# derivative action of isometries on hyperbolic 3-space (upper half-space model)

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 $$\mathcal{H}^3=\{\zeta=z+tj : z\in\mathbb{C}, 0 We have orientation preserving isometries $$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).

• 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. – reuns Feb 6 '19 at 5:52

## 1 Answer

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

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