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Example text
Consider the k -vector space V A spanned by G-translates of the ti . 4 that V is finite-dimensional. Without loss of generality we may asA sume now that (t1 : : : tn ) is a basis of this space. Let : S = k T1 : : : Tn ] be the surjective homomorphism defined by T i ti . The group G acts on S linearly by g (Ti ) = T , where g ( t ) = t ij j i ij j . Let I be the kernel of . G It is obviously G-invariant. We obtain that A = (S=I )G . 5, AG is integral over S G =I S G . Since we have shown already that S G is finitely generated, we are almost done (certainly done in the case when G is linearly reductive).
Recall that a normal ring is a domain integrally closed in its field of fractions. Let K be the field of fractions of A. It is clear that the field of fractions L of AG is contained in the field K G of G-invariant elements of K . We have to check that the ring AG is integrally closed in L. Suppose x L satisfies a monic equation xn + a xn;1 + + a = 0 2 with coefficients ai is verified. 1 0 2 AG . Since A is normal, x 2 A \ K G = AG and the assertion Bibliographical notes The proof of the Gordan–Hilbert Theorem follows the original proof of Hilbert (see [47]).
Since the standard tableau functions are linearly independent, we obtain that F I r0 m. 1. The equations PI J = 0 defining the Grassmannian Gr(r m) are called the Pl¨ucker equations. 3 implies that the Pl¨ucker equations describe the basic relations between the bracket functions. This result is sometimes referred to as the Second Fundamental Theorem of Invariant Theory. Now we are in business and finally can compute something. We start with the case r = 2. Let us write any degree d homogeneous standard tableau in the form 2 a1 a2 3 66 a112 a223 77 = 64 ..
