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Conjecture: Let be $f$ a modular form of weight $k$ and $j$ a strictly positive integer, then the set $f,f',...,f^{(j)}$ is $\mathbb C$-linearly independent in $A$.

Is that conjecture true or false? Do you know a counterexample or a proof?

Notation: Let $\Pi=\{x+iy\in \mathbb C|y>0\}$ be the upper half-plane, $Hol(\Pi)=\{f:\Pi\to\mathbb C|f \text{ is holomorphic}\}$, $M_k:=M_k(SL_2(\mathbb Z))$ be the space of modular forms of weight $k$ for $SL_2(\mathbb Z)$, $M_*:=\bigoplus_kM_k$, and finally let $A=Span(f^{(j)}|f\in M_* \text{ and } j\in \mathbb N)$ be the subalgebra of $Hol(\Pi)$ span (over $\mathbb C$) by the elements $f^{(j)}$ with $f\in M_*$ and $j\in \mathbb N$, where $f^{(0)}=f$, $f^{(1)}=f'=\frac{1}{2\pi i}\frac{df}{dz}$,...

In advance thank you very much for your answers.

Manuel

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  • $\begingroup$ This is false for two reasons -- the constant function is a weight zero modular form, and the zero function is a modular form of any weight :P But away from those case it's surely true, and one can perhaps prove it by observing that the derivative of $f$ is some more generalised version of a modular form, but has weight $k+2$ not $k$, and so on. $\endgroup$
    – user30035
    Commented Mar 19, 2013 at 21:38
  • $\begingroup$ That's right! thanks for your advice. $\endgroup$
    – MMa
    Commented Mar 20, 2013 at 19:30

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The only functions $f$ for which $f,f ',\ldots, f^{(j)}$ are linearly dependent are linear combinations of exponential functions, as one learns in a basic course on ordinary differential equations.

But maybe you meant to ask instead about whether $f$ and its derivatives are algebraically independent. Then it turns out that there always exists such an algebraic dependence. The reason is that $f$ and all its derivatives are quasimodular forms, and the ring of quasimodular forms has transcendence degree 3 over $\mathbb C$. So $f$ always satisfies some nonlinear third order differential equation. See e.g. Zagier's chapter in "The 1-2-3 of Modular Forms", section 5.

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    $\begingroup$ linear combinations of exponential functions ... No, You need linear combination of polynomial times exponential. $\endgroup$ Commented Mar 19, 2013 at 13:43
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    $\begingroup$ Of course, maybe Manuel is asking for the proof that modular forms cannot be written that way. $\endgroup$ Commented Mar 19, 2013 at 13:45
  • $\begingroup$ That's exactly what I need, but I already have a solution for that. Anyway thank you very much. $\endgroup$
    – MMa
    Commented Mar 20, 2013 at 19:35

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