Cayley’s Hyperdeterminant and Principal Minors

June 25, 2009

Just when you think you must know everything about a matematical structure you suddenly start to find out all sorts of new things about it. This has been happening to me recently with Cayley’s Hyperdeterminant  and I plan to blog about some of the things here.

First off is a surprise relationship between the principal minors of a 3×3 symmetric matrix and the hyperdeterminant that has been reported recently in arXiv:math/0604374 by Olga Holtz and Bernd Sturmfels (and possibly others).

The principal minors of a matrix are the determinants you are left with when you eliminate rows and columns of the original matrix passing through diaginal elements. The principal minors for a 3×3  matrix

          ( a   x   y )
    M =   ( u   b   z )
          ( v   w   c )

are

a111 = 1
a211 = a
a121 = b
a112 = c
a221 = ab – xu
a212 = ac – yv
a122 = bc – wz
a222 = det(M)

These are eight numbers which naturally fall on the corners of a cube to form a hypermatrix A, so what is its hyperdterminant? This can be worked out by hand to yield the answer,

det(A) = (yuw – xzv)2

In particular, if the matrix is symmetric the hyperdeterminant is zero. [The same is true for any symmetric matrix which is transformed by pre or post multiplication by a diagonal matrix]

Why is this so interesting? Well, the symmetry of a 3×3 matrix is preserved under SO(3) transformations applied simulataneously to rows and columns. The determinant a222 is invariant (of course a111 is trivially invariant). The principal minors appear in the diagonal elements of the inverse of M. So if the eight components of the hypermatrix are augmented with the six off-diagional elements of the matrix amd its inverse, then we seem to have a sytem of 14 dimensions with some extra symmetry. 

With a small adjustment we can extend this to a general 2x2x2 hypermatrix with components aijk and with a zero hyperdeterminant. Working in reverse we now define

a = a211
b = a121
c = a112
d = a222

k = a122
l = a212
m = a221
n = a111

then

x2 = ab – mn
y2 = ac – ln
z2 = bc – kn

This produces a symmetric matrix

          ( a   x   y )
    M =   ( x   b   z )
          ( y   z   c )

We can also add

r2 = da – lm
s2 = db – mk
t2 = dc – kl

And this gives a second symmetric matrix

          ( k   t   s )
    N =   ( t   l   r )
          ( s   r   m )

Multiplying these matrices together (and resolving sign ambiguities) we get a tidy relation

MN = nd I

The extra six components also have a simple interpretation. Since the hypermatrix has zero hyperdetermiant we know that it has a system of three vectors which act like the zero eignevectors of the system. The six components are the components of these vectors.

So what we have found is that the linear system consisting of the 8 components of a singular hypermatrix and the 6 components of its zero eigenvectors forms a system which has the SL(2) x SL(2) x SL(2) symmetry of the hypermatrix and also some extra SO(3) symmetries. So what is the system and what is its full symmetry?

If you know a bit about excpetional alegebras you can probably at least guess the answer to this question. The two symmetric 3×3 matrices can be regraded as elements in the Jordan algebra J3R. Then the matrices M and N and the two numbers n and d can be used to build a six dimensional matrix

    F =   ( nI   M  )
          ( N    dI )

This can be regarded as an element of a Freudenthal Triple System (FTS) which has dimention 14. Its automorphism group is a symplectic group (Sp(6) I think) and the relation MN = nd I is equivalent to det(F) = 0.

As far as I can tell this relationship is only valid between singular hypermatrices and singular elements of the FTS. It does not extend to non-singular elements in any way (please correct me if you think otherwise). I think it is of particular interest concerning the problem of Diophantine quadruples  since I can now think of these as special cases of an FTS as well as a hyperdeterminant.

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