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The standard metric on $RP^n$ is usually defined to be the metric that locally looks like the metric on $S^n$. But as a differentiable manifold (and not just as a set), $RP^n$ is not a subset of $S^n$, it is a quotient. So there is no natural map $RP^n\to S^n$, but rather a map $S^n\to RP^n$.

The standard metric on $RP^n$ would then be a sort of "push-forward metric", but there is no such a thing, metrics can only naturally be pulled back.

What am I missing?

glS
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geodude
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    I'm not sure if this is how it is done, but note that the quotient $S^n \to RP^n$ is a local diffeomorphism. So what you can do is for $x,y \in T_pRP^n$ define $(x,y)p= (x_1,y_1){p_1} + (x_2,y_2){p_2}$ with ${p_1, p_2}$ the preimage of $p$, and the vectors $x_i,y_i$ the unique preimages of $x,y$ in $T{p_i}S^n$. – s.harp Jan 25 '16 at 12:12
  • @s. harp, note that $p_1$ and $p_2$ has the same tangent space(consider $S^n$ to be embeded in $\mathbb{R}^n$). So there is no need to take sum of the two. – Kaushik Mar 18 '17 at 17:10
  • related: https://math.stackexchange.com/q/134836/173147 – glS May 25 '21 at 01:13

1 Answers1

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The point is that the metric on $S^n$ is invariant under the action of the group $\mathbb{Z}_2$, so it can be pushed down to the quotient. More explicitly, assume that we endow $S^n$ with a standard round metric and let $\pi \colon S^n \rightarrow \mathbb{RP}^n$ be the quotient map. For $p \in S^n$, we denote $\pi(p)$ by $[p]$. Let $A \colon S^n \rightarrow S^n$ be the antipodal map. Then $A$ is an isometry of $S^n$ and we have $\pi \circ A = \pi$.

The metric on $\mathbb{RP}^n$ is defined as

$$ \left< v, w \right>_{[p]} := \left< \left( d\pi|_p \right)^{-1}(v), \left( d\pi|_p \right)^{-1}(w) \right>_p. $$

This is well-defined because we have

$$ \left< \left( d\pi|_p \right)^{-1}(v), \left( d\pi|_p \right)^{-1}(w) \right>_p = \left< \left( d \left( \pi \circ A \right)|_p \right)^{-1}(v), \left( d \left( \pi \circ A \right)|_p \right)^{-1}(w) \right>_p = \left< \left( dA|_{p} \right)^{-1} \left( \left( d \pi|_{-p} \right)^{-1}(v) \right), \left( dA|_{p} \right)^{-1} \left( \left( d \pi|_{-p} \right)^{-1}(v) \right) \right>_p \\ = \left< \left( d\pi|_{-p} \right)^{-1}(v), \left( d\pi|_{-p} \right)^{-1}(w) \right>_{-p} $$

where we used the fact that $dA|_p$ is an isometry.

More generally, assume you have some Riemannian manifold $(M,g)$ with a Lie group $G$ acting on $M$ smoothly, freely and properly by isometries. Then the quotient $M / G$ is a manifold and it has a naturally and uniquely defined Riemannian metric that turns the map $\pi \colon M \rightarrow M / G$ into a Riemannian submersion. If $G$ is zero dimensional, as in your case, the map $\pi$ is a local isometry.

levap
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    Clear, and right to the point. Thank you! – geodude Jan 25 '16 at 13:11
  • How is the metric on $S^n$ invariant under the action of the group $\mathbb Z_2$? Antipodal points go from maximally far to distance zero! – Snared Jun 30 '23 at 05:45
  • @Snared the round metric in a sphere is a Riemannian metric, not a classical Hausdorf metric. That means it doesn’t tell you the distance between finitely separated points of the space. Instead it tells you an inner product on the tangent spaces, which you might think of as distance between infinitesimally separated points. The antipodal map does join antipodal points, but it doesn’t change infinitesimal distances. It is a local isometry. You get a finite separation Hausdorf metric from a Riemannian metric by integrating. – ziggurism Sep 30 '23 at 20:18