transcendence of periods of CM elliptic curves

Question

Let $E$ be an elliptic curve over $\overline{\mathbb{Q}}$ defined by a Weierstrass equation $$ y^2=4x^3+g_2x+g_3. $$ Then $H^1_{dR}(E/\overline{\mathbb{Q}})$ is spanned by the classes of the differential forms $\omega_1=\frac{dx}{y}$ and $\omega_2=x\frac{dx}{y}$ and "the" period matrix of $E$ consists of the four numbers $$ \int_{\gamma_i} \omega_j $$ where $\gamma_1, \gamma_2$ are generators of $H_1(E(\mathbb{C}), \mathbb{Q})$.

Theorem. The degree of transcendence of the field spanned by these numbers is at least $2$.

I am trying to understand why this implies that $\int_\gamma \omega_1$ and $\int_\gamma \omega_2$ are algebraically independent whenever $E$ has complex multiplication. This will follow if one shows that the ratio between periods attached to the same differential form is algebraic.

For $\omega_1$ I can see what's going on: by uniformization, $$ E(\mathbb{C}) \simeq \mathbb{C} / \mathbb{Z} \oplus \mathbb{Z} \tau $$ for some $\tau$ in the upper half plane. Complex multiplication forces $\tau$ to be imaginary quadratic. The differential form $\omega_1$ corresponds to $dz$ on the complex torus and the periods become $\int_0^1 dz=1$ and $\int_0^\tau dz=\tau$, so the quotient is $\tau$ which is algebraic.

What about the other differential form? I know that $\omega_2=x \frac{dx}{y}$ corresponds to $\mathcal{P}(z)dz$ where $\mathcal{P}$ is the Weierstrass function attached to the lattice. Then the periods $\int_{\gamma_i} \omega_2$ are minus the periods of $\zeta$, the primitive of $-\mathcal{P}$. Why their quotient is algebraic in this situation?

Answer 1

For the elliptic curve $E$ in the original post, we have two periods $\lambda_i = \int_{\gamma_i} \frac{dx}{y}$, $i=1,2$, and quasi-periods $\eta_i = \int_{\gamma_i} \frac{x\,dx}{y}$, $i=1,2$.

Theorem (Schneider 1936): Assume that $E$ is defined over $\overline{\mathbb{Q}}$. I.e., $g_2$, $g_3 \in \overline{\mathbb{Q}}$.

1. Each of $\lambda_1$, $\lambda_2$, $\eta_1$, $\eta_2$ is transcendental over $\mathbb{Q}$.
2. If $E$ does not have CM, then $\tau := \lambda_1/\lambda_2$ is transcendental over $\mathbb{Q}$.
3. If $E$ has CM, then $\mathbb{Q}(\tau)$ is an imaginary quadratic field.

Schneider's results were improved over the years, starting with work of Baker, Coates, and Masser, to results on $\overline{\mathbb{Q}}$-linear combinations of periods, quasi-periods, and elliptic logarithms. Parts 2 and 3 of the following theorem should settle the original question.

Theorem (Masser 1975): Again assume that $E$ is defined over $\overline{\mathbb{Q}}$. Let $V$ be the $\overline{\mathbb{Q}}$-linear span of the six numbers $1$, $\pi$, $\lambda_1$, $\lambda_2$, $\eta_1$, and $\eta_2$.

1. If $E$ does not have CM, then $\dim_{\overline{\mathbb{Q}}} V = 6$.
2. If $E$ has CM, then $\dim_{\overline{\mathbb{Q}}} V = 4$.
3. If $E$ has CM, then the three numbers $\eta_1$, $\eta_2$, and $\lambda_1$ are linearly dependent over $\overline{\mathbb{Q}}$, and the ratio $\eta_1/\eta_2$ is transcendental over $\mathbb{Q}$ if and only if $g_2g_3 \neq 0$.

A couple of further comments:

• Masser's results appear in his book "Elliptic Functions and Transcendence," Springer Lecture Notes 437, 1975.
• For further accounts of these types of results and their history, I highly recommend Waldschmidt's articles "Transcendence of periods: the state of the art," Pure Appl. Math. Q. 2 (2006), no. 2, part 2, 435-463, and "Elliptic functions and transcendence," Surveys in number theory, Dev. Math. 17, Springer, 2008, pp. 143-188. Both are available on his web page.
• The theorem alluded to in the original post, stating that the transcendence degree of $\overline{\mathbb{Q}}(\lambda_1,\lambda_2,\eta_1,\eta_2)$ over $\overline{\mathbb{Q}}$ is at least $2$ (with or without CM), is due to G. V. Chudnovsky (1976).

Answer 2

"This will follow if one shows that the ratio between periods attached to the same differential form is algebraic." This cannot be right. Assume that the periods of $\omega _2$ are $\eta $ and $\alpha \eta $, with $\alpha $ algebraic; then the ratio of the periods for $\omega _1+\omega _2$ is $\alpha +\dfrac{\tau -\alpha }{\eta +1} \cdot\ $ If it is algebraic, either $\alpha =\tau $ or $\eta $ is algebraic; in both cases this contradicts Legendre relation $(\alpha -\tau )\eta =2\pi i$.

Answer 3

Consider the space $V$ of $\mathbb{\bar{Q}}$-linear homomorphisms $H^{1}(E(\mathbb{C}),\mathbb{Q})\otimes_{\mathbb{Q}}\mathbb{\bar{Q}}\rightarrow H_{dR}^{1}(E/\mathbb{\bar{Q}}{})$ commuting with the actions of the field of complex multiplication. This can be regarded as an algebraic variety of dimension 2 over $\mathbb{\bar{Q}}$. Now $V(\mathbb{C})$ contains a canonical element $p$ whose matrix is the period matrix. It follows that $\mathbb{\bar{Q}}(p){}$ has transcendence degree at most $2$ (1.7 of Deligne's notes), and therefore exactly 2. The rest is linear algebra.