In my last post I showed 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 preliminary to allow us to construct the discriminant for the quartic which we will need to understand the all important Schlafli hyperdterminant for a 2x2x2x2 hypermatrix.

Historically the subject developped the other way round. discriminants of polynomials were studied first and by time Cayley graduated Boole had a general theory and new the degree of any multivariable discriminant. Cayley took up the challenge of looking at more general invariants and this discovered hyperdeteriminants. he spent much ofthe rest of his life buried in massive algebraic manipulations working out polynomial invariants to higher and higher orders.

We have seen a polynomial P(x) of degree n can be represented using a symmetric tensor

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

When we apply a linear transorm S to the vector x we generate a transformation on the underlying tensor where each index is transformed using the same matrix thus keeping it symmetric. The hyperdeterminant of the underlying tensor is invariant under special linear transforms applied more generally to any index individually, and the symmetric transformation with a matrix of determiant one is a special case of this. Since the hyperdeterminant is a power of the discriminant we can draw the conclusion that the discriminant is also actually an invariant of SL(2). Since we have come to this conclusion in a very roundabout way (and there were holes in the argument), it is worth trying to verify it directly from the definition of the discriminant.

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 and a_{n }is the leading coefficient. We apply a special linear transform (x,y) -> (ax+by, cx+dy) where ad-bc = 1. This induces a transformation of the form

F -> G such that G(ax+by, cx+dy) = F(x,y)

This in turn implies a transform of the polynomial coefficients and its roots. We need to check that the expression for the discriminant is invariant under this transformation.

The argument of the polynomial transforms rationally

x/y -> (ax+by)/(cx+dy) = (a(x/y)+b)/(c(x/y)+d)

So the set of roots is transformed by

r_{i }-> (ar_{i} + b)/(cr_{i}+d)

Using this we can verify that

r_{i} – r_{j }-> (r_{i} – r_{j })/[(cr_{i}+d)(cr_{j}+d)]

We also need to know how the leading coefficient transforms

a_{n }= F(1,0) -> G(1,0) = F(d,-c) = (-c)^{n}P(-d/c)

using the factorisation P(x) = a_{n}Π(x-r_{i}) we arrive at

a_{n } -> a_{n}(-c)^{n}Π(-d/c-r_{i}) = a_{n}Π(cr_{i}+d)

It is now just a matter of substitution of these transformations back into the definition of the discriminant to check that the exponent of the leading coefficient is correctly choosen so that the powers of (cr_{i}+d) cancel leaving the discriminant invariant.

Furthermore there is an extra surprise that we can get from this. Using simialr expressions in terms of the polynomial roots other invariants can sometimes be constructed. For example, in the case of the quartic we can define a new invariant as

V = a_{4}^{2}( (r_{1}-r_{2})^{2}(r_{3}-r_{4})^{2 }+ (r_{1}-r_{3})^{2}(r_{2}-r_{4})^{2 }+ (r_{1}-r_{4})^{2}(r_{2}-r_{3})^{2} )

This will give us an invariant of degree two in the coefficients. We see more about this later.

October 9, 2008 at 2:56 pm

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