What is the right measure for the upper half-plane?

Question

I think my question depends on the context of the problem so let me give you some background. I want to measure subsets of $\{(x,y)|~y>0, x\in \mathbb{R}\}$. I'm not involved with differential geometry. So what is the right measure for the upper halfplane. If one identifies it with the hyperbolic plane I saw that one possibility is $\frac{1}{y^2}dxdy$ but on the other hand this is a subset of $\mathbb{R}^2$ so with the Lebesgue measure we can also measure subsets.

The motivation behind it is as follows. We want to integrate a function on some subset of $\mathbb{R}^3$ and use cylindrical coordinates, assume there is no dependence on the angle. Then the integral becomes and integral on some subset of the upper halfplane. For me, it seems natural to measure things with the Lebesgue measure.

Is there a "natural" way to decide which measure to use?

Answer 1

Maybe I'm missing something - But if you want the value of the integral to be meaningful with respect to your original problem, you don't get to choose the measure.

If you have a function, given in Euclidean coordinates, it is likely this function is meaningful, as a measurable function, with respect to the Lebesgue measure.

If you want to integrate under different coordinates, that's fine as well - you have to use the Jacobian matrix to calculate the "fine" you have to pay.

This "fine" is actually the Lebesgue measure, given in polar coordinates. If we look at it from a Differential-Geometric point of view, you've just described the same volume form under different coordinates.

Answer 2

Yes, the most natural measure is proportional to ${dxdy\over y^2}$. This measure is invariant by the isometries of ${\bf H}$, just as the Lebesgue measure on the euclidean plane ${\bf R}^2$ is invariant by the transformations of the plane that preserves the norm.

The isometries of ${\bf H}$ that preserve the orientation are the homographies $z \mapsto {az+b \over cz+d}$ with $ad-bc = 1$ and $a,b,c,d \in {\bf R}$. You can check that these transformations preserve the measure $${dx\wedge dy\over y^2} = {i \ dz \wedge d\bar{z} \over 2 Im(z)^2}.$$