How can we continuously deform a height 1 formal group law into a height 2 formal group law?

273 Views Asked by At

A Quick Review:

The complex elliptic curve $\mathbb{C}/(\mathbb{Z} + \tau \mathbb{Z})$ may be rewritten using the exponential, $\text{exp(}{2 \pi i \tau}) =: q$ as $\mathbb{C}^\times/q^\mathbb{Z}$ .

The Tate-Jacobi theorem states:

There is a canonical formal group law associated to the complex elliptic curve $C^\times/q^{\mathbb{Z}}$ whose coefficients lie in the subring $\mathbb{Z}[E_4(q), E_6(q)]$ of $\mathbb{C}$, where the Eisenstein series $E_4$ and $E_6$ are convergent power series in $q$. After adjoining $q$, this formal group law becomes canonically isomorphic to the multiplicative one.

If I understand correctly, this theorem allows us to, among other things, continuously deform the multiplicative formal group law into a height 1 elliptic curve formal group law.

If we are to view the curve $\mathbb{C}^\times/q^\mathbb{Z}$ as a continuous family of groups fibered over the interval $[0,1)$

  • at $q=0$, we have the punctured complex plane, $\mathbb{C}^\times$, i.e., $\mathbb{G}_m$
  • for each $q$ such that $0 < |q| < 1$, we have ordinary elliptic curves

This seems analogous to the deformation of a supersingular elliptic curve to an ordinary one:

$$y^2 = 4x^3 + tx^2 + 2x$$

  • at $t = 0$ we have a supersingular elliptic curve, $y^2 = x^3 + x$ (up to rescaling of $x$) over Spec $W(F_9)[t]$
  • at $t \neq 0$, $t \in$ Spec $W(F_9)[t]$, we have ordinary elliptic curves

Here's my question: Do we have a "Tate-Jacobi-like" theorem for such a height 2 to height 1 family of elliptic curves (for some subring of a finite field, rather than a subring of $\mathbb{C}$)?

That is, how can we continuously deform a height 2 formal group law into a height 1 formal group law (in a somewhat canonical way)?

1

There are 1 best solutions below

0
On BEST ANSWER

Here’s a worm’s-eye view, perhaps not much help to you. You may always start with a supersingular curve in characteristic $p$, like $p=13$, where $y^2=x^3+x+4$ is supersingular. I choose $13$, because neither $y^2=x^3-x$ nor $y^2=x^3+1$ is supersingular there. Then you may always take your constant supersingular curve and jiggle it algebraically, here you could take $y^2=x^3+(1+t)x+4$ over $\Bbb F_{13}[t]$, to get a curve that’s ordinary for all but a finite number of values of $t$, among them $t=0$. Then, if you wish, you can take $y^2=x^3+(1+t)x+4$ as a curve over $W(\Bbb F_{13})[t]$. But I think you probably want something less ad hoc.