One way to understand an unfamiliar concept is to relate it to a more familiar one. According to the Google unpopularity test, hyperdeterminants are 400 times more unpopular than discriminants, but there is a close relation between the two, so lets see how that works.

The discriminant of the quadratic polynomial ax^{2} + bx + c is the familiar quantity

Δ = b^{2} – 4ac

When D > 0 the quadratic equation ax^{2} + bx + c = 0 has two real roots. When D < 0 it has no real roots and when D = 0 it has one double root.

The discriminant is related to the determinant of a 2 x 2 matrix M

M = | ( | a | b/2 | ) |

b/2 | c |

Then Δ = -4det(M).

Discriminants can be defined for higher order polynomials as well. The general definition of the discriminant for a polynomial

P(x) = a_{n}x^{n} + … + a_{1}x + a_{0}

is given by

Δ = a_{n}^{n-2} |
∏ | (r_{i} – r_{j})^{2} |

i < j |

where r_{i}for r = 1,…,n are the roots of P(x) = 0 including multiplicities. It is clear from this expression that the discriminant is zero iff the polynomial has multiple roots. For a specific n it is possible to expand the product into a symmetric polynomial in the roots. By a general theorem any symmetric polynomial can be expressed in terms of elementary symmetric polynomials that are equal to the coefficients on P(x). In this way the familiar expressions for discriminants as polynomials of degree 2n-2 in the coefficients can be constructed. E.g. for cubics we get

P(x) = ax^{3} + bx^{2}+ cx + d

Δ = b^{2}c^{2} – 4ac^{3} – 4b^{3}d – 27a^{2}d^{2}+ 18abcd

Just as the discriminant of a quadratic is the determinant of a symmetric matrix, the discriminant of a cubic is given by Cayley’s hyperdterminant of a symmetric hypermatrix A with components as follows

a_{111} = a

a_{211} = a_{121} = a_{112} = b/3

a_{221} = a_{212} = a_{122}= c/3

a_{222} = d

By applying the formula for Cayley’s Hyperdeterminant it is not difficult to verify that for cubics

Δ = -27 det(A)

How can we understand this relationship and what is the general result? Suppose then we have the polynomial of n^{th} degree P(x) = a_{n}x^{n} + … + a_{1}x + a_{0 }and we construct a symmetric hypermtrixT of rank n where a component with n-r indices equal to 1 and r indices equal to 2 given by

t_{1..12..2} = a_{r}/C_{r}^{n}

where C_{r}^{n} is a binomial coefficient

Using a vector **x** = (x, y) , then you can check that

P(x/y) = y^{-n} F(**x**) where F(**x**) = Σ t_{i…k} x^{(i)}…x^{(k)}

The form F(**x**) has a singular point where ∂F/∂**x = 0**which is equvalent to P'(x/y) = 0 and P(x/y) = 0 i.e the polynomial has a double root. So F(**x**) is singular iff Δ = 0.

Now recall the definition of a hyperdeterminantwhich is a discriminant for a more general multilinear form for a non-symmetric hypermatrix T given by

F(**x**_{1},…,**x**_{n}) = Σ t_{a…h} x_{1}^{(a)}…x_{n}^{(h)}

Of course this hyperdeterminant is equally well defined for the case in point where T is symmetric. Suppose then that Δ = 0 which implies that the form F(**x**) has a singular point **x = ξ**, but by comparing derivatives we find that the multilinear form F(**x**_{1},…,**x**_{n}) is also singular at the point where **x**_{i }= **ξ**for each i. It then follows from the definition that the hyperdeterminant must be zero. So we have that Δ = 0 => det(T) = 0 from which we conclude that det(T) = Δ Q for some polynomial Q in the coefficients of P. Of course if the degree of Δ matches the degree of det(T) as it does in the case of quadratics and cubics then Q must be a constant.

Actually we can go a little further. if det(T) = 0 then it has a singular point at say **x**_{i }= **ξ**_{i }, but since T is symmetric any permutation of the vectors gives another singular point **x**_{i }= **ξ**_{σ(i)}. In general the hypermatrixdoes not have multiple singular points and what we actually have here is just one with **x**_{i }= **ξ.** This means that the form F(**x**) is singular and Δ = 0. So the implication goes both ways. det(T) = 0 <=> Δ = 0 . Does this mean that det(T) and Δ must be equal up to a constant? Not quite. What actually happens is that

det(T) = KΔ^{m}

for some constant K and intger exponent m. Actually we have not quite proven this because we have assumed that Δ itself is not a power of some polynomial in which case m could be rational, however I think the result is correct.

So there is a link between polynomial discrminants of one variale and hyperdterminants on 2n hypermatrices. Of course this result can also be generalised to discriminants for multivariable polynomials which are similarly related to larger symmetric hypermatrices, but that’s not really the point. the importnant hting to see is that hyperdterminants are just generalisations of the familiar discriminants. Discriminants have been much more heavily studied and generalised in other ways. They an be applied in agebraic geometry and used to understand concepts such as ramification. Putting hyperdterminants in the same context should help us see more clearly where they belong.

I’ll finish by mentioning one last interesting consequnce of this result. Since the hyperdeterminant is a constant times a power of the discriminant it follows that the degree of the hyperdeterminant is a multiple of the degree of the discriminant. In other words the degree of the hyperdeterminant of format 2^{n} is a multiple of 2n -2. This may not sound much but it is not obvious at this stage what the degrees of the hyperdeterminants are so this is worth knowing.

October 9, 2008 at 11:01 am

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