How to compute dimensions of modular forms?

Question

I am reading Iwaniec's book on modular forms. After having proven the valence formula, he deduces constraints on the possible zeros. For instance, for the modular group and $k=4$, he deduces that a nonzero modular form needs to have a unique zero, simple, at $\rho$. I do not understanjd how to conclude from there. Are holomorphic functions entirely determined by their zeros and asymptotics? (this would be true on the entire plane by Liouville I guess, but what about the upper half-plane?).

Answer 1

Short answer: In the book it's not mentioned that $f - c G_4$ lives in $M_4$; and then you apply the contrapositive of the valence formula. Detailed proofs:

Let $M_k$ denote the $\def\C{\mathbb{C}}\C$-vector space of weight $k$ modular forms of full level. Assume the valence formula.

If you say you can't see how to conclude the case $k=4$, shouldn't you complain already for $k=0$? (In the book, the author says $m_f(z) = 0$ for all $z$ implies that $f$ is constant. How so? Why shouldn't there be a nonconstant modular function without zeros and poles?)

Proposition 1. $M_0 = \C \cdot 1$ are the constant functions.

Proof. Note "$\supseteq$" holds. Now take any $f \in M_0$. We need to show that $f$ is constant. Hmm, which value do we want to show that it takes everywhere? Well, pick your favorite point $x_0$ in the upper half-plane (or $x_0= \infty$), and we want/need to show $\forall z : f(z) = f(x_0)$. In other words, we need to show $g := f - f(x_0) = 0$. But recall that the constant function $z \mapsto f(x_0)$ is an element of $M_0$, and $M_0$ is a vector space; hence $g \in M_0$. So we have:

1. $g \in M_0$
2. $g$ is zero at at least one point (namely at $x_0$)

But then it follows from (the contrapositive of) the valence formula that $g = 0$. (If $g \neq 0$ then $g$ would have order $\geq 1$ at $x_0$, which contradicts the valence formula.) $\square$

(Btw note that we have used "$\supseteq$" to prove "$\subseteq$", which is a general strategy for those kind of proofs.)

For $M_4 = \C \cdot G_4$ one can either use a similar idea or reduce it to $M_0 = \C \cdot 1$. In the book you mention, the author takes the first approach, I prefer the second. Explicitly:

Proposition 2. $M_4 = \C \cdot G_4$.

Proof option 1. Note "$\supseteq$" holds. Let $f \in M_4$. Deduce from the valence formula that $f$ has a simple zero at all orbit points of $\rho$ and no other zeros and poles, just as $G_4$. Pick your favorite point $x_0 \neq \rho$ and consider $g := f - \frac{f(x_0)}{G_4(x_0)}G_4$. We want to show $g = 0$. But we have:

1. $g \in M_4$
2. $g(x_0) = 0$

Again, if $g \neq 0$ this would be a contradiction to the valence formula. $\square$

Note that for other $k$ one possibly must be a bit careful with choosing $x_0$, which is why I prefer:

Proof option 2. Note "$\supseteq$" holds. Let $f \in M_4$. Deduce from the valence formula that $f$ has a simple zero at all orbit points of $\rho$ and no other zeros and poles, just as $G_4$. Hence $f/G_4$ has no zeros and no poles. Also observe that $f/G_4$ is weight $0$ invariant! Hence $f/G_4 \in M_0$. By Proposition 1, $f/G_4 = c \in \C$ is constant, i.e. $f = c \cdot G_4$. $\square$

Note that:

• in option 1 we looked at the difference of $f$ and what we claim it is, and showed that it's equal to $0$
• in option 2 we basically looked at the quotient of $f$ and what we claim it is, and showed that it's equal to $1$

With such sort of function spaces both approaches have their right of existence. You want to use the first if you have "additive" information about the functions, and option 2 if you have "multiplicative" information about the functions. Since knowing orders of zeros and poles is "multiplicative" information, option 2 is the more natural choice.