The map from $SU(2) \times SU(2)$ to $SO(4)$

Question

Lie group $SO(4)$ is doubly covered by $SU(2) \times SU(2)$, I want to know the map from $SU(2) \times SU(2)$ to $SO(4)$.

The map from $SU_{2}$ to $SO(3)$ is $\begin{pmatrix} \alpha & \beta \\-\overline{\beta} & \overline{ \alpha} \end{pmatrix} \longrightarrow \begin{pmatrix} \frac{1}{2}(\alpha^{2}-\beta^{2}+\overline{\alpha}^{2}-\overline{\beta}^{2} & \frac{i}{2}(-\alpha^{2}-\beta^{2}+\overline{\alpha}^{2}+\overline{\beta}^{2} & -\alpha\beta-\overline{\alpha}\overline{\beta}\\\frac{i}{2}(\alpha^{2}-\beta^{2}-\overline{\alpha}^{2}+\overline{\beta}^{2} & \frac{1}{2}(\alpha^{2}+\beta^{2}+\overline{\alpha}^{2}+\overline{\beta}^{2} & -i(\alpha\beta-\overline{\alpha}\overline{\beta})\\ \alpha\overline{\beta}+\overline{\alpha}\beta & i(-\alpha\overline{\beta}+\overline{\alpha}\beta)& \alpha\overline{\alpha}-\beta\overline{\beta} \end{pmatrix}.$ Therefore what is the image of

$(\begin{pmatrix} \alpha & \beta \\-\overline{\beta} & \overline{ \alpha} \end{pmatrix},\begin{pmatrix} \alpha' & \beta' \\-\overline{\beta}' & \overline{ \alpha}' \end{pmatrix})\in SU_{2}\times SU_{2}$

Answer 1

Here's a sketch of the construction; I'll leave the (more or less straightforward) algebraic details to the reader.

Let $\Bbb V$ be the $4$-dimensional (real) vector space of complex $2 \times 2$ matrices $z \in M(2, \Bbb C)$ satisfying $$z^* = J z^{\top}J^{-1}, \qquad J := \pmatrix{0&-1\\1&0} .$$

Now, realize $SU(2)$ as the group of matrices $g \in M(2, \Bbb C)$ that satisfy $g^*g = \Bbb I$, and define an action of $SU(2) \times SU(2)$ on $\Bbb V$ by $$(g, h) \cdot z := g z h^* .$$ This action preserves the positive-definite bilinear form $$\langle z, w \rangle := \operatorname{Re}\operatorname{tr}(z w^*)$$ on $\Bbb V$ and so defines a map $SU(2) \times SU(2) \to SO(\Bbb V) \cong SO(4)$; this is the desired double cover.

Computing in a basis for $\Bbb V$---the basis $\left(\pmatrix{1&0\\0&1}, \pmatrix{i&0\\0&-i}, \pmatrix{0&1\\-1&0}, \pmatrix{0&i\\i&0}\right)$ is both convenient for computation and (up to an overall rescaling) orthonormal with respect to the bilinear form---gives an explicit formula analogous to that for the map $SU(2) \to SO(3)$ in the question statement.

For more details and similar constructions of the other sporadic coverings among Lie groups (both complex and real), see Paul Garrett's useful notes Sporadic isogenies to orthogonal groups [pdf].

Answer 2

In physics, this construction is called the spinor map, or the Weyl representation, or the nonlinear (chiral) σ model.

Consider the unit determinant 2×2 matrix mapped out of the 3-vector $\vec r \equiv (x,y,z)$,
$$ X = \begin{pmatrix} \sqrt{1+\vec r ^2} + z & x - iy \\ x + iy & \sqrt{1+\vec r ^2} - z \end{pmatrix}= \sqrt{1+\vec r ^2}~ 1\!\!1 +x\sigma_x +y\sigma_y+z\sigma_z , $$ and your unimodular ones $$ L=\begin{pmatrix} \alpha & \beta \\-\overline{\beta} & \overline{ \alpha} \end{pmatrix},\qquad R=\begin{pmatrix} \alpha' & \beta' \\-\overline{\beta}' & \overline{ \alpha}' \end{pmatrix}, $$ so $LXR$ is also unimodular. You may painstakingly do the multiplication and compare entries, to specify the 3×3 matrix M sending $\vec r \equiv (x,y,z)$ to the 3-vector specified by $LXR$. Each element of M will be linear in the unprimed (L) and also primed (R) elements.

Normally, in physics, this is done in a smoother, more efficient language involving the Lie algebra, and the simple expression of group elements in terms of Pauli matrices, $\exp(i \vec r \cdot \vec{\sigma}) = 1\!\! 1\cos{r} + i (\vec r \cdot \vec{\sigma}) \sin{r}/r $ , but this procedure should be equivalent.