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Suppose $RP^3$ is the real 3-dimensional projective space, prove the rotation group $SO(3)$ is homeomorphic to $RP^3$.

hbghlyj
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math112358
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2 Answers2

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Each rotation in $\Bbb R^3$ is characterized by an "oriented axis" $v\in S^2$ and an angle $\varphi\in [0,\pi]$ and the only relations are $(v,\pi)=(-v,\pi)$ and $(v,0)=(w,0)$ for each $v,w\in S^2$. If you represent $\Bbb RP^3$ as a 3-ball of radius $\pi$ with identified antipodal points $v\cdot \pi=-v\cdot \pi$ for each $v\in S^2$, then the map $SO(3)\to \Bbb RP^3$ just maps $(v,\varphi)$ to $[v\cdot \varphi]$. The angle $\varphi\,\,\mathrm{mod}\,2\pi$ depends continuously on the rotation and the axis $v$ depends continuously on the rotation whenever $\varphi\neq 0$.

Peter Franek
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  • Hi Peter. I posted a question concerning your answer...https://math.stackexchange.com/questions/3815819/the-proof-of-rp3-cong-so3-by-peter-franek really appreciate it if you could clarify... – No One Sep 06 '20 at 03:13
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The following proposition is incredibly useful here:

Proposition The homomorphism $R: S^{3} \rightarrow SO(3)$ is surjective with $\ker{R}=\{\pm 1\}$ equal to the centre of $S^{3}$. In particular, every matrix of the form \begin{pmatrix} a^2+b^2-c^2-d^2 & 2(bc-ad) & 2(bd + ac)\\ 2(bc+ad) & a^2-b^2+c^2-d^2 & 2(cd-ab) \\ 2(bd-ac) & 2(cd+ab) & a^2-b^2-c^2+d^2 \end{pmatrix} for some $\{a, b, c, d\} \in \mathbb{R}^{4}$ with $a^2+b^2+c^2+d^2=1$. The cosets of $\{\pm1\}$ in $S^{3}$ are simply pairs of antipodal points. Each pair determines a line in $\mathbb{R}^{4}$. the proposition above then determines that $SO(3)=\mathbb{RP}^{3}$ as topological spaces.

Aside.
You can also regard $\mathbb{RP}^{3}$ as the quotient of a solid ball in $\mathbb{R}^{3}$ by identifying antipodal points on the boundary. Every element in $SO(3)$ is a rotation about some axis by some angle $\theta \in [-\pi, \pi]$

blancket
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  • made 2 edits to fix mistakes – wlad May 14 '19 at 16:11
  • @manandlaptop Well spotted, thank you! –  May 15 '19 at 09:49
  • I think the 2nd row in your matrix should be $$[2(bc+ad), a^2-b^2+c^2-d^2, 2(cd-ab)]$$. – Yuval Apr 01 '22 at 05:34
  • The answer here comes with a god given map. Showing SU(2) is three sphere is easy. SU(2) is three dimensional and it act its lie algebra by adjoint action and this gives a representation of SU(2) which image in GL(R,3) is exactly SO(3). You can show that kernel of homomorphism is exactly {I,-I}. This should probably give the map in this answer or something similar to it. I made the calculations many years ago but they are on my old computer when I have them I'll add to here. – erolbarut May 16 '22 at 05:54
  • Is your answer supposed to read "There is a homomorphism $R$" instead of "The homomorphism $R$", and then "every matrix in $\operatorname{SO}(3)$ is of the form" instead of "every matrix of the form"? – LSpice May 24 '25 at 20:28