How to define a Riemannian metric in the projective space such that the quotient projection is a local isometry? 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?
 A: 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.
A: You are defining $\mathbb{P}^n = \mathbb{S}^n / p\sim A(p)$.  Let's assume that you have verified to your satisfaction that this action is "nice" enough to produce a smooth quotient manifold.  
Let's consider any point $p\in\mathbb{P}^n$. The only thing we know about $p$ is that it's an equivalence class $p = [x] = \{x_1, x_2\}$ which are related by $x_i = A(x_j)$.  The tangent spaces $T_{x_1}\mathbb{S}^n$ and $T_{x_2}\mathbb{S}^n$ are related by $dA$, and via this identification we can consider either of them a model for $T_p\mathbb{P}^n$.  So if we want to put a metric on $T_p\mathbb{P}^n$, there's really only one thing we can do: pick the metric on one of the tangent spaces $T_{x_i}\mathbb{S}^n$.
Now we worry!  We've made a choice of identification.  Does it agree with all the other possible choices we could have made?  In fact, yes, precisely because $A$ is an isometry.  The metric on $T_{x_j}\mathbb{S}^n$ pulled back by $dA$ is exactly what we would have picked if we had chosen $T_{x_j}\mathbb{S}^n$ to define our metric.
This strategy works whenever $Y = X/\Gamma$, where $X$ is a Riemannian manifold and $\Gamma$ is a discrete group acting freely on $X$ by isometries.  Then the quotient map is a covering and the covering neighborhoods are all isometric, so the choice of metric is natural.  
The general case $Y = X/G$, $X$ Riemannian and $G$ is any group acting by isometries, is more complicated.
