Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the book "Modular Forms" by Miyake one finds the definition of some obscure 'thing'. He calls it a metric on $\mathbb{H} = \{z \in \mathbb{C} : \operatorname{Im}(z) > 0 \}$. The following is defined:

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

Now the first thing that bothers me is that he writes a dependency of '$z$' while $dx$ and $dy$ are processes that require whole neighbourhoods. Anyhow: for a function $\phi : [0,1] \to \mathbb{H}$ that is continuous and smooth up to a finite number of points in $[0,1]$ he defines the length of this function as $$l_{\mathbb{H}}(\phi) = \int_0^1 \sqrt{x'(t)^2 + y'(t)^2}/y(t) \, dt$$

In the unit disc $B := \{ z \in \mathbb{C} : |z| \leq 1\}$ he defines another thing which gives rise to the definition of 'length' as $$d_B s^2 = \frac{4(dx^2 + dy^2)}{(1-|w|^2)^2}$$

$\mathbb{H}$ is 'holomorphically' related to the disc by $z \mapsto \rho.z$ where $\rho = \begin{pmatrix} 1 & i \\ 1 & -i \end{pmatrix}$ and $\rho$ acts as linear fractioinal transformation, i.e. $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}.z = \frac{az + b}{cz + d}$$

Now Miyake states two things and my question is: why do they hold?

1) He says that this $ds$ is invariant under the action of $GL_2^+(\mathbb{R})$ on the upper half plane. I guess that what he means is that if we have a curve $\phi$ as above and $\alpha \in GL_2^+(\mathbb{R})$, the curve $$\psi(t) := \alpha \circ \phi(t)$$ has the same length as $\phi$ in this $ds$ length-measurement.

2) Miyake states that the new $d_Bs^2$ is the push forward of the old one, i.e. $$l_B(\phi) = l_{\mathbb{H}}(\rho^{-1} \circ \phi)$$ where $l_B(\phi) = \int_0^1 4 \frac{\sqrt{x'(t)^2 + y'(t)^2}}{1 - |\phi(t)|^2} dt$ is the measurement with respect to the new thing.

I have seen some 'proofs' of these facts and they use strange relations between the $dx, dy, dz$ that i cannot understand, for example: Miyake says that for $dz = dx + i \, dy$ and $d\overline{z} = dx - i \, dy$ we have $dx^2 + dy^2 = dz \, d\overline{z}$ and then he rewrites the length as $l(\phi) = \int_0^1 ds(\phi(t))$ which does not make any sense in my eyes because: i can accept that $ds$ is some kind of differential operator that takes functions in two real variables and sends them to a function on $\mathbb{R}$ but $\phi$ is no such function: it takes only one variable, so i beg you: if you answer the question and use such relations then please describe them in a formal precise manner...

Thank you very much,

Fabian Werner, Germany

share|cite|improve this question
$ds$ is a differential form. Saying this is invariant under the action of $GL_2^+(\mathbb{R})$ means that for each $g\in GL_2^+(\mathbb{R})$, we have $g^*ds=ds$ – matgaio May 14 '12 at 16:49
I do get the impression you don't want an answer to a question but you want to complain about that book (which I don't know). The formulae you wrote down look all quite familiar, while not too modern (regarding notation) to me. The metric on the upper half plane is definitely not an obscure thing. Some people like the notation, others not. I don't know how concise these are defined in the book you are reading. If the definitions are not stringent enough for you I'd suggest you search for some other book on the topic and read some basic text on differential geometry to learn about metrics. – user20266 May 14 '12 at 17:13
I am sorry, i did not mean to sound so agressive. Indeed: i have no clue about diff. geometry: i only know a 'metric' as a function that measures the distance in a topological sense. I do not say that the metric itself is obscure but the way it was defined was very unclear to me (and the other books i have found do it in the same way :( ) – Fabian Werner May 14 '12 at 18:04
@Matgaio Just a pedantry: $\omega=ds^2$ is a differential form. – Giuseppe Tortorella May 14 '12 at 19:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.