Let $E_2, E_4,$ and $E_6$ be the first three Eisenstein series. There are well-known formulas due to Ramanujan for the derivatives of these quantities:

\begin{align*} DE_2 &= \frac{1}{12}(E_2^2-E_4) \\ DE_4 &= \frac{1}{3}(E_2E_4-E_6) \\ DE_6 & = \frac{1}{2}(E_2E_6-E_4^2), \end{align*} where $(Df)(\tau):=\frac{1}{2\pi i}f'(\tau)$. A noticeable feature of these identities is that, for each $k=2,4,6$, there are constants $a_k, b_k$ such that $$DE_k - a_kE_2E_k = b_kE_{k+2}.\tag{*}$$ I was wondering if a formula of this type held for higher $k$. Now, I know that given a modular form $f\in\mathcal{M}_k$ with $k\geq 4$, the so-called Serre derivative $\vartheta f = Df - \frac{k}{12}E_2f$ has the property $\vartheta f\in\mathcal{M}_{k+2}$. If $\mathcal{M}_{k+2}$ happens to be one-dimensional, then a formula of the type (*) indeed holds with $a_k=k/12$. This means the minimal $k$ for which such a formula might fail is $k=10$ (so that $\mathcal{M}_{k+2} = \mathcal{M}_{12}$ is two-dimensional). In principle I know one can calculate a Fourier expansion for $\vartheta E_{10}$ and then write it in the basis of $\mathcal{M}_{12}$ by comparing coefficients, but I am having a hard time doing the calculation.

Even if (*) turns out to be false, I wonder if a similar-looking recurrence relation might still hold. Perhaps this is well-known result (I know very little about modular forms), but I couldn't find anything in my cursory glance of the literature.

So, I essentially have four questions:

  1. Does (*) hold for $k=10$?
  2. What about for higher $k$?
  3. Even if (*) is false, is there still some generalized recurrence formula on the derivatives?
  4. And if the answer to the previous question is "yes," where might I find the details?
  • 1
    $\begingroup$ For $k \le 20$ it suffices to check the first two coefficients (if $h$ of weight $< 24$ has a double zero at $i\infty$ then $\Delta^{-2} h$ is a weight $< 0$ modular form so it is $0$) $\endgroup$
    – reuns
    Feb 11 at 11:42

1 Answer 1


I conjectured that $D E_k - a E_{k+2} = \sum_m b_m E_m E_{k+2-m}$ for some integers $a$ and $b_m$ depending on $k$ assuming a basis for $\mathcal{M}_{k+2}$ and then solved for those integers. My numerical results include: $$ D E_8 = (E_2 E_8 - E_{10})2/3, $$ $$ D E_{10} = (735 E_2 E_{10} - 44E_6 E_6 - 691 E_{12})/882, $$ $$ D E_{12} = (E_2 E_{12} - E_{14}), $$ $$ D E_{14} = (3773 E_2 E_{14} - 156E_6 E_{10} - 3617 E_{16})/3234, $$ $$ D E_{16} = (39787 E_2 E_{16} + 4080 E_4 E_{14} - 43867 E_{18})4/119361, $$ $$ D E_{18} = (5307907 E_2 E_{18} - 244188 E_6 E_{14} - 5063719 E_{20})3/10615814, $$ $$ D E_{20} = (1890862519 E_2 E_{20} - 90714360 E_6 E_{16} - 1800148159 E_{22})5/5672587557. $$ Other similar expressions can be found but I have no proofs. The first case where more than two nonzero terms are needed in the finite summation is when $k=22$ and more than three nonzero terms are needed when $k=34$.

  • $\begingroup$ This is a start! Are you just using that $E_mE_{k+2-m}$ form a basis for $\mathcal{M}_{k+2}$ to get the decomposition, or did you have a more explicit method for getting the coefficients? $\endgroup$
    – 4plus4man
    Feb 12 at 23:14

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