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.


FW, Germany


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.