Suppose we have an elliptic curve E with a point $P$ of order $5$ over a field of characteristic $0$. Denote $E'$ the curve $E/\langle P\rangle$. Now let $x$ (resp. $x'$) be a function on $E$ (resp. $E'$) with a double pole at $O_{E_1}$ (resp. $O_{E'}$) and let $f : E \rightarrow E'$ be an isogeny of degree 5 with kernel $\langle P\rangle$. Now let $U$ be the function of degree $5$ such that $x' \circ f = u \circ x$ ($x$ and $x'$ define a map to $\mathbb{P}^1$). We let $C_f$ denote the hyperelliptic curve over $K = k(t)$ where $T$ is a free parameter, defined by the affine equation

$$ C_f: y^2 = u(x) - T $$

If we have the following family of elliptic curves with $E_U : y^2 +(1-U)xy -Uy = x^3 -Ux^2$ with $(0,0)$ a point of order $5$. Then by Vélu's formulae (Jacques Vélu, Isogénies entre courbes elliptiques. C. R. Acad. Sci. Paris Sér. A-B 273, 1971) we can get an explicit description of the morphism $f$. Projecting to the $x$ coordinate then gives us the following family of hyperelliptic curves.

$$ C_5(U,T) : Y^2 = (1-Z)^3 +UZ((1-Z)^3 +UZ^2 - Z^3(1-Z)) - TZ^2(Z-1)^2. $$

I'm trying to see that this family is two dimensional (i.e. it isn't a one dimensional family in disguise), I've tried using the Igusa invariants, but that gets very messy. Does anyone have a solution?

  • 3
    $\begingroup$ I changed $<P>$ to $\langle P\rangle$. That is standard usage. $\endgroup$ Sep 26 '13 at 19:46
  • $\begingroup$ This might be a silly question, but what do you mean by "is two dimensional"? Do you mean that it is not a one-dimensional family in disguise? $\endgroup$ Sep 26 '13 at 21:38
  • $\begingroup$ Yes, exactly. I edited it. $\endgroup$
    – Han-Solo
    Sep 26 '13 at 21:58
  • $\begingroup$ I'm also not quite sure what happens behind the scenes when you say "we get a family". How is $C_5$ related to $E_U$ and $C_f$? Please write down the morphism. $\endgroup$ Sep 27 '13 at 1:10

I guess I found the answer: We want to show that the family of curves

$ C(U,T) : Y^2 = (1-Z)^3 + UZ((1-Z)^3 + UZ^2 - Z^3(1-Z)) - TZ^2(Z-1)^2 $

covers all isomorphism classes of hyperelliptic curves of genus $2$ with real multiplication by $[\frac{1+\sqrt{5}}{2}]$. We construct this family as follows: Take an ellliptic curve $E$ with a rational point $P$ of order $5$. Pick a point $T$ on $E$ and considering the image $H$ of the set $\{T, T+P, T+2P, T+3P, T+4P\}$ under the map $x_1: E \rightarrow \mathbb{P}^1$, where $x_1$ is a function on $E$ with a double pole at $O_E$. Now the curve $C(E_T)$ is the double covering of $\mathbb{P}^1$ ramified at the points of $H$ and the image of $O_E$ under $x_1$. Because any elliptic curve with a point of order $5$ can be written as

$ E: y^2 + (1-U)xy - Uy = x^3 - Ux^2, $

we write $C(E_T) = C(U,T)$.

First we recall a criterion due to Humbert.

Suppose that $\pi$ is a map from $C$ onto a plane conic $Q$ with branch points $P_1,P_2,P_3,P_4,P_5$ and $P_6$. Then $C$ has real multiplication by $\mathbb{Z}[\frac{1 + \sqrt{5}}{2}]$ if and only if for some ordering of the branch locus there is a plane conic $R$ passing through $P_6$ which is tangent to each of the edges of the pentagon $P_1P_2P_3P_4P_5$.

Suppose we have a curve $C$ with real multiplication by $\mathbb{Z}[\frac{1 + \sqrt{5}}{2}]$. Consider the two conics $Q,R$ obtained from the above criterion. We construct the elliptic curve $E$ by $E = \{(P,L) \in Q \times R^* | P \text{ lies on } L \}$. It is clear from the construction that the point $(P_1,L_1)$ for some $P_1 \in L_1$ has order $5$ on the curve $E$.

Now suppose we have an elliptic curve $E$ with a point $P$ of order $5$. We project to a conic $Q$ and consider the image of $H = \{T,T + P, T + 2P, T + 3P, T + 4P\}$ for some point $T \in E$. Now take a pentagon $G$ spanned by the image of $H$ and construct a conic $R$ inscribed to the sides of $G$. This conic $R$ will intersect $Q$ in some point, so by the criterion above this is equivalent to a hyperelliptic curve with real multiplication by $\mathbb{Z}[\frac{1 + \sqrt{5}}{2}]$.

These two constructions are inverse to eachother, so we've showed that any hyperelliptic curve of genus $2$ with real multiplication by $\mathbb{Z}[\frac{1 + \sqrt{5}}{2}]$ can be written in the above form.

  • $\begingroup$ Then you still have to show that the space of genus 2 curves with real mult. by $Z[(1+\sqrt{5})/2]$ is two dimensional, right ? $\endgroup$
    – Cantlog
    Oct 3 '13 at 16:22

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