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Let $q=e^{2\pi i\tau}$ and $$E_2(\tau) = 1 - 24 \sum_{n=1}^\infty\frac{nq^n}{1-q^n}$$ be the Eisenstein Series of weight $2$ and let $E_2^*(\tau) = E_2(\tau) - \frac{3}{\pi\cdot Im(\tau)}$ be the corresponding almost holomorphic modular form. Then let $$\eta(\tau)=e^{\pi i\tau /12}\cdot\prod_{n=1}^\infty(1-q^n)$$ be the Dedekind $\eta$-Function.

Question: How can I prove the following statement from here:

Let $\tau$ be any Complex Multiplication point. By basic theorems of complex multiplication, if you choose a suitable period $\omega(\tau)$, $E_4(\tau)/\omega(\tau)^4$, $E_6(\tau)/\omega(\tau)^6$, and $\sqrt{D}E_2^*(\tau)/\omega(\tau)^2$ (with $E_2^*(\tau)=E_2(\tau)-3/(\pi \cdot Im(\tau))$ and $D$ the discriminant of $\tau$) will be algebraic numbers of known degree, and if you choose $\omega(\tau)=\eta(\tau)^2$, they will even be algebraic integers.

Partial Solution for $E_2^*$:

Francois Brunault pointed out that this statement can be found in Prop. 5.10.6 on p. 202 of Cohen/Strömberg's book Modular Forms: A Classical Approach. Unfortunately, the proof of this proposition starts with "we only prove algebraicity, not the integrality". Who can help with proving the integrality? Since I am no expert in complex multiplication, I am looking for a rather detailed answer, or for a reference.

Complete Solution for $E_4$ and $E_6$:

The statement that $\frac{E_4}{\eta^8}$ is an algebraic integer follows from $$\left(\frac{E_4}{\eta^8}\right)^3 = j(\tau)$$ and the statement that $\frac{E_6}{\eta^{12}}$ is an algebraic integer follows from $$\left(\frac{E_6}{\eta^{12}}\right)^2 = j(\tau)-1728$$ with the absolute invariant $j(\tau)=\frac{1728E_4^3}{E_4^3-E_6^2}=\frac{E_4^3}{\eta^{24}}$ which is an algebraic integer (see Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Theorem 6.1, p. 140).

EDIT: Complete Solution for $E_2^*$:

The answer of Michael Griffin (see below) can now be found in more details in the appendix of this arXiv-preprint.

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  • $\begingroup$ Looking back through your question, you have a few identities that are not quite correct that you might want to fix... second to last line in your definition of $j(\tau)$ you have $E_4^3-E_6^2 = 1728 \eta^{24}$ not $\eta^{24}$. The correct definition of $j(\tau)$ aligns with $\frac{E_4^3}{\eta^{24}}$. Similarly, $\frac{E_6^2}{\eta^{24}}=j(\tau) -1728$ not $j(\tau)-1$ $\endgroup$ – Michael Griffin Mar 20 at 16:44
  • $\begingroup$ oops! I corrected it $\endgroup$ – L. Milla Mar 21 at 5:16
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I have seen this statement about $E_2^*$ tossed around off-handedly by experts a number of times, but never seen a complete proof referenced.

The tools to prove it are (mostly) in Masser's "Elliptic functions and transcendence", Appendix 1. There, Masser gives a formula for certain non-holomorphic modular functions. One of these fomrulas (Lemma A3) can be re-written as $$E_2^*(\tau)\left(\frac{\pi}{\omega_1}\right)^2=-\frac{3S}{\sqrt{D} ~\tau},$$ where $(\omega_1,\omega_2)$ are choices of periods of a CM elliptic curve with rational equation, $D$ is the discriminant of the CM point $\tau=\frac{\omega_1}{\omega_2}$, and $S$ is a sum of division points on the curve. If $\tau$ satisfies the reduced, integral quadratic, $C\tau^2+B\tau+A$, then Masser points out that by a theorem of Baker, $(AC)^2\wp$ is an algebraic integer. Moreover, the norm of $\tau$ is $A/C$, and so it's clear the only additional primes that could divide the denominator are divisors of $AC$.

On page 118 of Masser, he offers formulas for the function $$\gamma(\tau)=\frac{E_2^*(\tau)E_4(\tau)}{6E_6(\tau) j(\tau)}-\frac{7j(\tau)-6912}{6j(\tau)(j(\tau)-1728)}$$ at CM points, in terms of singular moduli of $j(\tau)$. The Gross-Zagier formula (which gives a factorization for the norm of differences of singular moduli) can then be used to show that no primes that split in the CM field can divide the denominator. Any prime dividing $A$ or $C$ is either split or ramified in the CM field. None of the split primes can divide the denominators. A more careful use of the Gross-Zagier formula shows that the ramified primes appear no more than expected, and so $\sqrt{D}E_2^*(\tau)\left(\frac{\pi}{\omega_1}\right)^2$ must be an algebraic integer.

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  • $\begingroup$ Thank you for this great answer! Looking into the details of Baker's Theorem, one can see that $AC\wp$ is also an algebraic integer (without the square). Thus with Masser's Appendix it follows that $(AC)^2\cdot\sqrt{D}E_2^*(\tau)/\omega(\tau)^2$ is an algebraic integer which is enough for my means - and so I don't need the Gross-Zagier-Formula. $\endgroup$ – L. Milla Mar 20 at 8:34
  • $\begingroup$ @L.Miller Great! $\endgroup$ – Michael Griffin Mar 20 at 16:38
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The statement about $E_2^*(\tau)$ is Proposition 5.10.6 in Cohen-Strömberg, Modular forms: a classical approach.

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  • $\begingroup$ Thank you! Unfortunately the proof of 5.10.6 begins with "we only prove algebraicity, not the integrality". How can one prove the integrality? $\endgroup$ – L. Milla Oct 4 '18 at 6:26
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    $\begingroup$ @L.Miller In principle, you can follow the proof of Prop 5.10.6 to get the minimal polynomial $\sqrt{D} E_2^*(\tau)$. Its coefficients are universal homogeneous polynomials in $E_4,E_6$ so it is a matter of checking whether these universal polynomials have integral coefficients (note that the writing as $P(E_4,E_6)$ is not unique, so you may need to change it). $\endgroup$ – François Brunault Oct 4 '18 at 10:52
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    $\begingroup$ @L.Miller A precision: it is not hard to show as an exercise that the algebra of modular forms with integral coefficients is $\mathbb{Z}[E_4,E_6,\Delta]$ (hint: starting with $f \in M_k(\mathbb{Z})$, substract a suitable monomial in $E_4$ and $E_6$, and then divide by $\Delta$ to reduce the weight). So you just need to check that the modular forms which appear are in $\mathbb{Z}[[q]]$ which shouldnot be too hard. Another reference is Zagier's article in the 1-2-3 of modular forms. $\endgroup$ – François Brunault Oct 4 '18 at 11:23
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    $\begingroup$ I should add that the purported strategy in my above comment doesn't work as stated since the modular forms which appear do not always have integral coefficients. I think that further knowledge of the theory of CM is needed $\endgroup$ – François Brunault Feb 7 at 21:33

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