TITLE: How to define a Riemannian metric in the projective space such that the quotient projection is a local isometry?
QUESTION [12 upvotes]: Let $A: \mathbb{S}^n \rightarrow \mathbb{S}^n$ be the antipode map ($A(p)=-p$) it is easy to see that $A$ is a isometry, how to use this fact to induce a riemannian metric in the projective space such that the quotient projection $ \pi: \mathbb{S}^n\rightarrow \mathbb{P}^n$ is a local isometry?

REPLY [7 votes]: Note first that the projective space, $\mathbb{P}^n$, is the quotient space,
$$\mathbb{P}^n=\mathbb{S}^n/ \Gamma$$ 
where, $\Gamma= \{A,Id\}$. We obtain an atlas for $\mathbb{P}^n$ simply choosing a neighborhoods $U$ of $\mathbb{S}^n$ such that $\pi|U$ is injective, where $\pi$ is the canonical projection. In this way, $\pi|U=A|U.$
Now fix $p$ in $\mathbb{P}^n$ and let $u,v\in T_p \mathbb{P}^n$. Define, 
$$<u,v>_p=<d(\pi|U)^{-1}u,d(\pi|U)^{-1}v>_{\pi^{-1}(p)}.$$
with this metric the canonical projection  is an  local isometry as defined by what has already been discussed above.