$j$-invariant for a singular curve

Question

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.

Answer 1

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 $\infty$. $\square$