The last post about multiplicative hyperdeterminants used a form of multiplication of hypermatrices where the last direction of one hypermatrix is contracted over the first direction of another. We can write the product using ordinary product notation so A multiplied by B is just AB. This generalises multiplication of matrices and multipication of matrices with vectors. For two vectors it gives the scalar product.
A more general product can be defined if we contract over more than one of the hypermatrix directions. I will write A(p)B to mean the product of hypermatrices A and B where the last p directions of A are contracted with the first p directions of B in reverse order. When a hypermatrix A of rank r is multiplied by a hypermatrix B of rank s the product A(p)B is a hypermatrix of rank r+s-2p. It is only defined if the dimensions of the last p directions of A match the dimensions of the first p directions of B. If A and B are multi-linear operators on vector spaces then the last vector spaces that A operates on must be the dual spaces of the first vector space that B operates on. When a vector space is transformed by a non-singular linear operator, its dual is transformed by the inverse of the operator.
The problem we want to look at now is how to invert a hypermatrix so that we can use it to reverse a multiplication. In other words, if C = AB we want to solve for A in terms of C and B. We can do this if we have a hypermatrix D with the same format as B, which inverts B in a suitable fashion. Given any matrix D with the same format as B (of rank r) we can form r matrices by contracting over all but one pair of matching directions. If all these matrices are a non-zero multiple of the identity matrix we call D an inverse of B. In particular we need B(r-1)D = xI and D(r-1)B = yI, but these are just 2 out of r ways of doing the contraction. Notice that we say “an” inverse rather than “the” inverse because such an inverse may not exist, and if it does it may not be unique (even up to a factor).
Such an inverse can be used to invert the multiplication C = AB to give the solution A = (1/x) C(r-1)D. This does not require all the properties of the inverse, it just requires B(r-1)D = xI. But our more specific definition of inverse is interesting because there is a means to construct such inverses from hyperdeterminants and other invariants.
The term inverse is justified because if D is an inverse of B then B is an inverse of D. This follows immediately from the symmetry in the definition. Furthermore the usual inverse of a non-singular matrix is also an inverse in the hypermatrix sense, but so is any scalar multiple of the inverse. Why not fix the normalisation? Take the trace of B(r-1)D = xI and D(r-1)B = yI and you get B(r)D = xd1 = ydr (where di is the dimension of the hypermatrix in the ith direction) and similarly for the other directions of the hypermatrix. If the hypermatrix is hypercubical, so that all the directions have the same dimension, then we could fix z = y =1, but that wont work for more general hypermatrix formats. Instead we would have to fix x = 1/d1, y = 1/dr etc., but that would be inconsistent with the inverse of a matrix. The best compromise is to not fix the normalisation at all.
Another plus for this definition is that the set of all inverses (with zero included) forms a vector space. The dimension of this vector space is an important attribute of the hypermatrix format and is not easy to calculate even for simple formats. What makes this interesting is that we can form inverses using invariants. This generalises the expression for the inverse in terms of its determinant, which is,
A-1 = (1/detA) (d/dA) detA
If A is a hyperdeterminant of rank r and K(A) is an invariant of degree m, then the inverse of A with respect to K is defined as
InvK(A) = (1/K(A)) (d/dA) K(A)
If we contract A with this inverse over all directions we find that
A(r)InvK(A) = m
This result is true for any homogeneous multivariate polynomial of degree m, but we can use the invariance property of K(A) to derive the stronger result that this inverse is an inverse in the sense defined previously. To see this consider a linear transform of A when its last direction is transformed by a linear operator δA = AE where E is small.
δK(A) = Tr( (d/dA)K(A) (r-1) δA ) = K(A) Tr( InvK(A) (r-1) AE ) + O(E2)
But by the transformation rule for invariants we get
K(A) -> K(A) det(I+E)m/d
=> δK(A) = (m/d) K(A) Tr(E) + O(E2)
By bringing these together we conclude that
InvK(A) (r-1) A = (m/d)I
With similar results for the other directions, which confirms that InvK(A) is an inverse of A.
That is all I have to say about inverses for now, but I’ll leave you with a few questions to consider. Firstly, I have not demonstrated that InvK(InvK(A)) = A. How can this be proved in general? Secondly, the ring of invaraints for a given format is finitely generated. Given a set of generators, we get a set of inverses. Are these linearly independent and do they generate the full vector space of inverses? I dont really know the answers to these questions so if you do know something, please leave a comment about it.