Presentation on theme: "Hyperdeterminants Fort Collins, 5/11/05. Let A be a n n matrix det A:= S n ( )a 1 (1)... a n (n) Its basic property is Det(A)=0 the linear."— Presentation transcript:
Hyperdeterminants Fort Collins, 5/11/05
Let A be a n n matrix det A:= S n ( )a 1 (1)... a n (n) Its basic property is Det(A)=0 the linear system Ax 0 has a solution x 0
Brief history 2 2 case is easy and goes back to antiquity around 1650, in Europe and in Japan, some particular linear systems were solved, leading to particular cases of the determinant in 1833, with the paper of Cayley, the theory takes the modern form
det is also characterized by 1) det is linear on each row 2) det(A)=0 if two rows of A are equal 3) det(I)=1 (normalization) Practical computation through Gaussian elimination needs properties 1), 2), 3).We add multiples of a row to another row until we find a triangular form.
Challenge Take a 2 2 2 hypermatrix or a 3 2 2 hypermatrix Is there any useful function analogous to det ?
In the multidimensional setting there are parallel slices instead of rows and columns Vertical slices Lateral slices Horizontal slices
First remark: linearity on slices is too strong a request, it cannot be preserved in the multidimensional setting
Look at the connection with systems Let A be a (k 0 +1) (k 1 +1) ... (k p +1) hypermatrix, regard it as a multilinear map A : k 1 +1 ... k p +1 k 0 +1 A is called degenerate if x 1 0,…, x p 0 such that A(x 1 ... x p ) 0
Second remark: the theory of hyperdeterminants is easier over , in contrast with the usual theory of determinants, which works equally well over every field. On , to check if a hypermatrix is degenerate is a subtle question, depending on real solutions of polynomials. On , to check if a hypermatrix is degenerate is much easier. For example all 2 2 2 hypermatrices are degenerate.
Theorem (over ) If k 0 k i all hypermatrices are degenerate. If k 0 k i the degenerate hypermatrices form an irreducible subvariety of codimension k 0 k i 1. The case k 0 k i is called of boundary format. 3 2 2 is of boundary format Let us consider hypermatrices of format (k 0 +1) (k 1 +1) ... (k p +1)
Definition In the boundary format case, degenerate hypermatrices form a hypersurface. The equation defining it is called the hyperdeterminant and it is denoted by Det. Above definition is correct up to a constant. We will soon normalize the function Det. In the n n case Det = det
Summary Let k 0 k i (boundary format). Let A be a (k 0 +1) (k 1 +1) ... (k p +1) hypermatrix. Det(A)=0 A(x 1 ... x p ) 0 has a nonzero solution This notion is invariant for the natural action of GL(k 0 +1) GL(k 1 +1) ... GL(k p +1)
By elementary invariant theory, there are constants n i such that, for any g i GL(k i +1) Det(A g i ) = Det(A) det(g i ) n i Theorem deg(Det) = We will prove the theorem in the 3 2 2 case, where an explicit formula is available, which goes back to Cayley. Cayley defined hyperdeterminants in The modern references are Gelfand, Kapranov, Zelevinsky, Hyperdeterminants, Adv. in Math., 96, (1992) 1994 Birkhauser book by the same authors
Consider a 000 a 001 a 010 a 011 a 100 a 101 a 110 a 111 a 200 a 201 a 210 a 211 and call A 00, A 01, A 10, A 11, the four 3 3 minors. Then Det(A) = A 01 A 10 A 00 A 11 A 00 A 11 A 10 A 01
Swapping two vertical slices, hyperdet does not change Because for g GL(3) Det(A g) = Det(A) det(g) 2
Swapping two horizontal slices, hyperdet changes its sign Because for g GL(2) Det(A g) = Det(A) det(g) 3
Boundary format hypermatrices have a well defined diagonal a i 0 …i p is on the diagonal if i 0 = i 1 +…+ i p
Computation of hyperdet of a 3 2 2 diagonal hypermatrix is interesting. a 000 a 001 a 010 a 011 a 100 a 101 a 110 a 111 a 200 a 201 a 210 a 211 Only the red elements are nonzero, hence A 01 = a 000 a 110 a 211 A 10 = a 000 a 101 a 211 A 00 = A 11 = 0 Then Det(A) = A 01 A 10 A 00 A 11 = a a 101 a 110 a 211 2
We can visualize the four diagonal elements In general, Det of a diagonal hypermatrix is given by the product of diagonal elements, each one with a certain exponent which can be determined by combinatorics.
Normalization of Det Let I be the matrix with 1 on the diagonal and 0 otherwise. Put Det(I) = 1
Is it possible to compute Det with Gaussian elimination on slices, by reducing to diagonal or triangular case? Answer: Unfortunately only in a few cases, namely when the format is 2 n (n+1) Natural question
Indeed, a nondegenerate A can be reduced to triangular form if and only if A is not a stable point according to Mumford GIT(Ancona-O. Advances in Geom. 1(2001) ). A can be reduced to diagonal form if and only if the stabilizer of A in the group SL(k 0 +1) ... SL(k p +1) is isomorphic to *. The stabilizer of A is always contained in SL(2), and the equality holds if and only if A can be reduced to I. Vandermonde matrices can be generalized in the multidimensional setting. They can always be reduced to I. Their hyperdeterminant is analogous to the usual square case.
The Binet-Cauchy formula det(AB)=det(A)det(B) can be generalized to the multidimensional setting, of course with certain exponents (Dionisi-O., J. Algebra 259 (2003) ).