Invariant Theory - Hilbert's Theorems

Hilbert's Theorems

Hilbert (1890) proved that if V is a finite dimensional representation of the complex algebraic group G = SLn(C) then the ring of invariants of G acting on the ring of polynomials R = S(V) is finitely generated. His proof used the Reynolds operator ρ from R to RG with the properties

  • ρ(1) = 1
  • ρ(a + b) = ρ(a) + ρ(b)
  • ρ(ab) = a ρ(b) whenever a is an invariant.

Hilbert constructed the Reynolds operator explicitly using Cayley's omega process Ω, though now it is more common to construct ρ indirectly as follows: for compact groups G, the Reynolds operator is given by taking the average over G, and non-compact reductive groups can be reduced to the case of compact groups using Weyl's unitarian trick.

Given the Reynolds operator, Hilbert's theorem is proved as follows. The ring R is a polynomial ring so is graded by degrees, and the ideal I is defined to be the ideal generated by the homogeneous invariants of positive degrees. By Hilbert's basis theorem the ideal I is finitely generated (as an ideal). Hence, I is finitely generated by finitely many invariants of G (because if we are given any – possibly infinite – subset S that generates a finitely generated ideal I, then I is already generated by some finite subset of S). Let i1,...,in be a finite set of invariants of G generating I (as an ideal). The key idea is to show that these generate the ring RG of invariants. Suppose that x is some homogeneous invariant of degree d > 0. Then

x = a1i1 + ... + anin

for some aj in the ring R because x is in the ideal I. We can assume that aj is homogeneous of degree d − deg ij for every j (otherwise, we replace aj by its homogeneous component of degree d − deg ij; if we do this for every j, the equation x = a1i1 + ... + anin will remain valid). Now, applying the Reynolds operator to x = a1i1 + ... + anin gives

x = ρ(a1)i1 + ... + ρ(an)in

We are now going to show that x lies in the R-algebra generated by i1,...,in.

First, let us do this in the case when the elements ρ(ak) all have degree less than d. In this case, they are all in the R-algebra generated by i1,...,in (by our induction assumption). Therefore x is also in this R-algebra (since x = ρ(a1)i1 + ... + ρ(an)in).

In the general case, we cannot be sure that the elements ρ(ak) all have degree less than d. But we can replace each ρ(ak) by its homogeneous component of degree d − deg ij. As a result, these modified ρ(ak) are still G-invariants (because every homogeneous component of a G-invariant is a G-invariant) and have degree less than d (since deg ik > 0). The equation x = ρ(a1)i1 + ... + ρ(an)in still holds for our modified ρ(ak), so we can again conclude that x lies in the R-algebra generated by i1,...,in.

Hence, by induction on the degree, all elements of RG are in the R-algebra generated by i1,...,in.

Read more about this topic:  Invariant Theory