A little background knowledge. We know that the imaginary quaternions $\mathrm{Im}\mathbb{H} = \mathrm{span}\{\sigma_1, \sigma_2, \sigma_3\}$and $\mathbb{R}^3$ are lie algebra isomorphims, i.e.
$$
(\mathrm{Im} \mathbb{H}, [ , ]) \simeq (\mathbb{R}^3, \times),
$$
where $\sigma_\alpha$ denotes the Pauli-matrices. The identification
$$
X = -i\sum\limits_{\alpha=1}^{3}X_\alpha \sigma_\alpha \in \mathrm{Im}\mathbb{H} \longleftrightarrow X = (X_1,X_2,X_3) \in \mathbb{R}^3
$$
provides us with the following matrix reprensentation
$$
X = \begin{pmatrix}-iX_3&-iX_1-X_2\\-iX_1 + X_2&iX_3\end{pmatrix}.
$$
Now i have a matrix
$$
U = \frac{2u}{\beta^2}\begin{pmatrix}-i(u+u^{-1})&-i\bar{a}\\-ia&i(u+u^{-1})\end{pmatrix} \in \mathrm{Im}\mathbb{H},
$$
where $u$ is a positive real-valued function, $a$ a complex-valued function and $\beta^2 = 2 + \left|a\right|^2 + u^2 + u^{-2}$.
I am having a hard time to transform $U$ into a vector in $\mathbb{R}^3$ by using this identification from above.
I know that $X_3 = u + u^{-1}$. But I don't know how to obtain $X_1$ and $X_2$, respectively , since we have different entries on the off-diagonals. In the end I want to calculate the norm of the desired vector in $\mathbb{R}^3$. The desired result should be
$$
\left|U\right|_{\mathbb{R}^3} = \frac{2u}{\beta}
$$
Any suggenstions?
Thanks in advance :)
By visual comparison, $X_3 = \dfrac {2u} {\beta^2} (u + u^{-1})$. If $a = A + \Bbb i B$, then
$$- \Bbb i X_1 + X_2 = \frac {2u} {\beta^2} (- \Bbb i a) = \frac {2u} {\beta^2} [- \Bbb i (A + \Bbb i B)] = - \Bbb i A \frac {2u} {\beta^2} + B \frac {2u} {\beta^2} \ ,$$
which gives $X_1 = A \dfrac {2u} {\beta^2} = \Re (a) \dfrac {2u} {\beta^2}$ and $X_2 = B \dfrac {2u} {\beta^2} = \Im (a) \dfrac {2u} {\beta^2}$.
Therefore, the isomorphism is
$$U \mapsto \frac {2u} {\beta^2} \big( u + u^{-1}, \Re (a), \Im (a) \big) \ .$$