Proving algebraically that $\mathbb RP ^3\cong SO(3,\mathbb R)$ I am giving a simple introductory course on algebraic geometry and I plan to mention that 
$$\mathbb RP ^3\cong SO(3,\mathbb R).$$
I know a rather simple proof of this  using the fact that $\mathbb RP^3$ is topologically a $3$-dimensional ball with an identification of opposite points on the boundary.
I would like to know if one can prove this fact in some symmetric "algebraic way".
$\bf Added.$ Giulio Bresciani made a good remark, that $SO(3,\mathbb R)$ is naturally an affine variety, and since $\mathbb RP ^3$ is projective, they are not isomorphic. In order to rectify this situation, let as compactify $SO(3,\mathbb R)$ by adding to it points on infinity (an empty set), by adding to $\mathbb R^9$ (where $SO(3,\mathbb R)$ sits) the space $\mathbb RP^8$. 
The new question is then: is there some nice birational map from $\mathbb RP^3$ to $SO(3,\mathbb R)$, which is as well a diffeo on the set of points.
 A: Correct me if I'm wrong, but I think they are NOT isomorphic, no more than $\mathbb{RP}^1$ and the unitary circle in the real plane. In fact, $\operatorname{SO}(3,\mathbb{R})$ is affine, and $\mathbb{RP}^3$ is not. You have non constant global functions on $\operatorname{SO}(3,\mathbb{R})$, for example the coordinates of the matrix.
A: The 'isomorphism'
$$\mathbb{RP}^3≅SO(3,\mathbb{R})$$
is true as a manifold morphism (actually a smooth manifold morphism), but not as an algebraic morphism. This means that the diffeomorphism between the two varieties can't be expressed locally by rational functions. On $\mathbb{RP}^3$ we have the standard structure given by $4$ charts choosing one of the $4$ coordinates to be $1$ (the change of coordinates between the charts is algebraic, hence this all make sense), while we can see $SO(3,\mathbb{R})$ as a smooth closed subvariety of the space of $3\times3$ matrices, $\mathbb{R}^9$ (all the equations defining $SO(3,\mathbb{R})$, $AA^T=\operatorname{Id}$ and $\operatorname{det}(A)=1$ are polynomial). They can't be isomorphic because the only algebraic functions on $\mathbb{RP}^3$ are the constants, while on $SO(3,\mathbb{R})$ we have the coordinates that are not constant.
