Let $\mathcal{H}^2=\{z=x+iy\in\mathbb{C} : y>0\}$ be the upper half-plane and let $g(z)=\frac{az+b}{cz+d}$, $a,b,c,d\in\mathbb{R}$, $ad-bc=1$, be an orientation preserving isometry of $\mathcal{H}^2$. If $z\in\mathcal{H}^2$ and $v\in\mathbb{C}$, then $PSL_2(\mathbb{R})$ acts on the tangent bundle by $$ g\cdot(z,v)=\left(g(z),g'(z)\cdot v\right)=\left(\frac{az+b}{cz+d},\frac{v}{(cz+d)^2}\right). $$
I want a similar formula for the action of $PSL_2(\mathbb{C})$ on the tangent bundle of $$ \mathcal{H}^3=\{\zeta=z+tj : z\in\mathbb{C}, 0<t\in\mathbb{R}\}\subseteq\mathbb{H}. $$ We have orientation preserving isometries $g(\zeta)=(az+b)(cz+d)^{-1}$, where the multiplication is quaternionic. What I want to know is the formula for the derivative, where the tangent space is identified with $\{z+tj\in\mathbb{C}+\mathbb{R}j\}\subseteq\mathbb{H}$, i.e. is it given by multiplation by a quaternion? I starting writing things in coordinates by I gave up and wondered if someone had a reference or had done it themselves before. Thanks.
Edit: I don't know if this is correct, but I obtained $$ g'(\zeta)\cdot\eta=(a-g(\zeta)c)\eta(c\zeta+d)^{-1} $$ for the derivative of $g$ at $\zeta$ applied to a tangent vector $\eta\in\mathbb{R}+\mathbb{R}i+\mathbb{R}j$. Is this correct? Does it define a group action (as it should)? At the very least, the derivative of a translation is the identity and the derivative of $\text{diag}(a,a^{-1})$ is $a^2z+|a|^2tj$ for $\eta=z+tj$ which seems correct (rotation and scaling).
Suppose $\zeta$ is a function whose derivative is $\eta$. First, let's compute
$$ (c\zeta+d)^{-1}(c\zeta+d)=1 $$
$$ \big[(c\zeta+d)^{-1}\big]'(c\zeta+d)+(c\zeta+d)^{-1}(c\eta)=0 $$
$$ \big[(c\zeta+d)^{-1}\big]'=-(c\zeta+d)^{-1}(c\eta)(c\zeta+d)^{-1}. $$
Apply this in differentiating $(a\zeta+b)(c\zeta+d)^{-1}$:
$$ (a\eta)(c\zeta+d)^{-1}-(a\zeta+b)(c\zeta+d)^{-1}(c\eta)(c\zeta+d)^{-1} $$
$$ =\big[a-(a\zeta+b)(c\zeta+d)^{-1}c\big] \eta \,(c\zeta+d)^{-1} $$
So your edit is correct. But the first part can be simplified with $(xy)^{-1}=y^{-1}x^{-1}$:
$$ a-a(\zeta+a^{-1}b)(\zeta+c^{-1}d)^{-1} $$
$$ a-a\big((\zeta+c^{-1}d)+(a^{-1}b-c^{-1}d)\big)(\zeta+c^{-1}d)^{-1} $$
$$ -a(a^{-1}b-c^{-1}d)(\zeta+c^{-1}d)^{-1}=(ad-bc)(\zeta c+d)^{-1} $$
Note $a,b,c,d$ commute with each other but not $\zeta$ (generally). Thus
$$ [\begin{smallmatrix}a & b \\ c & d \end{smallmatrix}](\zeta,\eta)=\big(\, (a\zeta+b)(c\zeta+d)^{-1}\, , \, (\zeta c+d)^{-1}\eta(c\zeta+d)^{-1} \, \big). $$