# Prove that $(\mathfrak{su}(2))^* \cong \mathfrak{sb}(2)$

Let $\mathfrak{su}(2)$ be the Lie algebra with basis elements $$e_1=\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} , \quad e_2=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} , \quad e_3=\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} ,$$ and $\mathfrak{sb}(2)$ to be the Lie algebra with basis elements $$f_1=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} , \quad f_2=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} , \quad f_3=\begin{pmatrix} 0 & i \\ 0 & 0 \end{pmatrix} .$$

I am currently calculating the coadjoints of two general elements of $\mathfrak{su}(2)$ and two general elements of $\mathfrak{sb}(2)$ and trying to show that Equation (ii'') on page 10 holds.

Is this the correct method, or is there a simpler way to tackle this?

Yes, there is a simpler way to tackle this, namely in the same way as the author tackles it for $\mathbb{sl}_2(\mathbb{C})$ on page $6$, and on page $12$, $13$, using the Killing form. The result is very similar; the brackets for $(\mathbb{sl}_2(\mathbb{C}))^*$ are $[f_1,f_2]=\frac{1}{4}f_2$, $[f_1,f_3]=-\frac{1}{4}f_3$ and $[f_2,f_3]=0$. This is, up to a scalar, just the same as $\mathfrak{sb}(2)$.