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This is an exercise from the Kirillov's book on Lie Algebras and Lie Groups.

Define a basis in $\mathfrak{su}(2)$ by $$ i\sigma_1 = \begin{pmatrix} 0& i\\ i& 0 \end{pmatrix},\quad i\sigma_2 = \begin{pmatrix} 0 &1\\ −1 &0 \end{pmatrix},\quad i\sigma_3 = \begin{pmatrix} i&0\\ 0 &−i\end{pmatrix} $$ Show that the map $\varphi: SU(2) → GL(3,\mathbb{R})$ sending $g$ to the matrix of $\mathrm{Ad} ~g$ in the basis $i\sigma_1,i\sigma_2,i\sigma_3$ gives a morphism of Lie groups $SU(2) → SO(3,\mathbb{R})$. (I have solved this one.)

Compute explicitly the map of tangent spaces $\varphi_{\ast}: \mathfrak{su}(2) →\mathfrak{so}(3,\mathbb{R})$ and show that $\varphi_{\ast}$ is an isomorphism.

Prove that the associated Lie algebra homomorphism $$\mathfrak{su}(2)\rightarrow\mathfrak{so}(3) $$ is isomorphism.

It is well-known, but how to prove it?

Especially, I want to know this isomorphism explicitly.

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So, if $U\in SU(2)$, then $\varphi(U)=\operatorname{Ad}U$. So, what is $\varphi(e^{ti\sigma_1})$? Since\begin{align}\varphi(e^{ti\sigma_1})(i\sigma_1)&=\begin{pmatrix}\cos t&i\sin t\\i\sin t&\cos t\end{pmatrix}\begin{pmatrix}0&i\\i&0\end{pmatrix}\begin{pmatrix}\cos t&-i\sin t\\-i\sin t&\cos t\end{pmatrix}\\&=i\sigma_1,\end{align}since\begin{align}\varphi(e^{ti\sigma_1})(i\sigma_2)&=\begin{pmatrix}\cos t&i\sin t\\i\sin t&\cos t\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}\cos t&-i\sin t\\-i\sin t&\cos t\end{pmatrix}\\&=\begin{pmatrix}-i\sin(2t)&\cos(2t)\\-\cos(2t)&i\sin(2t)\end{pmatrix},\end{align}and since\begin{align}\varphi(e^{ti\sigma_1})(i\sigma_3)&=\begin{pmatrix}\cos t&i\sin t\\i\sin t&\cos t\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}\cos t&-i\sin t\\-i\sin t&\cos t\end{pmatrix}\\&=\begin{pmatrix}i\cos(2t)&\sin(2t)\\-\sin(2t)&-i\cos(2t)\end{pmatrix},\end{align}we just have to differentiate this at $t=0$, in order to obtain $\varphi_*$. Then we get that $\varphi_*(i\sigma_1)(i\sigma_1)=0$, that $\varphi_*(i\sigma_1)(i\sigma_2)=\left(\begin{smallmatrix}-2i&0\\0&2i\end{smallmatrix}\right)=-2i\sigma_3$, and that $\varphi_*(i\sigma_1)(i\sigma_3)=\left(\begin{smallmatrix}0&2\\-2&0\end{smallmatrix}\right)=2i\sigma_2$.

Now, do the same thing in order to compute $\varphi_*(i\sigma_2)$ and $\varphi_*(i\sigma_3)$.

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  • $\begingroup$ @LeoGardon You're welcome. $\endgroup$ – José Carlos Santos Sep 22 '17 at 15:22

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