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What is the geometric meaning of $\omega=dx/(2y+a_1x+a_3)$ for an elliptic curve?

This question is an adjunct to MO Q1 on formal laws and L-series, which motivated Q2. Q1 (Silverman) and Darmon (pg. 6) state:

The invariant holomorphic differential form (Neron differential) attached to an elliptic curve is

$\omega=dx/(2y+a_1x+a_3)$.

(Ancilliary question: Relation to Weierstrass's elliptic functions?)

I'd like to broaden the question as a community wiki to ask, "What are some interesting manifestations of this one-form in various families of elliptic curves?"

E.g., J. Hoffman in Topics in Elliptic Curves and Modular Forms gives for the Jacobi quartic family of elliptic curves

$\omega=dx/(1+2\kappa x^{2}+x^{4})^{1/2}=\sum_{n=0}^{\infty}L_{n}(\kappa)x^{2n}dx$

with $L_{n}(\kappa)$ the Legendre polynomials.

edited Feb 24 2012 at 9:51

Tom Copeland

community wiki

4 revisions

Tom, the link to Weierstrass's elliptic functions seems to be broken. Also, do $a_1$ and $a_3$ come from the Weierstrass form of the equation? (Unfortunately, I don't have an answer to your question) – BR Dec 4 2011 at 6:22

There should be a square root in the denominator, no? – Mariano Suárez-Alvarez Dec 4 2011 at 6:48

Mariano, I don't think so (see p.6 of Darmon's article, and I checked Silverman's book in case that was a typo). – BR Dec 4 2011 at 7:28

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I'm not entirely sure exactly what kind of answer you are looking for, but for me, the invariant differential represents the fact that any algebraic group has trivial tangent bundle. See e.g. mathoverflow.net/questions/73824. The invariant differential is a choice of non-zero global section. In fact any non-zero scalar multiple of what you have written down will also be an invariant differential. – Daniel Loughran Dec 4 2011 at 8:41

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Depends on what level you want an answer. Your $\omega$ is only a Neron differential if you've started with the id component of a Neron model (a minimal Weierstrass equtation). The explanation in your answer below is okay over $\mathbb{C}$, but not over fields of charac 2 or 3. As noted, over a field, $E$ is a group variety of dim 1, so has a trivial (co)tangent bundle, so has a (unique up to scalar) everywhere holomorphic nonvanishing diff'l form. That's your $\omega$. A key fact is its translation invariance, $\tau_P^*\omega=\omega$ ($\tau_P$ = translation by $P$). – Joe Silverman Feb 21 2012 at 12:28

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Tom Copeland · Accepted Answer · 2012-02-24 09:57:10Z UTC

A paper by John Tate (pg. 1 and 2) gives a clear derivation of the diff. form:

Reparametrize the elliptic curve

$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4 x+a_6$

with $p(z)=x+(a_1^2+4a_2)/12$ and $p^{'}(z)=2y+a_1x+a_3$ to obtain

$(p^{'})^2=4p^3-g_2p-g_3$, defining the Weierstrass elliptic fct., and

$\omega=dp(z)/p^{'}(z)=dz=dx/(2y+a_1x+a_3)$.

Per Dan's comment, a coordinate transformation of $x=u^2x^{'}+r$ and $y=u^3y^{'}+su^2x^{'}+t$
leaves $\omega^'=u\omega$.

Given $\sigma=p(z)$ and the inverse $z=p^{-1}(\sigma)$,

$dz=(p^{-1}(\sigma))^{'}d\sigma=(p^{-1}(\sigma))^{'}p^{'}(z)dz$, so

$(p^{-1}(\sigma))^{'}=1/p^{'}(z)$ and $dz=d\sigma/p^{'}(z)=\omega$.

The amplitwist interpretation of differentiation and inversion presented by Tristan Needham in his book Visual Complex Analysis provides a geometric interpretation of these differential relations.

Consider as an analogy $P(\theta)=sin(\theta), P^{'}(\theta)=cos(\theta), and P^2+(P^{'})^2=1$.

Geometric picture of invariant differential of an elliptic curve

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