I had my final exam and the professor gave this question which I don't really know how to do.
The question: Let $p \in \mathbb{Z}^+$ and let $\mathbb{Z}_p$ act on $\mathbb{C}^2$ by $(z,w) \mapsto (e^{2 \pi i/p}z,e^{2 \pi i /p} w)$. This induces an action on $S^3 \hookrightarrow \mathbb{C}^2$ and the resulting quotient $S^3/\mathbb{Z}$ is referred to as the lens space $L(p,1)$. Show that this lens space is a smooth manifold and computs its de Rham cohomology groups.
I've seen the definition before but that's about it. I vaguely remembered that I should probably show something about the group action being properly discontinuous to show that $\pi:S^3 \to L(p,1)$ is a regular covering map. But before even that, how do I know $\pi$ is a quotient map and whether that says $L(p,1)$ is a manifold, let alone smooth?
Certainly $L(p,1)$ should be compact and connected as $S^3$ is. I believe it's orientable because multiplication by a complex number is a holomorphism (so orientation preserving) and in $\mathbb{C}^2$, it seems natural to say that a map is holomorphic if and only if its coordinate functions are holomorphic.
Thus, $H^0_{dR}(L(p,1) = H^3_{dR}(L(p,1)) = \mathbb{R}$. Also, $S^3$ is the universal cover of $L(p,1)$ and its deck transformations are $\pi_1(L(p,1)) = \mathbb{Z}_p$. Then, since $H^1_{dR}(L(p,1)) \cong H^1(L(p,1), \mathbb{R})$ by de Rham's theorem and $H_1(L(p,1), \mathbb{R}) = H_1(L(p,1),\mathbb{Z}) \otimes_\mathbb{Z} \mathbb{R} = \text{Abel}(\pi_1(L(p,1)) \otimes \mathbb{R} = 0$, $H^1_{dR}(L(p,1)) = 0$. By Poincare duality, $H^2_{dR}(L(p,1))=0$ as well. So the de Rham groups of $L(p,1)$ seem to be the same as $S^3$.
My questions:
- How do we show that $L(p,1)$ is a smooth manifold?
- How do we show it is orientable?
- Is my reasoning for the de Rham groups correct?
I apologize for being rather wordy. Thank you in advance.