Petersson inner product

Remarks

The function $f \overline{g} y^k$ and the measure $dx\, dy / y^2$ are invariant under $\mathrm{SL}_2(\Z)$, so any fundamental domain gives the same value. A cusp form decays like $e^{-2\pi y}$ as $y \to \infty$, so the integral converges. On the space $S_k$ of cusp forms, $\langle \cdot, \cdot \rangle$ is a positive definite Hermitian inner product. Example: $\langle \Delta, \Delta \rangle > 0$, since $\Delta$ has no zeros on $\mathbb{H}$, so the integrand $\abs{\Delta}^2 y^{12} \cdot y^{-2}$ is positive on $\mathcal{F}$. Non-example: for $f = g = E_4$ the integral diverges, because $\abs{E_4}^2 y^4 \cdot y^{-2}$ tends to infinity like $y^2$ as $y \to \infty$; this is why one form must be a cusp form. For a congruence subgroup $\Gamma$, the integral is taken over a fundamental domain of $\Gamma$, often divided by the index of $\Gamma$ in $\mathrm{SL}_2(\Z)$.