Hyperderminants as Invariants

October 1, 2008

Hyperdeterminants generalise determinants by extending their properties as a discriminant to multi-dimensional hypermatrices. If you ask a mathematician what a determinant is he probably will not reply that it is a type of discriminant despite the fact that its name reflects this property. He might define it by its multiplicative properties, or he might refer to its geometric interpretation as a scale factor for volumes when mapped by a linear transform. Sometimes he might also define it as a group invariant. If he is more constructive in outlook he would probably just give a formula based on alternating tensors. In this post and some later ones we will look at how some of these properties of determinants generalise to hyperdeterminants.

If you tranform a aquare N x N matrix M using a simialrity transform represented by an invertible matrix S so that

M → M’ = S-1MS

then the determinant is invariant, det(M’) = det(M). So the determinant is an invariant under the action of the general linear group GL(N) on the matrix, but it is not the only invariant. Another example of an invariant under this action is the trace of the matrix. However, the determinant actually has a bigger symmetry group. We can multiply the matrix by different matrices on both sides

M → M’ = TMS

then det(M) = det(M’)det(S)det(T) and the transformation is a symmetry if det(S)det(T) = 1. in that case we can divide S and T by their own determinants making them special linear transforms (ie. of determinant one) without changing the transformation. So the full symmetry group of the determinant is actually SL(N) ⊗ SL(N) and unlike the invariants of GL(N) any other invariant of SL(N) ⊗ SL(N) under the same action on the matrix will be a function of the determinant only.

How does this invariance property generalise to hyperdeterminants. A hypermatrix of rank n can be transformed by applying a linear transform to any of the n vector spaces from which it is formed. We can guess that the symmetry group for a hypermatrix of dimension N1 x … x Nnis going to be SL(N1) ⊗ … ⊗ SL(Nn). To prove it we can look at how the linear transformations affect the definition of the hyperdeterminant as a discriminant for a multilinear form F(x1,…,xn) where xi is an element of the vector space of dimension Ni. When the vector x1is transformed by a matrix non-singular S a new multilinear form FoS is generated

(FoS)(x1,x2,…,xn) = F(Sx1,x2,…,xn)

You can verify that FoS is singular iff F is singular and from the definition of the hyperdeterminant it follows that det(FoS) = det(F) P(S) for some polynomial P in the components of S. If we transform again using another non-singular transformation T we get (FoS)oT and det((FoS)oT) = det(FoS) p(T) = det(F) P(S) P(T), but also (FoS)oT = Fo(TS) and det(Fo(TS)) = det(F) P(TS), from which we deduce that P(TS) = P(T) P(S). From our knowledge of matrix algebra we know that the only multiplicative homomorphisms from the general linear group to the non-zero real numbers are powers of the determinant. I.e. P(S) = det(S)K

The exponent K can be determined by matching degrees of the expressions. If D is the degree of the hyperdeterminant and we know that the determinant has degree N1 then we find K = D/N1 and

det(FoS) = det(F) det(S)D/N1

Finally we can conclude that if S is a special linear transform then the hyperdeterminant is invariant under the transformation of S. This combined with similar arguments applied to the other vector arguments of the multilinear form gives the result that the hyperdeterminant is an invariant of the group SL(N1) ⊗ … ⊗ SL(Nn).

As an example, Cayley’s hyperdeteriminant is an invariant of the special linear group cubed SL(2)3

As a byproduct of this argument we now know that the exponent K = D/N has to be an integer therefore Ni divides D for each i, so D must be a multiple of the least common multiple of the set {Ni ,1≤i≤n}. This can be useful because in general it is hard to determine what the value of D is for a given hyperdeterminant format.

Advertisements

3 Responses to “Hyperderminants as Invariants”


  1. […] Filed under: Uncategorized — superstrings @ 10:07 am 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 […]


  2. […] how polynomial discriminants are related to hyperdeterminants. Previously I also showed that hyperdeterminants are invariants . Now I am going to put these together and look at  discriminants as invariants. This is a […]


  3. […] That much was already familiar to me, but David also remarked that this hyperdeterminant has a multiplicative property that generalises the well known property of matrices i.e. det(AB) = det(A)det(B). This was something I had not really appreciated before. The product of two hypermatrices is taken to be the contraction of an index from one hypermatrix with an index from the other. E.g. the product of an mn hypermatrix with an mk hypermatrix is an mn+k-2 hyprmatrix. The discriminant hypermatrix has a similar multiplicative property for boundary format hypermatrices but not for more general formats. One other case where a multiplicative property works is the product of a hypermatrix with a matrix as I demonstrated when looking at the hypermatrix as an invariant. […]


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: