Confusion about definition Petersson product

Question

I'm taking a course on modular forms, but my background in analysis is not that strong (I have taken complex analysis and measure theory before however). Therefore I'm a bit confused about the definition of the Petersson inner product.

Let $F$ be the usual fundamental domain for the SL$_2(\mathbb Z)$-action on the upper-half plane, then for cusp forms $f, g$ of weight $k$ we defined (with $z = x + iy$) $$ \langle f, g \rangle = \int_F f(z) \overline{g(z)} y^k \frac{dx dy}{y^2}. $$ One of the things I am confused by is putting the factor $1/y^2$ seperately. Why is this? If I instead write $\int_F f(z) \overline{g(z)} y^{k-2} dx dy,$ does this mean something else, or is this not a well-defined expression?

We also spoke about the "hyperbolic form", which as I understand it is the measure $\nu$ defined by $$ \nu(S) = \int_S \frac{1}{y^2} dx dy. $$ So, maybe what is meant is that we are integrating with respect to this measure, so $\langle f, g \rangle = \int_F f(z) \overline{g(z)} y^k d\nu(z)$. Is this something different from just integrating with respect to the Lebesgue measure and adding a factor $y^{-2}$?

Thanks in advance for any help and explainations.

Answer 1

This is a good, natural question. The point is that the measure $\mu(z) = \frac{dx dy}{y^2}$ is the unique (up to multiple) Haar measure here. This measure is invariant under the isometries of the upper half-plane, i.e. it's invariant under $\mathrm{SL}(2, \mathbb{R})$. This makes it very natural for modular forms, as modular forms are also invariant under (subgroups of) $\mathrm{SL}(2, \mathbb{Z})$.

In particular, this is how you can be certain that $\int_F f(z) d\mu(z)$ is independent of the choice of fundamental domain $F$.

There is one more oddity that might come up later. When performing integrals, especially when interacting with Fourier expansions, it is sometimes natural to perform the $x$ integral first. This leaves $\frac{dy}{y^2}$. If the integration has led to the region $\int_0^\infty F(y) \frac{dy}{y^2}$ for some function $F$, then this is typically written $\int_0^\infty F(y)y^{-1} \frac{dy}{y}$. Why? Because now we're integrating over $\mathbb{R}^+$, which has Haar measure $dy/y$.