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Starting on some links on MSE -- e.g. $RP^3\cong SO(3)$ it is clear to me the intuitive reason why $SO(3)$ is identified with $RP^3$.

From a formal point of view, how is $SO(3)$ topologized ? Can we define it applying quotient topology to $S^3$ ball ?

CarloC
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    Just the subspace topology of $SO(3)\subset M_{3\times 3}(\mathbb R)\simeq\mathbb R^9$. – Just a user Apr 03 '24 at 07:28
  • Ok, so we topologize $SO(3)$ via subspace topology from $\mathbb R^9$. Then it can be shown that such a topological space is actually homeomorphic to $RP^3$ -- i.e. with $B^3$ ball with antipodal points on the ball's surface identified. – CarloC Apr 03 '24 at 08:26
  • Can you provide explicitly an homemorphism between $SO(3)$ (topologized via subspace topology from $\mathbb R^9$) and $RP^3$ ? Thanks. – CarloC Apr 03 '24 at 09:14
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    The question you link contains a link to another question, which has an answer that produces an explicit homeomorphism. And there's more in other questions linked to that one in turn. – Thorgott Apr 03 '24 at 11:33

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Just a few words about rigor.

As said in the comment, the topology on $SO(3)$ should be the subsapce topology induced from the inclusion $$SO(3)\subset M_{3\times 3}(\mathbb R)\simeq \mathbb R^9$$

As a subspace, it's still Hausdorff. It's easy to see it's both closed and bounded, hence compact. Now given the explicit mapping from $S^3$ to $SO(3)\subset\mathbb R^9$ which is clearly continuous (all entries are just polynomials), and one can check that it's surjective. Therefore as both of $S^3$ and $SO(3)$ are compact Hausdorff, this is a quotient map.

(To drive the formula, we can combine the idea from the linked post and the Rodrigues' rotation formula. Maybe there is an easier way - such as using $SU(2)$ as the bridge mentioned in the comment, but anyway I won't bother here.)

Just a user
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  • How can we show that $SO(3)$ as subspace of $\mathbb R^9$ is closed and bounded ? – CarloC Apr 03 '24 at 20:31
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    It's closed because it can be described by polynomial equations. Bounded is also easy: every row has Euclidean norm $1$, hence bounded. – Just a user Apr 04 '24 at 01:16
  • As far as I can tell, the explicit mapping results in quotient map from $S^3$ to $SO(3)$. Which is the explicit homeomorphism between $SO(3)$ and $RP^3$ ? – CarloC Apr 04 '24 at 08:03
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    This shows there is a homeomorphism between a quotient of $S^3$ and $SO(3)$, and by definition, the quotient is also $\mathbb RP^3$. – Just a user Apr 04 '24 at 09:39