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show that the 2-sphere is a Lie group. I do not know with which binary action is the 2-sphere a Lie group? and with which binary action is the $\mathbb{R^3}-{0}$ a Lie group?

2-sphere is equal with $\{(x,y,z)\in \mathbb{R^3}; x^2 +y^2+ z^2 =1 \}$

user
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  • s-sphere means the 2-dimensional surface in $\mathbb{R^3}$ – user Jun 23 '14 at 09:20
  • See for example this question or the highest rated Related question on the list on the right margin. – Jyrki Lahtonen Jun 23 '14 at 09:28
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    The spheres $S^n$ with their standard smooth structure are Lie groups only in dimensions $1$ and $3$, $S^2$ is not a Lie group under any binary operation. – Peter Franek Jun 23 '14 at 10:10
  • Why? Can you introduce a reference for your answer?Is the $\mathbb{R}^3−{0}$ a lie group? – user Jun 23 '14 at 10:49
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    No, $\mathbb{R}^3\setminus{0}$ is not a Lie group either. The reason for $S^2$ is that if it were a Lie group, you could take a tangent vector $v$ in $T_e S^2$ (the tangent space in the identity element) left multiplication $L_g: x\mapsto g\cdot x$ would take it to a nonzero vector $(L_g)_∗(v)$ in the tangent space of any element $g$. But there doesn't exists a continuous nowhere zero vector fields on $S^2$, see the Hairy ball theorem. – Peter Franek Jun 23 '14 at 12:30
  • The reference to Hairy ball theorem is for example in Milnor, Topology from differentiable viewpoint, Chapt. 6, or Hatcher, Algebraic Topology, Thm 2.28. The fact that $S^1$ and $S^3$ are the only Lie group spheres is harder. You might have a look here.. – Peter Franek Jun 23 '14 at 12:31
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    ..to be honest, I don't know how to prove the $\mathbb{R}^3\setminus{0}$ case. Would be interested, if somebody more experienced comments.. – Peter Franek Jun 23 '14 at 13:19

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