lectures-3-math207

# lectures-3-math207 - 3 Hilbert-Mumford type theorems 3.1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3 Hilbert-Mumford type theorems. 3.1 Basics on group actions. 3.1.1 Algebraic group actions. Let X be an algebraic variety and let G be an algebraic group both over C . Then an (algebraic group) action of G on X is a morphism & : G & X ! X satisfying: 1. &(1 ;x ) = x ( 1 denoting the identity element of G ) for all x 2 X . 2, &( gh;x ) = &( g; &( h;x )) for all g;h 2 G and x 2 X . We will denote such an action by gx . The set Gx is called the orbit of x . Our main example is G a Z-closed subgroup of GL ( n;F ) , X = F n and gx is the matrix action of G on F n . More generally, we de&ne a regular representation of an algebraic group, G , to be a group morphism & : G ! GL ( V ) where V is a &nite dimensional vector space over C . We denote it ( &;V ) . One more bit of notation the isotropy group of x 2 X is the set f g 2 G j gx = x g and it will be denoted G x . We will con&ne our attention to the case when G is irreducible (that is connected in the Z-topology). Lemma 1 Let G be irreducible and act on an algebraic variety, X . Let x 2 X and let Y be the Z-closure of Gx . Then 1. Y is irreducible. 2. Gx is Z-open in Y: 3. There is a Z-closed G orbit in Y . 4. Y is the S-closure of Gx . Proof. We have seen in Theorem 26 of 1 : 3 : 4 that Gx has interior in Y . Since y ! gy de&nes an automorphism of Y we see that Gx is a union of open subsets of Y . This proves 2. Since Gx is the image under a morphism of an irreducible variety it is irreducible. As it is dense in Y , Y is irreducible. Let Z = Y ¡ Gx . Then Z is closed and G-invariant. If Z = ; then Y is closed. If not let V = Gz be an orbit of minimal dimension in Z: If W is the closure of V then W ¡ V is closed in W and since W is irreducible dim( W ¡ V ) < dim W . Thus the dimension of any orbit in W ¡ V would be lower than the minimum possible. Thus we must have V is closed proving 3. We note that 4 is an immediate consequence of 2 and Theorem 20 of 1.1.2.7. 1 Proposition 2 Let G be an a¢ ne algebraic group acting on an irreducible a¢ ne variety X . Then there exists an imbedding of & : X ! C n as a Z-closed subset of C n for some n and an algebraic group homomorphism, ¡ : G ! GL ( n; C ) such that & ( gx ) = ¡ ( g ) & ( x ) for all x 2 X and g 2 G . Proof. We may assume that X & C m is Z-closed, Let f i = x i j X for i = 1 ;::;m ( x i the standard coordinates on X ). Let the action of G on X be given by F: Then F & O ( X ) is a subalgebra of O ( G ¡ X ) = O ( G ) ¢ C O ( X ) . We also note that F & ( f )(1 ;x ) = f ( x ) also if we set ¢ ( g ) f ( x ) = f ( g ¡ 1 x ) then F & ¢ ( g ) f ( h;x ) = F & f ( hg ¡ 1 ;x ) . This implies that the linear span of f F & ¢ ( g ) f j g 2 G g is &nite dimensional for each f 2 O ( X ) . Thus we see that the linear span, W , of f ¢ ( g ) f i j g 2 G;i = 1 ;::;m g is &nite dimensional. Let u 1 ;:::;u d be a basis of W . Then the map & ( x ) = ( u 1 ( x ) ;:::;u d ( x )) de&nes an isomorphism of X into C d . We assert that & ( X ) is S-closed in C n...
View Full Document

{[ snackBarMessage ]}

### Page1 / 17

lectures-3-math207 - 3 Hilbert-Mumford type theorems 3.1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online