The group $\mathrm{PSL}_2(\mathbb{R})$ acts on $\mathbb{H}$ via Möbius transformations, that is \begin{align*} g=\begin{pmatrix} a & b \\ c & d\end{pmatrix}:z\mapsto \frac{az+b}{cz+d}. \end{align*} $\mathbb{H}$ may realised as a smooth manifold with its natural chart $\varphi:\mathbb{H}\to\{(x,y)\in\mathbb{R}^2:y>0\}$ by $\varphi(z)=(\Re(z),\Im(z))$. An element $g\in\mathrm{PSL}_2(\mathbb{R})$ thus may be considered a smooth map $g:\mathbb{H}\to\mathbb{H}$ and hence induces an action through its differential on the tangent bundle namely $Dg:T\mathbb{H}\to T\mathbb{H}$. It is widely known (see for example Ergodic Theory with a view towards Number Theory by Einsiedler and Ward) that this action is given by \begin{align*} Dg(z,v)=\left(g(z),g'(z)v\right). \end{align*} where $g'(z)$ would be the derivative of the transformation, namely $\frac{1}{(cz+d)^2}$. I have tried computing the action in coordinates namely if $v=v^1\frac{\partial}{\partial x}\Big|_z+v^2\frac{\partial}{\partial y}\Big|_z$, then I believe its image under the derivative action should be \begin{align*} \left(v^1\frac{\partial u}{\partial x}\Big|_{\varphi(z)}+v^2\frac{\partial u}{\partial y}\Big|_{\varphi(z)}\right)\frac{\partial}{\partial x}\Big|_{g(z)}+\left(v^1\frac{\partial v}{\partial x}\Big|_{\varphi(z)}+v^2\frac{\partial v}{\partial y}\Big|_{\varphi(z)}\right)\frac{\partial}{\partial y}\Big|_{g(z)}, \end{align*} where \begin{align*} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}&=\frac{|cz+d|^2-2c^2y^2}{|cz+d|^4},\\ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}&=\frac{2y(c^2x+cd)}{|cz+d|^4}. \end{align*} As I understand it, the result should have simply been \begin{align*} \frac{1}{(cz+d)^2}\left(v^1\frac{\partial}{\partial x}\Big|_{g(z)}+v^2\frac{\partial}{\partial y}\Big|_{g(z)}\right). \end{align*} If there is a mistake in the working/a much simpler method of computation of the derivative action, I would appreciate the help.

  • $\begingroup$ $g'(z) = \frac{1}{(cz+d)^2}$ not $\frac{1}{|cz+d|^2}$. Think "complex derivative", not "Jacobian derivative". $\endgroup$
    – Lee Mosher
    Feb 22, 2019 at 2:49
  • $\begingroup$ @LeeMosher Thank you, I have edited this above now. This does not affect the computation in coordinates though so I'm still unsure on where the error lies? $\endgroup$
    – user298173
    Feb 22, 2019 at 12:00

1 Answer 1


Your calculations are correct. The point is that $g'(z)\in\mathbb{C}$, so in your last line you did not express all terms in coordinates. For $z=x+iy$, $$ g'(z) = \frac{(cx+d)^2-c^2y^2}{|cz+d|^4} - i\frac{2cy(cx+d)}{|cz+d|^4} $$ Note that $(cx+d)^2-c^2y^2 = |cz+d|^2-2c^2y^2$. Let $v=v_1+iv_2$. If we compute the real and imaginary part of the complex multiplication $g'(z)v$ we see that $$ g'(z)v= \frac{v_1(|cz+d|^2-2c^2y^2) + v_2\cdot 2cy(cx+d)}{|cz+d|^4} +i \frac{-v_1\cdot 2cy(cx+d) + v_2(|cz+d|^2-2c^2y^2)}{|cz+d|^4} $$ With your notation, this means $$ g'(z)v = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ -\frac{\partial u}{\partial y} & \frac{\partial u}{\partial x} \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} $$ which is the multiplication of the Jacobian of $g$ with $(v_1,v_2)^T\cong v$.


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