Covering spaces of Lens spaces

Question

Let $L(p,q)$ be the Lens space with composite $p$, say $p=ab$.

What is the cyclic covering space of $L(p,q)$ induced from the quotient group homomorphism from $\mathbb{Z}/p$ to $\mathbb{Z}/a$?

Answer 1

I'm not sure exactly what it means to be "induced from a homomorphism". But I take it to mean "the covering space corresponding to the kernel of that homomorphism."

With this is mind, I claim that the covering is $L(b,q)$. Note that $\gcd(b,q) \leq \gcd(p,q) $ since $b\mid p$. Using the fact that $\gcd(p,q) = 1$, this implies $\gcd(b,q) = 1$ as well, so that $L(b,q)$ is well defined.

To save on typing, let $X = S^3$, $Y = L(p,q)$, and $G = \pi_1(Y) = \mathbb{Z}/p\mathbb{Z}$.

Note that $G$ acts on $X$ via Deck transformations. Further, the action is properly discontinuous ("nice") so the orbit space $X/G$ naturally has the structure of a smooth manifold.

Now, if $\pi:X\rightarrow Y$ denotes the projection, then clearly $\pi(g\ast x) = \pi(x)$ for any $x\in X$ and $g\in G$, so $\pi$ descends to a map from $X/G$ to $Y$ and it's not too hard to show that this map is, in fact, a diffeomorphism.

Let $H\subseteq G$ be the unique subgroup isomorphic to $\mathbb{Z}/b\mathbb{Z}$. By restricting the action of $G$ on $X$ to $H$, we still get a "nice" action of $H$ on $X$, so the quotient $X/H$ is a smooth manifold. Just as in the previous case, one can show that $X/H\cong L(b,q)$.

Finally, there is a natural map from $X/H\rightarrow Y$ given by sending $xH$ to $xG$. Said a different way, to go from $X$ to $X/H$, you do a little bit of quotienting, and then to go from $X/H$ to $Y(\cong X/G)$, you do all the rest of the quotienting. By barehands, one can prove that this map $X/H\rightarrow X/G$ is a covering. In fact, the fiber is naturally $G/H\cong \mathbb{Z}/a\mathbb{Z}.$