No linear relation amongst modular forms of different weights when looking at them as holomorphic functions

Question

It's just normally stated that it's simple to see this fact in most books, but it does not appear so simple to me.

Answer 1

Perhaps the best way to convince yourself of this is to think of a modular form of weight $k$ as a function on lattices $\Lambda \subset \mathbb C$, which is homogeneous of weight $-k$ for the action of $\mathbb C^\times$, i.e. $f(\lambda \Lambda ) = \lambda^{-k} f(\Lambda)$. This is related to the definition in terms of the variable $\tau$ by $f(\tau) = f(\mathbb Z + \tau \mathbb Z)$; then $f\left(\frac{a\tau + b}{c \tau + d}\right) = f(\mathbb Z + \frac{a\tau + b}{c \tau + d} \mathbb Z) = (c\tau+d)^k f(\mathbb Z + \tau \mathbb Z) = (c\tau+d)^k f(\tau)$ is the usual transformation rule.

If $f_0, \dots, f_k$ are modular forms, not all zero, and such that $f_i$ has weight $i$, then there exists a lattice $\Lambda$ for which $f_1(\Lambda), \dots, f_k(\Lambda)$ are not all zero. But if

$$\sum f_i = 0,$$

then for any $\lambda \in \mathbb C^\times$ we have

$$\sum f_i(\lambda \Lambda) = 0,$$

i.e.

$$\sum \lambda^{-i} f_i(\Lambda) = 0.$$

If we pick $k+1$ distinct elements $\lambda_0, \dots, \lambda_k \in \mathbb C^\times$, the $(k+1) \times (k+1)$ matrix $(\lambda_i^{-j})_{0 \leq {i},j \leq k}$ is invertible (being a Vandermonde matrix); and hence the system of equations

$$\sum \lambda_j^{-i} f_i(\Lambda) = 0$$

implies that $f_i(\Lambda) = 0$ for all $i$, contradicting the choice of $\Lambda$. Hence no such linear relation is possible.