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).

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    $\begingroup$ Look at $[z:1] \mapsto [az+b:cz+d]$ for $z \in \mathbb{C}, [z:w] \in P^1(\mathbb{C})$. Then $[a(z+v)+b:c(z+v)+d] = [(a(z+v)+b)(c(z-v)+d):(cz+d)^2-c^2v^2]$ $ = [ (az+b)(cz+d) + av(cz+d)-cv (az+b):(cz+d)^2] + O(v^2) $ $=[ (az+b)(cz+d) + v:(cz+d)^2] + O(v^2) =[ (az+b)(cz+d)^{-1} + v(cz+d)^{-2}:1] + O(v^2)$ whence $\partial_z [az+b:cz+d] = ((cz+d)^{-2},0) \in T_{[(az+b)(cz+d)^{-1}:1]}P^1(\Bbb{C})$. For $z \in\mathbb{H},[z:1]=[zw:w]\in P^1(\Bbb{H})$ the idea is the same except things don't commute so you need to replace $c(z-v)+d$ by the correct thing and be careful when expanding the LHS. $\endgroup$ – reuns Feb 6 '19 at 5:52

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

$$ [\begin{smallmatrix}a & b \\ c & d \end{smallmatrix}](\zeta,\eta)=\big(\, (a\zeta+b)(c\zeta+d)^{-1}\, , \, (\zeta c+d)^{-1}\eta(c\zeta+d)^{-1} \, \big). $$


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