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Recently in algebraic topology I was working with a certain graded ring $R$ equipped with an elliptic curve $C$. Now completion at the identity gives a 1-dimensional formal group $G$. This induces a module structure of the Lazard-Ring $\mathrm{MU}_*$ (or if simpler $\mathrm{BP}_*$ since I am ultimately interested in prime 2).

My problem now is that I only have an explicit description of the formal coordinate $-x/y=T+...$, and not of the elliptic curve or the formal group law (similarly to the scenario in Rezk notes for math 512). Is there any way I can now get the module structure? I know that the action of the $v_i$'s is given by the coefficients of the $[2]$-series, but I do not know a way to get this out of the coordinate.

I would also be happy if someone could give me a way to describe the formal group law or the Weierstraß equation.

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  • $\begingroup$ The formulae are spelled out in Silverman's "Arithmetic of elliptic curves" chapter IV. They are implemented in many place, for instance in sage do: sage: R.<a,b> = QQ[]; K = R.fraction_field(); E = EllipticCurve(K,[a,b]); Ehat = E.formal_group(); Ehat.mult_by_n(2, prec = 30) to get a series that starts like $2\,t - 12\,a\,t^5 - 54\,b\,t^7 + 44\,a^2\,t^9$. Not sure if that is what you are looking for. $\endgroup$ Commented Nov 3, 2023 at 12:57
  • $\begingroup$ Dear @ChrisWuthrich. If I understand your code right you start with an elliptic curve given in Weierstraß form and you extract Ehat out of it, but I need the opposite direction. I have Ehat and want E. $\endgroup$
    – Reihe27
    Commented Nov 3, 2023 at 13:25

1 Answer 1

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As far as I understand the question (I know elliptic curves, but I don't know what MU and BP are), the task is to express the coefficients of the Weierstrass equation given the series of multiplication by $2$ in its formal group. This can be done by the following: Suppose $[2](t) = 2\,t + c_2\, t^2 + c_3\, t^3 + \cdots$ and $E$ is given by a Weierstrass equation $y^2 + a_1\, xy + a_3\, y = x^3 +a_2\, x^2 + a_4\, x+ a_6$. Then

$\begin{align*} a_1 &= -c_2 \\ a_2 &= -c_3/2 \\ a_3 &= (a_1a_2-c_4)/7 \\ a_4 &= (2a_2^2-6a_1a_3-c_5)/12 \\ a_6 &= (-8a_1^3a_3 - 2a_2^3 - 14a_1a_2a_3 - 8a_1^2a_4 + 4a_3^2 + 4a_2a_4 - c_7)/54\end{align*}$

In particular, it is in short Weierstrass form if and only if $c_2=c_3=c_4=0$, in which case $A=-c_5/12$ and $B=-c_7/54$.

This follows from computing with sage as indicated above in the comment the series $[2](t)$ to get $ 2t - a_1 t^2 - 2a_2t^3 + (a_1a_2 - 7a_3)t^4 + (2a_2^2 - 6a_1a_3 - 12a_4)t^5 + (-a_1a_2^2 - 7a_1^2a_3 - 2a_2a_3 - 6a_1a_4) t^6 + (-8a_1^3a_3 - 2a_2^3 - 14a_1a_2a_3 - 8a_1^2a_4 + 4a_3^2 + 4a_2a_4 - 54a_6)t^7 + O(t^8) $

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