Cayley’s Hyperdeterminant

September 29, 2008

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 aijkx(i)y(j)z(k)

This has a stationary point when simulataneously

Σ aijkx(i)y(j) = 0
Σ aijkx(i)z(k) = 0
Σ aijky(j)z(k) = 0

Consider a matrix M(z) with components mij = Σ aijkz(k). Then in matrix notation the last two conditions read xTM(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

x2TM(z1) = x1TM(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*2y*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*1y*1z*1 + f2 x*2y*2z*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.

Advertisements

14 Responses to “Cayley’s Hyperdeterminant”


  1. […] an example, Cayley’s hyperdeteriminant is an invariant of […]


  2. […] features of hyperdeterminants. In fact there are two independent ways to make the link, one with Cayleys 2×2×2 hyperdeterminant and another with the 2×2×2×2 hyperdeteriminant. I’ll start with the […]


  3. […] and often we are interested in properties of small hypermatrices. An example already came up in the construction of cayley’s hyperdeterminant. For the 2×2×2 hypermatrix A we demonstrated that if the hyperdeterminant is not zero then the […]


  4. […] October 11, 2008 We analysed Cayley’s 2×2×2 hyperdeterminant earlier and now it’s time to look at Schläfli’s 2×2×2×2 hyperdeterminant. I’ll […]


  5. […] task of working out the degree of the hyperdeterminant of a given format. We have already seen that Cayley’s hyperdeterminant for a 2×2×2 hypermatrix is of degree 4 and the hyperdeterminant of size 2×2×2×2 is of […]

  6. David Glynn Says:

    Cayley published his first hyperdeterminant, that I call det_0, in 1843, not 1845. This is the most natural generalization of the ordinary determinant to higher dimensions. To get an invariant, symmetric in the dimensions (directions?), the number of these should be even. (For characteristic two fields odd dimensions are also ok.) The paper is A. Cayley, On the theory of determinants, Cambr. Phil. Trans. viii (1849), (starting at page 75 of Cayley’s Collected Papers, Vol. 1).
    It is hidden in the section section of the paper. Available as a download online from e.g. 2020ok.com/books/48/the-collected-mathematical-papers-of-arthur-cayley-47248.htm
    It is dated 1843 in the collected papers,
    not 1849, so he submitted it in 1843, at maybe age 21. The hyperdeterminant det_0 is (Binet-Cauchi) multiplicative wrt the inner product of hypercubes. The Det of 1845 is not, except for the boundary format case (see Dionisi and Ottaviani, arXiv:math/0104281v1).
    det_0 is only mentioned very briefly in the “hyperdeterminant” book, but is more important that is realised there.
    For example, the 3d version of det_0 can be considered to be an invariant of m mXm matrices over a field, having degree m in the n^3 coefficients. In general, det_0 of an m^n hypercube has degree m in the m^n variables.

  7. superstrings Says:

    David, thanks for your comment. It has always been the later hyperdeterminants that arose in my work, but I should look at the ealier one too.

    I see you also work on coding theory which is related to hyperdeterminants. I would like to find out more about that.

  8. David Glynn Says:

    Hi Superstrings, Hyperdeterminants are related to codes via hypercubes in various ways. For example, an (additive) quantum code corresponds to an additive set of states, each of which is basically a self-dual quantum code [[n,0,d]]_p (of 0 dimenions), and which can be represented by a single p^n hypercube over C (or “n-cube of side p”) (p is prime, usually 2). Then a hyperdeterminant gives an invariant of a quantum state useful in physics, ie. “entanglement measures”. e.g. Cayley’s Det (1845), the quartic function, determines the union of various orbits for 3-qubit binary states (2^3 cubes).
    The most common codes are linear over a finite field GF(q). They correspond to sets of points in projective space over GF(q). There are invariants of sets of points that lead to (or come from) the basic hyperdeterminants. These invariants can be produced by the process of “complete polarization” from the basic determinant function. Cayley’s first hyperdeterminant is basically a symmetric invariant of a set of n n^k matrices when k is even, but is antisymmetric when n is odd. It is a kind of n-ary product of hypercubes. People haven’t used hyperdeterminants in coding theory much, but there are connections with certain kinds of functions that enumerate codewords, and so on. I think that geometrical applications of
    hyperdeterminants are promising. For example, the incidence structure (points vs lines) of the Euclidean plane is related to a 4-cube over R that is the product of two 3-cubes, one representing the additive structure of R, and the other the multiplicative structure.


  9. […] I have started to get interested in. David recently left some comments on my blog article about Cayley’s Hyperdeterminants  so I think this is a good moment to continue this blog with some posts in which I will try to […]


  10. […] partial answer to this question is that the expression is a special case of Cayley’s hyperdeterminant for a 2×2×2 hypermatrix with […]

  11. latorre Says:

    Is there an explicit construction for the hyperdeterminant of a tensor 3x3x3x3 ?
    That corresponds to a quantum state of four qutrits. There is explicit construction of
    a global measure of entanglement.
    I understand the rank of such an object would be 36.

    cheers, ji.

    • PhilG Says:

      latorre, that is a good question. Actually 36 is the degree of the 3x3x3 hyperdeterminant. The degree of the 3x3x3x3 hyperdeterminant is 810. I worked that out using the formula cited at https://hyperdeterminant.wordpress.com/2008/10/18/the-degree-of-hyperdeterminants/. You should check it if you need to know that, I could have missed some terms.

      I have not seen any way to explicitly construct a hyperdeterminant larger than 2x2x2x2. I think the 3x3x3x3 hyperdeterminant will have more terms than there are atoms in the universe, but that does not rule out a construction that you could make some sense of. I dont think such a construction is known.


  12. It’ѕ aweѕοme to vіsit thіs web paցe ɑnd reading the viiews of aⅼl
    friends concerning this paragraph, while I aam aⅼso keen of getting knowledge.


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: