Earlier I showed you a surprising way in which Cayley’s 2x2x2x2 hyperdeterminant shows up in the theory of elliptic curves. Now we are ready to see how Schläfli’s 2x2x2x2 hyperdeterminant is related to elliptic curves in a more basic way.

Suppose then that you have a 2x2x2x2 hypermatrix with components t_{ijkl} taking values from the integers. I’ll pose the following problem for you: Find me three non-zero pairs of integers x^{j}, y^{k}, z^{l }such that

Σ t_{ijkl}x^{j }y^{k }z^{l} = 0

At first site you mght think this is going to be pretty easy since the problem is linear in each undetermined variable, but dont forget that there are two equations corresponding to i = 1,2. To solve it just peel off the vectors one at a time, firstly the matrix components M_{ij }=_{ }Σ t_{ijkl}y^{k }z^{l }must permit a solution to the equation

Σ M_{ij}x^{j }= 0

So we need det(M) = 0. Now think of M as being derived from a 2x2x2 tensor

A_{ijk }=_{ }Σ t_{ijkl}z^{l }=> M_{ij }=_{ }Σ A_{ijk}y^{k}

det(M) is now a quadratic form in the variables y^{k}. The condition for it to have integer roots is that the discriminant must be a square, but the discriminant is the same one that we identified earlier as Cayley’s hyperdeterminant . So we need

t^{2}= det(A)

But A is linear in z^{l }and the hyeprdeterminant is a quartic so we are left with a diophantine equation to solve of the form t^{2 }= av^{4}+ bv^{3}u + cv^{2}u^{2}+ dvu^{3}+ eu^{4}. Although elliptic curves are more commonly seen in a form where the right hand side is a cubic, this form is also an elliptic curve and can be reduced to the cubic form. So solving the original problem actually reduced to solving an elliptic curve based on Cayley’s Hyperdeteriminant and standard methods apply.

But we cango one step further. The quartic equation is the same one that arose when we were constructing Schläfli’s Hyperdeterminant for our 2x2x2x2 hypermatrix t. From the theory of elliptic curves we know that the nature of the elliptic curve is determined by its J-invariant given by

J = g_{2}^{3}/Δ = S^{3}/det(t)

where S is the octic invariant of the hypermatrix. It is remarkable enough that Schläfli hyperdeterminant is associated with the J-invariant of an elliptic curve but even more surprising is the significance that this implies the the degree of the hyperdeterminant which we showed was 24, the dimension of the Leech lattice.

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This entry was posted on October 11, 2008 at 2:55 pm and is filed under Uncategorized.

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