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We can realize $\mathbb{RP}^1$ as $S^1 \subset \mathbb{R}^2$ defined by the set $$ \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \} $$ Does there exist a nice description of $\mathbb{RP}^n$ as a subset of $\mathbb{R}^m$ similar to the one for the $n=1$ case?

54321user
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    If you don't mind about optimal dimension, I think the veronese embedding of degree 2 would yield such a set of defining equations for any $n$. – cjackal May 26 '17 at 06:29
  • Sure I don't care. Do you want to write it out explicitly for $\mathbb{RP}^2$? I will give you the check – 54321user May 26 '17 at 06:31
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    If you'll forgive a good-natured nitpick: The projective line is "really" the circle $x^{2} - x + y^{2} = 0$, i.e., the polar graph $r = \cos\theta$. The unit circle is the sphere; the $2$-to-$1$ map implemented by the polar coordinates parametrization is constant-speed (!), hence an isometric covering, see diagram at Flag manifold to Complex Projective line. – Andrew D. Hwang May 26 '17 at 11:38

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Caveat. This is far from optimal.

Let me try to embed $\mathbb{R}P^2$. We know from veronese embedding that $[x:y:z]\mapsto (x^2,y^2,z^2,xy,yz,zx)\in\mathbb{R}^6$ is an embedding.

(Here, I normalized the projective coordinates so that $x^2+y^2+z^2=1$.)

Then, if we write the generic coordinates in $\mathbb{R}^6$ to be $(x_{00},x_{11},x_{22},x_{01},x_{12},x_{02})$, the image lies in the locus of the equations $x_{ij}x_{kl}=x_{ik}x_{jl}, \forall i,j,k,l$. (Let's say for convenience that $x_{ij}=x_{ji}$ here.)

And as we normalize the coordinates, we have another identity: $x_{00}^2+x_{11}^2+x_{22}^2=1$.

Now, the usual proof of showing the image of the veronese embedding is the locus of the above relations can be applied to show these relations are all.

cjackal
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