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).

  • 1
    $\begingroup$ 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 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.