Let $V$ be a finite dimensional vector space over $\mathbb{C}$. Let $G = \text{SL}\_2(\mathbb{Z})$, say, and $\rho : G \to \text{GL}\_\mathbb{C}(V)$ a representation. Let us take a fixed basis $b\_1, ..., b\_n$ of $V$. We write functions $F : \mathbb{H} \to V$ as $\sum\_i F\_i b\_i$. We call a function $F : \mathbb{H} \to V$ 'weakly' (or whatever people call it) modular of some weight $k \in \mathbb{Z}$ for $G$ iff. for every $g \in G$, 


$F|\_g = \rho(g) F$


where $F|\_g = \sum\_i F\_i|\_g b\_i$ and $f|\_g = (c \tau + d)^{-k} f\left(\frac{a \tau + b}{c\tau + d} \right)$ where $g = \begin{pmatrix}a & b \\ c & d \end{pmatrix}$. The space of all of these functions will be denoted by $W\_k(\rho)$. Now let $\rho\_1, ..., \rho\_l$ be pairwise different representations [in the sense that for every $i < j$, there exist $g \in G, v \in V$ such that $\rho\_i(g) v \neq \rho\_j(g) v$. Is it true that, after viewing every space $W\_k(\rho\_i)$ as a subspace of the space of all functions from $\mathbb{H}$ to $V$, then


$$ \sum\_i W\_k(\rho\_i) = \oplus\_i W\_k(\rho\_i) $$


?? I.e. are vector valued modular forms w.r.t. different representations linearly independent?


What I figured out already:


If there is an element $g$ such that $\rho\_i(g) - \rho\_j(g)$ is invertible for all pairs $i < j$, then one can show the assertion with the same method as one shows that eigenvectors w.r.t. different eigenvalues are linearly independent.


In my case of certain Weil representations on a discriminant form, this is not always true.


Thanks,


FW, Germany



Modular Forms w.r.t. different representations linearly independent?
No, they are not. Consider $\rho\_1$ and $\rho\_1 \oplus \rho\_2$. It the $\rho\_j$ are pairwise disjoint, i.e., $Hom\_G(\rho\_j , \rho\_l)= \{0 \}$ for $j \neq l$, then this is sufficient. This can be expressed in terms of matrix coefficients, but that's an inequality of functions. 


