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I see that people say there is no Lie group structure on $S^2$. But $S^2$ can be identified with $SU(2)/U(1)$ by the Hopf fiberation. Since $U(1)$ is also a normal subgroup in $SU(2)$, can't you associate the quotient group structure to $S^2$?

What is wrong with this association? Thanks!

José Carlos Santos
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rice_cake
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    That copy of $U(1)$ is not normal in $SU(2)$. – Max May 17 '20 at 22:39
  • @Max What do you mean by "that copy"? I thought elements of U(1) commute with every element of SU(2), so it ought to be normal? – rice_cake May 17 '20 at 22:49
  • There are many subgroups of $SU(2)$ that are isomorphic to $U(1)$, but none of them are normal. There is a copy of $U(1)$ in $U(2)$ which is central- that's probably what you have in mind. My suggestion would be to write out explicitly the Hopf fibration to see what's going on. – Max May 17 '20 at 22:55
  • @Max Thanks a lot. This helps a lot. – rice_cake May 18 '20 at 02:05

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The group $U(1)$ is not a normal subgroup of $SU(2)$. For instance,\begin{multline}\begin{bmatrix}\frac12+\frac i2&-\frac12+\frac i2\\\frac12+\frac i2&\frac12-\frac i2\end{bmatrix}.\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}.\begin{bmatrix}\frac12+\frac i2&-\frac12+\frac i2\\\frac12+\frac i2&\frac12-\frac i2\end{bmatrix}^{-1}=\\=\begin{bmatrix}\cos(\theta)+i\sin(\theta)&0\\0&\cos(\theta)-i\sin(\theta)\end{bmatrix}\notin U(1).\end{multline}

And people don't just say that there is no Lie group structure on $S^2$. They prove it.

José Carlos Santos
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    Thanks very much. I was initially thinking $U(1)$ as the group of scalar multiplications, but the determinant of those matrices is not even one, so not even in $SU(2)$, so I was certainly wrong. But then if you take $U(1)$ to be the $SO(2)$ subgroup in $SU(2)$, it would not be normal, as you showed. Also thanks for the link for the proof. It is very enlightening. – rice_cake May 18 '20 at 02:09