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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.

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  • $\begingroup$ Quaternions would give you an operation of a 'line' in $\mathbb R^4$ on a suitable $\mathbb R^3$ $\endgroup$
    – Blah
    Jan 21, 2014 at 12:59
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    $\begingroup$ This question seems to be causing some confusion because you have not made it clear whether you are talking about isomorphisms of manifolds or algebraic varieties. (The tags suggest the latter, but you need to be explicit.) $\endgroup$
    – user64687
    Jan 21, 2014 at 14:40
  • $\begingroup$ I'm not sure that SO3 is a projective variety, as it's defining equation in $ℝ^9$ is not homogeneous ($A^TA-I=0$ has three non-homogeneous factors) $\endgroup$
    – rewritten
    Jan 22, 2014 at 10:08

2 Answers 2

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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.

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    $\begingroup$ And by "affine" you mean "non compact" (in the manifold picture). $\endgroup$ Jan 21, 2014 at 14:12
  • $\begingroup$ Yes, I think that "non compact" is the nearest intuitive concept. $\endgroup$ Jan 21, 2014 at 14:13
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    $\begingroup$ @MartinBrandenburg: I think you are asking a little too much of those quotation marks... $\endgroup$
    – user64687
    Jan 21, 2014 at 14:18
  • $\begingroup$ This is a good point. At the same time I want a positive answer to my question. So I will change it a bit. I will consider $\mathbb SO(3,\mathbb R)$ as projective and not affine, i.e. I want to compactify $\mathbb R^9$ where $SO(3,\mathbb R)$ is sitting, to an $\mathbb RP^9$ by adding points at infinity. Indeed a circle in $\mathbb R^2$ can be also looked at as a conic in $\mathbb RP^2$. Such a conic is of course isomorphic to $\mathbb RP^1$. $\endgroup$
    – agleaner
    Jan 21, 2014 at 16:30
  • $\begingroup$ @agleaner: I don't know if it works, at least from a classical point of view. For how my professor defined things when I took my first AG course, it doesn't change anything if you add some points to $\mathbb{R}^9$, there is still an obvious isomorphism between the new and the old SO3. Things are different if you look this from the point of view of schemes (but I don't think you want to do these things in an introductory course). If you talk about schemes, when you projectivize you are adding some (complex) points to SO3 changing it's structure, and this new SO3 is "compact" (proper). $\endgroup$ Jan 22, 2014 at 23:45
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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.

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    $\begingroup$ $\mathbb{RP}^3$ has inherent algebraic structure en.wikipedia.org/wiki/Algebraic_geometry_of_projective_spaces $\endgroup$ Jan 21, 2014 at 14:11
  • $\begingroup$ Yes, you are right. And the quaternions provide an explicit description of it. Still, I don't think that it would be good for an introductory course in AG. That said, I would downvote my answer too if I could. $\endgroup$
    – rewritten
    Jan 21, 2014 at 14:24
  • $\begingroup$ Well, if edited, it is a good first example that algebraic geometry$\neq$topology $\endgroup$ Jan 21, 2014 at 14:27
  • $\begingroup$ After the editing the answer is better, but it is still missing the point. Both spaces have an algebraic structure, and the fact that one of them is a Lie group or that neither of them has critic points is not relevant. You can compare them, and the result of the comparison is that they are not isomorphic with respect to the algebraic structure. Still, it's not only a matter of Zariski topology. For example, with the Zariski topology the real line is homeomorphic to the unitary circle, but they are not isomorphic as algebraic varieties. $\endgroup$ Jan 22, 2014 at 2:21
  • $\begingroup$ @GiulioBresciani please edit, I'll mark the answer as community. $\endgroup$
    – rewritten
    Jan 22, 2014 at 9:59

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