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Let $E$ be an elliptic curve. I want to consider its degeneration to the union of two projective lines $C:=\mathbb{P}^1 \cup_{x,y} \mathbb{P}^1$ attaching at two points $x,y$. The involution $-1$ on $E$ can "degenerates" to an involution on $C$ that preserve each component $\mathbb{P}^1$ and swaps $x$ and $y$.

My question is, can the translation by a 2-torsion point on $E$ degenerate to an involution on $C$? If this is possible, what kind of action will it be?

More generally is it true that for any translations $t_1, t_2$ on $E$, there exists a degeneration $C$ of $E$, such that all $t_1, t_2, -1$ degenerate to some action on $C$?

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$X=C$, right? $\ $ –  Allen Knutson Jun 3 '13 at 12:41
    
Yes, thank you for the correction. –  taryn Jun 3 '13 at 16:41

2 Answers 2

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Think of the group $E$ as degenerating to the group $C_{reg} \cong {\mathbb G}_m \times {\mathbb Z}/2$. In that group, the $2$-torsion is ${\mathbb Z}/2 \times {\mathbb Z}/2$, where one generator rotates each component of $C$ by $-1$, and the other one switches the two components.

EDIT: I'm very disturbed by the fact that $E$ can first degenerate to a one-nodal elliptic curve, whose group is only ${\mathbb G}_m$, i.e. with $2$-torsion only ${\mathbb Z}/2$. I retract my claim that one should think of $C_{reg}$ as a group, instead of a $({\mathbb G}_m \times {\mathbb Z}/n)$-space, as in S. Carnahan's answer. In that version, it becomes clear that the ${\mathbb Z}/n$ does indeed permute the nodes.

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Thank you for the answer. That's what I expected, but cannot really convince myself. Just to make sure, the second involution switches the two nodes, right? –  taryn Jun 3 '13 at 16:44
    
You can think of $C_{reg}$ as a group, because "generalized elliptic curves" come with distinguished identity sections to the smooth locus. –  S. Carnahan Jun 3 '13 at 23:48
    
But evidently not distinguished 2-torsion sections? What's bugging me is that I originally thought "let's follow the 2-torsion into the degeneration, and see that it becomes the 2-torsion in $C_{reg}$" but that doesn't make if I go to one node first. –  Allen Knutson Jun 4 '13 at 2:16

Concerning your general question, any finite order group of translations can be degenerated into an action on some Néron $n$-gon. An $n$-gon is made by taking $n$ copies of $\mathbb{P}^1$, parametrized by $\mathbb{Z}/n\mathbb{Z}$, and gluing the $\infty$ point of the $i$th copy transversely to the $0$ point of the $i+1$st copy for each $i$. An $n$-gon has a natural action of $\mathbb{Z}/n\mathbb{Z} \times \mathbb{G}_m$, where the $\mathbb{G}_m$ action fixes the nodes and the $\mathbb{Z}/n\mathbb{Z}$ rotates the polygon, so varying $n$ lets you encompass any choice of torsion. I think the Deligne-Rapoport article in Springer Lecture Notes 349 (Modular functions of one variable III) is the basic reference for these Néron polygons and the degeneration of torsion groups.

If you have a pair of translations that generate an infinite group, then your base field necessarily has torsion-free elements in the multiplicative group. In particular, things become easier, because you can use a "1-gon" nodal curve, with an action of $\mathbb{G}_m$, and send certain torsion-free elements to your favorite non-roots-of-unity.

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Thank you for the detail answer. –  taryn Jun 3 '13 at 16:46

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