Relations $\mathbb{H}^{*}$ and $\operatorname{GL}_{2}(\mathbb{C})$

Question

i find a homomorphism between quaternions minus the origin $\mathbb{H}^{*}$ and $\operatorname{GL}_{2}(\mathbb{C})$, where $\psi: \mathbb{H}^{*}\to \operatorname{GL}_{2}(\mathbb{C})$, $\psi(z,w)=\begin{bmatrix} z & -w\\ \bar{w} & \bar{z}\end{bmatrix}$. The fact that $\operatorname{GL}_{2}(\mathbb{C})$ is a Lie group and $\psi$ is a Group homomorphism implies that $\mathbb{H}^{*}$ is a Lie group? Whats is the relation between $\mathbb{H}^{*}$ and $\operatorname{GL}_{2}(\mathbb{C})$?

Answer 1

As Lee Mosher points out in the comments, we can see that $\mathbb{H}^*$ is a Lie group directly from the definition of Lie group (and of $\mathbb{H}^*$). However, your homomorphism shows (because it is injective and smooth!) that $\mathbb{H}^*$ is a Lie subgroup of $GL(2, \mathbb{C})$. (For Lie Mosher: every quaternion can uniquely written as $z + wj$ with $z, w \in \mathbb{C}$, where $\mathbb{C}$ is interpreted as the subalgebra of $\mathbb{H}$ spanned by $1$ and $i$.)

We can identify the subgroup a bit more precisely. Inside $GL(2, \mathbb{C})$ is sitting the subgroup $G$ of matrices with positive real determinant. (This is a subgroup since the positive real numbers are closed under multiplication and det is multiplicative.) Now let $A$ be such a matrix and $d$ be the positive squareroot of its determinant. Obviously $A$ can be written $A = DB$ where $D$ is the diagonal matrix both whose entries are $d$ and $B$ is the determinant 1 matrix $B = (1/d)A$. It follows that we have a decomposition of Lie groups $G = \mathbb{R}_{>0} \times SL(2, \mathbb{C})$.

Now inside $SL(2, \mathbb{C})$ there is the famous subgroup $SU(2)$, so inside $G$ we get the subgroup $H = \mathbb{R}_{>0} \times SU(2)$.

It is easy to check that your $\psi$ is an isomorphism of Lie groups between $\mathbb{H}^*$ and this subgroup $H$.

Could we see the isomorphism more directly on the $\mathbb{H}$ side? Well... Most sources would start by describing an isomorphism between the elements of norm 1 in $\mathbb{H}$ and $SU(2)$. The conclusion about all of $\mathbb{H}^*$ then follows from the trivial observation that any element $x \in \mathbb{H}^*$ can be written $x = |x|q$ where $q = x/|x|$ has norm 1.

However, the interesting step, constructing an iso between the norm 1 quaternions and $SU(2)$ is basically just giving your $\psi$!

The most interesting thing I could further say about this is that under this $\psi$ the square of the Euclidian norm on $\mathbb{H}$ is mapped to the determinant on $Mat(2, \mathbb{C})$. (Yes, I extended $\psi$ here to all of $\mathbb{H}$ but in the obvious way: sending 0 to 0.)