Say I have a plane cubic $f(x,y,z) \subset \mathbb{C}^3$ and I identify it with an elliptic curve by setting $z=1$ and end up with (perhaps after a change of variables) something of the form \begin{equation}\label{e} x^3+a_2x^2+a_4x+a_6=0. \end{equation} Is the $j$-invariant for this curve even defined? A normal form is \begin{equation} y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 \end{equation} and the $j$-invariant is given by a simple formula in the coefficients. But this formula assumes the curve is in normal form and in particular that the coefficient of the $y^2$ term is 1. It seems to me that the first curve is not in normal form.

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    $\begingroup$ There are no $y$'s at all in your first equation. Is that what you intended? $\endgroup$ – David E Speyer Oct 16 '11 at 16:04
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    $\begingroup$ Yes, and it's the source of my confusion. That "curve" would just be the union of three lines, correct? So if the $j$-invariant is defined at all I thought it would be some special value. $\endgroup$ – Derek Allums Oct 16 '11 at 19:23
  • $\begingroup$ I guess I could be more concise in my original question: is the $j$-invariant defined only when the curve can be put in normal form? Or does this normal form just happen to lend itself to easy computation of the $j$-invariant? $\endgroup$ – Derek Allums Oct 18 '11 at 0:48

There is no good way to define a $j$-invariant for this cubic.

As you have recognized, this cubic consists of three lines, and is thus singular. Properly speaking, the $j$-invariant is only defined for smooth cubics.

However, the situation is worse than that. The homogenous cubic $xyz$ has zero locus in $\mathbb{P}^2$ consisting of three lines. Morally, though, there is a sense in which $xyz$ has $j$ invariant $\infty$. Namely, let $F_t(x,y,z)$ be a family of cubics, depending on a parameter $t$, such that $F_0=xyz$ and such that $F_t$ is smooth for $t$ near $0$ but nonzero. Then $\lim_{t \to 0} j(F_t)$ will be $\infty$. I don't see how to give a low-tech proof of this, but the high tech reason is that $xyz=0$ is a semi-stable curve. The reason stable and semi-stable curves are so-named is because the behavior of invariants such as $j$ is continuous, or "stable", when they are perturbed.

By contrast, your curve consists of three lines through a single point; the technical term is that the lines are concurrent. In the $(x,y)$ coordinate chart, you have $3$ vertical lines, so they all pass through the point at vertical $\infty$. (In homogenous coordinates, the point $(0:1:0)$.) I claim that, for any number $j_0$, there is a family $F_t$ of cubics such that, for $t \neq 0$, the cubic $F_t$ is smooth with $j$-invariant $j_0$, and such that $F_0$ is your cubic.

Proof: First note that the symmetry group of $\mathbb{P}^2$ can take any three concurrent lines to any other three concurrent lines. So, I just need to build a family $F_t$ where $F_0$ is some triple of concurrent lines, and then apply a symmetry of $\mathbb{P}^2$ to that family.

Let $G$ be a cubic with $j$-invariant $j_0$. By making a change of coordinates if necessary, we can arrange that $G$ meets the line at $\infty$ in three distinct points. Then set $F_t(x,y,z) = G(x,y,tz)$. For $t \neq 0$, the cubic $F_t$ is isomorphic to $G$. When $t=0$, the cubic $F_t$ is the three lines joining the origin to the intersections of $G$ with the line at $\infty$. $\square$

  • $\begingroup$ Are you sure the curve $xyz = 0$ is semi stable? I thought the singularities of a semi stable curve should be ordinary. This means that there can only be two irreducible components going through a singular point, right? In your example there are 3 curves going through the singular point, no? $\endgroup$ – Insti Oct 27 '11 at 14:16
  • $\begingroup$ @Insti: I have converted your answer to a comment; answers should be reserved for posts that answer the question. But because you do not have 50 reputation points yet, you can only comment on your own questions and answers. So, you didn't do anything wrong; the "add comment" button will only appear for you once you gain 50 points. By the way, here is an explanation of reputation points. $\endgroup$ – Zev Chonoles Oct 27 '11 at 14:27
  • $\begingroup$ You are right that there should only be two components through any singularity. And there are! The components of $xyz=0$ are the lines $x=0$, $y=0$ and $z=0$. The form a triangle, with vertices at $(0:0:1)$, $(0:1:0)$ and $(1:0:0)$. $\endgroup$ – David E Speyer Oct 27 '11 at 15:10
  • $\begingroup$ Great answer. Thank you. $\endgroup$ – Derek Allums Oct 27 '11 at 16:05
  • $\begingroup$ I don't see how the equations xyz =0 gives a triangle. it's just the three coordinate axes, right? For example....ow wait, are we talking projective here? $\endgroup$ – Bana Oct 27 '11 at 16:18

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