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Let $\mathbb H^2 = \{(x, y) \in \mathbb{R}^2 : y> 0\}$ and consider the $2$-form in $\mathbb H^2$ defined by $$\varphi = \dfrac{dx \wedge dy}{y^2}.$$

Show that $\varphi$ is invariant $(T^*(\varphi)= \varphi)$ under the transformation $T$ from $\mathbb H^2$ to $\mathbb H^2$ given by $$T(z) = \dfrac{az + b}{cz + d}$$

where $z \in \mathbb H^2 \subset \mathbb{C}$ and $a, b, c, d \in \mathbb{R}$ with $ad - bc \neq 0$.

My attempt was to try to express $T(x, y)$ and calculate $dx $ and $ dy $ to make $ dx \wedge dy $, and calculate the pullback. I couldn't, the expression for $T$ was strange.

azif00
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Karly
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2 Answers2

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Write it in terms of $z = x + iy$ and $\overline{z} = x - iy$.

$$\frac{dx \wedge dy}{y^2} = -2i \cdot \frac{dz \wedge d\overline{z}}{(\overline{z} - z)^2}$$

Then plug in and simplify. You will find that

$$\frac{d\left(\frac{az + b}{cz+d}\right) \wedge d\left(\frac{a\overline{z} + b}{c\overline{z} + d}\right)}{\left( \left(\frac{a\overline{z} + b}{c\overline{z} + d}\right) - \left(\frac{a {z} + b}{c {z} + d}\right)\right)^2} = \frac{dz \wedge d\overline{z}}{(\overline{z} - z)^2}$$

Didier
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Flounderer
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  • I think there are two -minor- typos: first, $\mathrm{d}z\wedge \mathrm{d}\bar{z} = -2i \mathrm{d}x\wedge \mathrm{d}y$, and second, $(\bar{z} - z)^2 = -4y^2$. Hence there seems to be missing some "$\frac{i}{2}$" or something in your first equality. This of course does not change the validity of your argument. – Didier Mar 26 '21 at 08:07
  • Oh yes, you're right, I forgot about the squaring. – Flounderer Mar 26 '21 at 08:36
  • Now it's my bad, because of ambiguity: the $\frac{i}{2}$ term should be on the LHS! Hence, on the RHS it is more a $-2i$ term. – Didier Mar 26 '21 at 08:42
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As we know, every such transform is generated by 3 elements: $$ f_1(z) = z + a$$ $$ f_2(z) = \dfrac{1}{z} $$ $$ f_3(z) = az $$ So you need to check that you's form is invariant under this 3 transforms, what is not difficult

NeoFanatic
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