In "M. Kaneko and D. Zagier, A generalized Jacobi theta function and quasimodular forms, Prog. Math. 129, 165-172 (1995)" there is a proposition stating essentially that $E_2$, $E_4$ and $E_6$ are algebraically independent.

Unfortunately there is no proof there. I know how to show the algebraic independence of $E_4$ and $E_6$.

How does the proof for $E_2$ work?

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    $\begingroup$ Dear Jonas, Since $E_2$ is not a modular form, but has a "near-modular" transformation law under $\tau \mapsto -1/\tau$, it shouldn't be hard to deduce the independence by writing down a putative algebraic relation between $E_2$, $E_4$, and $E_6$ (so this is a polynomial in $E_2$ whose coefficients are sums of modular forms of various weights which vanishes) and then applying $\tau \mapsto -1/\tau$ to the relation and seeing what happens. Regards, $\endgroup$ – Emerton Mar 27 '13 at 1:25
  • $\begingroup$ Dear Emerton Well that's the obvious approach yes. I didn't find it so "easy". Can you be more specific? Is there a trick? For $E_4$ and $E_6$ one evaluates at $\frac{-1}{\tau+m}$ for enough $m$'s to write it as a Vandermonde matrix applied to a vector of forms to show that one can assume homogenous weights wlog (after that it is much easier). I find it a bit harder to find such an invertible matrix for $E_2$ (resp. to show that it is invertible). $\endgroup$ – Jonas Mar 27 '13 at 11:37
  • $\begingroup$ Dear Jonas, To expand slightly on Emerton's comment, $E_4$ and $E_6$ are modular forms, so in particular $E_4(-1/\tau)=\tau^4 E_4(\tau)$ and $E_6(-1/\tau)= \tau^6 E_6(\tau)$. On the other hand $E_2$ is not a modular form, it is only quasimodular. It obeys $E_2(-1/\tau)= \tau^2 E_2(\tau)- 6 i \tau/\pi$. No algebraic combination of $E_4,E_6$ can transform this way under $\tau \rightarrow -1/\tau$ so $E_2$ is algebraically independent of $E_4,E_6$. $\endgroup$ – Jeff Harvey Mar 27 '13 at 17:47
  • $\begingroup$ Dear Jonas, For $E_4$ and $E_6$ one uses that $\tau \mapsto -1/\tau$ takes $f(\tau)$ to $\tau^k f(\tau)$ for a weight $k$ modular forms, and this implies that in any sum of modular forms of various weights, the different weight terms are linearly independent (because they multiply by different powers of $\tau$). I don't really see where Vandermonde matrices come into it (although maybe that's one way of making the previous sentence rigorous, although there are other ways that I find easier). Now if you add in $E_2$, it is a bit more complicated, but shouldn't be too bad. Regards, $\endgroup$ – Emerton Mar 27 '13 at 19:19
  • $\begingroup$ P.S. Here is how I would do the $E_4$ and $E_6$ case: let $f_{k_1} + \cdots f_{k_n} = 0$ be a linear dependence between modular forms of increasing weights $k_1 < k_2 < \cdots < k_n$. Now evaluate at $-1/\tau$, and divide by $\tau^{k_1}$, to get $f_{k_1} + \tau^{k_2 - k_1} f_{k_2} + \cdots \tau^{k_n - k_1} f_{k_n} = 0.$ Subtracting the two equations, you get a linear dependence relation with fewer terms, and a well-organized induction should finish. $\endgroup$ – Emerton Mar 27 '13 at 19:22

Here is a reference for the algebraic independance of $E_2,E_4,E_6$ over $\mathbf{C}$ :

MR2186573 (2007a:11065) Martin, François ; Royer, Emmanuel . Formes modulaires et périodes. (French) [Modular forms and periods] Formes modulaires et transcendance, 1--117, Sémin. Congr., 12, Soc. Math. France, Paris, 2005.

See Lemme 117 p. 80. The proof is along the lines suggested by Emerton.

  • $\begingroup$ Nice to know that this book is available for free. $\endgroup$ – Chandan Singh Dalawat Mar 28 '13 at 10:26
  • $\begingroup$ Thanks a lot! :-) However I would not say that it is similar or as simple as suggested by Emerton. In fact it uses a different description for the space of quasimodular forms than polynomials in $E_2$, $E_4$ and $E_6$ (in fact the equality is shown much later on p. 83 Corollary 121). When trying to show the direct sum property the function is in essence evaluated at $\frac{-1}{\tau+m}$ for infinitely many $m$ ("closer to the Vandermonde idea"). So all in all the proof is rather complicated. I'm just saying this in my defense. I really do/did appreciate Emerton's effort to help! Best $\endgroup$ – Jonas Mar 28 '13 at 10:31

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