3

As a training problem I tried to explicitly compute the isomorphism between the Lie algebra of $S^3$ and $\mathbb{R}^3$ with the cross product. I used the following notations: Let $U=\{x=(x_0,x_1,x_2,x_3)\in S^3)\ |\ x_0>0\}$ and let $\varphi:U\to \mathbb{R}^3,\ x\mapsto (x_1,x_2,x_3)$. After a fairly long calculation, I arrived at the following: if we identify $\operatorname{Lie}(S^3)$ with $T_{e}S^3$ in the usual way, then we have $$ \left[\frac{\partial}{\partial\varphi_i}|_e,\frac{\partial}{\partial\varphi_{i+1}}|_e\right]=2\frac{\partial}{\partial\varphi_{i+2}}|_e $$ where $\varphi=(\varphi_1,\varphi_2,\varphi_3)$ and the indices are taken modulo 3. What surprises me about this is the presence of the factor of $2$, which makes it so that the searched isomorphism $\operatorname{Lie}(S^3)\cong\mathbb{R}^3$ is given by $$ f:\operatorname{Lie}(S^3)\cong\mathbb{R}^3\\ \frac{\partial}{\partial\varphi_1}|_e\mapsto (2,0,0),\ \frac{\partial}{\partial\varphi_2}|_e\mapsto (0,2,0),\ \frac{\partial}{\partial\varphi_3}|_e\mapsto (0,0,2), $$ but intuitively I would have thought that in the isomorphism there is no such factor of $2$. Have I just done an error somewhere in my calculations? Or is it indeed true that by some coincidence there is a factor of $2$ in the isomorphism?

Thanks a lot!

Redundant Aunt
  • 12,473

1 Answers1

3

The Lie algebra of $S^3$ is $\mathfrak{su}(2)$, and the Lie algebra for $\Bbb R^3$ with the cross product is $\mathfrak{so}_3(\Bbb R)$. And of course there is a Lie algebra isomorphism $f\colon \mathfrak{su}(2)\rightarrow \mathfrak{so}_3(\Bbb R)$ with $f(\sigma_i/2)=e_i$, where $\sigma_i$ are the Pauli matrices - see here.

Dietrich Burde
  • 140,055