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Is there a simple way to see that $SO(4) \simeq SO(3)\bigotimes SO(3) /\mathbb{Z}_2$ ?

DanielC
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ahala
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    I've never come up with an intuitive way to see this. Far easier is to see that $SO(4) = S^3 \times SO(3)$, the argument being this: take a $4 \times 4$ matrix in $SO(4)$ and send it to its first column (an element of $S^3$). That's a fibration, and the fiber over the "north pole" (1,0,0,0) is just the set of 4x4 orthogonal matrices whose first col is (1,0,0,0), i.e., matrices whose lower right 3x3 matrix is in SO(3). So you have an SO(3)-bundle over $S^3$. Is it trivial? Yes! Because the clutching function is in $\pi_2(SO(3))$, which is $0$. – John Hughes Dec 09 '13 at 22:07
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    I think this will mostly answer your question: http://math.stackexchange.com/questions/3646/recovering-the-two-su2-matrices-from-so4-matrix – Seub Dec 09 '13 at 22:21
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    This question may also be helpful. – Jim Belk Dec 09 '13 at 22:51

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Since $SO(3)$ is related to $SU(2)$ I recommend this paper : More on the Isomorphism $SU(2)\otimes SU(2) \simeq SO(4)$ by Kazuyuki Fujii , Hiroshi Oike and Tatsuo Suzuki.

draks ...
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