Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 29 / 6 / c

For positive integer $k$, the equivalence is the bounded invariant norm characterization of cusp forms. The three conditions below work together: meromorphicity and the bound remove interior poles; the transformation law produces a periodic local function at every modular cusp; the same bound makes that function tend to zero at the modular cusp. Conversely a cusp form has exponentially decaying Fourier expansions of a modular form, and finitely many compact core regions control its invariant norm everywhere. These implications are proved under the corresponding labelled items.

Without a positive-weight convention, weight zero is a genuine exception. The function $f=1$ satisfies all three printed conditions at $k=0$ for every congruence subgroup, but is not a cusp form. Indeed a weight-zero cusp form would descend to a holomorphic function on the compact curve $X(\Gamma )$ and hence be constant, and vanishing at a modular cusp would make it zero. Thus the literal unrestricted claim needs the qualification $k>0$.

If negative integer weights are included, both sides contain only the zero function. To check the bounded side, each modular cusp translate has at most power growth in $y$, hence at most a power of $\log (1/|q_h|)$. For a negative Laurent coefficient of order $m$, Cauchy's coefficient estimate bounds its absolute value by $C{r}^m(\log (1/r){)}^{|k|/2}\to 0$, so there is no negative coefficient and every modular cusp is holomorphic. If such a nonzero negative-weight form existed, the product of its slash translates over $\Gamma \backslash S{L}_2(\mathbb{Z})$ would be a nonzero holomorphic level-one modular form of negative weight. Right multiplication permutes the translates, and all modular cusp factors are bounded, so that product really is a level-one form. Its existence contradicts the valence formula. This also excludes negative-weight modular cusp forms. The nonzero counterexample is specifically weight zero.

A cusp form is holomorphic in the upper half-plane by definition, so it is meromorphic there. In the reverse direction suppose $f$ is meromorphic andOn a small closed disk around an interior point, $y$ is bounded above and away from zero. Hence $|f|$ is bounded there by a fixed constant. A pole of positive order would make $|f|$ unbounded along points approaching its center, a contradiction. Every possible pole is therefore removable, and $f$ is holomorphic on $\mathbb{H}$. Meromorphicity alone would not establish this conclusion; the bounded invariant norm is being used here.

The transformation condition is necessary by the definition of a cusp form. For sufficiency, fix $\sigma \in S{L}_2(\mathbb{Z})$ representing any modular cusp, and set $g=f{|}_k\sigma$. By item (i), $g$ is holomorphic in the upper half-plane. Choose an actual period $h$ with $\sigma T^h{\sigma }^{-1}=\gamma \in \Gamma$. Associativity of slash operations and the given transformation law giveThus $g(\tau +h)=g(\tau )$ and $g$ is a single-valued holomorphic function on a punctured disk with coordinate $q_h=e^{2\pi i\tau /h}$. The calculation in part (b) transports the original uniform bound to this translate:This puts every modular cusp in exactly the situation needed in item (iii). No assumption that a single modular cusp represents all of $\Gamma$'s modular cusps is made.

First suppose $f$ is a cusp form. Choose finitely many representatives ${\sigma }_j$ for $\Gamma \backslash S{L}_2(\mathbb{Z})$; their translates of the standard modular region cover a fundamental region for $\Gamma$. On the compact truncated part of each translate, the continuous function $|f(\tau )|y^{k/2}$ is bounded. On its tail, the corresponding modular cusp expansion of $g_j=f{|}_k{\sigma }_j$ has zero constant term, so $g_j=O(e^{-2\pi y/h_j})$ uniformly in the real part. Therefore $y^{k/2}|g_j|$ is bounded there and tends to zero. Part (b) identifies this expression with $\varphi ({\sigma }_j\tau )$. There are only finitely many representatives; taking the maximum of these bounds and using $\Gamma$-invariance proves that $\varphi$ is bounded on all of $\mathbb{H}$.

Conversely assume all three conditions and $k>0$. Items (i) and (ii) show that at each modular cusp a holomorphic periodic translate $g$ satisfies $|g(\tau )|\le B{y}^{-k/2}$. In its punctured-disk coordinate,The Riemann removable singularity theorem extends $g$ holomorphically to zero with $g(0)=0$. Equivalently its Laurent expansion has no negative powers and no constant term. Since this holds at every modular cusp, $f$ is a cusp form. Hence the three conditions are sufficient as well as necessary in positive weight, with the weight-zero exception explained in the root solution.