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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

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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.

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