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I am given the following manifold $N=\{(x,y)\in \mathbb{R}^2, y>0\}$ with metric:

$$ds^2=\frac{dx^2+dy^2}{y^2}$$

There is a suggestion to take $z=x+iy$ and consider the transformations:

$$z\to z+c\,, \quad z\to cz\,,\quad z \to \frac{z}{cz+1}\,,\quad c\in \mathbb{R}$$

I have to find three independent Killing vector fields.

I think the idea is to define a flow from those transformations and then compute the Killing vectors associated to that flow. Is there any systematic way to do this?

egreg
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Joaquin Liniado
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1 Answers1

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You have the right idea. First, define \begin{align} \varphi_1(t,z) &= z + t & \varphi_2(t,z) &= e^{2t}z & \varphi_3(t,z) &= \dfrac{z}{tz +1} \end{align} They correspond to the matrices in $PSL_2(\mathbb{R})$, generating $1$-parameter subgroups \begin{align} A_1 &=\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}& A_2 &=\begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix}& A_3 &=\begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} \end{align} Show they are flows acting by isometry on the hyperbolic plane. Then, compute $X_i(z) = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi_i(t,z)$. They will be Killing vector fields.

For example, fix $z$ in the hyperbolic plane. Then the vector field $X_1(z)$ at $z$ is the tangent vector in $T_z\mathbb{H} \simeq \mathbb{C}$ $$ X_1(z) = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi_1(t,z) = 1 = 1 + i\cdot 0 = \partial_x $$

Didier
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  • Thank you very much for your answer. On the last step you say that the Killing vector fields will be given by $\frac{d}{dt} \varphi_i(t,z)$. But for example, $\frac{d}{dt} \varphi_1(t,z)=z+1$. How do I get the vector field? – Joaquin Liniado Dec 24 '20 at 11:30
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    I added a line to highlight how to compute $X_1$. You did some error in the differentiation. Moreover, recall that you are working in the identity chart, so coordinates have to be translated into right vector fields while differentiating. – Didier Dec 24 '20 at 11:38