It seems that most introductory books on elliptic curves simply state the definition of the j-invariant of an elliptic curve without giving any background on how that definition was conceived. Of course, for moduli reasons, it is clear why one might want such an invariant, but the actual formula has always seemed quite mysterious to me. Does anyone know of a nice self-contained source that explains the definition of the j-invariant?

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    $\begingroup$ The complete definition, or just the definition up to a constant? Matt has given a nice explanation for why one should care about the j-invariant up to a constant, but the precise form of the j-invariant has to do with its number-theoretic properties and is not easy to explain in an introduction. On the other hand books on elliptic curves don't want to give you the wrong definition! So there is a dilemma here. $\endgroup$ – Qiaochu Yuan Feb 12 '11 at 9:27

Actually, Ravi Vakil's notes give a great reason (modulo the strange constant out front). This explanation is given somewhere in the Foundations of Algebraic Geometry notes here http://math.stanford.edu/~vakil/preprints.html#coursenotes. I'm just going off memory, so don't attribute any errors I make to him.

Notice that once you prove that every elliptic curve has an affine model given by $y^2=x(x-1)(x-\lambda)$, then you know the $j$-invariant has to be independent of the different $\lambda$ you get by permuting.

You can just explicitly work it out that the six choices of lambda then are $\lambda, \frac{1}{\lambda}, 1-\lambda, \frac{1}{1-\lambda}, \frac{\lambda}{\lambda -1}, \frac{\lambda -1}{\lambda}$.

Some obvious first choices for an invariant with respect to all of these is to multiply them all together. You get $1$, oops, that isn't a good invariant since it not only doesn't depend on the choice of $\lambda$, but also is independent of curve. So you could also try adding them all, oops again, they come in pairs each adding to $1$, so they add to $3$.

So let's try the next best thing which is to sum the squares. If you check, this is exactly the $j$ invariant (but without the constant which is stuck in for characteristic $2$ reasons).

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    $\begingroup$ In other words, we have an action of the symmetric group of degree $3$ on the field of rational functions on $\lambda$ and we look at the invariant subfield. We know from Lüroth's theorem that that subfield is the field of rational functions of some rational function of $\lambda$. Standard ideas of invariant theory help us find that function. Moreover, in this way we see why it is enough to consider the usual invariant: all others can be deduced from it. $\endgroup$ – Mariano Suárez-Álvarez Feb 12 '11 at 5:18

The $j$-invariant has the following classical interpretation.

Consider a model $E\subset{\Bbb P}^2$ of the elliptic curve (one knows that $E$ is a cubic). Let $P\in E$. There are 4 lines through $P$ that are tangent to $E$ and one can show that the set of cross-ratios $c$ of these 4 lines are independent of $P$. Then $$ j=\frac{(c^2-c+1)^3}{c^2(c-1)^2} $$ is invariant under the 24 permutations of the 4 tangents (which give up to 6 different values of the cross-ratio) and is the $j$-invariant of the elliptic curve, up to normalization.

When the curve $E$ is given in Legendre form $y^2=x(x-1)(x-\lambda)$ the formula for the $j$-invariant is obtained taking $P$ the point at infinity and the 4 tangents the line at infinity and the lines $x=0$, $x=1$ and $x=\lambda$.


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