# Algebraic independence of $P,Q,R$ or $E_2,E_4,E_6$ over $\mathbb C(z)$

Let $$P,Q,R$$ be the Fourier series of the Eisenstein series $$E_2,E_4,E_6$$, that is,

$$P(q)=1-24\sum_{n=1}^{\infty}\sigma_1(n)q^n,$$

$$Q(q)=1+240\sum_{n=1}^{\infty}\sigma_3(n)q^n,$$

$$R(q)=1-504\sum_{n=1}^{\infty}\sigma_5(n)q^n.$$

I heard that $$P,Q,R$$ are algebraically independent over $$\mathbb C(q)$$, due to Mahler, but I cannot find a reference. Could you help me if you know the reference?

Meanwhile, I found 'On Algebraic Differential Equations Satisfied by Automorphic Functions', by Mahler, 1969. I do not think that this paper addresses this topic, but if it does, why does it imply the algebraic independence of $$P,Q,R$$?

• The problem for just $Q$ and $R$ is settled at mathoverflow.net/questions/282554/… – Gerry Myerson Oct 30 '18 at 11:55
• What about $\mathbb C (z)$? – LWW Oct 30 '18 at 12:05
• @GerryMyerson OP's question is about independence over $\mathbb{C}(z)$, not $\mathbb{C}$, so a harder question. It may be that using the functional equation for the Eisenstein series will reduce to the easier problem but it's not obvious. A harder question, that probably gives an answer to OP's question, is the Mahler-Manin conjecture, proved by K. Barré-Sirieix et al., (Invent. Math. 124 (1996), 1-9). That paper might contain a proof or a reference to OP's question. – Felipe Voloch Oct 30 '18 at 16:08
• Voloch // why Mahler-Manin conjecture implies the question?? I didn't understand. – LWW Oct 31 '18 at 2:25
• If you find $z$ such that $P(z),Q(z),R(z)$ are all algebraic, then by Mahler-Manin, $z$ is transcendental. This wouldn't happen if the functions were algebraically dependent. Have a look at the comment after Theorem 4 of Waldschmidt's Bourbaki talk n° 824. I think it completely answers your question. – Felipe Voloch Oct 31 '18 at 4:53

## 1 Answer

I think the follwing is the answer of this question. It is not depend on the above comments because I didn't understand them. This is just my approach. (But not fully my idea because it depends on a strong proposition which is already known, and the remaining part is just a corollary.)

Proposition. Let $$f$$ be a modular form of weight $$k$$, defined over $$\mathbb Q^{\mathrm{alg}}$$. If $$f$$ is non constant, then the functions $$f, Df$$ and $$D^2 f$$ are algebraically independent over the function field $$\mathbb C(q)$$.

The proof is in the book, 'Introduction to Algebraic Independence Theory', Y.V.Nesterenko and P.Philippon, (Eds), Ch.1 prop 1.1. Here $$q=e^{2\pi i \tau}$$ and $$D=\frac{1}{2\pi i}{\tau}=q\frac{d}{dq}$$.

Let $$\Delta = E_4^3-E_6^2$$, cusp form of weight 12. Due to Ramanujan, we have the following formulas :

$$DP=\frac{1}{12}(P^2-Q), \quad DQ=\frac{1}{3}(PQ-R), \quad DR=\frac{1}{2}(PR-Q^2).$$

Thus we have $$D\Delta = P\Delta$$. Take $$D$$ in both side and do some elementary algebra, we have

$$Q=13\frac{(D\Delta)^2}{\Delta^2}-12\frac{D^2 \Delta}{\Delta}.$$

By the proposition, $$\{\Delta, D\Delta, D^2\Delta\}$$ is algebraically independent over $$\mathbb C(q)$$, so $$\{Q^3-R^2,P,Q\}$$ is algebraically independent over $$\mathbb C(q)$$, so $$\{P,Q,R^2\}$$ is algebraically independent over $$\mathbb C(q)$$. This is equivalent to say that $$R^2$$ is transcendental over $$\mathbb C(q)[P,Q]$$, and this implies that $$R$$ is transcendental over $$\mathbb C(q)[P,Q]$$, i.e. $$\{P,Q,R\}$$ is algebraically independent over $$\mathbb C(q)$$.

• It's worth mentioning Zagier's generalization of Jacobi and Ramanujan's D-algebraic dependence, see people.mpim-bonn.mpg.de/zagier/files/doi/10.1007/… – Igor Pak Nov 4 '18 at 20:06
• Pak // Could you let me know where can I find those contents? I tried to find it in the textbook you refered but fail... – LWW Nov 4 '18 at 22:46
• Don't remember the exact place. Start with p. 49 for D-alg eq. but also see short section on p. 85 for alg independence. – Igor Pak Nov 4 '18 at 22:52