Cayley’s Hyperdeterminant

Hyperdeterminants were discovered by the English mathematician Arthur Cayley in 1845. He worked out the formula for the most simple hyperdeterminant that is not a determinant, i.e. the 2x2x2 hyperdeterminant. Although he also looked at larger examples we now refer to this specific case as Cayley’s hyperdeterminant. If you fancy reading Cayley’s original work “On the Theory of Linear Transformations” you will find it in  “The collected mathematical papers of Arthur Cayley Vol 1” which can be downloaded from http://www.archive.org/details/collmathpapers01caylrich .

The 2x2x2 array A has 8 components which we refer to either using index notation of individual letters 

a₁₁₁ = a
a₁₁₂ = b
a₁₂₁ = c
a₁₂₂ = d
a₂₁₁ = e
a₂₁₂ = f
a₂₂₁ = g
a₂₂₂ = h

Then Cayley’s hyperdetermiant written exactly as he did himself is

det(A) = a²h² + b²g² + c²f² + d²e² – 2ahbg – 2ahcf – 2ahde – 2bgcf – 2bgde – 2cfde + 4adfg + 4bech

It is useful to see how this is derived from the definition of the hyperdeterminant. Recall that we start with a form

F = Σ a_ijkx⁽ⁱ⁾y^(j)z^(k)

This has a stationary point when simulataneously

Σ a_ijkx⁽ⁱ⁾y^(j) = 0
Σ a_ijkx⁽ⁱ⁾z^(k) = 0
Σ a_ijky^(j)z^(k) = 0

Consider a matrix M(z) with components m_ij = Σ a_ijkz^(k). Then in matrix notation the last two conditions read x^TM(z) = 0 and M(z)y = 0, and this can be realised iff det(M(z)) = 0. Since det(M(z)) is quadratic in the components z^(k) it can also be written as a symmetric quadratic form in the two components u = z⁽¹⁾, v = z⁽²⁾ 

det(M(z)) = (ag – ce) u² + (ah + bg – cf – de) uv + (bh – df) v²

The nature of the solutions depends on the discriminant of this equation

disc. = (ah + bg – cd -de)² – 4(ag – ce)(bh – df)

Which can be expanded to confirm that it is Cayley’s hyperdeterminant given above. If this is non-zero then there are two solutions of det(M(z)) = 0 up to scale factors over the complex numbers, but if it is zero there is just one. Dealing with the non-zero case first we label the two soltuions as z₁ and z₂ . The corrseponding eigenvectors of M(z) will be x₁, x₂, y₁, y₂, labelled in reverse so that

x₂^TM(z₁) = x₁^TM(z₂) = 0

M(z₁)y₂ = M(z₂)y₁ = 0

It is also possible to work through similar arguments by starting with the application of x or y instead of z. The discriminants in these cases are again the same hyperdeterminant which we know to be non-zero and the same vectors are the solutions. It follows that x₁ is distinct from x₂ and y₁ is distinct from y₂

From this we can see that M(z₁) = f x*₂⊗y*₂ where the star operator is a dualization defined by (u,v)* = (v,-u) (so that z.z* = 0), and f is a non-zero scale factor. Likewise we find that the original array can be written as the sum of two products

A = f₁ x*₁⊗y*₁⊗z*₁ + f₂ x*₂⊗y*₂⊗z*₂

Where f₁and f₂ are non-zero. Once we know that the hypermatrix can be written in this form it is easy to see that we can satisfy any two of the three equations for a singular point at a time, but not all three at once. So if the hyperdeterimant is non-zero then the hypermatrix is non-singular. I’ll leave it as an excersize to prove the converse.