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?

  • $\begingroup$ Briefly, the existence of complex multiplications forces relations between the periods, and if the two you note were algebraically dependent, this would force the transcendence degree to be one. The discussion of these things at the end of Section 1 of Deligne's notes "Hodge Cycles on Abelian Varieties" may be helpful (available by googling). $\endgroup$
    – abz
    Nov 1, 2013 at 18:45
  • $\begingroup$ Thanks for your comment anon, but you are not really answering my question. I would like to know why the integrals $\int x\frac{dx}{y}$ over two generators $\gamma_1$ and $\gamma_2$ of the homology of $E(\mathbb{C})$ are algebraically dependent. I don't remember this is adressed in Deligne LNM 900. $\endgroup$
    – legrel
    Nov 1, 2013 at 19:19
  • $\begingroup$ Not directly, but the answer is obvious from the setup there. $\endgroup$
    – abz
    Nov 1, 2013 at 20:12
  • $\begingroup$ could you explain why it is obvious? $\endgroup$
    – legrel
    Nov 1, 2013 at 20:32
  • $\begingroup$ Proposition 1.6 proves indeed that the transcendence degree is $\leq$ the dimension of the Mumford-Tate group, which in that case is 2. But I guess there is a simpler proof, just using the function $\zeta$. Or there isn't? $\endgroup$
    – legrel
    Nov 1, 2013 at 20:43

3 Answers 3


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).
  • $\begingroup$ Isn't $\lambda_1=\int_0^1dz=1?$ $\endgroup$
    – shenghao
    Jun 10, 2015 at 3:54
  • $\begingroup$ @shenghao: If $E$ has period lattice $\mathbb{Z} + \mathbb{Z}\tau$, then Schneider's Theorem implies that one of the $g_2$, $g_3$ must be transcendental. If $j(\tau)$ is algebraic, then $E$ is isomorphic to a curve defined over $\overline{\mathbb{Q}}$, so there is some non-zero $\lambda \in \mathbb{C}$ so that the elliptic curve associated to $\mathbb{Z}\lambda + \mathbb{Z}\lambda\tau$ is defined over $\overline{\mathbb{Q}}$. In this case, we can take $\lambda_1=\lambda$ and $\lambda_2 = \lambda\tau$, and both numbers are transcendental. $\endgroup$ Jun 17, 2015 at 23:25
  • 1
    $\begingroup$ Maybe my confusion arises from the difference in terminology: the torus $\mathbb C/\mathbb Z+\mathbb Zi$ is not said to be ``defined over $\mathbb Q$", because of the transcendency of its $g_2,$ right? Even though it is biholomorphic to the algebraic curve $y^2=x^3-x$ base changed from $\mathbb Q$ to $\mathbb C.$ $\endgroup$
    – shenghao
    Jun 24, 2015 at 4:03

"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 $(\alpha -\tau )\eta =2\pi i$.

  • $\begingroup$ I'm really confused... Then how do you explain that the degree of transcendence of the periods of a CM elliptic curve is at most 2? $\endgroup$
    – legrel
    Nov 2, 2013 at 11:22
  • $\begingroup$ $1$ and $\tau$ are algebraic, so it is certainly $\leq 2$. $\endgroup$
    – abx
    Nov 2, 2013 at 11:30
  • $\begingroup$ I think I see where the mistake is. The elliptic curve is uniformized by $\mathbb{Z} \int_{\gamma_1} \omega_1 \oplus \mathbb{Z} \int_{\gamma_2} \omega_1$. Normalizing the lattice one gets $\mathbb{Z} \oplus \mathbb{Z} \tau$ with $\tau=\frac{\int_{\gamma_1} \omega_1}{\int_{\gamma_2} \omega_1}$ and CM forces this $\tau$ to be algebraic. So the periods $\int_{\gamma_1} \omega_1$ and $\int_{\gamma_2} \omega_1$ are algebraically dependent. But not algebraic! What happens with the other periods? $\endgroup$
    – legrel
    Nov 2, 2013 at 12:36
  • $\begingroup$ Well, they should be one more algebraic dependence relation, but why should that imply that it involves only the two other periods? $\endgroup$
    – abx
    Nov 2, 2013 at 13:09
  • $\begingroup$ I am confused my your confusion; you have said that the transcendence degree of the field generated by the four numbers $1,\tau$ and $\int _0^1P(z)dz$ and $\int _0^{\tau} P(z)dz$ is two; by assumption, $1,\tau$ are algebraic. So the other two periods are algebraically independent and are transcendental numbers. $\endgroup$ Dec 2, 2013 at 16:01

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.


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