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 Post subject: disjoint circles
PostPosted: Wed, 12 May 2010 17:53:53 UTC 
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We all know that S^3 can be split into a disjoint union of circles by viewing S^3\subseteq\mathbb{C}^2 as the unit sphere.

How about \mathbb{R}^3?

What if 'circles' mean congruent circles?

(These results are probably well-known among certain circles, pun intended.)

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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