Earlier we established that hyperdeterminants are invariants. The next obvious step is to look at how to construct the invariants and see what they can tell us about hyperdeterminants.
When Cayley discovered hyperdeterminants there was not very much known about the theory of invariants. Today the subject is taught routinely to undergraduates of mathematics and physics in some form. I am going to base this post on the knowledge of invariants as traditionally taught to physicists as tensor analysis without the more abstract approaches that mathematicians prefer.
From this standpoint our hypermatrix is a tensor T of rank n which we may write using indices over its components Tij…k. Even if you prefer not to think of tensors as a system of components, the notation can be regarded as an abstract way of labelling the vector spaces. Indices corresponding to a dual space and written as superscripts and the Einstein summation convention for contraction over indices is a way of indicating which spaces act on other spaces. I will assume that the notation is familiar. I still find this clearer and less constrained than the alternative notations devised by mathematicians including Sweedler notation.
From the theory of tensor analysis we know that all invariants can be constructed using tensor products and contractions over invariant tensors. In the present case where the relevant group is SL(N) the invariant tensors are the alterating tensors εi…k with N indices which are antisymmetric in any pair of indices. The components of this tensor are 0 or ±1 and are uniquely determined. This notation for the alternating tensor is commonly called the Levi-Civita symbol.
For the hypermatrix Tij…k, the subscript indices label different vector spaces which might be of different dimensions. To construct the invariants we need an alternating tensor for each of these spaces and we must be careful to contract over indices in the same space. In this way we can construct all invariants of the tensor and since the hyperdeterminant is an invariant we know that it must be possible to represent it in this way. As an example, the determinant of a 3×3 matrix M would be written
det(M) = 1/6 εijkεlmnMilMjmMkn
All hyperdeterminants are invariants of this form but the converse is not true. A simple counterexample would be that the square of the determinant which is also obviously invariant. More generally, the sums and products of invariants are also invariants and in fact the collection of all polynomial invariants forms a ring which is graded over the natural numbers by dint of the degree of the polynomial.
Hilbert proved a fundamental theorem of invriants called the Hilbert Basis Theorem that tells us that this ring is always finitely generated. In other words, for any hypermatrix format a finite number of invariants can be used to generate all the invariants. In the case of 2x2x2 hypermatrices, Cayley’s hyperdeterminant is sufficient to generate the entire ring of SL(2)3invariants, but in general the hyperdeterminant is not enough. For example, if Tijklis a 2x2x2x2 hypermatrix, then εimεjnεkpεlqTijklTmnpqis an invariant of degree two but it is not the hyperdeteriminant which we will see later is actaully a much more complex object of degree 24.
In general the problem of finding a complete set of generators of a ring for a given format of hypermatrix (and any relations between them) is an arduous task except for the simplest cases. Furthermore it is not even easy to form the correct invriant expression for a hyperdeterminant in a systematic way. However I’ll finish this exercise by giving a treatment for the case of Cayley’s hyperdeterimant.
For a 2x2x2 hypermatrix the only independent expression that might give an invariant of degree two is
X = εimεjnεkpTijkTmnp
But we can commute the components, relabel indices and alternate the tensors …
X = εimεjnεkpTijkTmnp = εimεjnεkpTmnpTijk = εmiεnjεpkTijkTmnp = -εimεjnεkpTijkTmnp = -X
So this attempt to form an invariant just gave us an expression that must be identically zero. By the way the similar invariant we cited for the 2x2x2x2 format above is not identically zero. As a general rule we must always be careful in this game to check that any expression we give is not zero and it is not always obvious. It can also happen that two apparently different expressions give the same result or even that a complicated expression can factorise into a product of two simpler ones unexpectedly. Beware these pitfalls.
There is no way of forming invariant expressions of odd degree for the 2x2x2 hyperdeteriminant. We can see this because the alternating tensor has two components. Recall also that how we showed that the dimension of the vector spaces must divide the order of the determinant when we demonstrated that it was an invariant. This result can be extended to any invariant. So the lowest degree for invariants on this case is quartic and the possible expressions can be quickly reduced to these four
W = εaeεimεbjεfnεcpεgkTabcTefgTijkTmnp
X = εaeεimεbfεjnεcpεgkTabcTefgTijkTmnp
Y = εaeεimεbjεfnεcgεkpTabcTefgTijkTmnp
Z = εaeεimεbjεfnεckεgpTabcTefgTijkTmnp
You can see that the first expression is more symmetrical than the other three so does this mean that W is the hyperdeterminant which we know to be symmetrical? There are relationships between expressions involving alternating tensors e.g
εaeεim + εaiεme + εamεei =0
That this expression is identically zero can be verified by noticing that it is antisymmetric under exchange of any two indices which is not possible for a rank four tensor in two dimensions.
using this identity we can show that
W – X + Y = W – Y + Z = W – Z + X = 0
which implies that
W = 0 and X = Y = Z
So our guess based on symmetry was wrong and in fact the hyperdeterminant is given by the other expressions.