Group action of $\text{SL}(2, \mathbb{C})$

Question

Let's denote upper-half space model of hyperbolic 3-space identified with quaternions as $$\text{H}^{+}_3 = \biggr\{z+wj\in \mathbb{C}\oplus \mathbb{R}^{+}j\biggr\}\tag{1}$$

Then a group action $\rho : \text{SL}(2,\mathbb{C})\times \text{H}^{+}_{3}\rightarrow \text{H}^{+}_{3}$ seems to be defined via linear fractional transformations, i.e

$$\rho(M, z+wj) \mapsto M(z+wj) = (\alpha(z+wj)+\beta)(\gamma (z+wj)+\delta)^{-1} = z^{*}+w^{*} j \in \text{H}^{+}_{3}\tag{2}$$

Where $$z^{*}=\frac{(\alpha z+\beta)(\bar{\gamma} \bar{z}+\bar{\delta})+\alpha \bar{\gamma}w^{2}}{|\gamma z+\delta|^{2}+|\gamma|^{2} w^{2}} , \space\space w^{*}= \frac{w}{|\gamma z+\delta|^{2}+|\gamma|^{2} w^{2}}\tag{3}$$

Does this induce a group action $\rho : \text{SL}(2,\mathbb{C})\times \mathcal{L}^2(\text{H}^{+}_3)\rightarrow \mathcal{L}^2(\text{H}^{+}_3)$ through the following map?

$$\rho(M, f)(z+wj)\mapsto f(M^{-1}(z+wj))\tag{4}$$

If so, can we further define the induced Lie algebra action $d\rho : \mathfrak{sl}(2, \mathbb{C})\times \mathcal{L}^2(\text{H}^{+}_3)\rightarrow \mathcal{L}^2(\text{H}^{+}_3)$ through the following map?

$$\begin{align}d\rho(m, f)(z+wj)\mapsto \frac{d}{dt}\biggr|_{t = 0}\rho(e^{tm}, f)(z+wj) &= \frac{d}{dt}\biggr|_{t = 0}f(e^{-tm}(z+wj))\end{align}\tag{5}$$

Where $z+wj\in \text{H}^{+}_3$.

Answer 1

By the OP's request, I collect my comments into an answer.

1. The usual left-action of $\mathrm{SL}_2(\mathbb{C})$ on $\text{H}^{+}_3$ is defined by quaternionic right-division: $MP:=(\alpha P+\beta)(\gamma P+\delta)^{-1}$. Here $P=z+wj$ is a Hamiltonian quaternion with $k$-component zero, and so is $MP$. This left-action induces a right-action on functions $\text{H}^{+}_3\to\mathbb{C}$: the image of the function $P\mapsto f(P)$ is $P\mapsto f(M P)$. It also induces a left-action on functions $\text{H}^{+}_3\to\mathbb{C}$: the image of the function $P\mapsto f(P)$ is $P\mapsto f(M^{-1} P)$. These group actions (on functions $\text{H}^{+}_3\to\mathbb{C}$) give rise to Lie-algebra actions (on smooth functions $\text{H}^{+}_3\to\mathbb{C}$) in the usual way. Note that smooth and compactly supported functions are dense in $\mathcal{L}^2(\text{H}^{+}_3)$.

2. Both the left and the right action are unitary, because $\mathrm{SL}_2(\mathbb{C})$ is a unimodular group, hence it has a two-sided Haar measure. The Lie-algebra action is as follows. An element $m\in\mathfrak{sl}_2(\mathbb{C})$ maps the function $P\mapsto f(P)$ to the function $P\mapsto\frac{\partial}{\partial t}f(e^{-t m}P)\mid_{t=0}$. Of course this action is only defined for special functions $P\mapsto f(P)$, e.g. smooth and compactly supported functions are fine.

3. The space $\text{H}^{+}_3$ can be naturally identified with $\mathrm{SL}_2(\mathbb{C})/\mathrm{SU}_2(\mathbb{C})$; this identification comes from the left action of $\mathrm{SL}_2(\mathbb{C})$ on $\text{H}^{+}_3$. Hence a natural measure on $\text{H}^{+}_3$ is induced by a right Haar measure on $\mathrm{SL}_2(\mathbb{C})$. However, this right Haar measure is also a left Haar measure, whence the natural measure on $\text{H}^{+}_3$ is in fact invariant under the left action of $\mathrm{SL}_2(\mathbb{C})$. So this left action induces a unitary action on $\mathcal{L}^2(\text{H}^{+}_3)$.