Projective plane - equation in $\Bbb R^5$ I've just read that the projective plane $\Bbb RP^2$ is a surface equivalent to the set of $(x,y,u,v,w) \in \Bbb R^5$ such that
$$u^2=xy,v^2=x(1-x-y),w^2=y(1-x-y)$$
How is it true? I don't see how to get this result, even looking at questions like these : Give an explicit embedding from $\mathbb{R}P_2$ to $\mathbb{R}^4$, Injective map from real projective plane to $\Bbb{R}^4$, embedding projective plane in 4-space?.
Thank you!
 A: I'm not sure what is meant by "equivalent", but it is not as smooth manifolds.  More specifically, the set $$X=\{(x,y,u,v,w): u^2 = xy, v^2 = x(1-x-y), w^2 = y(1-x-y)\}$$ is not a smoothly embedded submanifold of $\mathbb{R}^5$.
To see this, consider the point $p = \left(0,\frac{1}{2},0,0,\frac{1}{2}\right)\in X$.  I claim that $T_p X$ contains the three vectors $(0,1,0,0,0)$, $(0,0,1,1,0)$ and $(0,0,1,-1,0)$.  Thus, if $T_p X$ is a vector space, it's a least 3-dimensional.  On the other hand, we will find a path $\alpha(t)$ in $X$ connecting $p$ to another point $q\in X$ where the tangent space is $2$-dimensional.  Thus, on a single connected component of $X$ we find it's $2$ dimensional and at least $3$ dimensional, a contradiction.
So, let's argue that $T_p X$ contains those three vectors.
To do that, let $\gamma(t) = (0,t,0,0,\sqrt{t-t^2})$.  Then one can easily verify that $\gamma\left(\frac{1}{2}\right) = p$, $\gamma(t)\in X$ for all $t$ it's defined, and $\gamma'\left(\frac{1}{2}\right) = (0,1,0,0,0)$.
For the other two, consider $\alpha_{\pm}(t) = \left( 2t^2, \frac{1}{2}, t, \pm t\sqrt{1-4t^2}, \sqrt{\frac{1}{4}-t^2}\right)$.  Then $\alpha_{\pm}(0) = p$, and $\alpha_{\pm}(t)\in X$ whenever it's defined, and $\alpha'(0) = (0,0,1,\pm 1,0)$.
Finally, for either choice of $\alpha$, note that for small non-zero $t$, all components are non-zero.  So $\alpha$ is a path in $X$ connecings $p$ to a point where are coordinantes are non-zero.
Now, one can compute the jacobian:  One gets $\begin{bmatrix}-y & -x & 2u & 0 & 0\\ 2x + y-1 & x & 0 & 2v & 0 \\ y & 2y+x-1 & 0 & 0 & 2w\end{bmatrix}$.  In particular, if $u$, $v$, and $w$ are all non-zero, this matrix has full rank.  Thus, at such points, the tangent space of $X$ is $2$-dimensional.
