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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?
4plus4man
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    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 $0$) – reuns Feb 11 '23 at 11:42

2 Answers2

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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 $k=34$.

Somos
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  • 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? – 4plus4man Feb 12 '23 at 23:14
  • The results in your answer can be deduced using the recurrence relation given by Ramanujan. I have added some details as an answer. – Paramanand Singh Jan 09 '25 at 16:09
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In his paper On certain arithmetical functions (Transactions of the Cambridge Philosophical Society, XXII, No.9, 1916, 159 – 184), Ramanujan described a very ingenious approach to his version of Eisenstein series and their derivatives and I describe the same below.

He uses the following notation in the paper: \begin{align} S_r(q) & := - \frac{B_{r+1}}{2(r+1)}+\sum_{n\geq 1}\frac{n^rq^n}{1-q^n}=- \frac{B_{r+1}}{2(r+1)}+\sum_{n\geq 1}\sigma_r(n) q^n\tag{1}\\ \Phi_{r,s}(q)&:=\sum_{m\geq 1}\sum_{n\geq 1}m^rn^s x^{mn} =\sum_{n\geq 1}n^s\sigma_{r-s}(n)q^n\tag{2} \end{align} where $B_r$ are Bernoulli numbers defined via $$\frac{x} {e^x-1}=\sum_{r\geq 0}B_r\cdot\frac{x^r}{r!}\tag{3}$$ The Eisenstein series $E_{2k}$ is related to function $S_{2k-1}$ via $$E_{2k}(\tau)=-\frac{4k}{B_{2k}}S_{2k-1}(q)\tag{4}$$ where $\tau$ is a complex number with positive imaginary part and $q=\exp(2\pi i\tau) $ so that $|q|<1$. One can observe that $\Phi_{r, s} =\Phi_{s, r} $ and further $$S_r(q) =-\frac{B_{r+1}}{2(r+1)}+\Phi_{0,r}(q)\tag{5}$$

The link between differential operators related to $\tau$ and $q$ is given by $$q\frac{d} {dq} \equiv \frac{1}{2\pi i} \frac{d} {d\tau} \tag{5}$$ and above is equivalent to $D$ operator in question. Next we can note that $$q\frac{d} {dq} \Phi_{r, s} (q) =\Phi_{r+1,s+1}(q)\tag{6}$$ and hence $$\Phi_{r, s} (q) =\left(q\frac{d} {dq} \right) ^r \Phi_{0,s-r}(q)\tag{7}$$ What Ramanujan does next is simply beyond imagination and he springs up two trigonometric series identities out of the blue: $$ \left(\frac{1}{4}\cot\frac{\theta}{2}+\sum_{n\geq 1}\frac{q^n}{1-q^n}\sin n\theta\right) ^2=\frac{1}{16}\cot^2\frac{\theta}{2}+\sum_{n\geq 1}\frac{q^n}{(1-q^n)^2}\cos n\theta+\frac{1}{2}\sum_{n\geq 1}\frac{nq^n}{1-q^n}(1-\cos n\theta) \tag{8}$$ $$\left(\frac{1}{8}\cot^2\frac{\theta}{2}+\frac{1}{12}+\sum_{n\geq 1}\frac{nq^n} {1-q^n}(1-\cos n\theta)\right)^2=\left(\frac{1}{8}\cot^2\frac{\theta}{2}+\frac{1}{12}\right) ^2+\frac{1}{12}\sum_{n\geq 1}\frac{n^3q^n}{1-q^n}(5+\cos n\theta) \tag{9}$$ Ramanujan proved the simpler identity $(8)$ in his paper (the same proof is available in this answer) and left the proof of the more difficult identity $(9)$ for readers as an exercise.

Equating coefficients of even powers of $\theta$ on both sides of equation $(9)$ gives rise to the following identity $$\frac{(n-2)(n+5)}{12(n+1)(n+2)} S_{n+3}(q)=\binom{n} {2}S_3(q)S_{n-1}(q)+\binom{n} {4}S_5(q)S_{n-3}(q) +\dots + \binom{n} {n-2}S_{n-1}(q)S_{3}(q)\tag{10}$$ where $n>2$ is even and $$\binom{n} {r} =\frac{n!} {r! (n-r)!} $$ is the usual binomial coefficient. This gives rise to a relation between sums $S_r(q) $ for odd $r$ and Ramanujan says that $$S_r(q) =\sum K_{m, n} Q^m(q) R^n(q) \tag{11}$$ where $m, n$ are non-negative integers with $4m+6n=r+1$ and $r$ is odd with $r>1$ and $K_{m, n} $ are rational numbers and $$Q(q) =240S_3(q)=E_4(\tau)$$ and $$R(q) =-504S_5(q)=E_6(\tau)$$ and for completeness $$P(q) =-24S_1(q)=E_2(\tau)$$ Ramanujan says that the formula $(11)$ can be proved via induction. This allows us to express Eisenstein series as polynomials in $E_4,E_6$. His presentation avoids elliptic function theory, modular forms and uses just plain algebraic manipulation.

Next Ramanujan equates coefficients of $\theta^n$ for even $n\geq 2$ in $(8)$ and gets the following identity $$\frac{n+3}{2(n+1)}S_{n+1}(q)-\Phi_{1,n}(q)=\binom{n} {1}S_1(q)S_{n-1}(q)+\binom{n} {3}S_3(q)S_{n-3}(q) +\dots + \binom{n} {n-1}S_{n-1}(q)S_{1}(q)\tag{12}$$ and he says that one can establish the following formula via induction $$\Phi_{1,s}(q)=\sum K_{l, m, n} P^l(q) Q^m(q) R^n(q) \tag{13}$$ where $l, m, n $ are non-negative integers with $l\leq 2$ and $2l+4m+6n=s+2$ and $s\geq 2$ is even. This gives us the derivatives of $\Phi_{0,s-1}$ or essentially of Eisenstein series $E_s$. In particular one gets the relations mentioned in question by putting $n=2,4,6$ in equation $(12)$.

Using equation $(7)$ Ramanujan concludes further that $$\Phi_{r, s} (q) =\sum K_{l, m, n} P^l(q) Q^m(q) R^n(q) \tag{14}$$ where $$2l+4m+6n=r+s+1$$ and $r, s$ are non-negative integers of different parity (ie $r-s$ is odd) and $l-1$ does not exceed $\min(r, s) $.

Ramanujan gave explicit expressions for $\Phi_{r, s} $ as polynomials in $P, Q, R$. As an example involving small numbers we have $$144\Phi_{1,14}=Q(3Q^3+4R^2-7PQR)$$ where we have dropped the variable $q$ to reduce typing effort. This can be translated in terms of Eisenstein series by noting that $$\Phi_{1,14}=D\Phi_{0,13}=DS_{13}=-\frac{B_{14}}{28}DE_{14}=-\frac{1}{24}DE_{14}$$ and hence $$DE_{14}=\frac{E_4(7E_2E_4E_6-4E_6^2-3E_4^3)}{6}$$