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According to

Algebraic variety and real manifold

lots of manifolds can be realized as the real points of an affine variety. It is obvious how to do this for the sphere. How about for slightly more complicated manifolds like projective space? In particular, does anyone have an example of a affine algebraic variety whose real points are the real projective plane? Is the minimal dimension (number of affine coordinates) here five?

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    See https://math.stackexchange.com/questions/2158230/embedding-rp2-into-r4 for instance. – KReiser Feb 18 '22 at 02:17
  • @KReiser thanks for the comment! The link you provided gives a lovely regular map whose image is $ \mathbb{R}P^2 $. That's awesome! Do you have in mind a particular set of polynomials in four variables such that the affine subvariety of $ \mathbb{R}^4 $ that they define is the image of the regular map you linked to? Or is there some well known way, given a regular map, of finding polynomials such that the affine subvariety they define is the image of that regular map? If so I would love to know about that! – Ian Gershon Teixeira Feb 18 '22 at 22:18
  • Sure, I've explained how one might start discovering these things and provided a calculation of the explicit equations for that particular embedding below. – KReiser Feb 20 '22 at 09:42

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Since $\Bbb RP^n$ is the quotient of $S^n$ under the group action by $\Bbb Z/2$ identifying antipodes, there's a reasonably easy solution. Because $k[V/G]=k[V]^G$, where the superscript denotes invariants, we can just calculate the invariants of the group action. The group action on $\Bbb R[x_1,\cdots,x_n]/(1-\sum x_i^2)$ is by $x_i\mapsto -x_i$, which has algebra of invariants generated by the degree-2 polynomials. This is clearly a finitely-generated $\Bbb R$-algebra, so choosing generators and computing relations, one can find the variety you request.

In the case $n=2$, you get the coordinate algebra $$\Bbb R[x_1^2,x_1x_2,x_1x_3,x_2^2,x_2x_3,x_3^3]/(x_1^2+x_2^2+x_3^2-1)$$ which can be written as $\Bbb R[y_1,\cdots,y_6]/I$ for $y_1=x_1^2$, $y_2=x_2^2$, $y_3=x_3^2$, $y_4=x_1x_2$, $y_5=x_1x_3$, $y_6=x_2x_3$ where $I$ is generated by $y_1+y_2+y_3-1$ and all the appropriate quadratic relations of the form $y_iy_j-y_ky_l$ like $y_1y_2-y_4^2$. (These are somewhat annoying to write out, please forgive me for not doing all of them.) This is not the minimal embedding, though - for instance, making the substitution $y_3=1-y_1-y_2$, we can get down to 5 variables, showing that $\Bbb RP^2$ can be represented as the real points of some variety in $\Bbb R^5$.

This isn't great for trying to find the minimal dimension, though - we're embedding $\Bbb RP^n$ in $\Bbb R^{\binom{n+1}{2}-1}$. We know that for any smooth affine variety of dimension $n$, there's an embedding in affine space of dimension $2n+1$ - projecting from a point which isn't on any line through our variety with target a hyperplane produces a map which restricts to an isomorphism on our variety but drops the dimension of the ambient space by 1. (This is roughly the analogue of the Whitney embedding theorem from topology.) In the case $n=2$, it happens that we can actually project one further time: this is related to some very special properties of the Veronese surface. Topological properties show we can't do any better: $\Bbb RP^2$ does not embed in $\Bbb R^3$ for orientability reasons.

The link from the comments gives an example of a polynomial map $\Bbb R^3\to\Bbb R^4$ by $(x,y,z)\mapsto (x^2-y^2,xy,xz,yz)$ which when restricted to $S^2\subset\Bbb R^3$ gives a double covering of its image and therefore a polynomial embedding of $\Bbb RP^2$ in to $\Bbb R^4$. You've asked how to calculate the equations for the image. The techniques for this belong to an area called elimination theory. Briefly, what we're looking to do is to compute the inverse image of the ideal $(x^2+y^2+z^2-1)$ under the ring map $\Bbb R[a,b,c,d]\to \Bbb R[x,y,z]$ sending $a\mapsto x^2-y^2$, $b\mapsto xy$, $c\mapsto xz$, $d\mapsto yz$. After a little computer assistance, we find that the ideal in question is $$(a^2d + 4b^2d + c^2d + d^3 - 2bc + ad,\\ bc^2 - acd - bd^2,\\ 2b^2c - abd + cd^2 - bd,\\ abc + 2b^2d + c^2d - bc,\\ a^2c + c^3 + 2abd - cd^2 - ac,\\ a^2b + 4b^3 + acd + 2bd^2 + cd - b,\\ a^3 + 4ab^2 + ac^2 + ad^2 + c^2 - d^2 - a).$$ I hope one additional thing you take from this is that sometimes looking for explicit equations can be less convenient than knowing you have a variety by other means.

KReiser
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  • Nice description, I have tried to find others maps and to find the the equations of their image. Give a look here if you are interested: https://math.stackexchange.com/questions/4627844/the-projective-plane-as-a-smooth-surface-in-a-4-dimensional-space?noredirect=1#comment9773712_4627844 However, it seems that I only need 6 equations instead of 7, as your computation suggested. Is there some problem? – Federico Fallucca Feb 04 '23 at 21:28