Completion along zero section of an elliptic curve.

I am trying to understand the intuition that I should have about the formal group of an elliptic curve. Say that I have an elliptic curve $E\to \text{Spec} R$ for some ring $R$, with section $0\colon \text{Spec} R\to E$. My first question is: when I hear speaking about the "completion of $E$ along $0$", should I think that such a thing is the formal scheme whose underlying topological space is $0(\text{Spec} R)$ and whose sheaf of rings is the completion of $\mathcal O_E$ with respect to the ideal sheaf defining $0(\text{Spec} R)$ in E? And what is the relation of this object with the formal group of $E$? My second question is: say that I have a nowhere vanishing differential $\omega \in H^0(E,\Omega_{E/R}^1)$. I somehow have this idea (but I can't understand how true is it) that completion along the zero section tells us about "Taylor expansion" of $\omega$. How does one formalize that? Also, is the sheaf $\Omega_{E/R}^1$ always globally isomorphic to $\mathcal O_E$? or is it just invertible? Thank you in advance if you're willing to help me!

Solution 1:

Most of this post is a quilt of examples and comments others have offered me when asked about formal schemes, which I've found very helpful. Hopefully, they'll help you.

This is just following up on Sebastian's answer to explain why your intuition is right: the formal completion of E along the "0"-section is like the Taylor series expansion of E about the origin.

As you know, Taylor series are usually of the form $f(x+y) = f(x) + yP_1(x) + y^2P_2(X) + ...$, where $P_i(x) = \frac{f^{(i)}(x)}{i!}$. However, this uses the exponential, which fails to make sense, even as a formal power series, in characteristic $p$.

In some sense, 'completion at a point $t=0$' gives you the Taylor series of a map in the algebraic-geometric setting, but I'd like to stress that it's more general than this. Instead of a plain old power series $\text{A}[[t]]$, we'll get a topologized power series with the $t$-adic topology, $\text{Spf } A[[t]]$.

Spf is the "formal spectrum of a ring," its worth noting explicitly that its topology is different from Spec: $$\text{colim }(\text{Spec } A[t]/t^n) =: \text{Spf } A[[t]]$$ $$\text{Spec } (\text{lim } A[t]/t^n) =: \text{Spec } A[[t]]$$

Note that a formal group law is the formal spectrum of a power series ring.

Keep in mind that a formal scheme can be thought of as an algebraic replacement of tubular neighborhoods (when we consider the formal completion of a subvariety inside the ambient variety).

Let's look at an elliptic curve $C \to \text{Spec } A$.

$\text{Spec } A[t]/t^2$ is like truncating the Taylor series after the information given by the first derivative, it's the the first infinitesimal neighborhood of $\text{Spec }A$.

Similarly, $\text{Spec } A[t]/t^3$ is the second infinitesimal neighborhood, and we need a slightly larger infinitesimal neighborhood to take the second derivative.

I want you to imagine:

• $\text{Spec } A$ as a line, sitting on $t=0$,
• $\text{Spec }A[t]/t^2$ as an infinitesimal normal bundle sticking out of that line
• $\text{Spec }A[t]/t^3$ as a slightly fatter infinitesimal normal bundle sticking out of that line, etc.

In the case of a formal group law, we want to write down all of the derivatives, so we take the colimit of $\text{Spec } A[t]/t^n$, which is by definition the formal scheme $\text{Spf } A[[t]]$.

Zariski locally, the completion of $C$ is isomorphic to the formal scheme, $\text{Spf }A[[t]]$.

For example: Zariski locally, over $\text{Spec } A$, we can put our elliptic curve in Weierstrass normal form $y^2 + a_1 xy = x^3 + a_2 x^2 + a_4 x + a_6$ (outside of characteristic 2 and 3, we can take $a_1 = 0$).

In this form, the variable $z = x/y$ is a uniformizer at the identity ($y$ has a pole of order 3 at the identity, $x$ has one of order 2, so $x/y$ has zero of order 1). So it's reasonable to expect $\text{Spf }A[[z]]$ to be the formal completion along the identity.

Solution 2:

Regarding the first question: You're right. If we have an elliptic curve over a base scheme $\operatorname{Spec}(R)$, the zero section corresponds to a surjective map $\mathcal{O}_E \to \mathcal{O}_{\operatorname{Spec}(R)}$ whose kernel is the ideal sheaf $I$ defining $\operatorname{Spec}(R)$ in $E$. The formal completion of $\mathcal{O}_E$ with respect to $I$ gives the formal completion, which is by definition the formal group of the elliptic curve $E$. (Note that $E$ is an abelian group)

Regarding the Taylor expansion I'm not sure, but be aware that if you have a nowhere vanishing differential $\omega$ as you say, this is a nowhere vanishing global section of the line bundle $\Omega_{E/R}^1$. Thus the line bundle is trivial and hence globally isomorphic to $\mathcal{O}_E$.