z_1e^{2\pi i / p},z_2 e^{2\pi i q / p})$ the orbit of the point $(w_1,w_2)$ is the set $$O_1=\{(w_1e^{2\pi i k / p},w_2 e^{2\pi i kq / p})\,|\, k=0, \dots, p-1\}.$$

Now we study orbit of $(w_2,w_1)$, the under the action given by $(z_1,z_2)\mapsto (z_1e^{2\pi i / p},z_2 e^{2\pi i q' / p})$. This orbit is $$O_2=\{(w_2e^{2\pi i k' / p},w_1 e^{2\pi i k'q' / p})\,|\, k'=0, \dots, p-1\},$$ which, by substituting $k'=kq \bmod p$, can be rewritten as $$O_2=\{(w_2e^{2\pi i kq / p},w_1 e^{2\pi i k / p})\,|\, k=0, \dots, p-1\}.$$ Thus the map $\phi \colon S^3 \rightarrow S^3$ given by $\phi(z_1,z_2)=(z_2,z_1)$ carries the orbit $O_1$ of $(w_1,w_2)$ under the first action onto the orbit $O_2$ of $(w_2,w_1)$ under the second action, thus inducing a diffeomorphism between $L(p,q)$ and $L(p,q')$ where $qq'\equiv 1 \bmod p$.

The case of $-q$ is similar. Consider the orbit of $(w_1,\overline{w_2})$, the under the action given by $(z_1,z_2)\mapsto (z_1e^{2\pi i / p},z_2 e^{-2\pi i q / p})$. This orbit is $$O_3=\{(w_1 e^{2\pi i k' / p}, \overline{w_2} e^{-2\pi i k'q / p})\,|\, k'=0, \dots, p-1\}.$$ Thus $O_3$ is the image of $O_1$ under the map $\psi\colon S^3\rightarrow S^3$ given by $\psi(z_1,z_2)=(z_1,\overline{z_2})$. Thus $\psi$ induces a diffeomorphism between $L(p,q)$ and $L(p,-q)$. Here I'm using $\overline{z}$ to denote complex conjugation.