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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?

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    $\begingroup$ 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 $T_{p_i}S^n$. $\endgroup$ – s.harp Jan 25 '16 at 12:12
  • $\begingroup$ @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. $\endgroup$ – Kaushik Mar 18 '17 at 17:10
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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.

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  • $\begingroup$ Clear, and right to the point. Thank you! $\endgroup$ – geodude Jan 25 '16 at 13:11

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