One of the connections that makes hyperdeterminants very interesting is their link with elliptic curves. Elliptic curves are connected to may other things of interest including the proof of Fermat’s lat theorem and monstrous moonshine and their unexplored link with elliptic curves may throw new light on these areas.
I wanted to introduce this link as early as possible in the blog so I’ll do it now before I have mentioned a number of other basic features of hyperdeterminants. In fact there are two independent ways to make the link, one with Cayleys 2x2x2 hyperdeterminant and another with the 2x2x2x2 hyperdeteriminant. I’ll start with the former.
An elliptic curve is a name for an equation that is usually put into this form
y2 = px3 + qx2 + rx + s
A general problem associated with elliptic curves is to find solutions in rationals when p,q,r and s are given rationals. There is a group theoretic element to the solutions in that they form an abelian group under a certain operation. I am going to assume that you know enough about these things already.
For this excersize I am going to look at a specific class of elliptic curves which have a solution at x = 0 and for which the cubic side of the equation fully factorises. This means we can write it in the form
y2 + 4(bx – n)(cx – k)(ex – m) = 0, where nkm = t2
My aim is to try and rewrite this in a form that is as symmetric as possible, so i’ll satisfy the last condition by setting n = fd, k = dg, m = fg, so that t = fdg and the equation can be written concisely as
y2 + 4(bx – fd)(cx – dg)(ex – fg) = 0
This gives us a symmetry corresponding to S3 the symmetric group on three elements. The symmetry is realised by an action on the “known” coefficients by applying the same permutation to (b,c,e) and to (g,f,d). Of course, our elliptic curve originally had four coefficients so by increasing the number to six we must have introduced a two parameter redundancy. This symmetry is generated by transformations with parameters r and s as follows
b -> rb, f -> rf, c -> c/r, g -> g/r
b -> sb, d -> rd, e -> e/s, g -> g/s
These are transformations on the known coefficients and there does not appear to be any symmetry between the two unknown parameters x and y. However, it is also possible to make elliptic curves symmetrical in two unknowns by writing them in this form
u2v2 + 2αuv + 2βu + 2γv + δ = 0
In this form the equation is symmetrical under exchange of the two unknowns u and v if the fixed coefficients β and γ are also swapped. The standard form for an elliptic curve can be recovered by completing the square with respect to the equation as a quadratic in u. This leads us to rewrite it as
(uv2 + αv + β)2 + 2γv3 + (δ -α2)v2 – 2αβv – β2 = 0
This form can be identified with our original elliptic curve if we make the equations
x = v
y = uv2 + αv + β
4bce = 2γ
4(bcfg + bedg + cefd) = α2-δ
4fdg(bg + cf + ed) = -2αβ
2fdg = β
These can then be inverted to give as the coefficients α, β, γ and δ in terms of b,c,d,e and f. Substituting back into the equation for u and v we get
u2v2 – 2(bg + cf + ed)uv + 4fdgu + 4bcev + (bg + cf + ed)2 – 4(bcfg + bedg + cefd) = 0
expanding and rearranging gives
u2v2 + b2g2 + c2f2 + e2d2 – 2bguv – 2cfuv – 2eduv – 2bcfg – 2bedg – 2cefd – 4fdgu + 4bcev = 0
It is clear from this expression that we now have more symmetry than we were originally looking for. We set up an S3 symmetry acting on the coefficients and a Z2 symmetry in the varaibles, but we now also have symmetries in the expression that interchange the coefficients and the variables. And of course you will recognise the expression as Cayley’s hyperdeterminant in the cube formed by the eight variables. The symmetry group of the expression is therefore at least the symmetry of the cube which is isomorphic to S4. In fact, as we shall see next, the hyperdetermiant has a much larger symmetry even than that because the continuous symmetries have also expnaded. We shall look at what the fullsymmetry is shortly.