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INTRODUCTION TO ALGEBRAIC GROUPS AND INVARIANT THEORY
IVAN V. ARZHANTSEV

Preface
These notes originate in a series of eleven lectures and eight seminars given at the Eberhard Karls Universit Tubingen in the Spring of 2007. The material here is at based on special courses and seminars given by Professor Ernest Vinberg, Dmitri Timashev and myself at the Faculty of Mechanics and Mathematics of Moscow State University. I am grateful to everyone who attended my lectures in Tubingen for their patience and critical remarks. I am especially thankful to Professor Dr. Jurgen Hausen for invitation and useful discussions, and to Devrim Celik for technical support.

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ALGEBRAIC GROUPS AND INVARIANT THEORY

iii

Contents
Preface 1. A ne Algebraic Groups 1.1. Examples and rst properties Exercises to subsection 1.1 1.2. Connected components and homomorphisms Exercises to subsection 1.2 1.3. Actions and representations of algebraic groups Exercises to subsection 1.3 1.4. Homogeneous spaces Exercises to subsection 1.4 1.5. The tangent algebra Exercises to subsection 1.5 1.6. Algebraic tori Exercises to subsection 1.6 1.7. Jordan decompositions Exercises to subsection 1.7 1.8. Solvable groups. Maximal tori and Borel subgroups Exercises to subsection 1.8 1.9. Reductive groups Exercises to subsection 1.9 2. Invariant Theory 2.1. Finite generation Exercises to subsection 2.1 2.2. The quotient morphism and categorical quotients Exercises to subsection 2.2 2.3. Rational invariants. Rosenlicht's Theorem Exercises to subsection 2.3 3. Appendix One: Some Facts from Algebraic Geometry 4. Appendix Two: Some Facts on Lie Algebras 5. Hints and Solutions to Exercises References i 1 1 4 5 8 9 14 15 19 21 26 27 31 33 38 39 43 45 50 51 51 54 55 59 61 65 67 69 71 87

ALGEBRAIC GROUPS AND INVARIANT THEORY

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1. Affine Algebraic Groups 1.1. Examples and rst properties. Let K be an algebraically closed eld, and put V = Kn . The elements of the polynomial algebra K[x1 ; : : : ; xn ] can be viewed as K-valued functions on V . They form the algebra O(V ) of regular functions on V . An algebraic subvariety X V is the set of solutions of a system of polynomial equations on V with the algebra of regular functions O(X ). By de nition, functions in O(X ) = O(V )=I(X ) are restrictions of elements of O(V ) to X , and I(X ) = {f O(V ) : f |X 0}. There are natural notions of morphisms and isomorphisms between algebraic subvarieties. An a ne variety is an isomorphism class of algebraic subvarieties.

De nition 1.1.1. An a ne algebraic group is a group G equipped with a structure of an a ne variety such that the multiplication map : GвG G, (g1 ; g2 ) = g1 g2 and the inverse map i : G G, i(g) = g-1 are morphisms of a ne varieties.
of all invertible (n в n)-matrices over K. It has a structure of an a ne variety. 2 Namely, take the vector space Kn +1 with coordinates (t; aij ); i; j = 1; : : : ; n. Then 2 GL(n) may be realized as a subset of Kn +1 de ned by the equation det(aij )t = 1. 1 ]. This implies O(GL(n)) = K[aij ][ det The multiplication map is given as (A; B ) = AB = C , and if A = (aij ), B = (bij ), C = (cij ), then cij = n=1 aik bkj . As we know from the course of linear k -1)i+j algebra, i(A) = A-1 = (dij ) with dij = (det(A) mj i , where mj i are minors of A. This shows that the maps : GL(n) в GL(n) GL(n) and i : GL(n) GL(n) are polynomial, thus GL(n) is an a ne algebraic groups.

Example 1.1.2 (The general linear group). Consider the group GL(n; K) = GL(n)

Remark 1.1.3. In this course we consider only a ne algebraic groups, so the
adjective "a ne" will be sometimes omitted.

Proposition 1.1.4. Let G GL(n) be a subgroup. Assume that G is a (closed) subvariety in GL(n). Then G is an algebraic group.
Proof. The multiplication and inverse maps for G are the restrictions to GвG and G of the corresponding maps for GL(n). Clearly, such restrictions are morphisms.
A closed (in Zariski topology) subgroup of GL(n) is called a linear algebraic group. As we shall prove later (see Theorem 1.3.20 and Remark 1.3.21), any a ne algebraic group admits a realization as a closed subgroup of some GL(n). So, the notion "a ne algebraic group" and "linear algebraic group" turn out to be synonyms. On the other hand, some authors use the term "linear algebraic group" in order to x the matrix realization of a group G.

Example 1.1.5 (Classical Groups). Besides GL(n), there are four other series of
algebraic groups that are called classical. (i) The special linear group SL(n) consists of the matrices of determinant 1 in GL(n). It is clearly a subgroup de ned by the equation det(aij ) - 1 = 0 in GL(n). Thus SL(n) is an algebraic group. Since it is a hypersurface in the space of matrices Mat(n в n; K), the dimension of SL(n) equals n2 - 1. (ii) Let q : V в V K be a bilinear symmetric non-degenerate form on the space V = Kn . De ne a group O(q) := {A GL(n) : q(Av1 ; Av2 ) = q(v1 ; v2 ) for any v1 ; v2 V }:

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IVAN V. ARZHANTSEV

Let Q be a symmetric (n в n)-matrix associated with q, i.e. q(v1 ; v2 ) = T v1 Qv2 for any v1 ; v2 V . Then T q(Av1 ; Av2 ) = (Av1 )T Q(Av2 ) = v1 AT QAv2 ; and A O(q) if and only if AT QA = Q. The last equality may be considered as a system of quadratic equations on aij , where, as usual, A = (aij ). This shows that O(q) is an algebraic group. It is called the orthogonal group associated with the form q. For example, if v1 = (x1 ; : : : ; xn ), v2 = (y1 ; : : : ; yn ), and the form q is de ned as q(v1 ; v2 ) = x1 y1 + · · · + xn yn , then Q = E and O(q) = {A GL(n) : AT A = E }. In this case we denote O(q) as O(n). (iii) The equation AT QA = Q implies det(A) = ±1 (Q is non-degenerate !). De ne the special orthogonal group as SO(q) = {A O(q) : det(A) = 1}: Clearly, SO(q) is a subgroup in O(q), and this subgroup is proper (see Exercise 1.1.11). (iv) Assume that char K = 2. Let V = K2n , ! : V в V K be a bilinear skew-symmetric non-degenerate form on V , and be the associated skewsymmetric (2n в 2n)-matrix. De ne the symplectic group as Sp(!) = {A GL(2n) : !(Av1 ; Av2 ) = !(v1 ; v2 ) for any v1 ; v2 V } = = {A GL(2n) : AT A = }: Again for the standard skew-symmetric form !(v1 ; v2 ) = x1 y2 - x2 y1 + · · · + x2n-1 y2n - x2n y2n-1 we reserve the notation Sp(!) = Sp(2n). Example 1.1.6 (Finite groups). Recall that any nite set X with n elements admits a canonical structure of an a ne algebraic variety (over K). This variety has n irreducible one-point components and the algebra of regular functions O(X ) is the direct sum of n copies of the eld K: O(X ) = K · · · K. In particular, any K-valued function on X is regular, and any map X Y to another a ne variety Y is a morphism. This shows that any nite group G has a canonical structure of an a ne algebraic group. Example 1.1.7 (Additive and multiplicative groups). The additive group Ga is the a ne line K1 with group low (x; y) = x + y and i(x) = -x. The multiplicative group Gm is the a ne open subset Kв K with (x; y) = xy, i(x) = x-1 . Clearly, they are commutative one-dimensional algebraic groups. The group Gm may be realized as GL(1), but for a matrix realization of Ga one needs 2 в 2-matrices: 1 c ; cK : 01 Proposition 1.1.8. Let G1 and G2 be a ne algebraic groups. Then the direct product G1 в G2 has a canonical structure of an a ne algebraic group.

Proof. If X1 ; X2 ; Y1 ; Y2 are a ne varieties and 1 : X1 Y1 , 2 : X2 Y2 are morphisms, then : X1 в X2 Y1 в Y2 , (x1 ; x2 ) = (1 (x1 ); 2 (x2 )) is again a morphism. This shows that the multiplication map and the inverse map: : (G1 в G2 ) в (G1 в G2 ) G1 в G2 ; ((g1 ; g2 ); (g1 ; g2 )) = (g1 g1 ; g2 g2 ); - - i : G1 в G2 G1 в G2 ; i(g1 ; g2 ) = (g1 1 ; g2 1 ) are morphisms.

ALGEBRAIC GROUPS AND INVARIANT THEORY

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Proposition 1.1.8 allows us to construct many examples of algebraic groups. In particular, the direct product T k = Gk = Kв в · · · в Kв (k times) is a commutative m algebraic group called an algebraic torus. Note that, if ti is the coordinate in the ith copy of Kв , then O(T k ) = K[t1 ; t-1 ; : : : ; tk ; t-1 ] is the algebra of Laurent 1 k polynomials. We nish this subsection with some further examples of linear algebraic groups that will be used frequently in this course. Example 1.1.9 (Some other matrix groups). Let D(n), B (n) and U (n) be the subgroups of diagonal, upper triangular, and upper triangular unipotent matrices in GL(n) respectively:
D(n) =
0 : : : 0

::: 0

0

: : : :

: : : :

: : : :

0 0 0

; B (n) =

0 : : : 0

::: 0

: : : :

: : : :

: : : :

; U (n) =

:

1 0 :: 0

1 ::: 0

: : : :

: : : :

: : : :

1

:

Note that these subgroups are de ned in GL(n) by equations of the form: aij = 0 or 1 for some i; j; so they are algebraic subgroups.

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IVAN V. ARZHANTSEV

Exercises to subsection 1.1. T T Exercise 1.1.10. Check that if A; B Mat(n в n) and v1 Av2 = v1 B v2 for any v1 ; v
Kn

, then A = B .

2

Exercise 1.1.11. Show that the group O(q) contains matrices A1 and A2 with det(A1 ) = 1 and det(A2 ) = -1. Exercise 1.1.12. Check that SO(2) is commutative, but O(2) is not. Exercise 1.1.13. Prove that for any bilinear symmetric non-degenerate form q on V = Kn the subgroup O(q) (resp. SO(q)) is conjugate to O(n) (resp. to SO(n)) in GL(n). Moreover, for any bilinear skew-symmetric non-degenerate form ! on V = K2n the subgroup Sp(!) is conjugate to Sp(2n) in GL(2n). Exercise 1.1.14 (*). Prove that det(A) = 1 for any A Sp(!). Exercise 1.1.15 (*). Show that SL(2) = Sp(2), but Sp(2n) is a proper subgroup of SL(2n) for any n > 1. Exercise 1.1.16. Show that GL(n; R) is not a closed subgroup in GL(n; C). Exercise 1.1.17. Find the linear span of SL(n) in the vector space Mat(n в n). Exercise 1.1.18. Find the center of a) GL(n); b) SL(n); c) B (n); d) U (n). Exercise 1.1.19. Calculate the dimension of a) D(n); b) B (n); c) U (n). Exercise 1.1.20. Describe elements of nite order in Ga and Gm . Exercise 1.1.21. Do all elements of nite order form a subgroup in GL(n) ? Exercise 1.1.22. Find a matrix realization for Gk в Gs . m a Exercise 1.1.23. Let A be a nite-dimensional K-algebra. Prove that the automorphism group of A is an a ne algebraic group. Exercise 1.1.24 (*). Let G be a group with a structure of an a ne algebraic variety such that the multiplication map : G в G G is a morphism. Prove that the inverse map i : G G is a morphism. Exercise 1.1.25 (**). Is the group SL(2) isomorphic to A3 as an a ne variety ?

ALGEBRAIC GROUPS AND INVARIANT THEORY

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1.2. Connected components and homomorphisms. Let G be an algebraic group. Our aim is to show that the group structure on G imposes strong restrictions on its geometry. The reason for this is "homogeneity" or "equal status of points". More precisely, with any element g G one associates two mappings: Lg : G G; Lg (x) = gx; Rg : G G; Rg (x) = xg: By de nition of an algebraic group, these mappings are morphisms. Moreover, Lg 1 (resp. Rg 1 ) is inverse to Lg (resp. Rg ), so Lg and Rg are automorphisms of the variety G. For any x; y G the map Lyx 1 sends x to y. In particular, the automorphism group of the variety G acts on G transitively. Lemma 1.2.1. Let G be an algebraic group. Then the variety G is smooth.
- - -

Proof. Let x G and y sends y to x, so x is also Lemma 1.2.2. If H is closure H is an algebraic

be a smooth point of G. Then the automorphism Lxy 1 a smooth point. an (abstract) subgroup of an algebraic group G, then its subgroup of G.
-

Proof. Since for any h H the automorphism Lh : G G preserves H , it also preserves H . This implies H H H . So for any g H one has H y H . But H y coincides with the closure of the subset H y and is contained in H . This proves that H H H . The map i is an automorphism of G and, if it preserves a subset H , then it preserves its closure H . So H is a subgroup in G, and by Proposition 1.1.4 this subgroup is algebraic.
Recall that a subset Z of a variety X is called local ly closed if Z is open in its closure Z . Lemma 1.2.3. Let G be an algebraic group and H G a local ly closed subset that is an (abstract) subgroup of G. Then H is closed.

Proof. By Lemma 1.2.2, for any g H the coset gH is contained in H . Moreover, gH is open in H as a translate of an open subset H H . But H is dense in H , so H gH = . If x H gH , then g = xh-1 for some h H , thus g H .

Theorem 1.2.4. Let G be an algebraic group and e G the unit. Then
(i) irreducible components of G coincide with connected components; (ii) the connected component G0 containing e is a normal subgroup of a nite index in G, and al l other connected components are cosets of G by G0 .

Proof. Let G0 ; : : : ; Gk be the irreducible components of G. By de nition, there exists g G0 \ (G1 · · · Gk ). The image of g under any automorphism of the variety G also belongs to a unique irreducible component of G. This shows that any element of G lies in a unique irreducible component, so Gi Gj = for any i = j , and (i) is proved. Since the variety G0 в G0 is irreducible, so is its image (G0 в G0 ). Then (G0 в G0 ) is contained in an irreducible component of G. But e (G0 в G0 ), and thus (G0 в G0 ) = G0 . The same arguments show that i(G0 ) = G0 , so G0 is a subgroup. Now consider the morphism : Gi в G0 G, (g; h) = ghg-1 . The variety Gi в G0 is irreducible, (g; e) = e, thus (Gi в G0 ) G0 , or ghg-1 G0 for any g G; h G0 , and G0 is normal in G. For any g Gi the automorphism Lg maps G0 isomorphically to some irreducible component of G. But Lg (e) = g, thus Lg maps G0 to Gi . Moreover, the number of

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irreducible components of any variety is nite, so G0 has a nite index in G, and the proof of (ii) is completed.

Lemma 1.2.5. Let G be a connected algebraic group and U; V G be two nonempty open subsets. Put U V := {uv : u U; v V }. Then G = U V .
{

Proof. g-1 : gG v V,

Since the map i is an g V } is open. By the intersection of two u U such that gv-1

automorphism of the variety G, the subset V -1 := Theorem 1.2.4, the group G is irreducible. For any open subsets gV -1 and U is non-empty, so there are = u, or g = uv.

Proposition 1.2.6. Let G be an algebraic group and {Mi G : i I } be a family
of subsets such that
1) e Mi for any i I ; 2) Mi are irreducible (in the induced topology); 3) Mi contains an open subset of Mi .

Then the subgroup H

G generated by the subsets Mi is closed and connected.

::: Proof. Consider Mi11:::ikk := 1 ::: k (Mi1 в · · · в Mik ), where 1 ::: k (g1 ; : : : ; gk ) := 1 ::: k 1 : : : g k with g1 i = ±1. Clearly, Mi1 :::ik satis es 1). Since the image and the k product of irreducible subsets are irreducible, we also have 2). Finally, take open subsets Ui Mi Mi . By Theorem 3.0.24, the image of ::: Mi1 в · · · в Mik Mi11:::ikk

contains an open subset of Mi and

1 ::: k 1 :::ik

, and we get 3). Note that
Mi
1 ::: k 1 :::ik

H=

::: ::: ::: ::: ::: ::: Mi11:::ikk ; Mj11:::jss Mi11:::ikkj11:::jss = Mi11:::ikk Mj11:::jss : ::: ::: ::: Set M := Mi11:::ikk with the maximal dim Mi11:::ikk . Then any Mj11:::jss is contained in ::: ::: Mi11:::ikkj11:::jss = M (Theorem 3.0.25), thus H = M . This shows that H is connected. By Lemma 1.2.2, H is a subgroup. Applying Lemma 1.2.5 to H and U = V an open subset of M , we get H = U U H , so H is closed.

Corollary 1.2.7. The groups GL(n) and SL(n) are connected.
Proof. As we know from linear algebra, any non-degenerate matrix is a product of elementary matrices. Thus one may apply Proposition 1.2.6 to irreducible curves Mij = {E + cEij : c K} (i = j ) and Mii = {E + ( - 1)Eii : Kв } which generate GL(n). In the case of SL(n) the curves Mij (i = j ) are su cient.

Remark 1.2.8. For GL(n), there is a simpler proof: it is irreducible as an open

subset of Mat(n в n). Corollary 1.2.9. Let G be a connected algebraic group. Then the commutant [G; G] is a closed connected subgroup of G.

Proof. Consider a morphism : G Proposition 1.2.6 to {Mi } = { (G)}.

в

G G, (x; y) = xyx-1 y-1 and apply

Now we come to the de nition of a morphism in the category of algebraic groups.

ALGEBRAIC GROUPS AND INVARIANT THEORY

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H is said to be a homomorphism if (g1 g2 ) = (g1 )(g2 ), i.e. is a homomorphism of abstract groups. Example 1.2.11. For any integral k the map A det(A)k is a homomorphism from GL(n) to Kв . On the other hand, the map C Cв , a ea is a homomorphism of abstract groups, but not a morphism. Theorem 1.2.12. Let : G F be a homomorphism of algebraic groups. Then Ker() G and Im() F are closed subgroups. Moreover, dim G = dim Ker() + dim Im():

De nition 1.2.10. Let G and F be algebraic groups. A morphism : G

Proof. Since Ker() = -1 (e), it is a closed normal subgroup of G. Assume that G is connected. The image Im() is a subgroup of F , and, by Theorem 3.0.24, it contains an open subset U of Im(). By Lemmas 1.2.2 and 1.2.5, Im() = U U = Im(). If G0 ; G1 ; : : : ; Gk are connected components of G and g1 G1 ; : : : ; gk Gk , then Im() = (G0 ) (g1 )(G0 ) · · · (gk )(G0 ) is closed. In order to prove the dimension formula, one may assume that (G) = F . Any ber of |G0 : G0 F 0 is a coset of G0 by Ker()0 := Ker() G0 . By Theorem 3.0.26, one has dim G0 = dim Ker()0 + dim F 0 . Clearly, dim G = dim G0 and dim F = dim F 0 . We claim that dim Ker() = dim Ker()0 . Indeed, the connected component Ker()0 is contained in G0 , thus dim Ker()0 dim Ker()0 dim Ker() = dim Ker()0 :

De nition 1.2.13. A homomorphism : G F is called an isomorphism if there is a homomorphism : F G such that = idF and = idG . Remark 1.2.14. If char K = 0, then by Theorem 3.0.27 any bijective homomorphism is an isomorphism. For char K = p > 0 this is not the case: one may consider the Frobenius homomorphism : Ga Ga , (x) = xp .

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IVAN V. ARZHANTSEV

Exercises to subsection 1.2. Exercise 1.2.15. Construct two non-isomorphic structures of an algebraic group on the
variety
A3

.

algebraic group.

Exercise 1.2.16 (*). Show that the variety

A1 \ {

0; 1} does not admit a structure of an

ponent of O(2) consists of elements of order two. Exercise 1.2.18. Let H be a subgroup of a nite index in G. Prove that H contains G0 . Exercise 1.2.19. Give an example of an algebraic group G such that the center Z (G) is nite, but Z (G0 ) is in nite. Exercise 1.2.20. By Theorem 1.2.4, for any algebraic group G the quotient group F = G=G0 is nite. Give an example of G which is not isomorphic to a semidirect product of F and G0 . Exercise 1.2.21. Let K be a eld. Prove that the polynomial det(aij )-1 K[a11 ; : : : ; ann ] is irreducible. Exercise 1.2.22. Calculate the commutant of a) GL(n); b) SL(n); c) B (n); d) U (n). Exercise 1.2.23. Let G GL(n) be a closed connected subgroup with a nite center Z (G). Prove that Z (G=Z (G)) = {e}. It is true for a non-connected G ? Exercise 1.2.24 (*). Let G be an algebraic group. Prove that [G; G] is a closed subgroup of G. Exercise 1.2.25 (*). Give an example of a homomorphism of Lie groups : G1 G2 such that Im() is not a (closed) Lie subgroup of G2 .

Exercise 1.2.17. Show that SO(2) is isomorphic to Gm , and the second connected com-

ALGEBRAIC GROUPS AND INVARIANT THEORY

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1.3. Actions and representations of algebraic groups. Let X be an algebraic variety and G an algebraic group. De nition 1.3.1. A morphism : G в X X is said to be an (algebraic) action, if it satis es the following properties: (i) (e; x) = x for any x X ; (ii) (g1 ; (g2 ; x)) = (g1 g2 ; x) for any g1 ; g
2

G, x X .

Further we abbreviate (g; x) as g · x. The fact that we have an action of G on X will be denoted as G : X , and a variety with a G-action will be called shortly a G-variety. (g; g1 ) = g1 g-1 , C (g; g1 ) = gg1 g-1 . De nition 1.3.3. Assume that a group G acts on a variety X . For any x X de ne its orbit Gx := {y X : y = g · x for some g G} and its stabilizer Gx := {g G : g · x = x}. Theorem 1.3.4. Let an algebraic group G act on a variety X . Then for any x X the stabilizer Gx is a closed subgroup of G and the orbit Gx is a smooth local ly closed subvariety of X . Moreover, dim G = dim Gx + dim Gx :
L

Example 1.3.2. There are three remarkable actions of a group G on itself. Namely,
(g; g1 ) = gg1 ,
R

Proof. Consider a morphism : G в X X в X; (g; x) = (x; g · x) - and the embedding rx : G G в X , rx (g) = (g; x). Clearly, Gx = rx 1 ( -1 (x; x)), thus Gx is closed. For the orbit morphism G0 X , g g · x, the image G0 x contains an open subset U of G0 x (Theorem 3.0.24). Then gi U is open in gi G0 x = gi G0 x = Gi x. This implies that Gx contains an open subset W = k=0 gi U of Gx = k=0 Gi x (here i i g0 = e). But then Gx = gG gW is open in Gx. Smoothness of Gx follows from transitivity of the G-action (by automorphisms) on Gx. The dimension of any ber of the orbit morphism G0 G0 x equals dim(G0 )x , and Theorem 3.0.26 implies dim G0 = dim(G0 )x + dim G0 x. Finally, one can easily check that dim G = dim G0 , dim Gx = dim(G0 )x , and dim Gx = dim G0 x.

Example 1.3.5. The group GL(n) acts on the space of symmetric n в n-matrices as (A; S ) AS AT . Non-degenerate symmetric matrices form an open orbit and the stabilizer of the unit matrix equals O(n). By Theorem 1.3.4, dim GL(n) = dim O(n) + n(n2+1) , thus dim O(n) = n(n2-1) . Corollary 1.3.6. For any action G : X there is a closed G-orbit.
Proof. The orbit Gx is open and dense in Gx. Hence dim(Gx \ Gx) < dim Gx and any orbit of the smallest dimension is closed in X .

Corollary 1.3.7. For any subset A G de ne the centralizer ZG (A) := {g G :
gx = xg for any x A}. Then ZG (A) is a closed subgroup of G.

Proof. For any a A the centralizer ZG (a) is the stabilizer of a G with respect to the action C (see Example 1.3.2). Thus ZG (a) is closed, and ZG (A) = aA ZG (a).

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IVAN V. ARZHANTSEV

Proposition 1.3.8. For any action G : X and any integer k the subsets
Xk := {x X : dim Gx k} and X k := {x X : dim Gx k} are closed.

Proof. Since the sets Xk and X k coincide for G and G0 , and G0 preserves irreducible components of X , one may reduce the proof to the case when G in connected and X is irreducible. Again consider : G в X X в X , (g; x) = (x; g · x). By Theorem 3.0.26, there exist an integer m and a non-empty open subset U (G в X ) such that
(i) for any point (x; y) U any component of the ber -1 (x; y) has dimension m; (ii) any component of any non-empty ber -1 (x; y) has dimension m. Note that -1 (x; y) = {(g; x) : g · x = y} gGx Gx . The image of the pro jec= = tion of U to the rst factor contains a non-empty open subset (Theorem 3.0.24). This shows that Xk = X for k m and there is an open subset W X which is contained in X \ Xm+1 . Since Xm+1 is a G-invariant subset, the open G-invariant subset GW also is contained in X \ Xm+1 . Consider the decomposition into irreducible components: X \ GW = Y (1) · · · Y (s). Any Y (i) is an irreducible G-variety with dim Y (i) < dim X . Arguing by induction on the dimension, one may assume that Y (i)k is closed for any k. On the other hand, for k > m one has Xk = Y (1)k · · · Y (s)k . Finally, X k = Xdim G-k (Theorem 1.3.4).

Proposition 1.3.9. Let an algebraic group G act on a variety X . Then the subset
X G of G- xed points in closed in X .

Proof. For any g G de ne a morphism g : X X в X , g (x) = (x; g · x). The set X g of g- xed points is the preimage of the diagonal in X в X , thus is closed. The set X G is the intersection of closed subsets: X G = gG X g : representation of an algebraic group G in the space V is a homomorphism : G GL(V ) of algebraic groups. Here V is said to be a ( nite-dimensional) rational G-module. A representation is called faithful, if Ker() = {e}.

De nition 1.3.10. Let V be a nite-dimensional K-vector space. A (rational)

De nition 1.3.11. Any representation : G

GL(V ) de nes an action G в V V , g · v = (g)v. Such actions are called linear.

:

Remark 1.3.12. A rational representation of GL(n) in a space V is a homomor-

phism : GL(n) GL(V ) such that the matrix entries of (A) are polynomials in 1 1 aij ; det(A) . The presence of det(A) motivates the term "rational".

Remark 1.3.13. Standard constructions of representation theory (restrictions to

invariant subspaces, quotient and dual representations, direct sums, tensor products, symmetric and exterior powers,...) allow to produce numerous rational Gmodules from given ones.

Lemma 1.3.14. Any rational representation : G

GL(V ) de nes a natural algebraic action on the projective space P(V ): g · [v] := [(g)v].

ALGEBRAIC GROUPS AND INVARIANT THEORY

11

Proof. Let x1 ; : : : ; xn be a coordinate system on V . The preimage of the open chart Ui : xi = 0 on P(V ) under the action map : G в P(V ) P(V ) is Ui = {(g; [v]) : xi ((g)v) = 0}. The inverse image of the coordinate function xj on Ui is xi (g)xj (v) x ((g)v) xj ( )(g; [v]) = j = k kj ; xi xi ((g)v) k ki (g )xi (v ) which is a regular function on Ui . This proves that is a morphism.
Now we came to actions of an algebraic group G on an a ne variety X . There is a natural action of G on regular functions: (g · f )(x) := f (g-1 · x); f O(X ); x X; g G: () Lemma 1.3.15. Formula () gives a wel l-de ned linear G-action on O(X ) by automorphisms of a K-algebra.

Warning: in general, the K-space O(X ) has in nite dimension, so at the moment
we speak only about linear actions of an abstract group.
-

Proof. For any g G the function g · f is a composition X X K of the automorphism Lg 1 and the function f , so g · f O(X ). Clearly, e · f = f and - - - (g1 · (g2 · f ))(x) = (g1 · f )(g2 1 · x) = f (g2 1 · (g1 1 · x)) = f ((g1 g2 )-1 · x) = (g1 g2 · f )(x), thus we have an action of an abstract group. Moreover, (g · (1 f1 + 2 f2 ))(x) = (1 f1 + 2 f2 )(g-1 · x) = = (1 f1 )(g-1 · x) + (2 f2 )(g-1 · x) = (1 (g · f1 ) + 2 (g · f2 ))(x); and g · (f1 f2 ) = (g · f1 )(g · f2 ) by the same arguments.

Theorem 1.3.16. The algebra O(X ) is a union
rational G-submodules W
iO

(X ).

i=1

Wi of a nite dimensional

Proof. The morphism nate algebras:

:GвX

X corresponds to a homomorphism of coordiK O

: O (X )

O

(G в X ) O(G) =
F (g · x) =

(X );

(F ) =

m i=1

fi hi

with

m i=1

fi (g)hi (x):

This proves that the linear span GF of the orbit GF is contained in h1 ; : : : ; hm , so is nite-dimensional. One may assume that h1 ; : : : ; hm form a basis in GF . Since this subspace is G-invariant, the following formula holds for any g G: (g · hi )(x) = hi (g
-1

·

x) =

m

j =1

pij (g-1 )hj (x):

for some pij O(G). There are regular functions pij O(G), such that pij (g) = pij (g-1 ). Thus an element g G acts in the space GF by the matrix (pj i (g)), and the G-module GF is rational. Finally, take a countable K-basis F1 ; F2 ; F3 ; : : : in i O(X ) and set Wi = j =1 GFj .

De nition 1.3.17. A G-module V is called rational if any element v contained in a nite-dimensional rational submodule. Corollary 1.3.18. The G-module O(X ) is rational.

V is

12

IVAN V. ARZHANTSEV

Let G be a group and X , Y be two G-sets. Recall that a map : X Y is called (G-)equivariant if (g · x) = g · (x) for all g G and x X . Moreover, if g · y = y for all y Y , then an equivariant map : X Y is called (G-)invariant. Theorem 1.3.19. Let G be an algebraic group acting on an a ne variety X . Then there are a rational nite-dimensional G-module V and a closed G-equivariant embedding I : X , V .

Proof. Since the algebra O(X ) is nitely generated, there is a nite-dimensional rational G-submodule U O(X ) which generates O(X ). Let V = U be the module dual to U . De ne the map I : X V , I (x) = lx U , where lx (u) := u(x). By de nition, lx (1 u1 + 2 u2 ) = (1 u1 + 2 u2 )(x) = 1 u1 (x) + 2 u2 (x) = 1 lx (u1 ) + 2 lx (u2 ); and the function lx is linear. In order to prove that I is a closed embedding it is su cient to check that the dual homomorphism I : O(V ) O(X ) is surjective. The regular functions on V form the symmetric algebra Sym(U ) of the space U . By de nition, for any u U one has I (u)(x) = u(I (x)) = u(lx ) = lx (u) = u(x). This shows that I (U ) coincides with the subspace U O(X ), thus generates O(X ), and the homomorphism I : O(V ) = Sym(U ) O(X ) is surjective. We need to check that I is equivariant. This is straightforward: I (g · x)(u) = lg·x (u) = u(g · x) = (g-1 · u)(x) = lx (g-1 · u) = (g · lx )(u) = (g · I (x))(u):

Theorem 1.3.20. Any a ne algebraic group admits a faithful representation.
Proof. Consider the action L of G on itself be left translations: g · g1 = gg1 . By Theorem 1.3.19, there are a rational nite-dimensional G-module V and an equivariant closed embedding I : G , V . Since the action L is e ective, the corresponding representation : G GL(V ) is faithful. Remark 1.3.21. It follows from the above theorem that any a ne algebraic group may be realized as a closed subgroup in some GL(n). Indeed, the homomorphism : G (G) is bijective and (G) is closed in GL(V ) (Theorem 1.2.12). If char K = 0, then automatically de nes an isomorphism between G and (G). In general, one may argue as follows: Take the unit e G and consider the orbit morphism : (G) I (G), ((g)) = (g) · I (e). Since I : G I (G) is an isomorphism, the composition I -1 : G G is the identity map. This shows that I -1 is inverse to .
Let an algebraic group G act on an algebraic variety X We are going to explain that the algebra of regular functions O(X ) is a rational G-module (with respect to the action de ned by (*)). Proposition 1.3.22. Let X and Y be algebraic varieties. Then O(X в Y ) O(X ) K O(Y ): =

Proof. Assume that X is a ne and Y = s=1 Yi be an a ne covering of Y . Take any i f O(X в Y ). Let fi be the restriction of f to X в Yi . Fix a K-basis g1 ; g2 ; : : : of O(X ). Then fi = k gk hik with hik O(Yi ). Since fi - fj is identically zero on the a ne open subset Yi Yj , and hik ; hj k O(Yi Yj ), one has 0 = k gk (hik - hj k ), thus hik and hjk coincide on Yi Yj . This shows that hik glue together to a regular function hk O(Y ), and f = k gk hk . So the natural homomorphism O(X ) вK O(Y ) O(X в Y ) is surjective. If some k gk hk maps to zero, then we may restrict this equality to any a ne chart X в Yi and get hk = 0 for all k.

ALGEBRAIC GROUPS AND INVARIANT THEORY

13

For an arbitrary variety X , x an a ne covering X = i Xi . Taking any f O(X в Y ), restricting it to all Xi в Y , and repeating the above arguments, we get the statement. Now the proof of Theorem 1.3.16 works also for: Theorem 1.3.23. Let G be an algebraic group. For any G-variety X the algebra O(X ) is a rational G-module.

14

IVAN V. ARZHANTSEV

GL(n) (A) is connected. Is it true for SL(2) ? Exercise 1.3.25. For any closed subset A G de ne the normalizer NG (A) = {g G : gxg-1 A for any x A}. Check that NG (A) is a closed subgroup of G. Exercise 1.3.26. Assume that there is an algebraic action of G on a vector space V . Show that this action is linear (in the sense of De nition 1.3.11) if and only if g · (1 v1 + 2 v2 ) = 1 (g · v1 ) + 2 (g · v2 ) for any g G; 1 ; 2 K; v1 ; v2 V : Exercise 1.3.27. Describe orbits of the tautological linear actions: (a) SL(n) : Kn ; (b) D(n) : Kn ; (c) B (n) : Kn ; (d) U (n) : Kn ; (e) O(n) : Kn ; (f ) SO(n) : Kn ; (g) Sp(2n) : K2n . Exercise 1.3.28. Following Example 1.3.5, calculate dim Sp(2n). Exercise 1.3.29 (*). Prove that the groups SO(n) and Sp(2n) are connected, and O(n) consists of two connected components. Exercise 1.3.30 (*). For any pair of groups listed below construct explicitly a surjective two-sheeted homomorphism: (a) SL(2) SO(3); (b) SL(2) в SL(2) SO(4); (c) Sp(4) SO(5); (d) SL(4) SO(6). Exercise 1.3.31. Let K be a eld. Prove that the polynomial det(aij ) K[a11 ; : : : ; ann ] is irreducible. Exercise 1.3.32. Let V be a nite-dimensional rational G-module. Prove that the natural G-action on the variety F (V ) of complete ags in V is algebraic. Exercise 1.3.33 (*). De ne an algebraic action of an algebraic group G on a prevariety X . Show that in this case the subset X G need not be closed in X . Exercise 1.3.34. Give an example of a non-linear algebraic action of the group Ga on Kn . Exercise 1.3.35. Let G be a nite group. Prove that any G-module is rational. Exercise 1.3.36. Assume that an algebraic group G acts on an irreducible variety X . Then the formula (g · f )(x) := f (g-1 · x) de nes a structure of a G-module on the eld of rational functions K(X ). Give an example where this module is not rational. Exercise 1.3.37 (*). Let G be an algebraic group acting of a quasia ne variety Y . Prove that there are an a ne G-variety X and an open equivariant embedding J : Y , X .

Exercises to subsection 1.3. Exercise 1.3.24. Prove that for any A GL(n) the centralizer Z

ALGEBRAIC GROUPS AND INVARIANT THEORY

15

1.4. Homogeneous spaces. Let G be an a ne algebraic group and H a closed subgroup of G. The set of left cosets G=H admits a natural transitive G-action: g · g1 H = gg1 H . The aim of this section is to introduce a structure of an algebraic variety on G=H such that the action above becomes algebraic. This structure is based on the following result. Theorem 1.4.1 (Chevalley's Theorem (1951)). Let G be an a ne algebraic group and H a closed subgroup of G. (1) There are a representation : G GL(V ) and a vector v V such that H = {g G : (g)v v }: (2) If the subgroup H is normal, then there is a representation : G GL(V ) such that H = Ker( ).

Proof. Denote by I(H ) the ideal of all functions f O(G) that are zero on the (closed) subvariety H G. As any ideal of O(G), I(H ) is nitely generated: I(H ) = (f1 ; : : : ; fs ). Consider the action G : G by left translations. Clearly, g H gH = H g · I(H ) = I(H ): Fix a nite-dimensional (rational) G-submodule W O(X ) containing f1 ; : : : ; fs and set U = W I(H ), dim U = k. We claim that g H if and only if g · U = U . Indeed, the subspace U contains f1 ; : : : ; fs and is contained in I(H ), thus the set of common zeroes of functions from U is H , and g · U = U gH = H . Conversely, gH = H implies g H . Let us recall a lemma from linear algebra (see Exercise 1.4.9). Lemma 1.4.2. Assume that A is a linear operator on a vector space W and U is a k-dimensional subspace of W . The operator A acts natural ly on the kth exterior power k W of W , and A preserves the subspace U W if and only if it preserves the line k U k W .
Let us x a basis u1 ; : : : ; uk of the subspace U . In order to prove (1), one should set V = k W and v = u1 · · · uk . Now we came to (2). By the above construction, the vector v in an eigenvector for H , and there exists a homomorphism 0 : H Kв such that h · v = 0 (h)v. Let us consider all homomorphisms i : H Kв such that there is a non-zero vi V with h · vi = i (h)vi for any h H . The subspaces Vi = {w V : h · w = i (h)w for any h H } form a direct sum in V . Denote this sum as: ~ V=
k i=0

Vi :

If w Vi , then h · (g · w) = g · ((g-1 hg) · w) = g · i (g-1 hg)w = i (g-1 hg)(g · w); ~ thus g · w belongs to some Vj . This implies that V is G-invariant. Then G acts ~ ) of linear operators on V : if : G GL(V ) is our ~ ~ naturally on the space L(V ~ ~ ), then g · C := (g)C (g)-1 . Since the group G representation and C L(V ~ ~ ~ ~ permutes the summands Vi in V , the subspace L0 := k=0 L(Vi ) of L(V ) is Gi invariant. Let us consider the representation : G GL(L0 ). We claim that Ker( ) = H . Indeed, elements of H are sent to operators that are scalar on any Vi . These are precisely the operators that commutes with each element of k=0 L(Vi ). i

16

IVAN V. ARZHANTSEV

On the other hand, if an element g G is sent to an operator which is scalar on any Vi , then, in particular, g preserves the line v , and, by assumption, belongs to H. Let us suppose for the rest of this section that char K = 0. Corollary 1.4.3. The set G=H of left cosets admits a unique structure of a quasiprojective algebraic variety such that the natural action G : G=H is algebraic.

Proof. Let us rstly introduce an algebraic structure on G=H . Consider the pro jective space P(V ) and a point [v] P(V ) corresponding to a pair (V ; v) constructed in Theorem 1.4.1 (1). (Since the case G = H is trivial, we may assume that v = 0.) By Lemma 1.3.14, the induced action G : P(V ) is algebraic, and the stabilizer of [v] coincides with H . The orbit G[v] is open in its closure (Theorem 1.3.4) and thus has a structure of a quasipro jective variety with an algebraic transitive G-action. The orbit map G P(V ), g g · [v] de nes a bijection G=H G[v], and induces a structure of a quasipro jective variety on G=H such that the natural action G : G=H is algebraic. Now assume that there exists another algebraic structure (G=H )1 on G=H . By assumption, there is a commutative diagram G z HH
G=H
zzz zz }zz

HH HH HH H$ / (G=H )1

;

where and are orbit morphisms and is a set-theoretical bijection. Suppose that G is connected. By Theorem 3.0.28, the map is a rational morphism, thus there exists an open subset W G=H with : W (W ) being isomorphism. Since G acts transitively on G=H and all maps are G-equivariant, we may produce a covering of G=H (by shifts of W ) such that extends as an isomorphism to any element of this covering: (g · w) = g · (g-1 · (g · w)). This shows that G=H and (G=H )1 are isomorphic. For non-connected G, one again may extend an isomorphism G0 =(G0 H ) (G0 =(G0 H ))1 to G=H (G=H )1 by translations.

Corollary 1.4.4. Let H be a closed normal subgroup of G. Then the quotient group
G=H has a unique structure of an a ne algebraic group such that the projection G G=H is an algebraic homomorphism.

Proof. Consider the representation : G GL(V ) from Theorem 1.4.1 (2). It de nes a bijective homomorphism G=H (G). Since (G) is closed in GL(V ) (Theorem 1.2.12), it has a structure of an algebraic group, and induces a structure of an algebraic group on G=H . If G=H admits a structure of an algebraic group such that the pro jection G G=H is algebraic, then we have a sequence of morphisms: G в G=H G=H в G=H G=H; (g; g1 H ) (gH; g1 H ) gg1 H; which shows that the left action G : G=H is algebraic. Thus the uniqueness follows from Corollary 1.4.3.

Proposition 1.4.5. Let G be an algebraic group and H a closed subgroup. Then
the projection morphism p : G

G=H is open.

ALGEBRAIC GROUPS AND INVARIANT THEORY

17

Proof. All bers of the morphism p : G G=H are isomorphic to H ; in particular, all their components have the same dimension. The variety G=H is smooth (e.g, normal). Thus for connected G Proposition 1.4.5 follows from Theorem 3.0.32. For arbitrary G, one should apply Theorem 3.0.32 to the restriction of p to a connected component of G, where all bers are isomorphic to H G0 .

Proposition 1.4.6. Let G be an algebraic group and H a closed subgroup. Then
O

(G=H ) = O(G)H := {f

O

(G) : f (gh) = f (g) for any g

G; h H }:

Proof. The dominant morphism p : G G=H corresponds to an embedding p : O(G=H ) O(G)H . On the other hand, any function f O(G)H de nes a commutative diagram:
z p zzz zz }zz

G AA

G=H

AA f AA AA / A1

:

If G is connected, then O(G=H ) (Corollary 3.0.29). For arbitrary G, these arguments prove that is regular on any irreducible (=connected) component of G=H , thus f O(G=H ). We nish this section with some examples of homogeneous spaces.

Example 1.4.7 (Grassmannians and Flag Varieties). The group GL(n) acts transitively on the set of k-dimensional subspaces of V = Kn (1 of the standard k-subspace e1 ; : : : ; ek is P (k; n) := A B : A GL(k); C GL(n - k); B 0C

k n). The stabilizer
в

Mat(k

(n - k)) :

So the homogeneous space GL(n)=P (k; n) is isomorphic to the Grassmannian Gr(k; n). Now consider the subgroup B (n) GL(n). It is the stabilizer of the standard complete ag {0} e1 e1 ; e2 · · · e1 ; : : : ; en = Kn in Kn . Since GL(n) acts transitively of the set of complete ags, we again have that GL(n)=B (n) is isomorphic to the ag variety F (n). It is well known that the varieties Gr(k; n) and F (n) are pro jective. By Theorem 1.3.4, we have: dim Gr(k; n) = n
2-

(k2 + (n - k)2 + k(n - k)) = k(n - k);
2-

n(n + 1) n(n - 1) = : 2 2 Example 1.4.8 (Homogeneous spaces for G = SL(2)). (1) Take G = S L(2) and H = B := {A B (2) : det(A) = 1}. In order to apply Chevalley's Theorem, consider the tautological SL(2)-module V = K2 and the rst standard vector e1 V . Clearly, B = {A SL(2) : A · e1 e1 }. Since SL(2) acts transitively on the set of lines in V , the homogeneous space SL(2)=B is isomorphic to the pro jective line P1 .

dim F (n) = n

(2) Take G = SL(2) and H = U := U (2). Again consider V = K2 and v = e1 , and note that U = {A SL(2) : A · e1 = e1 }: Thus, SL(2)=U is isomorphism to the orbit of e1 in V . This is a quasi-a ne (non-a ne !) variety K2 \ {0}.

18

IVAN V. ARZHANTSEV

(3) Finally, take G = SL(2) and H = T := {A D(2) : det(A) = a three-dimensional space of 2 в 2-matrices with trace zero, where conjugation: (A; C ) AC A-1 . Set 0 v = 1 -1 : 0 The stabilizer of v coincides with T , and the orbit Gv consists of eigenvalues 1 and -1. This orbit is de ned in V by the equation Thus SL(2)=T is an a ne quadric in A3 .

1}. Let V be SL(2) acts by

matrices with det(C ) = -1.

ALGEBRAIC GROUPS AND INVARIANT THEORY

19

component, and G0 acts transitive on any such component. Moreover, G=H is connected if and only if H intersects all connected components of G, or, equivalently, G0 H = G. Exercise 1.4.12 (*). Prove that a homogeneous space G=H is quasi-a ne if and only if there are a rational nite-dimensional G-module V and a vector v V such that H = {g G : g · v = v }: Exercise 1.4.13. Prove that a homogeneous space G=H is a ne if and only if there are a rational nite-dimensional G-module V and a vector v V such that H = {g G : g · v = v } and the G-orbit Gv is closed in V . Exercise 1.4.14. Assume that G=H is a ne and there are a rational nite-dimensional G-module V and a vector v V such that H = Gv . Is it true that the orbit Gv is closed in V ? Exercise 1.4.15. Give an example of a homogeneous space G=H such that O(G=H ) = K, but G=H is not pro jective. Exercise 1.4.16. Describe the variety SL(3)=U (3). Exercise 1.4.17. Let G be an algebraic group and H1 H2 G closed subgroups. Show that the map G=H1 G=H2 , gH1 gH2 is a morphism. Exercise 1.4.18 (*). Give an example of an algebraic group G and a closed subgroup H such that the bering p : G G=H is not locally trivial.

Exercises to subsection 1.4. Exercise 1.4.9. Prove Lemma 1.4.2. Exercise 1.4.10. Find a faithful representation of the group GL(n)=Z (GL(n)). Exercise 1.4.11. Prove that any irreducible component of G=H is also a connected

ALGEBRAIC GROUPS AND INVARIANT THEORY

21

1.5. The tangent algebra. From now on we shall assume that char K = 0. Consider the group G = GL(n). Since GL(n) Mat(n в n) is an open subset, the tangent space TE (G) at the unit matrix E may be identi ed with Mat(n в n). One may de ne a bilinear operation in this space: [A; B ] = AB - B A. It is easy to check that [·; ·] possesses the following properties: [A; B ] = -[B ; A]; [A; [B ; C ]]+[B ; [C; A]]+[C; [A; B ]] = 0 for any A; B ; C Mat(nвn): De nition 1.5.1. A K-vector space g with a bilinear operation [·; ·] : g в g g is called a Lie algebra, if the operation [·; ·] satis es the following conditions: (i) (Antisymmetry) [x; y] = -[y; x] for any x; y g; (ii) (The Jacobi Identity) [x; [y; z ]] + [y; [z ; x]] + [z ; [x; y]] = 0 for any x; y; z g. Remark 1.5.2. By analogy with the Leibnitz rule (f g) = f g + f g , the Jacobi Identity for Lie algebras may be rewritten as: [x; [y; z ]] = [[x; y]; z ] + [y; [x; z ]]: The space Mat(n в n) with the operation [A; B ] = AB - B A is an example of a Lie algebra. The standard notation for this Lie algebra is gl(n). A natural question arises: Why do we de ne an operation in TE (G) as AB - B A, but not via some other formula ? Below come some motivations for this de nition. (1) Let A GL(n) and LA : Mat(n в n) Mat(n в n), LA (B ) = AB (resp. RA (B ) = B A). Then LA (resp. RA ) is a linear operator and its di erential dLA : TE GL(n) TE GL(n) is given by dLA (X ) = AX (resp. dRA (X ) = X A). (2) Consider the inverse map i : GL(n) GL(n), i(B ) = B -1 . The differential di : TE GL(n) TE GL(n) may be calculated using the explicit formula for inverse matrix (Exercise 1.5.23). But now we prefer another arguments. Any vector X TE GL(n) is a tangent vector to a smooth curve d : K GL(n), (0) = E , i.e., dt |t=0 (t) = X . The image di(X ) is the tangent vector to the curve i( (t)) = (t)-1 . One may di erentiate the identity (t) (t)-1 = E at t = 0 and obtain X (0)-1 + (0)(di(X )) = 0, or di(X ) = -X . (3) Consider an inner automorphism aA : GL(n) GL(n), aA (B ) = AB A-1 . Here the di erential daA : TE GL(n) TE GL(n) is given as daA (X ) = AX A-1 , because aA is the composition of LA and RA 1 . We have an (algebraic) representation: Ad : GL(n) GL(Mat(n в n)); Ad(A)(X ) = AX A-1 : (4) Let us calculate the di erential ad := dE Ad : Mat(n в n) Mat(Mat(n в n)): d Take again a curve (t) with (0) = E and dt |t=0 (t) = X . Then for any Y Mat(n в n): d d |t=0 Ad( (t))(Y ) = | (t)Y (t)-1 = dt dt t=0 = X Y (0)-1 + (0)Y (-X ) = X Y - Y X = [X; Y ]: This proves that ad(X )(Y ) = [X; Y ]. De nition 1.5.3. A representation of a Lie algebra g is a linear map from g to some gl(m) such that ([x; y]) = (x) (y) - (y) (x) for all x; y g:
-

22

IVAN V. ARZHANTSEV

For any nite-dimensional Lie algebra g de ne a linear map ad : g gl(g), ad(x)(y) = [x; y]. Lemma 1.5.4. The map ad : g gl(g) is a representation of the Lie algebra g.

Proof. The statement follows from the Jacobi identity: ad([x; y])(z ) = [[x; y]; z ] = [[x; z ]; y] + [x; [y; z ]] = [x; [y; z ]] - [y; [x; z ]] = = (ad(x)ad(y) - ad(y)ad(x))(z ):

Proposition 1.5.5. Let G be a closed subgroup of GL(n).
(i) The subspace g := TE G TE GL(n) is a Lie subalgebra of gl(n). (ii) The structure of the Lie algebra on g = Te G does not depend on a (closed) embedding G GL(n).

Proof. Since for any g G the inner automorphism ag (of GL(n)) maps G to G, the tangent space g = TE G is an invariant subspace for the representation Ad : G GL(gl(n)). So the same holds for its di erential ad : g gl(gl(n)), and for any X; Y g one has ad(X )(Y ) = [X; Y ] g. This proves (i). Further, the maps ag : G G, Ad(g) : g g and ad(X ) : g g are de ned in the internal terms of the group G, thus the Lie bracket [X; Y ] := ad(X )(Y ) depends only on G itself.
Since any algebraic group G may be realized as a closed subgroup of some GL(n) (Theorem 1.3.20), the tangent space Te G possesses a (canonical) structure of Lie algebra. De nition 1.5.6. The representation Ad : G GL(g) (resp. ad : g gl(g)) is called the adjoint representation of an algebraic group G (resp. of a Lie algebra g). More generally, if H G is a closed subgroup (resp. h g is a Lie subalgebra) and Ad (resp. ad) is the adjoint representation of G (resp. of g), then the restriction Ad |H (resp. ad |h ) de nes a linear action H : g (resp. h : g). We shall call this action an adjoint action of H (resp. of h). Remark 1.5.7. Recall that a Lie algebra g is commutative if [x; y] = 0 for any x; y g. If an algebraic group G is commutative, then ag = id for all g G Ad(g) = E for all g G ad(x) = 0 for all x g, thus Lie(G) is a commutative Lie algebra. Example 1.5.8. Consider G = SL(n). In order to nd the tangent algebra, one should calculate the di erential of the map GL(n) Kв , A det(A) at the unit. Direct calculations (see Exercise 1.5.22) shows that dE (det)(X ) = tr(X ), thus the tangent algebra Lie(SL(n)) := sl(n) = {X gl(n) : tr(X ) = 0}: Now take G = SO(n). For any smooth curve : K SO(n), (0) = E , we need to d nd the tangent vector X = dt |t=0 (t): Di erentiating the identity (t)T (t) = E , T + X = 0. This implies that Lie(SO(n)) := so(n) is contained in the we get X space of skew-symmetric matrices. On the other hand, in the system AT A = E of n2 de ning equations for O(n) only n(n2+1) are pairwise di erent, and dim so(n) = dim SO(n)(= dim O(n)) n2 - n(n2+1) = n(n2-1) (cf. Example 1.3.5). Hence the algebra so(n) is the algebra of skew-symmetric n в n-matrices.

ALGEBRAIC GROUPS AND INVARIANT THEORY

23

g;

Below for algebraic groups G; H; : : : we denote their tangent algebras as Lie(G) = Lie(H ) = h; : : : . De nition 1.5.9. Let g and h be Lie algebras. A linear map : g h is said to be a homomorphism (of Lie algebras) if ([x; y]) = [(x); (y)] for all x; y g: Theorem 1.5.10. Let G and H be algebraic groups and : G H a homomorphism. Then its di erential d : g h is a homomorphism of Lie algebras.

Proof. Let L(g; h) be the space of linear maps from g to h. Consider the map : G L(g; h); (g)(Y ) = (d Ad(g))(Y ) for all Y g: The statement of the theorem follows from the calculation of d : g L(g; h) in two ways:

G
ag

/ /

H H

a(g)

G

d / h == == Ad(g) = Ad((g)) d = / h: g

g= =

d Namely, take a smooth curve (t), (0) = E in G with dt |t=0 (t) = X . Then ( (t))(Y ) = (d Ad( (t)))(Y ) d (X )(Y ) = (d ad(X ))(Y ) = d([X; Y ]):

On the other hand, ( (t))(Y ) = (Ad(( (t))) d(Y )) d (X )(Y ) = = ad(d(X ))(d(Y )) = [d(X ); d(Y )]:

Lemma 1.5.11. Let G be an algebraic group and H a closed subgroup. Then
TeH G=H g=h: =

Proof. Consider the surjection p : G G=H . By Theorem 3.0.33, for a generic g G the kernel of the di erential dg p : Tg G TgH G=H equals the tangent space to the ber. But G acts transitively on G and G=H and the map p is G-equivariant. This implies that all points have equal status, and thus Ker(dg p) = Tg (gH ) h = for any g G. On the other hand, G, H and G=H are smooth, and dim(G=H ) = dim G - dim H . So dg p is surjective, and TeH G=H = Im(de p) g=Ker(de p) = g=h: =

Lemma 1.5.12. Let : G

H be a homomorphism of algebraic groups. Then Lie(Ker ) = Ker d; Lie(Im) = Im(d):

Proof. We may applies arguments given above to the surjection G
Lie(
Hi ) =

(G).

Lemma 1.5.13. Let {Hi : i I } be a family of closed subgroups of G. Then
iI i I

Lie(Hi ):

24

IVAN V. ARZHANTSEV

Proof. Since any variety is Noetherian, there are elements H1 ; : : : ; Hk of our family with k=1 Hj = iI Hi . Moreover, we may reduce the proof to the case k = 2 by j induction. Consider the commutative diagram A of group homomorphisms and the induced commutative diagram B of di erentials:

G O
/

G=H O

1

g O h2

/ g=h1 O / h2 =h1 h2

H

2

/

H2 =H1 H2

:

By Lemma 1.5.11, the kernel of the lower arrow in diagram B equals Lie(H1 H2 ). Passing in diagram B by another way and applying Lemma 1.5.11 to the upper arrow, we get that the kernel is Lie(H1 ) Lie(H2 ).

De nition 1.5.14. Let G be an algebraic group. A Lie subalgebra
called algebraic if there exists a closed subgroup H

h Lie(G) is G with h = Lie(H ).

Warning: Not any Lie subalgebra h Lie(G) is algebraic, see Proposition 1.6.20. Theorem 1.5.15. Let G be an algebraic group. The map Lie : H Lie(H ) de nes
a bijection: {closed connected subgroups H

G} {algebraic Lie subalgebras h Lie(G)}:

Proof. By de nition 1.5.14, the map Lie is surjective. In order to prove injectivity, assume that Lie(H1 ) = Lie(H2 ). In particular, dim H1 = dim H2 . If H1 H2 , then H1 = H2 by Theorem 3.0.25. In the general case, one has Lie(H1 ) = Lie((H1 H2 )0 ) = Lie(H2 ) (Lemma 1.5.13), and H1 = (H1 H2 )0 = H2 .
Now consider a rational representation : G GL(V ) and its tangent representation := d : g gl(V ). For any subspace U V de ne NG (U ) = {g G : (g)U = U }:

Proposition 1.5.16. For any vector v V and any subspace U
Lie(Gv ) =
gv

V one has:

:= {x

g

: (x)v = 0}; Lie(NG (U )) =

n

g

(U ) := {x

g

: (x)U

U }:

Proof. Applying Lemma 1.5.12 to G G=Ker(), we may assume that : G GL(V ) is injective. In the case G = GL(V ), the subgroups GL(V )v and NGL(V ) (U ) and their tangent algebras may be easily described in the matrix form. This description implies that the statements are true. In the general case, (Gv ) = GL(V )v (G), (NG (U )) = NGL(V ) (U ) (G), and one applies Lemma 1.5.13 to the closed subgroup (G) GL(V ).

Theorem 1.5.17. Let : G

GL(V ) be a rational representation of a connected group G, and := d : g gl(V ) be the tangent representation. A subspace U V is G-invariant if and only if it is g-invariant. In particular, the representation is irreducible if and only if is irreducible.

ALGEBRAIC GROUPS AND INVARIANT THEORY

25

Proof. A subspace U is G-invariant (resp. g-invariant) if and only if NG (U ) = G (resp. ng (U ) = g). For a connected G, NG (U ) coincides with G if and only if their tangent algebras coincide (Theorem 1.5.15). By Proposition 1.5.16, Lie(NG (U )) = g if and only if ng (U ) = g. De nition 1.5.18. Let g be a Lie algebra. A subspace a g is called an ideal if [x; y] a for all x g; y a. Remark 1.5.19. A subspace a g is an ideal if and only if it is a g-invariant subspace with respect to the ad-representation. Proposition 1.5.20. Let H be a closed connected subgroup of a connected group G. The subgroup H is normal in G if and only if Lie(H ) is an ideal of Lie(G). Proof. A subgroup H is normal in G if and only if ag (H ) = H for all g G. We claim that this is equivalent to the condition "Lie(H ) is an Ad(G)-invariant subspace of Lie(G)". Indeed, if G preserves H , then it preserves its tangent space. Conversely, assume that ag (H ) = H for some g G. Then ag (H ) is another closed connected subgroup of G, and by Theorem 1.5.15 Lie(H ) = Lie(ag (H )) = Ad(g)(Lie(H )). Finally, the subspace Lie(H ) Lie(G) is G-invariant if and only if it is g-invariant (Theorem 1.5.17).
For a Lie algebra g, the center is de ned as z(g) := {x g : [x; y ] = 0 for all y g}: Moreover, set Z (g) := {g G : Ad(g)(x) = x for all x g}; and, for any x g, ZG (x) = Z (x) := {g G : Ad(g)x = x}; zg (x) = z(x) := {y g : [y; x] = 0}: Proposition 1.5.21. For any algebraic group G and any x Lie(G) = g one has Lie(Z (x)) = z(x); Lie(Z (g)) = z(g):

Proof. The rst statement follows directly from Proposition 1.5.16 applied to the adjoint representation Ad with v = x. For the second one, take a basis x1 ; : : : ; xn in g. Then Z (g) = n=1 Z (xi ), z(g) = i n z(xi ), and we may use Lemma 1.5.13. i=1

26

IVAN V. ARZHANTSEV

Exercises to subsection 1.5. Exercise 1.5.22. Consider the map det : GL(n)

, A det(A). Prove that dE (det)(X ) = tr(X ) for any X TE GL(n). Exercise 1.5.23. Calculate the di erential di : TE GL(n) TE GL(n) using the formula for inverse matrix in terms of minors. Exercise 1.5.24. Prove that a Lie algebra is commutative if and only if any its subalgebra is an ideal. Exercise 1.5.25. What is an n-dimensional representation of a commutative Lie algebra g? Exercise 1.5.26. Assume that G1 and G2 are connected algebraic groups such that the Lie algebras Lie(G1 ) and Lie(G2 ) are isomorphic. Is it true that G1 and G2 are isomorphic ? Exercise 1.5.27. Describe the tangent algebras of the groups Sp(2n), D(n), B(n) and U (n). Exercise 1.5.28. Prove that the following Lie algebras are isomorphic:
K
в

(a) (b) (c) (d)

sl(2) sl(2) sp(4) sl(4)

Exercise 1.5.29. Show that for any rational representation : G

so(3); = в sl(2) so = so(5); = so(6). =

(4);

GL(V ) any Ginvariant subspace U V is g-invariant, but for a non-connected G the converse is not true. Exercise 1.5.30. Let H be a closed subgroup of G. Prove that if H is normal in G, then Lie(H ) is an ideal of Lie(G). Give an example where the converse is not true. Exercise 1.5.31. Set ZG (g) = {h G : gh = hg}. Then Lie(ZG (g)) = {x g : Ad(g)x = x}: Exercise 1.5.32. Prove that a connected algebraic group is commutative if and only if its tangent algebra is commutative. Exercise 1.5.33. For a connected G, prove that Lie(Z (G)) = z(g). For any G, one has Lie(Z (G)) z(g), but the inclusion may be strict. Exercise 1.5.34. Let A be a nite dimensional algebra. By a derivation of A we mean a linear map D : A A such that D(ab) = D(a)b + aD(b) for any a; b A. Denote by Der(A) the subspace of all derivations in gl(A). (i) Check directly that Der(A) is a Lie subalgebra in gl(A). (ii) Let Aut(A) be the automorphism group of A, see Exercise 1.1.23. Prove that Lie(Aut(A)) = Der(A):

ALGEBRAIC GROUPS AND INVARIANT THEORY

27

1.6. Algebraic tori. This section is devoted to a special important class of algebraic groups, namely algebraic tori T m Kв в · · · в Kв (m times). The group = T m is connected, commutative, and has a lot of other important properties that are discussed below. We begin with the study of characters. De nition 1.6.1. Let G be an algebraic group. A character of G is a homomorphism : G Kв . The set X(G) of all characters has a structure of an Abelian the operation (1 + 2 )(g) := 1 (g)2 (g), with the neutral for all g, and with the inverse element (-)(g) := (g)-1 . Let G be the torus T m . Take an integer vector u = (a1 ; : : : an (algebraic) homomorphism u : T m Kв ; u ((t1 ; : : : ; tm )) = ta1 : : : 1 Conversely, Lemma 1.6.2. Any character : T m Kв has a form u group with respect to element 0 , 0 (g) = 1
; am ) Zm . It de nes t am : m

for some u

Zm

.

Proof. We start with m = 1. By de nition,
(t) =
k i=-j

ai t

i

is a Laurent polynomial with ak = 0. Since it does not have roots in Kв , the same is true for p(t) = tj (t). But any polynomial may be decomposed into linear factors, so p(t) = ak tk+j . The condition (1) = p(1) = 1 implies ak = 1 and nally (t) = tk . For m > 1, we have ((t1 ; : : : ; tm )) = ((t1 ; 1; : : : ; 1)) : : : ((1; : : : ; 1; tm )) = ta1 : : : tam m 1 for some ai Z. Corollary 1.6.3. The Abelian group X(T m ) is isomorphic to the lattice Zm .

Proof. The map : X(T m ) tions.

Zm

, (u ) = u, is bijective and respects the opera-

Now we shall describe automorphisms of T m . Lemma 1.6.2 shows that any homomorphism : T m T m has a form: ((t1 ; : : : ; tm )) = (ta11 : : : ta1m ; : : : ; tam1 : : : tamm ) m m 1 1 for some integer aij . Consider a matrix A = (aij ) Mat(n в n; Z), and denote the automorphism as A . Lemma 1.6.4. BA = B A for any A; B Mat(n в n; Z).

Proof. For any t = (t1 ; : : : ; tm ) T m , we have
(t
b11 a11 +···+b1m am1 1

:::t

BA (t) = b11 a1m +···+b1m amm m

; : : : ; tbm 1

1 a11 +···+bmm am1

= B (A (t)):

:::t

bm1 a1m +···+bmm amm m

)

28

IVAN V. ARZHANTSEV

The lemma implies that a homomorphism A is an automorphism if and only if the matrix A is invertible. So the group Aut(T m ) is isomorphic to GL(n; Z). Further, for any algebraic group G, an automorphism : G G acts on X(G) as ( · )(g) := (-1 (g)). In our case, let e1 ; : : : ; em be the standard basis of X(T m ) Zm . Since -1 = A 1 , we have = A -1 A ((t1 ; : : : ; tm )) = (tc11 : : : tc1m ; : : : ; tcm1 : : : tcmm ); m m 1 1 -1 where A = (cij ), and thus A (ei ) = ci1 e1 + · · · + cim em . This proves that A acts in the bases e1 ; : : : ; em via the matrix (AT )-1 . In particular, there is an isomorphism between Aut(T m ) and Aut(X(T m )). Let us summarize all these observations. Proposition 1.6.5. There are isomorphisms: Aut(T m ) Aut(X(T m )) GL(m; Z): = = Remark 1.6.6. The group GL(m; Z) does not have any reasonable structure of an a ne algebraic group. Remark 1.6.7. Any isomorphism T m (Kв )m de nes the standard basis in the = lattice X(T m ). Conversely, any basis e1 ; : : : ; em in X(T m ) de nes an isomorphism : T m (Kв )m , (t) = (e1 (t); : : : ; em (t)). =
-

Our next ob jective is the classi cation of rational T m -modules. De nition 1.6.8. An (algebraic) quasitorus is an algebraic group Q isomorphic to T m в A, where A is a nite Abelian group. Lemma 1.6.9. Elements of nite order are dense in Q.

Proof. Clearly, an element (t; a), t T m , a A has a nite order if and only if t does. So it is su cient to prove the statement for Q = T m . - - Let F (X1 ; X1 1 ; : : : ; Xm ; Xm1 ) be a Laurent polynomial such that F ( 1 ; -1 ; : : : ; m ; 1 N = 1 for some N > 0. We claim that F is identically zero. In0 for any i with i deed,
- (1) If m = 1, then F (X1 ; X1 1 ) has in nitely many roots in K, which is possible only for the zero polynomial; (2) Assume that m > 1. By the inductive hypothesis, for any element Kв of nite order, the polynomial - - - - F (X2 ; X2 1 ; : : : ; Xm ; Xm1 ) := F ( ; -1 ; X2 ; X2 1 ; : : : ; Xm ; Xm1 ) is identically zero. This implies that F does not depend on X1 (Exercise 1.6.21) and F is identically zero.

1 m
-

)=

Let us recall a de nition and two facts (see Exercises 1.6.25 and 1.6.26) from linear algebra. De nition 1.6.10. An element A Mat(n в n) is called semisimple (or diagonalizable) if it is conjugate to a diagonal matrix. Lemma 1.6.11. Any element of nite order in GL(n) is semisimple. Lemma 1.6.12. Let {Ai : i I } be a family of pairwise commuting semisimple operators on a nite-dimensional vector space. Then the operators Ai may be diagonalized simultaneously.

ALGEBRAIC GROUPS AND INVARIANT THEORY

29

Let V be a rational Q-module. For any character X(Q), de ne a subspace V = {v V : g · v = (g)v for all g Q}: Theorem 1.6.13. Let Q be a quasitorus. For any rational Q-module V one has: V V=
X(Q)

Proof. We start with a nite-dimensional case. Lemma 1.6.14. Any rational nite-dimensional Q-module is a direct sum of a one-dimensional submodules. Proof. Let : Q GL(V ) be a nite-dimensional rational representation. For any element g Q of nite order the operator (g) is semisimple (Lemma 1.6.11). Since Q is commutative, there exists a basis in V such that the operators (g) are represented by diagonal matrices for all elements g Q of nite order (Lemma 1.6.12). Thus the image of a dense subset of Q is contained in the subgroup D(n) of diagonal matrices in GL(n) (n = dim V ). But the subgroup D(n) is closed in GL(n), and this implies (Q) D(n).
Since any one-dimensional submodule of V is contained in some V , we have V = X(Q) V . The same holds for any rational Q-module, because any its element is contained in a nite-dimensional rational submodule. We claim that the sum X(Q) V is direct. Indeed, suppose that v1 + · · · + vk = 0 is a linear combination with non-zero vi Vi , where i are pairwise di erent, and k is the smallest possible. Take g Q with 1 (g) = 2 (g). Then 1 (g)v1 + · · · + k (g)vk = 0, and we may nd a shorter combination: (1 (g) - 2 (g))v2 + · · · + (1 (g) - k (g))vk = 0; a contradiction. Theorem 1.6.15. Any closed subgroup H T m is a quasitorus. Moreover, there are an isomorphism T m (Kв )m and positive integers d1 ; : : : ; ds (s m) such = that H = {(t1 ; : : : ; tm ) : td1 = · · · = tds = 1}: s 1

Proof. The torus T m is commutative, and H is a normal subgroup of T m . By Theorem 1.4.1 (2), there is a nite-dimensional rational representation : T m GL(k) such that H = Ker(). There are characters 1 ; : : : ; k such that the representation is equivalent to t diag(1 (t); : : : ; k (t)) (Theorem 1.6.13). This proves that H = {t T m : 1 (t) = · · · = k (t) = 1}. Now we need one more fact from a course of algebra. Proposition 1.6.16. Let B be a nitely generated subgroup of Zm . Then B is a free Abelian group. Moreover, there are a basis a1 ; : : : ; am of Zm , a basis b1 ; : : : ; bs of B (s n) and positive integers d1 ; : : : ; ds such that b1 = d1 a1 ; : : : ; bs = ds as .
Take a subgroup B generated by 1 ; : : : ; k in X(T m ) Zm . Clearly, = m : (t) = 1 for all B }: H = {t T Let us x the corresponding bases a1 ; : : : ; am and b1 ; : : : ; bs of X(T m ) and B respectively. Applying an automorphism of T m , one may assume that a1 ; : : : ; am is a standard basis of Zm . Then H = {(t1 ; : : : ; tm ) T m : td1 = · · · = tds = 1}; s 1

30

IVAN V. ARZHANTSEV

so H is isomorphic to where
Zr

is a cyclic group of order r. Corollary 1.6.17. A closed subgroup of a quasitorus is a quasitorus.

Zd1 в · · · в Zds в (Kв )m-s

;

Proof. Since Q T m в A and A is isomorphic to the direct product of l nite = cyclic groups, the quasitorus Q (as well as any its closed subgroup) is isomorphic to a closed subgroup of the torus T m в T l .
Now we came to the tangent algebra Lie(T m ) := t. At the rst glance, it is not an interesting ob ject: t is just an m-dimensional commutative Lie algebra. But we shall show now that it is naturally equipped with an additional structure, so called Q-form. The di erential of a character : T m Kв de nes a linear function de := d : t K that may be considered as an element of the dual space t . The equalities (1 + 2 )(t) = 1 (t)2 (t) and 1 (e) = 2 (e) = 1 imply d(1 + 2 ) = d1 + d2 . Let us denote the pairing between t and t as (·; ·), (v; l) = l(x). De ne a sublattice: t(Z) := {x t : (x; d) Z for all X(T m )}; and a Q-subspace: t(Q) := {x t : (x; d) Q for all X(T m )}: De nition 1.6.18. A subspace h t is called Q-de ned if h is the K-linear span of h t(Q), i.e., h = h t(Q) K : Lemma 1.6.19. A subspace h t is Q-de ned if and only if there are 1 ; : : : ; s X(T m ) with
h

=

s

i=1

Ker(di ):

Proof. There are 1 ; : : : ; s such that h t(Q) = {x t(Q) : (x; di ) = 0; i = 1; : : : ; s}: Then h t(Q) K = {x t : (x; di ) = 0; i = 1; : : : ; s}; and h coincides with h t(Q) K if and only if it may be de ned by a desired system of linear equations. Proposition 1.6.20. A subspace h t is an algebraic subalgebra of t if and only if h is Q-de ned. Proof. We know that any closed subgroup H T m is given as H = {t T m : 1 (t) = · · · = s (t) = 1 for some 1 ; : : : ; s or s H = Ker(i ):
i=1

X(T m )}

;

By Lemma 1.5.12, Lie(Ker(i )) = Ker(di ), and by Lemma 1.5.13, Lie(H ) = s i=1 Ker(di ): Now Lemma 1.6.19 completes the proof.

ALGEBRAIC GROUPS AND INVARIANT THEORY

31

for a K. Prove that if Fa (Y2 ; : : : ; Yn ) coincides with Fb (Y2 ; : : : ; Yn ) for in nitely many b K, then F (Y1 ; : : : ; Yn ) does not depend on Y1 . Exercise 1.6.22. Find the group of characters X(G) for G equals (a) GL(n); (b) SL(n); (c) O(2); (d) B (n); (e) a quasitorus; (f ) a nite group. Exercise 1.6.23. Describe the set of homomorphisms { : T m T r }. Exercise 1.6.24. Let A Mat(n в n) be a semisimple element and U V := Kn be an A-invariant subspace. Prove that A |U is a semisimple element of GL(U ). Exercise 1.6.25. Prove that any element of nite order in GL(n) is semisimple. Exercise 1.6.26. Let {Ai } be a family of pairwise commuting semisimple operators on a nite-dimensional vector space. Prove that the operators Ai may be diagonalized simultaneously. Exercise 1.6.27. Do semisimple elements form a subgroup of GL(n) ? Exercise 1.6.28. Prove that semisimple elements form a dense subset of Mat(n в n), but for n > 1 this subset is not open. Exercise 1.6.29. Let G be a commutative algebraic group such that G0 is a torus. Prove that G is a quasitorus. Exercise 1.6.30. Let Q be a quasitorus and : Q G be a surjective homomorphism. Prove that G is a quasitorus. Exercise 1.6.31. Let H be a closed subgroup of T m . Prove that any character of H may be extended to a character of T m . Exercise 1.6.32. In terms of Lemma 1.6.19, prove that h t(Q) K = {x t : (x; di ) = 0; i = 1; : : : ; s}: Exercise 1.6.33. Prove that the di erential de : Aut(T m ) GL(t) de nes an isomorphism between Aut(T m ) and the subgroup H := {A GL(t) : A(t(Z)) = t(Z)}:

Exercises to subsection 1.6. Exercise 1.6.21. Let F (Y1 ; : : : ; Yn ) be a polynomial and Fa (Y2 ; : : : ; Yn ) := F (a; Y2 ; : : : ; Yn )

ALGEBRAIC GROUPS AND INVARIANT THEORY

33

1.7. Jordan decompositions. De nition 1.7.1. An element A Mat(n в n) is called nilpotent if AN = 0 for some N , or, equivalently, all eigenvalues of A are 0. De nition 1.7.2. An element A GL(n) is called unipotent if A - E is nilpotent, or, equivalently, all eigenvalues of A are 1. Lemma 1.7.3. The subset Nil(n) Mat(n в n) of al l nilpotent elements and the subset Uni(n) GL(n) of al l unipotent elements are closed.

Proof. Let PA (x) = xn + a1 xn-1 + · · · + an be the characteristic polynomial of a matrix A. It is well known that the coe cients a1 ; : : : ; an depend on A polynomially. So the subvariety Nil(n) is de ne in Mat(n в n) by a1 (A) = · · · = an (A) = 0: The subvariety Uni(n) is E +Nil(n), thus it is closed in Mat(nвn) and in GL(n).
For any nilpotent A one may de ne correctly the exponent: A A2 A3 exp(A) := E + + + + :::: 1! 2! 3! Since for a nilpotent A Mat(n в n) one has An = 0, the map exp : Nil(n) GL(n) 2 A is a morphism. Moreover, the matrix 1! + A + : : : is nilpotent, so exp(A) is 2! unipotent, and exp sends Nil(n) to Uni(n). Conversely, for any B Uni(n) one may de ne the logarithm: B2 B3 ln(B ) = ln(E + B0 ) := B0 - 0 + 0 - : : : : 2 3 Lemma 1.7.4. The map exp : Nil(n) Uni(n) is an isomorphism of varieties.

Proof. The fact that exp and ln are inverse to each other follows from well-known identities of formal power series: exp(ln(x)) = x; ln(exp(x)) = x:
Let A GL(n). Since the intersection of any family of closed subgroups in GL(n) is again a closed subgroup, there exists the smallest closed subgroup containing A. Denote it by G(A). Assume that A Uni(n). For any t K de ne At := exp(t ln(A)). De nition 1.7.5. A closed subgroup G GL(n) is called unipotent if any element A G is unipotent. Proposition 1.7.6. Let A = E be a unipotent matrix. Then 1) {At : t K} is a one-dimensional unipotent subgroup in GL(n) isomorphic to Ga ; 2) G(A) = {At : t K}.

Proof. Note that A1 ; A2 Nil(n); A1 A2 = A2 A1 exp(A1 + A2 ) = exp(A1 ) exp(A2 ): This implies exp((t1 + t2 ) ln(A)) = exp(t1 ln(A)) exp(t2 ln(A)), and the map expA : K GL(V ), expA (t) = At is a homomorphism of algebraic groups. By Theorem 1.2.12, the subgroup expA (K) = {At : t K} is closed in GL(n).

34

IVAN V. ARZHANTSEV

Lemma 1.7.7. The group Ga has no proper closed subgroups.
Proof. Since Ga is one-dimensional and irreducible, any its proper closed subset is nite. But Ga contains no proper nite subgroups, because the only element of nite order in Ga is the unit.
Thus for 1) it is su cient to prove that {At : t K} is non-trivial. But this is clear, since it contains A = E . Moreover, G(A) {At : t K} by minimality, and, again by Lemma 1.7.7, G(A) = {At : t K}. Now we need some more facts from linear algebra. For any A Mat(n в n) consider the factorization of the characteristic polynomial: PA (x) = (x - 1 )k1 : : : (x - r )kr ; 1 ; : : : ; r K; k1 + · · · + kr = n: For V = Kn , de ne the root subspaces: V i := {v V : (A - i E )ki v = 0}: It is well known that V = V 1 · · · V r . Consider the semisimple operator As that acts on any V i as i E . The operator An := A - As is nilpotent. Moreover, if A is invertible, then As is invertible too, and Au := AA-1 is unipotent. Finally, s As commutes with A, An and Au . De nition 1.7.8. (i) Additive Jordan decomposition of a matrix A Mat(nв n) is a decomposition A = As + An , where As is semisimple, An is nilpotent, and As An = An As . (ii) Multiplicative Jordan decomposition of a matrix A GL(n) is a decomposition A = As Au , where As is semisimple, Au is unipotent, and As Au = Au As . The arguments given above explain that any A Mat(n в n) (resp. A GL(n)) possesses the additive (resp. multiplicative) Jordan decomposition. Proposition 1.7.9. For any A Mat(n в n) (resp. A GL(n)), the additive (resp. multiplicative) Jordan decomposition is unique.

Proof. Assume that A = As + An An A = AAn , we have As V i V i ; But As is semisimple, and there is V i = V1i · · · Vlii

is another decomposition. Since As A = AAs ,
An V i V i ; i = 1; : : : ; r: a decomposition with Vji := {v V i : As v = j v}:

The operators A and An preserve all Vji . Let vij be an eigenvector for An in Vji . Then Avij = (As + An )vij = j vij + 0vij = j vij Thus j = i for any i; j , and As = As , An = A - As = A - As = An . The same arguments work in the multiplicative case.

Corollary 1.7.10. An element A

Mat(n в n) is semisimple (resp. nilpotent, unipotent) if and only if A = As (resp. A = An , A = Au ). Lemma 1.7.11. For any A Mat(n в n) there is a polynomial F (x) K[x] such that As = F (A).

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35

Proof. By the Chinese remainder theorem there exists F (x) F (x) i (mod(x - i )ki ): Then F (A) |V i i E , thus F (A) = As .

K[

x] with

Corollary 1.7.12. For any B Theorem 1.7.13. Let G
As ; Au G.

Mat(n в n) the condition AB = B A implies As B = B As , An B = B An , Au B = B Au .

GL(n) be a closed subgroup and A

G. Then

Proof. By de nition, A G implies G(A) Lemma 1.7.14. G(As ) is a quasitorus.

G. One may suppose that As ; Au = E .

Proof. The element As is contained in a closed subgroup H conjugate to D(n) GL(n). Thus G(As ) H and G(As ) is a quasitorus (Theorem 1.6.15).

The intersection of G(As ) and G(Au ) consists of e, because all elements of G(As ) are semisimple, and all elements of G(Au ) are unipotent. Lemma 1.7.15. B C = C B for any B G(As ); C G(Au ).

Proof. Note that As ; Au ZGL(n) (As )ZGL(n) (Au ), thus ZGL(n) (Au ) by minimality. This proves that As Hs := {B G(As ) : B C = C B for all Au Hu := {C G(Au ) : B C = C B for all Again by minimality we have G(As ) = Hs , G(Au ) = Hu

G(As ); G(Au ) ZGL(n) (As ) C G(Au )};

.

B G(As )};

Denote by G(As ; Au ) the subgroup of GL(n) generated by G(As ) and G(Au ). We have proved that the map G(As ) в G(Au ) GL(n), (B ; C ) B C is an injective homomorphism. Thus G(As ; Au ) is a closed subgroup of GL(n) isomorphic to G(As ) в G(Au ), and A = As Au implies G(A) G(As ; Au ). Consider the pro jections s : G(A) G(As ), u : G(A) G(Au ). Since Ker(s ) G(Au ), and G(Au ) has no proper closed subgroups, there are two possibilities: (i) Ker(s ) = G(Au ) Au G(A) Au G As G; (ii) Ker(s ) = {e}. Then G(A) is a quasitorus and u (g) = E for any g (Proposition 1.7.16). But u (A) = Au , a contradiction. The proof of Theorem 1.7.13 is completed.

G(A)

and

Proposition 1.7.16. Let Q be a quasitorus. Then al l homomorphisms : Q
: Ga

Q are trivial.

G

a

Proof. Since the set F of elements of nite order is dense in Q (Lemma 1.6.9), and the only element of nite order in Ga is e, we have (F ) = {e}. The preimage -1 (e) is a closed subset of Q containing F , thus -1 (e) = Q. Conversely, (Ga ) is contained in Q0 , and by Proposition 1.6.15 (Ga ) is a torus. Let g = e, g (Ga ) be an element of order N . Then g = (u), and (N u) = e. But Ga has no proper closed subgroups, and Ker( ) = Ga .

Corollary 1.7.17. G(A) = G(As ; Au ).

36

IVAN V. ARZHANTSEV

Proof. By Theorem 1.7.13 As ; Au G(A), thus G(As ); G(Au ) G(A), and G(As ; Au ) G(A). On the other hand, A = As Au G(As ; Au ), and G(A) G(As ; Au ).

Theorem 1.7.18. Let G

GL(n), G1 GL(n1 ) be closed subgroups and : G G1 be a homomorphism of algebraic groups. Then for any A G one has (As ) = (A)s , (Au ) = (A)u .

Proof. The condition (A) (G(A)) GL(n1 ) implies G((A)) (G(A)). On the other hand, the closed subgroup -1 (G((A))) contains A, thus G((A)) = (G(A)). One knows that G(A) G(As ) в G(Au ); G((A)) G((A)s ) в G((A)u ); = = and may consider homomorphisms s : G(Au ) G((A)s ); u : G(As ) G((A)u ): By Proposition 1.7.16, they are trivial, and (G(As )) G((A)s ), (G(Au )) G((A)u ). Hence (As ) is semisimple and (Au ) is unipotent. Moreover, these elements commute, and (A) = (As )(Au ) is the multiplicative Jordan decomposition of A. By the uniqueness property, (As ) = (A)s , (Au ) = (A)u .

Corollary 1.7.19. Let G be an algebraic group and g G. The Jordan decomposition g = gs gu is wel l-de ned, i.e., does not depend of a closed embedding G GL(n).
Now we come to Jordan decomposition in the tangent algebra. Let G GL(n) be a closed subgroup and x g gl(n). De ne G(x) as the smallest closed subgroup in GL(n) such that x belongs to its tangent algebra. Clearly, G(x) G, G(x) is connected and dim G(x) 1 for any x = 0. If x is a nilpotent element of gl(n), then G(x) coincides with {exp(tx)}; if x is semisimple, then G(x) is a torus. Theorem 1.7.20. 1) x g xs ; xn g; 2) if : G G1 is a homomorphism, then d(xs ) = d(x)s , d(xn ) = d(x)n .

Proof. The proof is parallel to the proofs of Theorems 1.7.13 and 1.7.18, so we just sketch it. Let us take the stabilizer (in GL(n)) of the element xs under the adjoint action. Its tangent algebra is the centralizer of xs in gl(n), and thus contains xn . This shows that G(xn ) stabilizes xs . We claim that G(xn ) normalizes G(xs ). Indeed, if gG(xs )g-1 = G(xs ) for some g G(xn ), then Lie(gG(xs )g-1 ) = Lie(G(xs )) and xs Lie(gG(xs )g-1 G(xs )) = Lie(gG(xs )g-1 ) Lie(G(xs )); a contradiction with minimality of G(xs ). The same arguments show that G(xs ) normalizes G(xn ). Since G(xs ) G(Xn ) = {e}, we again have that G(xs ; xn ) := G(xs ); G(xn ) is a closed subgroup isomorphic to G(xs ) в G(xn ). Moreover, G(x) G(xs ; xn ). Using pro jections we get G(xn ) G(x) G, thus xn g and xs = x - xn g. Concerning 2), one easily checks that (G(x)) = G((x)); (G(xs )) G(d(x)s ); (G(xn )) G(d(x)n ): This implies that d(xs ) is semisimple, d(xn ) is nilpotent and by Lemma 1.5.10 they commute, thus form additive Jordan decomposition of d(x).

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37

Corollary 1.7.21. Additive Jordan decomposition x = xs + xn for an element x g = Lie(G) is wel l-de ned. Theorem 1.7.22. Let G GL(n) be a unipotent subgroup. Then any element x g is nilpotent and the map exp : g G is an isomorphism of algebraic varieties.
Proof. If x = xs + xn with xs = 0, then G(xs ) G, where G(xs ) is a torus, a contradiction. Thus the map exp : g GL(n) is de ned. Moreover, for any x g the subgroup G(x) = {exp(tx) : t K} is contained in G, hence exp : g G. Conversely, for any A G the subgroup G(A) = {At : t K} is contained in G. Set B = ln(A). The tangent vector to the curve tB t2 B 2 At = E + + + ::: 1! 2! at t = 0 is B . This shows that B g and ln sends G to g. Since G is a closed subset of GL(n), Lemma 1.7.4 completes the proof. Corollary 1.7.23. Any unipotent group G is isomorphic (as a variety) to an a ne space. In particular, G is connected. Proposition 1.7.24. Let G be a commutative unipotent group. Then exp : g G is an isomorphism of algebraic groups, where g is considered as an additive group of the underlying vector space. Proof. Since g is a commutative Lie algebra (Remark 1.5.7), any linear operators B ; C g commute, and exp(B + C ) = exp(B ) exp(C ). Corollary 1.7.25. Any commutative unipotent group of dimension m is isomorphic to (Ga )m .

38

IVAN V. ARZHANTSEV

Exercises to subsection 1.7. Exercise 1.7.26. Prove that any connected one-dimensional algebraic group is isomor-

phic either to Ga or to Gm . Exercise 1.7.27. Prove that the product and the sum of two commuting semisimple (resp. nilpotent) operators is again semisimple (resp. nilpotent). Prove that the product of two commuting unipotent operators is again unipotent. Exercise 1.7.28. Is the variety Nil(n) irreducible ? Exercise 1.7.29. Assume that A is a degenerate matrix and de ne a multiplicative Jordan decomposition of A as in De nition 1.7.8 (ii). Does such a decomposition always exist and is it unique ? Exercise 1.7.30. Prove that for any A Mat(n в n) there is a polynomial f (x) such that f (A) = As and f (0) = 0. Exercise 1.7.31. Find an element A GL(n) with G(A) = D(n). Exercise 1.7.32. What is the maximal possible dimension of a subgroup of the form G(A), where A GL(n) ? Exercise 1.7.33. Assume that all elements of Lie(G) are nilpotent. Prove that G0 is a unipotent group. Exercise 1.7.34. Let G1 and G2 be connected algebraic groups. Show that for a homomorphism of Lie algebras : Lie(G1 ) Lie(G2 ), the equalities (xs ) = (x)s and (xn ) = (x)n do not hold in general. In particular, not any homomorphism of tangent algebras may be "integrated" to a homomorphism of corresponding connected groups. Exercise 1.7.35 (*). Does there exist an algebraic group G such that all elements of G0 are unipotent, but there is a connected component gG0 of G consisting of semisimple elements ? Exercise 1.7.36 (*). Prove that any commutative algebraic group G is isomorphic to Gs в Gu , where Gs is a quasitorus and Gu is a commutative unipotent group. Exercise 1.7.37 (*). Give an example of a Lie subgroup G GL(n; C) and an element A G such that As ; Au G. =

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39

1.8. Solvable groups. Maximal tori and Borel subgroups. Theorem 1.8.1 (Borel's Fixed Point Theorem (1956)). Assume that a connected solvable algebraic group G acts on a complete (e.g., projective) variety. Then X G = .

Proof. We proceed by induction on the length of the derived series of G. For the basis, assume that G is commutative. Then a closed orbit Y in X (see Corollary 1.3.6) is isomorphic to G=H , where H is a closed (normal) subgroup of G, thus Y is a ne (Corollary 1.4.4). On the other hand, Y is irreducible (G is connected) and complete (as a closed subset of a complete variety). This proves that Y is a G- xed point. For a non-commutative G, consider the commutant [G; G]. We know that [G; G] is a proper closed connected subgroup of G (Corollary 1.2.9). Since [G; G] is normal in G, the group G=[G; G] acts on X [G;G] . By inductive hypothesis, the set X [G;G] is non-empty. By Proposition 1.3.9, X [G;G] is closed in X , thus complete. Again by inductive hypothesis, the set (X [G;G] )G=[G;G] = X G is non-empty.

Theorem 1.8.2 (The Lie-Kolchin Theorem (1948)). Let G be a connected solvable

algebraic group and : G GL(V ) be a rational representation. Then there is a non-zero vector v V such that (g)v = (g)v for some X(G) and any g G.

Proof. On may assume that V is nite-dimensional. The action G : P(V ) (see Lemma 1.3.14) has a G- xed point. This point corresponds to a G-invariant line L in V , and G acts on L via some character.
The Lie-Kolchin Theorem admits the following useful reformulation. Theorem 1.8.3. Let G be a connected solvable algebraic group and : G GL(V ) be a rational nite-dimensional representation. Then there is an element A GL(V ) such that ::: : : : -1 0 : A(G)A B (n) = : : : : : : : : : 0 0 :::

Proof. Since the variety F (V ) of complete ags in V is pro jective, the natural action G : F (V ) has a xed point. Now we should take a basis in V compatible with a G- xed ag.

Corollary 1.8.4. Let G be a connected solvable algebraic group and U (G) G be
the set of al l unipotent elements in G. Then U (G) is a closed normal subgroup of G and G=U (G) is a torus. Proof. We may assume that G GL(n) and even G B (n). Here U (G) = G U (n) is a closed normal subgroup in G and the natural homomorphism G B (n)=U (n) T n identi es G=U (G) with a connected subgroup of T n . =

Theorem 1.8.5. Let G GL(n) be a unipotent subgroup. Then there is A GL(n)
such that
AGA-1 U (n) =

1 01 ::: ::: 00

: : : :

: : : :

: : : :

1

:

40

IVAN V. ARZHANTSEV

Proof. We claim that G is solvable. In order to prove it, we use induction on dim G. If dim G = 1, then G = G(A) for any A G, A = e, and thus is commutative. If dim G > 1, take a maximal proper closed subgroup H G. It is connected (as a unipotent group) and solvable by inductive hypothesis. The Ad-representation de nes a linear action H : g=h. By the Lie-Kolchin Theorem, there is a non-zero eigenvector x + h g=h. But all elements of H are unipotent, and x + h is H xed. The condition Ad(H )x x + h implies ad(h)x h, or [x; h] h. The one-dimensional subgroup G(x) is not contained in H (x h) and normalizes H . = By maximality of H , the semidirect product G(x) H coincides with G. We have proved that H is normal in G and dim G=H = 1. Hence H and G=H are solvable, and so is G. Applying Theorem 1.8.3, we nd A GL(n) such that AGA-1 B (n). But all element of G are unipotent, so AGA-1 U (n). Corollary 1.8.6. Any unipotent group is solvable.
In fact, we may say more. Corollary 1.8.7 (of the proof of Theorem 1.8.5). If G is a unipotent group, then there is a sequence of subgroups {e} = G0 G1 · · · Gk = G; Ga . such that Gi =Gi-1 = Corollary 1.8.8. The subgroup U (n) is a maximal (with respect to inclusion) unipotent subgroup of GL(n). Corollary 1.8.9. X(G) = 0 for any unipotent group G.

Proof. Assume that : G Kв is a non-trivial character. Then there is a number i with (Gi-1 ) = 1 and |Gi being non-trivial, so is a non-trivial character of the group Gi =Gi-1 , a contradiction with Proposition 1.7.16.
Now we are ready to prove the structural theorem for connected solvable groups. Theorem 1.8.10. Let G be a connected solvable algebraic group. Then there is a subtorus T G such that G = T U (G).

Proof. By Corollary 1.8.4, the quotient G=U (G) is a torus H . Lemma 1.8.11. For any torus T m , there is t0 T m such that G(t0 ) = T m . Proof. By Theorem 1.6.15, it is su cient to nd an element t0 T m such that any character of T m is not equal to 1 at t0 . On may take t0 = (p1 ; : : : ; pn ), where p1 ; : : : ; pn are pairwise di erent (positive) prime integers.
Now take t0 H with G(t0 ) = H , consider the pro jection : G H and x g G with (g) = t0 . If g = gs gu is Jordan decomposition, then t0 = (gs )(gu ). Hence (gu ) = e, and one may assume that g = gs is semisimple. The subgroup G(g)0 is a torus of dimension dim H , because (G(g)0 ) can not be a proper subgroup of H . By dimension reasons, G(g)0 U (G) coincides with G, and G(g)0 is the desired subtorus T . De nition 1.8.12. Let G be an algebraic group. A subtorus T G is called a maximal torus of G if it is not contained in any other subtorus of G. Example 1.8.13. The subgroup D(n) is a maximal torus of GL(n).

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Proposition 1.8.14. Any two maximal tori of a connected solvable algebraic group
G are conjugate.

Proof. We know that G = T U (G). Lemma 1.8.15. Any semisimple element s G is conjugate to an element of T . Proof. Consider the homogeneous G-space G=T . An element s G is conjugate to an element of T if and only if s has a xed point on G=T . The pro jection := |U (G) : U (G) G=T is bijective, and thus is an isomorphism of varieties. By Theorem 1.7.22, the map := exp : Lie(U (G)) G=T is also an isomorphism. Moreover, the torus T acts on Lie(U (G)) via adjoint action, t exp(A)t-1 = exp(tAt-1 ), and thus the morphism is T -equivariant with respect to the action T : G=T , t · uT = tuT = tut-1 T . This shows that G=T may be identi ed with an a ne space Am with a linear T -action. Suppose that U (G) is commutative. Then exp : Lie(U (G)) U (G) is an isomorphism of groups (Proposition 1.7.24) and, via our identi cation G=T Am , = the U (G)-action on G=T is the action by parallel translations on Am . Finally, the group G acts on Am via a ne transformations, and thus preserves the nitedimensional subspace A of a ne functions in O(Am ). By assumption, the element s is semisimple, and there is a basis {f1 ; : : : ; fm+1 } in A consisting of s-eigenvectors. Renumbering, one may suppose that the linear parts of f1 ; : : : ; fm are linearly independent. The hyperplanes f1 = 0; : : : ; fm = 0 are s-invariant and their intersection is an s- xed point. If U (G) is not commutative, we proceed by induction. The commutant [U (G); U (G)] is a closed normal subgroup of G, and for the group G=[U (G); U (G)] T U (G)=[U (G); U (G)] = the element s[U (G); U (G)] is conjugate to an element of T . Thus for some g G, one has gsg-1 T [U (G); U (G)]. Again by inductive hypothesis, this element is conjugate to an element of T .
The condition s gT g-1 implies G(s) gT g-1 . This proves that any subgroup G(s) is conjugate to a subgroup of T . By Lemma 1.8.11, any subtorus in G is conjugate to a subtorus in T . (In particular, T is a maximal torus in G.) De nition 1.8.16. Let G be an algebraic group. A maximal connected solvable subgroup of G is called a Borel subgroup. Example 1.8.17. Theorem 1.8.3 implies that B (n) is a Borel subgroup of GL(n). Lemma 1.8.18. Let G be an algebraic group and B G be a Borel subgroup of maximal dimension. Then the homogeneous space G=B is projective.

Proof. By Theorem 1.4.1, there exist a rational G-module V and a non-zero v V such that B = {g G : g · v v }: Let F0 be a closed subvariety of the ag variety F (V ) consisting of complete ags with the rst element v . The subvariety F0 is B -invariant, and by Borel's Fixed Point Theorem B has a xed point F F0 . The stabilizer GF coincides with B , because B is the stabilizer of F 's rst element. We claim that the G-orbit of F is closed in F (V ). Indeed, any G-orbit in the closure of GF has smaller dimension.

42

IVAN V. ARZHANTSEV

On the other hand, the stabilizer of any point on F (V ) is solvable, and B has a maximal dimension among (closed) solvable subgroups of G. We have proved that GF G=B is closed in the pro jective variety F (V ), thus is = pro jective too. Theorem 1.8.19. Let G be an algebraic group. Then (1) any two Borel subgroups of G are conjugate; (2) any two maximal tori of G are conjugate.

Proof. Let B be a Borel subgroup of G and B0 be a Borel subgroup of maximal dimension. By Borel's Fixed Point Theorem, B has a xed point on G=B0 , or, equivalently, there is g G with gB g-1 B0 . By maximality of B , gB g-1 = B0 , and the rst statement is proved. Now take two maximal tori T1 and T2 in G. Since T1 and T2 are connected and solvable, there are Borel subgroups B1 and B2 with T1 B1 , T2 B2 . We know that gB1 g-1 = B2 for some g G. By Proposition 1.8.14, the torus T2 and gT1 g-1 are conjugate. Corollary 1.8.20. Let G be an algebraic group and B a Borel subgroup of G. Then the homogeneous space G=B is projective.

ALGEBRAIC GROUPS AND INVARIANT THEORY

43

B and B1 isomorphic as algebraic groups ? Exercise 1.8.22. Show by examples that the following conditions:

Exercises to subsection 1.8. Exercise 1.8.21. Let B and B1 be Borel subgroups in SL(2) and SO(3) respectively. Are

a) G is connected; b) G is solvable; c) X is complete are essential in Borel's Fixed Point Theorem. Exercise 1.8.23. Check that NGL(n) B (n) = B (n). Exercise 1.8.24. Let G be a solvable algebraic group and U (G) G be the set of all unipotent elements in G. Prove that U (G) is a closed normal subgroup of G. Is it true that G=U (G) is a quasitorus ? Exercise 1.8.25. Let G be a unipotent group and X an a ne G-variety. Prove that any G-orbit on X is closed. Exercise 1.8.26. A subgroup P of an algebraic group G is said to be parabolic if P contains a Borel subgroup of G. Prove that the homogeneous space G=H is pro jective if and only if H is parabolic. Exercise 1.8.27 (*). Classify up to conjugation parabolic subgroups of GL(n). Exercise 1.8.28 (*). Let G be an algebraic group acting on a variety X and B be a Borel subgroup of G. Assume that Y X is a closed B -invariant subset. Prove that GY := {g · y : g G; y Y } is closed in X . Exercise 1.8.29. Prove that any two maximal unipotent subgroups of an algebraic group G are conjugate. Exercise 1.8.30. Show that two maximal connected commutative subgroups of G need not be conjugated. Exercise 1.8.31 (*). Describe Borel subgroups and maximal tori in SO(n) and Sp(2n). Exercise 1.8.32 (**). Let G be a solvable algebraic group and H a closed subgroup of G. Prove that the homogeneous space G=H is a ne.

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45

1.9. Reductive groups. In the present section we rst introduce two concepts of the radical for algebraic groups.

Lemma 1.9.1. Let F be a solvable (resp. unipotent) algebraic group and : G

F be a surjective homomorphism. The group G is solvable (resp. unipotent) if and only if so is Ker.

Proof. The statement concerning solvability is standard. If s G is a semisimple element, then (s) is semisimple, so (s) = e and s Ker. This proves that G contains no semisimple elements (i.e., is unipotent) if and only if so does Ker.

Lemma 1.9.2. Let G be an algebraic group and H1 , H2 two normal closed connected solvable (resp. unipotent) subgroups of G. Then the subgroup H1 H2 possesses the same properties.
Proof. The image of the map H1 в H2 G, (h1 ; h2 ) h1 h2 satis es the conditions of Proposition 1.2.6, so H1 H2 is a closed connected subgroup of G. Clearly, it is normal. Applying Lemma 1.9.1 to the homomorphism
H1 H2 H1 H2 =H2 H1 =(H1 H2 ); = we get that H1 H2 is solvable (resp. unipotent).

De nition 1.9.3. The radical R(G) of an algebraic group G is the largest (closed) connected normal solvable subgroup of G. Remark 1.9.4. By Lemma 1.9.2, the subgroup R(G) is well-de ned. Moreover, R(G) = R(G0 ) because R(G) R(G0 ) and R(G0 ) is a normal subgroup of G. De nition 1.9.5. An algebraic group G is semisimple if R(G) = {e}. Lemma 1.9.6. If G is an algebraic group, then G=R(G) is semisimple.
{e}. Then -1 (R(G=R H 0 R(G)

Proof. Let : G G=R(G) be the pro jection. Assume that R(G=R(G)) = it has positive dimension. By Lemma 1.9.1, the preimage H := (G))) is a normal solvable subgroup of G of positive dimension, thus and (H ) is nite, a contradiction.

Proposition 1.9.7. An algebraic group G is semisimple if and only if its tangent
algebra g is semisimple. Proof. If r(g) = 0, then g contains a non-zero commutative ideal a: take the last non-zero element in the sequence of commutants of r(g). Then a z(zg (a)). On the other hand, the ideal z(zg (a)) is the tangent algebra to the normal algebraic subgroup H = {g G : Ad(g)y = y for all y zg (a)}: By Proposition 1.9.8, the group H 0 is commutative, and thus H 0 R(G).
Conversely, if R(G) = {e}, then it contains a connected commutative subgroup that is normal in G, and its tangent algebra is a commutative ideal of g.

Proposition 1.9.8. If G is connected and g = Lie(G) is a commutative Lie algebra,
then the group G is commutative.

46

IVAN V. ARZHANTSEV

Proof. Let gs (resp. gn ) be the set on semisimple (resp. nilpotent) elements in g. Since g is commutative, gs and gn are ideals in g. Let us x a closed embedding G GL(n). The map exp : gn G de nes an isomorphism between gn and some commutative unipotent subgroup Gu of G. Moreover, Gu is normal in G (because gn is an ideal). One may assume that gs is the set of diagonal matrices in g (Lemma 1.6.12). Then gs = Lie(G D(n)) = Lie(Gs ), where Gs := (G D(n))0 is a torus. Again, the subgroup Gs is normal in G. Since Gu Gs = {e}, the map : Gu в Gs G, (g1 ; g2 ) = g1 g2 is a injective homomorphism. Since any element of G admits multiplicative Jordan decomposition, is an isomorphism.
Now we introduce a product of normal subgroups of G which is "almost" direct. De nition 1.9.9. Let G be a connected algebraic group and H1 ; : : : ; Hk its closed connected subgroups. The group G is said to be an almost direct product of H1 ; : : : ; Hk (notation: G = H1 · H2 · ::: · Hk ), if (i) hi hj = hj hi for any hi Hi , hj Hj , i = j ; (ii) the homomorphism : H1 в · · · в Hk G; (h1 ; : : : ; hk ) = h1 : : : hk ; is surjective and Ker is a nite (central) subgroup of H1 в · · · в Hk . Example 1.9.10. The group GL(n) is an almost direct product of H1 = {E : Kв } and H2 = SL(n) with Ker = {( E ; -1 E ) : n = 1}. Lemma 1.9.11. Let G be a connected algebraic group and H1 ; : : : ; Hk be closed connected subgroups. Then G is an almost direct product of H1 ; : : : ; Hk if and only if g = h1 · · · hk (a direct sum of Lie algebras).

Proof. Assume that k = 2. If G = H1 · H2 , then h1 h2 = 0, h1 and h2 are ideals of g and the di erential of : H1 в H2 G de nes an isomorphism h1 h2 g. = Conversely, if g = h1 h2 , then H1 and H2 are normal subgroups of G with H1 H2 being nite. A commutator h1 h2 h-1 h-1 belongs to H1 H2 , and irreducibility of 12 - - H1 H2 H1 1 H2 1 implies that h1 h2 = h2 h1 . Finally, : H1 в H2 G is surjective, because of Theorem 1.2.12. For k > 2 we use induction. If G = H1 · ::: · Hk , then g = Lie(H1 · ::: · Hk-1 ) hk = h1 · · · hk :
Conversely, if g = at
h1 · · · hk

, then the subgroups Hi are normal in G, and, looking

H1 в H2 в · · · в Hk-1 H1 H2 :::Hk-1 ; one checks that Lie(H1 H2 :::Hk-1 ) = h1 h2 · · · hk-1 . Hence we obtain G = (H1 · :::: · Hk1 ) · Hk = H1 · ::: · Hk-1 · Hk :

largest normal unipotent subgroup of G. Remark 1.9.13. By Lemma 1.9.2, the subgroup Ru (G) is well-de ned and again Ru (G) = Ru (G0 ). De nition 1.9.14. An algebraic group G is reductive if Ru (G) = {e}. Example 1.9.15. Any nite group and any quasitorus are reductive.

De nition 1.9.12. The unipotent radical Ru (G) of an algebraic group G is the

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47

Theorem 1.9.16. Let G be an algebraic group. The fol lowing conditions are equivalent:
(1) (2) (3) (4) (5)
G is reductive; R(G) is a torus; G0 = T · S , where T is a torus and S is a connected semisimple subgroup; any nite-dimensional rational representation of G is completely reducible; G admits a faithful nite-dimensional completely reducible rational representation.

Proof. (1)(2) Since R(G) = T U (R(G)), where U (R(G)) is the set of all unipotent elements in R(G), the subgroup U (R(G)) is invariant under any automorphism of G, and thus U (R(G)) Ru (G). (2)(3) Set T = R(G) = R(G0 ) and S = [G0 ; G0 ]. Let V be a nite-dimensional G-module. Since T is a torus, there is a decomposition
V=

X(T

)

V ; V := {v V : t · v = (t)v}:

The subgroup T is normal in G, thus G permutes the summands V . On the other hand, the stabilizer of V is closed in G, so G0 preserves any V . In particular, taking a faithful representation , we see that T is contained in the center of G0 . Any element of S is a product of commutators, so it acts on any V with determinant 1. This implies that S T is nite. By Lemma 1.9.6, the group G=T is semisimple. Then Lie(G=T ) is semisimple and [Lie(G=T ); Lie(G=T )] = Lie(G=T ) (Corollary 4.0.40).

Lemma 1.9.17. Let F be a connected algebraic group with [Lie(F ); Lie(F )] =
Lie(F ). Then [F; F ] = F .

Proof. If [F; F ] is a proper subgroup, then F =[F; F ] is an Abelian group of a positive dimension, Lie(F =[F; F ]) = Lie(F )=Lie([F; F ]) is commutative, and thus [Lie(F ); Lie(F )] Lie([F; F ]).
We have proved that the group G0 =T coincides with its commutant, and the natural pro jection : S G0 =T is surjective, with Ker() being nite. This proves that S is semisimple (Lie(S ) = Lie(G0 =T ), use Proposition 1.9.7) and G0 = T · S . (3)(4) Take any nite-dimensional G-module V and again consider the decomposition V = X(T ) V with respect to T . Any subspace V is G0 -invariant. We claim that V is a completely reducible S -module. Indeed, Theorem 1.5.17 reduces the problem to representations of Lie algebras, where the statement is know as Weyl's Theorem (see Theorem 4.0.41). We have obtained that V is completely reducible G0 -module. Now let U be a G-invariant subspace of V and p : V U be a G0 -equivariant pro jection. For any g G de ne an operator pg (v) := g · p(g-1 · v). Fix a representative gi in any connected component of G.

Lemma 1.9.18. The operator
P :=

1 |G=G

0|

is a G-equivariant projection on U .

i

pgi : V V

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IVAN V. ARZHANTSEV

Proof. Clearly, P : V U , and it is su cient to check that P (u) = u for any u U . But 1 1 - - P (u) = gi · p(gi 1 · u) = gi · (gi 1 · u) = u: |G=G0 | i |G=G0 | i
Finally, the kernel of P is a G-invariant subspace complementary to U . (4)(5) Follows from Theorem 1.3.20. (5)(1) Let : G GL(V ) be a faithful completely reducible representation . u Suppose that Ru (G) = {e}. By Theorem 1.8.5, the subspace V R (G) = u (G) 0. But Ru (G) is normal in G, and V R is G-invariant. By complete reducibility, there is a complementary G-invariant subspace W : u V = V R (G) W: Theusubspace W is non-zero ( is faithful), and, again by Theorem 1.8.5, u W R (G) = 0, a contradiction with V R (G) W = 0.

Corollary 1.9.19. For a reductive G, R(G) is a central subtorus of G0 . Corollary 1.9.20. Al l classical groups are reductive.
Proof. Since the tautological representation of a classical group is completely reducible, we may apply (5). (In fact, the tautological representation is almost always irreducible, see Exercise 1.9.34.)

De nition 1.9.21. A connected algebraic group G is said to be simple, if Lie(G)
is a simple Lie algebra.

Warning: A simple algebraic group G may contain a proper normal subgroup:
take G = SL(2).

Proposition 1.9.22. For any reductive group G, one has
G0 = T · G1 · ::: · Gk ; where T is a torus and Gi are simple subgroups.

Proof. For the semisimple part S G0 , one has Lie(S ) = g1 · · · gk , where gi are simple ideals (Theorem 4.0.39). We claim that gi is an algebraic Lie subalgebra of Lie(S ). Indeed, the center of a simple Lie algebra is trivial, and gi = {x Lie(S ) : [x; y ] = 0 for all y gj ; j = i}: Thus gi = Lie(Si ), where Si := {g S : Ad(g)y = y for all y gj ; j = i}:
Lemma 1.9.11 implies that S = G1 · ::: · Gk , where Gi := Si0 , and, nally, G0 = T · G1 · ::: · Gk . We nish this section with complete reducibility of any rational G-module for a reductive G. Let = (G) be the set of isomorphism classes of rational nite-dimensional simple G-modules. For any rational G-module V one can de ne a submodule V as the sum of all simple submodules of V whose class is .

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49

De nition 1.9.23. The submodule V is called the isotypic component of type
of the module V .

Theorem 1.9.24. Let G be a reductive group and V a rational G-module. Then
V =

V :

Proof. Since any vector v V lies in a nite-dimensional submodule, Theorem 1.9.16 (4) implies that v V . Suppose that v1 + · · · + vk = 0, where non-zero vi belong to di erent Vi . There is a nite-dimensional G-submodule W that contains v1 ; : : : ; vk . Consider a decomposition W = W1 · · · Ws ; where Wi are simple G-modules. We claim that V W is a sum of some Wi . Indeed, if U is a simple G-submodule in W , then, by the Schur Lemma, U has zero pro jection on any Wi that is not isomorphic to U . This shows that the sum W = V W is direct, a contradiction. Remark 1.9.25. The decomposition V = V is called the isotypic decomposition of a rational G-module V .

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IVAN V. ARZHANTSEV

Exercises to subsection 1.9. Exercise 1.9.26. Classify all closed subgroups of positive dimension in SL(2) up to
conjugation.
· · · < ks = n of positive integers and the standard basis e1 ; : : : ; en in Kn . Let Pk1 ;:::;ks be the stabilizer in GL(n) of the ag {0} e1 ; : : : ; ek1 e1 ; : : : ; ek2 · · · e1 ; : : : ; eks = Kn : Calculate R(Pk1 ;:::;ks ) and Ru (Pk1 ;:::;ks ). Exercise 1.9.28. Prove that SO(4) is an almost direct product of two closed subgroups isomorphic to SL(2). Exercise 1.9.29. Let G be an algebraic group. Prove that G=Ru (G) is reductive. Exercise 1.9.30. Do there exist connected algebraic groups G1 and G2 such that G1 is reductive, G2 is not, and Lie(G1 ) Lie(G2 ) ? = Exercise 1.9.31. Show that a direct product, a normal subgroup and a quotient group of a reductive group is reductive. Exercise 1.9.32. Let G be a reductive group and : F G be a surjective homomorphism with nite kernel. Show that F is reductive. Exercise 1.9.33. Prove that Lie(R(G)) = r(g). Exercise 1.9.34. Let V be the tautological module of a classical group G (recall that G = GL(n); SL(n); O(n); SO(n); Sp(2n)). Prove that V is irreducible with the only exception G = SO(2). Exercise 1.9.35 (*). List all classical groups that are semisimple. Which of them are simple ? Exercise 1.9.36. Let H and N be closed connected subgroups of an algebraic group G. Assume that H normalizes N . Then the subgroup H N is closed in G and Lie(H N ) = Lie(H ) + Lie(N ). Exercise 1.9.37 (*). Assume that all elements of an algebraic group G are semisimple. Prove that G0 is a torus.

Exercise 1.9.27. Fix a sequence 1 k1 < k2 <

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51

2. Invariant Theory 2.1. Finite generation. Let G be an algebraic group and X a G-variety. Recall that the algebra of regular functions O(X ) is a rational G-module with respect to the action: (g · f )(x) := f (g-1 · x): G := {f O(X ) : g · f = f for all g G}. Clearly, O(X )G is a De ne O(X ) subalgebra in O(X ) and the restriction of an element f O(X )G to a G-orbit in X is a constant. The algebra O(X )G is called the algebra of invariants of the action G : X , and an element f O(X )G is called an invariant. Theorem 2.1.1 (Hilbert's Theorem on Invariants (1890)). Let G be a reductive algebraic group and X an a ne G-variety. Then the algebra of invariants O(X )G is nitely generated.

Proof. Consider the isotypic decomposition of the G-module O(X ): O(X ) = (G) O(X ) : If = 0 is a class of the one-dimensional G-modules with trivial G-action, then O(X )0 = O(X )G . Consider a G-invariant subspace O(X )G := =0 O(X ) : G O (X ) . Clearly, O(X ) = O(X ) G De nition 2.1.2. The pro jection R : O(X ) O(X )G along O(X )G is called the Reynolds operator on O(X ).
Let us collect some properties of the Reynolds operator. Lemma 2.1.3. (i) The map R is G-invariant; (ii) R(f f1 ) = f R(f1 ) for any f O(X )G , f1 O(X ); (iii) any G-invariant subspace of O(X ) is R-invariant.

Proof. (i) Since the subspaces O(X )G and O(X )G are G-invariant, we have R(g· f ) = R(f ).
+ - + - + - (ii) Set f1 = f1 + f1 , where f1 O(X )G , f1 O(X )G . Then f f1 = f f1 + f f1 , + O(X )G . Since for any simple G-submodule U O(X ) of type the subspace f f1 f U = {f u : u U } is also a simple G-submodule of the same type, one has O(X )G O(X )G O(X )G ; - + thus f f1 O(X )G , and R(f f1 ) = f f1 = f R(f1 ).

(iii) Let W O(X ) be a G-invariant subspace. The isotypic decomposition of W leads to W = W G WG , there W G O(X )G , WG O(X )G . Thus f = f + f with f W G , f WG , and R(f ) = f W . Theorem 2.1.1 in the particular case, where X is a rational algebra O(V ) is a polynomial algebra that has a natural (V )n , where O(V )n is the span of monomials of degree n. preserves homogeneous components. In particular, O(V )G = n0 O(V )G : n Put I1 := n>0 O(V )G . Then I := I1 O(V ) O(V ) is the ideal of O(V ) generated n by invariants of positive degree. By the Hilbert Basis Theorem, any ideal of O(V ) is nitely generated, thus I = (f1 ; : : : ; fk ). One may assume that the elements f1 ; : : : ; fk are homogeneous and belong to I1 . We come to the proof of G-module V . Here the grading O(V ) = n0 O Since G acts linearly, it

52

IVAN V. ARZHANTSEV

Lemma 2.1.4. f1 ; : : : ; fk generate the algebra O(V )G .
Proof. Take a homogeneous element F O(V )G . In order to prove that F K[f1 ; : : : ; fk ], we apply induction on deg F . The case deg F = 0 is obvious. If deg F > 0, then F I , thus F = a1 f1 + · · · + ak fk for some ai O(V ). By Lemma 2.1.3, one has F = R(F ) = R(a1 f1 ) + · · · + R(ak fk ) = R(a1 )f1 + · · · + R(ak )fk : Since deg fi > 0, we may suppose that R(ai ) are homogeneous invariants with deg R(ai ) < deg F . By inductive hypothesis, R(ai ) K[f1 ; : : : ; fk ], so F K[f1 ; : : : ; fk ].
Now we are ready to prove Theorem 2.1.1 for arbitrary a ne G-variety X . By Theorem 1.3.19, X may be realized as a closed G-invariant subvariety of some Gmodule V . The embedding X , V corresponds to the surjective G-equivariant restriction homomorphism p : O(V ) O(X ). Clearly, the restriction of p to O(V )G de nes a homomorphism p : O(V )G O(X )G .

Lemma 2.1.5. Let V be an a ne G-variety and X its closed G-invariant subvariety. Then the restriction homomorphism p : O(V )
GO

(X )G is surjective.

Proof. For any non-zero f O(X )G , the preimage of the line p-1 ( f ) is a Gsubmodule W O(V ) that is mapped surjectively on f . By the Schur Lemma, p sends any isotypic component of non-zero type in W to zero. Thus there is a G-invariant F W such that p(F ) = f .
Since O(V )G is nitely generated, so is its homomorphic image O(X )G , and Theorem 2.1.1 is proved. Unfortunately, the proof given above does generating set for O(X )G . It turns out to this section this a method that allows to cases. This method is called the method of Assume that S

not provide any algorithm for nding a be a very di cult problem. We nish nd a generating set in many important sections.

X is a closed subvariety. De ne

Z (S ) := {g G : g · s = s for all s S }; N (S ) := {g G : g · s S for all s S }:

Clearly, Z (S ) is a normal subgroup of the group N (S ), and the quotient group W = W (S ) := N (S )=Z (S ) acts (algebraically) on S . The surjection O(X ) O(S ) de nes a homomorphism : O(X )G O(S )W .

Lemma 2.1.6. Suppose that there is an open dense subset U any x U the orbit Gx intersects S . Then is injective.
Proof.

X such that for

f O(X )G ; (f ) = 0 f |S 0 f |U 0 f = 0:

Moreover, if f1 ; : : : ; fk O(X )G and (f1 ); : : : ; (fk ) generate O(S )W , then is an isomorphism. In particular, f1 ; : : : ; fk generate O(X )G .

ALGEBRAIC GROUPS AND INVARIANT THEORY

53

в n) with the adjoint G-action: A · M := AM A-1 . Take S to be the subspace of diagonal matrices. Here Z (S ) = D(n) and N (S ) is the group of monomial matrices. Indeed, the standard basis in Kn is the only basis proper for any operator from S , and N (S ) may only permutes basis vectors. This shows that W is isomorphic to the permutation group n and it acts on S permuting the diagonal entries. Let 1 ; : : : ; n be elementary symmetric polynomials in standard coordinates on S . It is well known that 1 ; : : : ; n is a generating set for O(S )W . In order to verify the condition of Lemma 2.1.6, one may take the open subset U of matrices with pairwise di erent eigenvalues. It is well known that any matrix from U is diagonalizable. Now let f1 ; : : : ; fn be the coe cients of the characteristic polynomial PA (x) = xn + f1 (A)xn-1 + · · · + fn (A); considered as polynomial functions on the space Mat(n в n). The restriction of fi to the subspace of diagonal matrices equals (-1)i i . This proves that O(Mat(n в n))GL(n) = K[f1 ; : : : ; fn ]: Moreover, we know that 1 ; : : : ; n generate O(S )W freely, and so do f1 ; : : : ; fn . Example 2.1.8. Let G = SL(n) and X be the space Sym(n) of symmetric n в nmatrices with the action: A · M := AM AT . It is natural to consider the line aE , a K as the subvariety S . Here Z (S ) = SO(n). If A N (S ), then AE AT = aE . Taking determinant, we get an = 1. Thus N (S ) = Cn Z (S ), where Cn = {diag(- ; ; : : : ; ) : n = -1}; and W Zn acts on S via multiplications by = 2 , n = 1. This proves that = O(S )W = K[F ], F (aE ) = an . Since any non-degenerate symmetric matrix is GL(n)-equivalent to E , its SL(n)orbit intersects S , and we may put U to be the set of non-degenerate symmetric matrices. The determinant det de nes a G-invariant function on X . Its restriction to S coincides with F , so O(Sym(n))SL(n) = K[det]:

Example 2.1.7. Consider G = GL(n) and X = Mat(n

54

IVAN V. ARZHANTSEV

Exercises to subsection 2.1. Exercise 2.1.9. Give an example of a non- nitely generated subalgebra in K[x1 ; x2 ]. Exercise 2.1.10. Check that R is the only G-invariant pro jection of O(X ) to O(X )G . Exercise 2.1.11. Assume that G is nite and f O(X ). Show that
R(F ) =
|G| gG

1

g · f:
Kn

Exercise 2.1.12. For the tautological action GL(n) :

), show that R(f ) = f ((0; : : : ; 0)). Exercise 2.1.13. Let f (x1 ; : : : ; xn ) K[x1 ; : : : ; xn ] be a homogeneous polynomial of degree k. Prove that there exist linear forms l1 (x1 ; : : : ; xn ); : : : ; lN (x1 ; : : : ; xn ) such that k k f = l1 + · · · + lN . Exercise 2.1.14. Let G be a nite group and V a nite-dimensional G-module. Prove that O(V )G is generated by homogeneous invariants of degree |G| (Noether's Theorem). Exercise 2.1.15. Using the method of sections, nd a generating set of the algebra of invariants for (a) the tautological action SO(n) : Kn ; (b) the tautological action Sp(2n) : K2n ; (c) the diagonal action SL(n) : Kn · · · Kn (s times, s n). Exercise 2.1.16. Find a generating set of the algebra of invariants for the linear actions: (a) Zn : K2 , · (x1 ; x2 ) = ( x1 ; x2 ), n = 1; (b) Zn : K2 , · (x1 ; x2 ) = ( x1 ; -1 x2 ), n = 1. Exercise 2.1.17. Find a generating set of the algebra of invariants for the linear action Kв : K4 ; t · (x1 ; x2 ; x3 ; x4 ) = (t3 x1 ; tx2 ; t-1 x3 ; t-2 x4 ):

and f

O(Kn

ALGEBRAIC GROUPS AND INVARIANT THEORY

55

2.2. The quotient morphism and categorical quotients. Let G be a reductive group and X an a ne G-variety. Assume that f1 ; : : : ; fk generate the algebra O(X )G . Consider a morphism: : X Kk ; (x) = (f1 (x); : : : ; fk (x)): Clearly, is constant on G-orbits. Moreover, the morphism : X (X ) does not depend on the choice of generators f1 ; : : : ; fk , because it may be realized as the morphism X Spec(O(X )G ) corresponding to the inclusion O(X )G O(X ).

De nition 2.2.1. The morphism : X
the inclusion O(X X. )G O

Y := Spec(O(X )G ) corresponding to (X ) is called the quotient morphism for an a ne G-variety

Let us summarize basic properties of the quotient morphism.

Theorem 2.2.2.

(1) is surjective; (2) if Z X is a closed G-invariant subvariety, then (Z ) is closed in Y ; (3) if {Z } is a family of closed G-invariant subvarieties of X , then
(Z ) = ( Z );
O

(4) if U Y is an open subset, then U : O(U ) phism.

(-1 (U ))G is an isomor-

Proof. (1) For any y Y consider the ideal Iy := {f O(Y ) : f (y) = 0} O(Y ) = O(X )G : Applying the Reynolds operator, we get (O(X )Iy )G Iy . In particular, O(X )Iy = O(X ). Let m be a maximal ideal of O(X ) with O(X )Iy m. Then (x) = y , where x X is a point corresponding to the maximal ideal m.
(2) Let : Z , X tion homomorphisms jective (Lemma 2.1.5). YZ := Spec(O(Z )G ), is a
O

be the closed G-equivariant embedding. The restric: O(X ) O(Z ) and : O(X )G O(Z )G are surThis proves that the morphism : YZ Y , where closed embedding,
o

(OZ )

O

(X ) O
Z

Z
Z

/

X

/Y Y (Z )G o O(X )G and (Z ) = (YZ ), because the quotient morphism Z for the a ne G-variety Z is surjective.

O

(3) Let I G. R(f1 ) + opposite Suppose that

I O(X ) be the ideal corresponding to Z . We claim that ( I Indeed, if F = f1 + · · · + fk ( I )G with fi I i , then F = R( · · · + R(fk ) with R(fi ) I Gi (Lemma 2.1.3 (iii)), and F I G. inclusion is obvious. that f O(X )G is identically zero on Z . Then there is m N fm ( I )G = I G;

)G = F) = The such

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IVAN V. ARZHANTSEV

and thus f vanishes on (Z ). Since both in Y , this proves the inclusion
(Z ) (

(Z ) and ( Z ):

Z ) are closed

The opposite inclusion holds set-theoretically. (4) First assume that U = Yf and
O

Y is a principal open subset. Here f1 O(U ) = (O(X )G )f = { m : f1 O(X )G } f

(-1 (U ))G = (O(X )f )G =

But then f2 O(X )G and O(U ) = O( Since any open U is a nite union of principal open subsets, the statement follows from O(U1 U2 ) = O(U1 ) O(U2 ) = O( -1 (U1 ))G O( -1 (U2 ))G = = (O(-1 (U1 ))
O

f2 f2 { k: k ff -1 (U ))G .

is G - invariant}:

(-1 (U2 )))G = O(-1 (U1 U2 ))G :

In general, invariants do not separate all G-orbits in X . The following corollary shows that they do separate closed orbits.
Y be the quotient morphism. For any y Y , the ber -1 (y) contains a unique closed G-orbit Oy . This orbit is contained in the closure of any orbit from -1 (y).

Corollary 2.2.3. Let : X

Proof. Since -1 (y) is closed and G-invariant, it contains a closed G-orbit (Corollary 1.3.6). Let O1 and O2 be two closed G-orbits in -1 (y). Then O1 O2 = , and by Theorem 2.2.2 (3), (O1 ) (O2 ) = . But (O1 ) = (O2 ) = y, a contradiction. Finally, Corollary 1.3.6 also implies that the closure of any orbit from -1(y) contains a closed G-orbit, and it should be Oy .

action: A · M := AM A-1 . It was shown in Example 2.1.7 that here Y = Kn and the map sends a matrix to the coe cients of its characteristic polynomial. A ber of is the set of matrices with xed eigenvalues (with multiplicities). Any ber contains a nite number of G-orbits which are parametrized by Jordan normal forms with the prescribed diagonal. In particular, the ber -1 (0; : : : ; 0) is the set Nil(n) of nilpotent matrices and n n n -1 ((- ; ; : : : ; (-1)n )) 1 2 n is the set Uni(u) of unipotent matrices. Moreover, if the discriminant of the polynomial xn + a1 xn-1 + · · · + an is non-zero, then the ber -1 (a1 ; : : : ; an ) consists of one G-orbit. Now we are going to prove an important universal property of the quotient morphism.

Example 2.2.4. Consider G = GL(n) and X = Mat(n в n) with the adjoint G-

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57

De nition 2.2.5. Let G be an algebraic group and X a G-variety. A G-invariant morphism : X Y is said to be a categorical quotient, if for any G-invariant morphism : X Z there exists a unique morphism : Y Z such that the following diagram is commutative: X@
Z Y Remark 2.2.6. It follows from the de nition that, if the categorical quotient exists, then it is unique (up to isomorphism), see Exercise 2.2.16.
~~~ ~~ ~~~ o

@@ @@ @@

The usual notation for the categorical quotient is : X X==G. Theorem 2.2.7. Let G be a reductive group and X an a ne G-variety. Then the quotient morphism : X Y is the categorical quotient.

Proof. Let : X Z be a G-invariant morphism. We de ne rstly the desired map as a map of sets: for any y Y put (y) = (-1 (y)). Clearly, it is the only way to make the above diagram commutative. But we need to explain that (-1 (y)) is one point. Indeed, the ber -1 (y) contains a unique closed G-orbit Oy , and any other orbit O -1 (y) contains Oy in its closure, thus (O) = (Oy ) = z for some z Z. Now we prove that is a continuous map. De nition 2.2.8. A G-invariant subset W X is said to be saturated if x X; w W; Gx Gw = x W: Lemma 2.2.9. A subset W X is saturated if and only if there is a subset W1 Y such that W = -1 (W1 ). Proof. The condition Gx1 Gx2 = is equivalent to "x1 and x2 are in the same ber of ". So, W is saturated if and only if it consists of bers of the quotient morphism.
Take an open subset U Z . Then -1 (U ) is open and saturated in X . Theorem 2.2.2 (2), the set D := (X \ -1 (U )) is closed in Y , thus -1 (U (-1 (U )) = Y \ D is open. Finally we check that is a morphism, i.e., for any open U Z and f O(U ) function (f ) lies in O( -1 (U )). Theorem 2.2.2 (4) implies (f ) O(-1 (U ))G = O(-1 ( -1 (U )))G = (O( -1 (U ))) (f ) O( By )= the
-1

(U )):

We nish this section with the following unexpected alternative: for a G-module V the quotient space V ==G is either an a ne space or a singular variety. Proposition 2.2.10. Let V be a nite-dimensional rational G-module. Then O(V )G is a polynomial algebra if and only if (0) is a smooth point on V ==G.

Proof. If O(V )G is a polynomial algebra, then V ==G = Spec(O(V )G ) is an a ne space, and any its point is smooth. Conversely, suppose that (0) V ==G is smooth. Let m O(V )G be the maximal ideal corresponding to (0). We know that T(0) V ==G (m=m2 ) . Hence, if n = dim V ==G, then dim m=m2 = n, and = there are homogeneous elements f1 ; : : : ; fn m whose images form a basis of m=m2 .

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Using induction on degree, one easily checks that f1 ; : : : ; fn generate the ideal m. By Lemma 2.1.4, the elements f1 ; : : : ; fn generate the algebra O(V )G . If there exists a non-zero polynomial F (X1 ; : : : ; Xn ) such that F (f1 ; : : : ; fn ) 0, then the transcendency degree of the eld of quotients QO(V )G is less then n, a contradiction with n = dim V ==G.

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that the algebra O(X )G is nitely generated. Give an example where the morphism : X Spec(O(X )G ) de ned by the embedding O(X )G O(X ) is not surjective. Exercise 2.2.12. Let : X Y be the quotient morphism from Example 2.2.4. Prove that the set of semisimple elements in any ber -1 (y) is the unique closed GL(n)-orbit Oy . Exercise 2.2.13. Let G be a nite group, X a G-variety and : X X==G the quotient morphism. Prove that is a nite morphism and any ber of is a G-orbit. Exercise 2.2.14. Let G be a reductive group and X an irreducible a ne G-variety. Show by an example that the components of a ber of the quotient morphism : X X==G may have di erent dimension. Exercise 2.2.15. Let : X X==G be the quotient morphism and U X==G be an open subset. Prove that |-1 (U ) : -1 (U ) U is the categorical quotient for the G-variety -1 (U ). Exercise 2.2.16. Prove that the categorical quotient is unique (up to isomorphism). Exercise 2.2.17. Let G be an algebraic group, X a G-variety and : X X==G the categorical quotient. Prove that the morphism is surjective. Exercise 2.2.18. Consider an action T m : Kn , t · (x1 ; : : : ; xn ) = (1 (t)x1 ; : : : ; n (t)xn ) with some 1 ; : : : ; n X(T m ). Prove that Kn ==T m is a point if and only if all i are non-zero and the cone generated by 1 ; : : : ; n in the space X(T m ) Z Q is strictly convex. Exercise 2.2.19 (Igusa's Criterion). Let G be a reductive group, X an irreducible a ne G-variety, and Y a normal irreducible a ne variety. Assume that there is a dominant G-invariant morphism : X Y such that codimY (Y \ (X )0 ) 2 and there exists open U Y such that -1 (y) contains a dense G-orbit for any y U . (Here (X )0 denotes the maximal open subset of the image (X ), see Theorem 3.0.24.) Prove that : X Y is the categorical quotient. Exercise 2.2.20. Set G = O(n) and X = Kn · · · Kn (s times, s n) with the diagonal G-action. De ne Fij (v1 ; : : : ; vs ) := q(vi ; vj ). Prove that O(X )G is a polynomial algebra generated by Fij . Formulate and prove the corresponding statement for G = Sp(2n). Exercise 2.2.21. Let G be an algebraic group and H G a reductive subgroup. Prove that the homogeneous space G=H is a ne. Exercise 2.2.22. Let G be an algebraic group, X a G-variety, and : X Z a Ginvariant morphism. We say that two points x; x X are equivalent if there is a sequence of points on X : x = x1 ; x2 ; : : : ; xk = x such that Gxi Gxi+1 = for i = 1; : : : ; k - 1. Prove that (x) = (x ). Exercise 2.2.23 (*). Consider an action T m : Kn+1 , t · (x1 ; : : : ; xn+1 ) = (1 (t)x1 ; : : : ; n+1 (t)xn+1 ) with some 1 ; : : : ; n+1 X(T m ), 1 = 2 . Prove that the corresponding action T m : Pn admits the categorical quotient with Pn ==T m being a point.

Exercises to subsection 2.2. Exercise 2.2.11. Let G be an algebraic group and X an a ne G variety. Suppose

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2.3. Rational invariants. Rosenlicht's Theorem. Let G be an algebraic group and X an irreducible G-variety. There is a natural G-action on the eld of rational functions K(X ): (g · f )(x) = f (g-1 · x). Consider the eld of rational invariants: K(X )G := {f K(X ) : g · f = f for all g G}: In this situation, there is no problem with nite generation. Proposition 2.3.1. Let K L be a nitely generated eld extension. Then for any K E L, the extension K E is nitely generated.

Proof. Let Y1 ; : : : ; Yr be a transcendency basis in E . It may be extended to a transcendency basis Y1 ; : : : ; Yr ; Yr+1 ; : : : ; Yk in L. Then K(Y1 ; : : : ; Yk ) L is a nite algebraic extension of some degree N . Any element E is algebraic over K(Y1 ; : : : ; Yr ). Let F (T ) (resp. H (T )) be the minimal polynomial of over K(Y1 ; : : : ; Yr ) (resp. K(Y1 ; : : : ; Yk )). Then H (T ) divides F (T ), thus H (T ) does not depend on Yr+1 ; : : : ; Yk , and F (T ) H (T ). In particular, the degree of over K(Y1 ; : : : ; Yr ) is bounded by N . Applying the Primitive Element Theorem, we get that the degree of E over K(Y1 ; : : : ; Yr ) is bounded by N . Corollary 2.3.2. For any algebraic group G and any irreducible G-variety X , the eld K(X )G is nitely generated over K.
The quotient of two regular invariants is a rational invariant, thus QO(X )G K(X )G . In general, we do not have an equality here. Example 2.3.3. Consider an action Kв : K2 , t · (x1 ; x2 ) = (tx1 ; tx2 ). In this case x1 O(X )G = K, but the function x2 is a non-constant rational invariant. Proposition 2.3.4. Assume that X is an irreducible a ne G-variety and one of the fol lowing conditions holds: (a) G is nite; (b) G is unipotent; (c) G is connected, X(G) = 0, O(X ) is factorial, and O(X )в = Kв . Then QO(X )G = K(X )G .

Proof. (a) If F

K

(X )G , then f1 ( f F= 1= f2

g = e g · f2 ) g G g · f2

QO(X )G :
2

(b) Take F = f1 K(X )G . By f2 contains a non-zero G-invariant h f g ·f F= 1= i 1= i f2 gi · f2 i (c) Assume that

Theorem 1.8.5, the linear span of the orbit Gf = i i (gi · f2 ), i Kв . Then (g · f ) (gi · f1 ) = i i i 1 QO(X )G : (gi · f2 ) i i (gi · f2 )

pa1 : : : pak G k F = 11 b : : : q bs K(X ) ; q1 s where p1 ; : : : ; pk ; q1 ; : : : ; qs are pairwise di erent primes. The G-action preserves this decomposition. On the other hand, G can not permute the factors (G is connected) and can not send a factor to an associated element (O(X )в = Kв and X(G) = 0). Hence all p1 ; : : : ; pk ; q1 ; : : : ; qs are G-invariants.

Now we came to the "best possible" notion of a quotient for a G-variety.

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De nition 2.3.5. Let G be an algebraic group and X an irreducible G-variety. A G-invariant morphism : X Y is said to be a geometric quotient, if the following conditions hold:
(G1) (G2) (G3) (G4)

is surjective; (y) is a G-orbit for any y Y ; is open; -1 G U : O(U ) O( (U )) is an isomorphism for any open U
-1

Y.

Notation: : X X=G. By Proposition 1.3.8 and Theorem 3.0.26, if there is a G-invariant morphism : X Y satisfying condition (G2), then all orbits in X are closed and have the same dimension. Unfortunately, the later conditions are not su cient for the existence of geometric quotient (Exercise 2.3.23). Proposition 2.3.6. Let G be an algebraic group and X an irreducible G-variety. If there is a surjective G-invariant morphism : X Y , where Y is a normal G-variety and -1 (y) is a G-orbit for any y Y , then is a geometric quotient.

Proof. Conditions (G1) and (G2) are included into assumptions. Condition (G3) follows from Theorem 3.0.32. Finally, (G4) follows from Corollary 3.0.29. Example 2.3.7. Let G = Kв and X = Kn \{0} with t·(x1 ; : : : ; xn ) = (tx1 ; : : : ; txn ). By Proposition 2.3.6, the morphism : X Pn-1 , ((x1 ; : : : ; xn )) = [x1 : ::: : xn ] is a geometric quotient. In particular, for reductive G and quasia ne X the variety X=G need not be quasia ne. Proposition 2.3.8. A geometric quotient : X X=G is the categorical one. Proof. Let : X

Z be a G-invariant morphism: XA
~~~ ~~ ~~~ o

De the one and

ne (y) = subset -1 has (f ) =

Z Y: ( (y)). The map is continuous, because for any open U Z (U ) = (-1 (U )) is open in Y (use (G3)). Finally, for any f O(U ) O(-1 (U ))G = (O( -1 (U ))) (use (G4)). Since is injective , the function (f ) is contained in O( -1 (U )).
-1

AA AA AA

Corollary 2.3.9. If a geometric quotient : X
(up to isomorphism).

X=G exists, then it is unique

We come to the main result of this section. Theorem 2.3.10 (Rosenlicht's Theorem (1956)). Let G be an algebraic group and X an irreducible G-variety. Then there is a non-empty open G-invariant subset U X which admits a geometric quotient : U U =G.

Proof. By Proposition 1.3.8, there is an open G-invariant subset U X such that all G-orbits on U are of the same dimension. Fix a generating set f1 ; : : : ; fm of K(X )G . Reducing U , one may assume that f1 ; : : : ; fm are regular on U . Moreover, f1 ; : : : ; fm are regular on gG gU , so U may be supposed to be invariant. Consider a subalgebra K[f1 ; : : : ; fm ] K(X )G and set Y := Spec(K[f1 ; : : : ; fm ]), The inclusion K[f1 ; : : : ; fm ] O(U ) de nes a dominant morphism : U Y .

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Reducing Y to a principal open subset, one may assume that Y is normal and is surjective. Consider a morphism : G в U U в U; (g; x) = (x; g · x) and two sets A := Im() = {(x; x ) U в U : x Gx}; B := U вY U = {(x; x ) U в U : (x) = (x )}: Since is G-invariant, we get A B . Lemma 2.3.11. A is dense in B .

Proof. Let V and V be open a ne subsets of U . It is su cient to prove that : W := -1 (V вY V ) V вY V ; (g; x) (x; g · x) is dominant. Since U is irreducible, the set V в V meets the diagonal U , and W is non-empty. The variety V вY V is a ne, and we shall check that : O(V вY V ) = O(V ) O(Y ) O(V ) O(W );

(

s i=1

ui vi ) (g; x)
s i=1 ui vi lies s hg (x) = ui i=1

s i=1

ui (x)vi (g · x)

is injective. Suppose that

in Ker( ). For any g (x)vi (g · x)

G the function

is a rational function on U . Since it vanishes on V g-1 V , it is identically zero. We claim that this condition implies that F = s=1 ui vi is a zero (rational) function i on U вY U . Firstly, one may assume that v1 ; : : : ; vs are linearly independent over K(U )G = (K(Y )). Indeed, if, say, v1 = c2 v2 + · · · + cs vs , ci K(U )G , then F coincides with s=2 (ui + ci u1 ) vi on U вY U . Now the statement follows from i Lemma 2.3.12. Let ui , vi , i = 1; : : : ; s be rational functions on U such that v1 ; : : : ; vs are linearly independent over K(U )G . If s=1 ui (g · vi ) = 0 for al l g G, i then u1 = · · · = us = 0.

Proof. We argue by induction on s. The case s = 1 is obvious. Suppose that s > 1. If u1 = 0, then
s

for all g; h G. Thus

i=1 s i=2

h · (ui u-1 )(g · vi ) = 0 1

(h · (ui u-1 ) - ui u-1 )(g · vi ) = 0: 1 1
1 K(

By inductive hypothesis, h · (ui u-1 ) = ui u-1 for all h G, so ui u- 1 1 1 linear dependence v1 + u2 u-1 v2 + · + us u-1 vs = 0 1 1 leads to a contradiction. Lemma 2.3.11 is proved.

U )G . The

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Further, the variety B = U вY U contains a dense irreducible subset A = Im(), thus B is irreducible. Lemma 2.3.13. Let B be an irreducible variety, W B a non-empty open subset and : B Z a dominant morphism. Then there is a non-empty open subset V Z such that -1 (z ) W is dense in -1 (z ) for any z V .

Proof. Let C1 ; : : : ; Ck be irreducible components of B \ W . Renumbering, one may assume that (C1 ); : : : ; (Cr ) are dense in Z , and (Cr+1 ); : : : ; (Ck ) are not. Set
V := Z \
k i=r+1

(Ci ):

Reducing V , we may assume that (i) the dimension of any component of any ber of the morphism : -1 (V ) V equals dim B - dim Z ; (ii) for any i = 1; : : : ; r the dimension of any component of any ber of the morphism |Ci : Ci V equals dim Ci - dim Z (Theorem 3.0.26). Since dim Ci < dim B , now any component of a ber -1 (z ) is not contained in C1 · · · Cr , and hence meets the open subset W . We are going to apply Lemma 2.3.13 to W = A0 , where A0 is the maximal open subset of A = Im(), and to the pro jection : B U , (x; x ) = x. Reducing Y , we may suppose that any ber of has a dense intersection with A. But -1 (x) = {x} в -1 ((x)); A -1 (x) = {x} в Gx; and all G-orbits in U are closed. This proves that for any y Y the ber -1 (y) is a G-orbit. Now Proposition 2.3.6 shows that : U Y is a geometric quotient. Corollary 2.3.14 (of the proof ). In notation of Theorem 2.3.10, for any generating set f1 ; : : : ; fm of K(X )G there is a non-empty open G-invariant subset U X such that al l fi are regular on U and for any two points x; x U the condition Gx = Gx is equivalent to fi (x) = fi (x ), i = 1; : : : ; m. Corollary 2.3.15. An action G : X has an open orbit if and only if K(X )G = K. Proposition 2.3.16. Assume that there is a non-empty open G-invariant subset W X and rational invariants f1 ; : : : ; fk that are regular on W , such that for any x1 ; x2 W the condition fi (x1 ) = fj (x2 ) for al l i; j = 1; : : : ; k implies Gx1 = Gx2 . Then f1 ; : : : ; fk generate K(X )G .

Proof. Extend f1 ; : : : ; fk to a generating set f1 ; : : : ; fk ; fk+1 ; : : : ; fm of K(X )G . By assumptions, there is a dominant morphism : W Spec(K[f1 ; : : : ; fk ]) whose bers are G-orbits. Reducing W , we may assume that there is a dominant morphism : W Spec(K[f1 ; : : : ; fm ]) with the same bers. By Theorem 3.0.28, the varieties Spec(K[f1 ; : : : ; fk ]) and Spec(K[f1 ; : : : ; fm ]) are birationally isomorphic.

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with QO(X )SL(2) = K(X )SL(2) . Exercise 2.3.18. Let G be a connected algebraic group and H G a closed subgroup. Check that the pro jection p : G G=H is a geometric quotient with respect to the right H -action on G. Exercise 2.3.19. For a reductive G and an irreducible a ne X , the quotient morphism : X X==G is a geometric quotient if and only if all G-orbits in X are of the same dimension. Exercise 2.3.20. For a nite G and an irreducible a ne X , the quotient morphism : X X==G is a geometric quotient. Exercise 2.3.21. Let G be reductive and X be irreducible a ne G-variety. Prove that the following conditions are equivalent: (i) QO(X )G = K(X )G ; (ii) there is a non-empty open V X==G such that for any y V the ber -1 (y) of the quotient morphism contains a dense G-orbit. Exercise 2.3.22. Let : X X=G be a geometric quotient. Prove that K(X )G = (K(X=G)). Is the same true for a categorical quotient ? Exercise 2.3.23 (*). Consider an action Kв : K2 \ {0}, t · (x1 ; x2 ) = (tx1 ; t-1 x2 ). Prove that here all orbits are closed and of dimension one, but the action does not admit geometric quotient.

Exercises to subsection 2.3. Exercise 2.3.17. Give an example of an SL(2)-action on an irreducible a ne variety X

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3. Appendix One: Some Facts from Algebraic Geometry

Theorem 3.0.24. Let : X

Y be a dominant morphism of irreducible algebraic varieties. Then (X ) contains an open subset of Y .

Theorem 3.0.25. Let X be an irreducible variety and Y
such that dim X = dim Y . Then X = Y .

X a closed subvariety

Theorem 3.0.26. Let : X

Y be a morphism of irreducible algebraic varieties. Then there is a non-empty open subset U Y such that for any y0 U (X ) any irreducible component of the ber -1 (y0 ) has dimension dim X - dim Y . Moreover, any irreducible component of any non-empty ber -1 (y), y Y , has dimension dim X - dim Y .

Theorem 3.0.27. Assume that char K = 0 and : X

Y is a bijective morphism between normal (e.g., smooth) varieties. Then is an isomorphism.

X , Y , Z are irreducible varieties with given dominant morphisms : X Y , : X Z such that there is a non-empty open subset U X with the fol lowing property: (x) = (x ) implies (x) = (x ) for any x; x U . Then there exists a rational (dominant) morphism : Y Z such that the diagram X@ @@ ~~~ @@ ~ @@ ~ ~~~ /

Theorem 3.0.28 (Factorization of a morphism). Assume that char K = 0, and

Y

Z
A1

is commutative.

Corollary 3.0.29. In notations of Theorem 3.0.28, assume that Z =
normal and codimY (Y (X )):
\

(X

)0 )

, Y is 2 (here (X ) is the maximal open subset in
0

If

Y : O(X ), then there exists O(Y ) with ( ) = .

X BB BB ~~~~ BB ~ BB ~~ ! ~ / A1

Theorem 3.0.30. [Hu75, Th. 4.5] Let : X Y be a dominant morphism of irreducible varieties, r = dim X - dim Y . Assume that for each closed irreducible subset Z Y al l components of -1 (Z ) have dimension r + dim Z . Then the morphism is open. Theorem 3.0.31. [Hu75, Th. 4.3] Let : X Y be a dominant morphism of irreducible varieties, r = dim X - dim Y . Then Y has a non-empty open subset U such that:
(i) U (X ); (ii) if W Y is an irreducible closed set which meets U , and if Z is a component of -1 (W ) which meets -1 (U ), then dim Z = dim W + r.

Theorem 3.0.32. (cf.[PV94, p. 187]) Let : X

Y be a dominant morphism of irreducible varieties such that the dimension of any component of any ber equals dim X - dim Y . Assume that Y is normal. Then is an open morphism.

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Theorem 3.0.33. Let : X Y be a dominant morphism between irreducible varieties. Then, for any x X , Tx (-1 )(x) Ker dx ; and there is a non-empty open subset U X such that the equality takes place for any x U .

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4. Appendix Two: Some Facts on Lie Algebras In this Appendix all Lie algebras are supposed to be nite-dimensional. We refer to [Hu72] for a detailed exposition of the sub ject. De nition 4.0.34. A Lie algebra g is called simple if it is not commutative and has no proper ideals. De nition 4.0.35. The commutant [g; g] of a Lie algebra g is the linear span of [x; y], x; y g. Clearly, [g; g] is an ideal of g. This de nition leads to the notion of a solvable Lie algebra. Remark 4.0.36. If g is simple, then [g; g] = g. De nition 4.0.37. The radical r(g) of a Lie algebra g is the largest solvable ideal of g. De nition 4.0.38. A Lie algebra g is called semisimple if r(g) = 0. Theorem 4.0.39. A Lie algebra g is semisimple if and only if it is isomorphic to a direct sum of simple Lie algebras: g = g1 · · · gk : Corollary 4.0.40. If g is semisimple, then [g; g] = g. Theorem 4.0.41 (G. Weyl). Any nite-dimensional representation of a semisimple Lie algebra is completely reducible.

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1.1.10. aij = eT Ae i 1.1.11. Let e1 ; : : : ; ei , i > 1. 1.1.12. S O(2) = 1.1.13.
A

j

= eT B ej = bij : i en be an orthogonal basis. Set A1 = E and A2 (e1 ) = -e1 , A2 (ei ) =

5. Hints and Solutions to Exercises

1.1.14.

1.1.15. 1.1.16. 1.1.17. 1.1.18.

1.1.19. 1.1.20. 1.1.21. 1.1.22.

a -b : a2 + b2 = 1 : For O(2), take -1 0 and 0 1 : ba 01 10 T q(v1 ; v2 ) = v1 Qv2 , where Q is symmetric non-degenerate, so there is a nondegenerate S with S T QS = E , or Q = (S -1 )T S -1 . Then O(q ) AT QA = Q AT (S -1 )T S -1 A = (S -1 )T S -1 S T AT (S -1 )T S -1 AS = E S -1 AS O(n): Similarly, any bilinear skew-symmetric non-degenerate form is equivalent to the standard one. (char K = 0) A Sp(2n) if and only if A preserves ! 2 V . Then A preserves n ! 2n V = det . Since ! is non-degenerate, n ! is a non-zero multiple of det. This proves that det(Av1 ; : : : ; Av2n ) = det(v1 ; : : : ; v2n ) for any v1 ; : : : ; v2n V , so det A = 1. In arbitrary characteristic, one may prove that Sp(2n) is generated by symplectic transections: u;a : V V ; u;a (v) = v + a!(v; u)v; u V ; a K; see (L.C. Grove, Classical Groups and Geometric Algebras, Grad. Studies in Math. 39, AMS, 2002) for details. For K2 , det coincides with the standard bilinear skew-symmetric non-degenerate form. For n > 1, E + E13 SL(2n) \ Sp(2n). The intersection of GL(n; R) with the subvariety of scalar matrices is not closed in Zariski topology. For n > 1 SL(n) = Mat(n в n), because E ; E + Eij (i = j ), and E - Eii + Eij - Ej i belong to SL(n). If A commutes with all matrices of a linear group G, then A commutes with all matrices of the linear span of G. But if A commutes with all Eij , then it is scalar, so Z (GL(n)) = {E : Kв }, Z (SL(n)) = {E : n = 1}. Similarly, Z (B (n)) = {E : Kв } and Z (U (n)) = {E + aE1n : a K}. dim D(n) = n, dim B (n) = n(n2+1) , dim U (n) = n(n2-1) . There are in nitely many elements Kв with N = 1 for some N . If char K = 0, then 0 is the only element of nite order in Ga . If char K = p > 0, then any non-zero element of Ga has order p. 0 No, if n > 1. For n = 2 consider the re ections 1 -1 and 1 -2 : 0 -1 0

1.1.23. Let e1 ; : : : invertible (ei ej ) where ck ij

diag(t1 ; : : : ; tk ; 1 a1 ; : : : ; 1 as ): 01 01 ; en be a basis of A and be the (bilinear) multiplication. Clearly, an linear map : A A is an isomorphism if and only if (ei ) (ej ) = for any i; j = 1; : : : ; n. Set (ei ) = l ali el and ei ej = k ck ek , ij are structural constants of the algebra A. Then ( ali el ) ( asj es ) = ck ( ark er ) ij
l s k r

is a system on polynomial equations on aij . 1.1.24. (char K = 0) Consider : G в G G в G, (g1 ; g2 ) = (g1 ; g1 g2 ). This morphism is bijective, thus it is an isomorphism, and -1 (h1 ; h2 ) = (h1 ; h-1 h2 ). Then 1 i(g) = p2 ( -1 (g; e)), where p2 : G в G G, p2 (g1 ; g2 ) = g2 . 1.1.25. I have only topological arguments: SL(2; C) is homotopy-equivalent to the real sphere S 2 .

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a b 2 = E. b -a 1.2.18. If H G0 is a proper (closed) subgroup, then dim H = dim H 0 = dim H G0 < dim G0 = dim G: 1.2.19. G = O(2). t0 0 1.2.20. G = { 0 t-1 s-1 -s : t; s Kв } : there are no elements of order 0 two in the second component. 1.2.21. One may assume that K is algebraically closed. Corollary 1.2.7 implies that the hypersurface de ned by det(aij )-1 is irreducible. In order to see that det(aij )-1 is not a proper power, consider the leading term of det(aij ) - 1 with respect to a monomial order. 1.2.22. Clearly, [GL(n); GL(n)] SL(n) and [B (n); B (n)] U (n). On the other hand, (E +Eij )(E +(t-1)Eii +(t-1 -1)Ej j )(E -Eij )(E +(t-1 -1)Eii +(t-1)Ejj ) = E +(1-t2 )Eij : Thus [GL(n); GL(n)] = [SL(n); SL(n)] = SL(n) and [B (n); B (n)] = U (n). If j > i + 1, then (E + Eii+1 )(E + Ei+1j )(E - Eii+1 )(E - Ei+1j ) = E + Eij : This implies that [U (n); U (n)] is de ned in U (n) by a12 = a23 = · · · = an-1n = 0. 1.2.23. Let g G \ Z (G). The map g : G G, g (h) = ghg-1 h-1 has an irreducible image which is not equal to {e}. Thus g (G) is not contained in Z (G), and gZ (G) Z (G=Z (G)). = In the disconnected case the statement is not true: one may take a nite group G, say, a non-commutative group of order 8. 1.2.24. Consider the subgroup H generated by xyx-1 y-1 ; x G0 ; y G. It is generated by subsets y (G0 ), where y (x) = xyx-1 y-1 , thus is closed and connected (Proposition 1.2.6). So it is su cient to check that H has a nite index in [G; G]. - - - Since g1 [g; h]g1 1 = [g1 gg1 1 ; g1 hg1 1 ] for any g; g1 G, h G0 , the subgroup H is normal in G. Further, [gH; hH ] = [g; h]H = H , and G0 =H is contained in the center Z (G=H ). This shows that Z (G=H ) has a nite index in G=H , and the group [G=H; G=H ] is nite. But [G; G]=H [G=H; G=H ], and H has a nite index in [G; G]. 1.2.25. Let G1 be ( ; +) and G2 a compact torus (R2 ; +)=Z2 . Consider a homomorphism R (a) = (a; 2a) + Z2 . Check that the intersection of Im() with the circle x1 Z in G2 is an in nite countable set.

1.2.15. (Ga )3 and U (3). 1.2.16. Denote X = A1 \ {0; 1}. It is su cient to prove that the automorphism group 11 Aut(X ) of the variety X does not act transitively on X. Since O(X ) = K[T ; T ; T -1 ], any automorphism Aut(X ) de nes an automorphism of the eld K(T ). But the automorphism group of K(T ) is isomorphic to PGL(2), or Aut(P1 ). Our should preserve {0; 1; } P1 . But any automorphism of P1 that xes 0, 1 and is the identity, so Aut(X ) is nite. (In fact, Aut(X ) 3 .) = a -b a + bi, i2 = -1. For the second component, 1.2.17. SO(2) Gm , b a

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73

1.3.24. Note that {B Mat(n в n) : AB = B A} is a subspace, and its intersection with an open subset GL(n) is irreducible. For SL(2), take A = 1 1 : 01 1.3.25. For any x A, g NG (A) if and only if f (gxg-1 ) = 0 for all f I(A). 1.3.26. If V is a rational G-module, then GвV V , (g; v) (g)v is a morphism. Conversely, suppose that G в V V is a morphism. By assumption, (g; v) (g)v, where (g) is a linear operator on V . Fixing a basis {ei } in V and restricting the action to G в {ei } V , we get that (g) depends on g algebraically. 1.3.27. (a) n > 1: {0} and Kn \ {0} (any non-zero vector may be included into a unimodular basis); (b) 2n orbits ( x zero coordinates); (c) n + 1 orbits ( x a place of the last non-zero coordinate); (d) (x1 ; : : : ; xn ) and (y1 ; : : : ; yn ) are in the same orbits k: xk = · · · = xn = yk = · · · = yn = 0, xk-1 = yk-1 = 0 (set formally xn+1 = yn+1 = 0, x0 = y0 = 1); (e) and (f ) n > 2: non-zero vectors v1 and v2 are in the same orbit q(v1 ; v1 ) = q(v2 ; v2 ). If q(v; v) = 0, then v may be included in an orthogonal basis. If q(v; v) = 0 then there is e Kn with q(e; e) = 1, q(e; v) = 1, so q(e; v - e) = 0, q(v - e; v - e) = -1, and we may include e and v - e in an orthogonal basis. For n = 2, the conditions q(v; v) = 0; v = 0 de ne two SO(2)-orbits that are permuted by O(2). (g) {0} and K2n \ {0} (any non-zero vector may be included into a symplectic basis). n 1.3.28. dim Sp(2n) = (2n)2 - 2n(22 -1) = 2n2 + n. Kв , Sp(2) = SL(2) are connected. Further, : SO(n) 1.3.29. We know that SO(2) = S n-1 , where S n-1 = {(x1 ; : : : ; xn ) : x2 + · · · + x2 = 1} is irreducible, and 1 n (g) = g · e1 . Any ber of is isomorphic to SO(n - 1), thus is connected. This shows that SO(n) is connected. Similarly, : Sp(2n) K2n \ {0}, (g) = g · e1 , and any ber of is isomorphic to Sp(2n - 2) в K2n-1 . Remark. If an orbit Gx and the stabilizer Gx are connected, then G is connected. Indeed, the stabilizer Gx is contained in G0 and di erent connected components of G correspond to di erent G0 -orbits in Gx. Such orbits are closed and have empty intersections, a contradiction with connectedness of Gx. 1.3.30. In all cases it is su cient to construct a homomorphism with kernel of order two (use Theorem 1.2.12 and compare dimensions). (a) SL(2) : {M Mat(2 в 2) : tr(M ) = 0}, A · M := AM A-1 , the action preserves det; (b) SL(2) в SL(2) : Mat(2 в 2), (A; B ) · M := AM B -1 , the action preserves det; (d) SL(4) : 2 K4 , the action preserves q(w1 ; w2 ) = w1 w2 ; (c) Sp(4) : V , where V is a 5dimensional complement to an invariant line in 2 K4 . 1.3.31. It is su cient to prove that the hypersurface det(aij ) = 0 is irreducible. But this hypersurface contains a dense GL(n) в GL(n)-orbit, where GL(n) в GL(n) : Mat(n в n), (A; B ) · M = AM B -1 . 1.3.32. Note that F (V ) is a closed G-invariant subvariety in
в P(2 в · · · в P(n-1

P

(V )

V)

V );

where n = dim V . 1.3.33. Take a line with a double point and the Z2 -action permuting the glueing lines. 1.3.34. Ga : Kn , a · v = v + a(1; 1; : : : ; 1) { no xed points. 1.3.35. The linear span of any orbit is nite-dimensional and any representation of a nite group is algebraic. 1.3.36. Take G = Gm , X = K1 . Prove that dim tx1 : t Gm = . +1 1.3.37. Suppose A O(Y ) is a nitely generated subalgebra de ning an embedding of Y as a dense open subvariety of an a ne variety X := Spec(A ). By Theorem 1.3.16, a generating set of A is contained in a nite-dimensional Ginvariant subspace. Hence the subalgebra A is contained in a nitely generated G-invariant subalgebra A O(Y ). Put X := Spec(A). There is a commutative

74

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diagram of dominant morphisms:
~~ ~~~ ~ J ~~ ~

YB B

X

One may nd f1 ; : : : ; fk A such that Y = i Xfi . Since A A O(Y ) O(Xfi ), all morphisms in the diagram become isomorphisms after localization at fi . Thus J (Y ) = i Xfi is open in X .

BB BB BB ! /X:

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75

1.4.9. It is su cient to check that the subspace U = u1 ; : : : ; uk W is uniquely de ned by ! := u1 · · · uk . We claim that U = {w W : ! w = 0}. Indeed, w U if and only if u1 ; : : : ; uk ; w are linearly dependent. 1.4.10. : GL(n) GL(Mat(n в n)), A · M := AM A-1 . 1.4.11. The group G=G0 acts transitively on the set of G0 -orbits in G=H , so these orbits are closed. Hence they are irreducible components with empty intersections. It implies that G=H is connected if such an orbit is unique, or G0 eH = G. 1.4.12. If G=H is quasia ne, use Exercise 1.3.37 and Theorem 1.3.19. Conversely, the orbit Gv is open in the a ne variety Gv. 1.4.13. Follows from Theorem 1.3.19. 1.4.14. No, consider Gm : K1 , or SL(2) : S3 K2 , v = e2 1 e2 . a t1 0 b : Here G=H P2 в P2 \ . 1.4.15. Take G = SL(3) and H = 0 t2 = 0 0 t-1 t-1 12 1.4.16. SL(3)=U (3) X \ (Y Z ), where X; Y ; Z K6 de ned by x1 y1 + x2 y2 + x3 y3 = 0 = x1 = x2 = x3 = 0 and y1 = y2 = y3 = 0 respectively : consider V = K3 (K3 ) , v = (e1 ; e ). 3 1.4.17. Use Theorem 3.0.28 with X = G. 1.4.18. Let G be a connected group and H a non-trivial nite subgroup. If there is an open U G=H such that p-1 (U ) U в H , then G contains an open reducible = subset.

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1.5.22.

@ det @ aij

|A=E

=

n ;(i)=j

sgn() a

1(1)

: : : ai(i) : : : an(n) |A=E =

ij

;

1.5.23.

and dE (det)(X ) =

i;j ij xij

= trX .

1.5.24.

1.5.25. 1.5.26. 1.5.27.

@ Aj i @ det A A @ detji @ akm det A - Aj i @ akm A| |A=E = A=E = @ ak m (det A)2 = - ki mj (1 - ij ) + ij km (1 - ik ) - ij km = - ik jm : In order to prove that [x; y] = 0, consider the one-dimensional subalgebras x and y . One may be interesting in a group-theoretical analog of this statement: A group G is commutative if and only if any its subgroup is normal. This statements is wrong: take G = Q8 = {±1; ±i; ±j; ±k}, where ghg-1 = ±h for any g; h G. It corresponds to a subspace U in Mat(n в n) consisting of pairwise commuting matrices together with a surjective linear map g U . Take G1 = Ga , G2 = Gm . They are not isomorphic, because Gm contains in nitely many elements of nite order, but Ga { only one. sp(2n) = {X Mat(2n в 2n) : X T + X = 0}, where is the skew-symmetric matrix, corresponding to the standard form (follow Example 1.5.8);
0 : : : 0

Lie(D(n)) =

::: 0

0

: : : :

: : : :

: : : :

0 0 0

; Lie(B (n)) =

0 : : : 0

::: 0

: : : :

: : : :

: : : :

; Lie(U (n)) =

:

0 0 :: 0

0 ::: 0

: : : :

: : : :

: : : :

0

(D(n) and B (n) are open in subspaces of Mat(n в n), and U (n) is a shift of a subspace). 1.5.28. Follows from Lemma 1.5.12 and Exercise 1.3.30. For (a), one may get an isomorphism directly. Set e = 0 1 , h = 00 1 0 , f = 0 0 . Then [h; e] = 2e, [h; f ] = -2f , [e; f ] = h. On the 10 0 -1 other hand, in so(3) we have x12 = E12 - E21 , x23 = E23 - E32 , x13 = E13 - E31 with [x12 ; x23 ] = x13 , [x23 ; x13 ] = x12 , [x13 ; x12 ] = x23 . Calculating the eigenvalues of ad(x12 ), we get an isomorphism e ix23 - x13 , h 2ix12 , f ix23 + x13 . 1.5.29. Any G-invariant subspace is G0 -invariant. For converse, take a nite group. 1.5.30. If H G, then H 0 G. For converse, take a nite group. 1.5.31. We may assume that G GL(n) is a closed subgroup. There is a linear action G : Mat(n в n), h · A = hAh-1 , which is the restriction of the adjoint representation for GL(n). Thus the tangent action is x · A = [x; A] for all x g. By Proposition 1.5.16, Lie(ZG (g)) = gg = {x g : [x; g] = 0} = {x g : gxg-1 = x} = {x g : Ad(g)x = x}: 1.5.32. If g is commutative, then Z (x) = G for any x g, because Lie(Z (x)) = z(x) = g (Proposition 1.5.21). This means that for any g G Lie(ZG (g)) = g (Exercise 1.5.31), and ZG (g) = G. 1.5.33. By de nition, Z (G) = gG ZG (g). Using the Noether property, one may assume that this is an intersection of nitely many subgroups, and Lemma 1.5.13 implies Lie(Z (G)) = {x g : Ad(g)x = x for all g g}: If G is connected, then, by Proposition 1.5.16, Lie(Z (G)) = {x g : ad(y)x = 0 for all y g} = z(g): For non-connected G, consider G = O(2). d 1.5.34(ii). Assume that D Lie(Aut(A)). Then D = dt |t=0 (t), where (t) is a smooth curve in Aut(A) with (0) = id. Fix a basis e1 ; : : : ; en in A. Let ( ij (t)) be d the matrix representing (t), ij (0) = ij . Then dt |t=0 ij (t) = dij , where

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77

D = (dij ). It is su cient to check that So, d d D(ei ej ) = |t=0 (t)(ei ej ) = dt dt d = |t=0 ( ki (t)ek )( lj (t)el ) = dt k l

D(ei ej ) = D(ei )ej + ei D(ej ) for any i; j .
|t=0

( (t)(ei ))( (t)(ej )) =
k;l ki

d | ( dt t=0

(t) lj (t)ek el ) =

=

Conversely, suppose that D Der(A) (K = C). Consider the (non-algebraic) curve t2 D 2 (t) := exp(tD) = E + tD + + ::: 2! d Clearly, dt |t=0 (t) = D. We have to show that exp(D) Aut(A), or D2 D2 D2 (E + D + + : : : )(ab) = (E + D + + : : : )(a)(E + D + + : : : )(b): 2! 2! 2! It is easy to prove by induction the "higher Leibniz rule": n nk Dn (ab) = D (a)Dn-k (b): k k=0 Then the desired equality follows from
11 k! (n-k)!

k

dki ek ej +

l

dlj ei el = D(ei )ej + ei D(ej ):

=

n1 k n!

:

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IVAN V. ARZHANTSEV

1.6.21. Consider F (Y1 ; : : : ; Yn ) as polynomial in Y2 ; : : : ; Yn . Then all its coe cients are constants. 1.6.22. Note that X(G) = X(G=[G; G]). This implies X(GL(n)) = X(T 1 ) Z; X(SL(n)) = = 0; X(B (n)) = X(T n ) Zn (Exercise 1.2.22). Moreover, [O(2); O(2)] = SO(2) = and X(O(2)) Z2 . For a quasitorus Q T m в A, we have X(Q) Zm в A. = = = Finally, X(G) G=[G; G] for a nite G. = 1.6.23. Hom(T m ; T n ) = X(T m )n . 1.6.24. Recall that A is semisimple if and only if its minimal polynomial A (T ) has no multiple roots. Note that A|U (T ) divides A (T ). 1.6.25. A (T ) divides T n - 1. 1.6.26. Restrict all Ai to eigenspaces Vj of A1 and use induction. 11 2 1.6.27. Consider 2 1 01 0 1. 1.6.28. The set U of matrices with pairwise di erent eigenvalues is de ned as Disc(PA (T )) = 0, where PA (T ) is the characteristic polynomial of A, and thus U is open. On the other hand, E + tE11 is semisimple only if t = 0. 1.6.29. The nite group G=G0 is commutative, thus G=G0 Zn1 · · · Zns . Let ti be = a representative of the component, corresponding to a xed generator ai Zni . Since tni G0 and the group G0 is divisible (i.e., for any t G0 and N N i there is h G0 with hN = t), one may assume that tni = e. Let F be the i subgroup of G generated by ti . Check that G G0 в F . = 1.6.30. Theorem 1.6.15 implies that (Q0 ) is a torus. Use Exercise 1.6.29. 1.6.31. Use Theorem 1.6.15. 1.6.32. Let Ax = 0 be a system of linear equations in Km . Assume that F K is a sub eld and A Mat(m в m; F ). Since rkK A = rkF A, KerK (A) and KerF (A) have the same dimension over K and F respectively. This proves that there is a basis of KerK (A) with elements in F n . 1.6.33. Let 1 ; : : : ; m be the standard basis of X(T m ): i ((t1 ; : : : ; tm )) = ti . Then di ((c1 ; : : : ; cm )) = ci , where (c1 ; : : : ; cm ) t and t(Z) = {(c1 ; : : : ; cm ) : ci Z}. a1 On the other hand, if ((t1 ; : : : ; tm )) = (ta11 : : : tm m ; : : : ; tam1 : : : tamm ), then d m 1 1 acts in t via matrix A = (aij ).

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79

1.7.26. Note that Lie(G) = x , where x is either semisimple or nilpotent, and G = G(x). 1.7.27. For semisimple elements, see Lemma 1.6.12; for nilpotents, use the binomial formula; reduce the unipotent case to the nilpotent one. 1.7.28. Yes, it contains a dense GL(n)-orbit. 1.7.29. One may reduce the questions to the case where A is nilpotent. The operator Au restricted to any proper space of As de nes a non-degenerate operator. Thus As = 0, and A = 0. In the last case a decomposition is not unique. 1.7.30. By the Chinese remainder theorem there exists f (x) K[x] with f (x) i (mod(x - i )ki ); f (x) 0 (modx): 1.7.31. See Lemma 1.8.11. 1.7.32. Any G(A) is a direct product of the quasitorus G(As ) and the unipotent subgroup G(Au ). Elements of G(As ) are diagonalizable simultaneously (Lemma 1.6.12), so dim G(As ) n. Moreover, dim G(Au ) 1. But all elements of G(As ) commute with Au , and if Au = E , then G(As ) is a proper subgroup of D(n). This proves that dim G(A) n, and the value n is attained by Exercise 1.7.31. 1.7.33. If x Lie(G), then G(x) G0 and thus exp(Lie(G)) G0 . By Lemma 1.7.4, exp(Lie(G)) is a closed irreducible subvariety of G0 of the same dimension, so exp(Lie(G)) = G0 . 1.7.34. Consider a non-zero map Lie(Ga ) Lie(Gm ). a 1.7.35. Take G = 0 -1 : 3 = 1; a K : 1.7.36. When G is connected, see the proof of Proposition 1.9.8. For arbitrary G, check that G0 is divisible and repeat arguments of Exercise 1.6.29. 1.7.37. t t t exp 0 t = e tet : t C 0e t or

t00 et 0 0 exp 0 0 t = 0 1 t 000 001

:t

C

:

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IVAN V. ARZHANTSEV

1.8.21. No, Z (B ) Z2 , but Z (B1 ) = {e}, see Exercise 1.2.23. = 1.8.22. a) Z2 в Z2 acts on P1 by [x1 : x2 ] [-x1 : x2 ] and [x1 : x2 ] [x2 : x1 ]; b) SL(2) acts on P1 transitively; c) G acts on G by left translations transitively. 1.8.23. Clearly, e1 is the only line xed by B (n). Taking V = e1 and applying induction, one checks that the standard complete ag is a unique B (n)- xed point on F (V ). Hence this point is also xed by NGL(n) B (n). 1.8.24. For any unipotent A G the irreducible curve {At : t K} connects A and E , thus any unipotent element is contained in G0 . We know that U is a closed normal subgroup of G0 . Since a conjugate to a unipotent element is again unipotent, U is normal in G and G=U is a nite extension of a torus. But it need not be commutative ! 1.8.25. Assume that for some x X the boundary Y := Gx \ Gx is non-empty. By Theorem 1.8.5, there is a non-zero G- xed vector f in the G-invariant ideal I(Y ) O(Gx). But then f is a constant on Gx, a contradiction. 1.8.26. If B P , then G=B G=P is surjective, thus G=P is complete and by Corollary 1.4.3 is pro jective. Conversely, if G=P is pro jective, then by Borel's Fixed Point Theorem, B has a xed point on G0 =(P G0 ). 1.8.27. Take the set I = {(12); (23); : : : ; (n - 1n)}, a subset J I , and consider a parabolic subgroup PJ GL(n) de ned by aij = 0, where i > j and {(j j + 1); : : : ; (i - 1i)} is not contained in J . In particular, P = B (n) and PI = GL(n). Prove that any connected parabolic is conjugate to some PJ . First, prove that any Lie subalgebra in gl(n) containing all diagonal matrices is spanned by some Eij . Then describe subalgebras which contain Lie(B (n)). Check that di erent PJ are not conjugate (use Exercise 1.8.23 and Theorem 1.8.19). Again, Exercise 1.8.23 and Theorem 1.8.19 imply that any PJ coincides with its normalizer in GL(n), thus any parabolic subgroup is connected. 1.8.28. Consider a morphism : G в X X , (g; x) = g-1 · x. Then Z := -1 (Y ) = {(g ; x) : g -1 x Y } is a closed subset. Consider the pro jections p : G G=B , p2 : G в X X and p2 : G=B в X X . There is a commutative diagram: GвX F
G=B
в

X

rr yrrr

pвidrrrr

r

p2

FF FF 2 p FF FF " /X

Since Y is B -invariant, one has Z = (p в id)-1 ((p в id)(Z )). Taking normalization, one may assume that X in normal. By Theorem 3.0.32, the image of (G в X ) \ Z is open in G=B в X , so (p в id)(Z ) is closed. Further Y := p2 ((p в id)(Z )) is closed in X , because G=B is complete. On the other hand, Y = p2 (Z ) = GY . 1.8.29. Any maximal unipotent group is solvable and coincides with the set of unipotent elements of some Borel subgroup. 1.8.30. Check that D(n) is a maximal connected commutative subgroup of GL(n). On the other hand, a commutative unipotent group is not conjugated to a subgroup of D(n). 1.8.31. Prove that for q((x1 ; : : : ; xn ); (y1 ; : : : ; yn )) = x1 yn + · · · + xn y1 the set of diagonal matrices in SO(q) is a maximal torus. For this and for a description of a Borel subgroup, describe Lie(SO(q)). Similar arguments work for Sp(!), where !((x1 ; : : : ; x2n ); (y1 ; : : : ; y2n )) = x1 y2n - x2n y1 + · · · + xn yn+1 - xn+1 yn : 1.8.32. One may assume that G is connected and H is connected (right action of a nite group preserves a neness). Then G = T U , H = T1 U1 , U1 U and, up to conjugation, T1 T . In particular, any character of H may be extended to G. There is a pair (V ; v), where V is a G-module and H is the stabilizer of v , where H acts by a character . Let be an extension of to G. Consider the tensor product of V with a one-dimensional G-module corresponding to the character

ALGEBRAIC GROUPS AND INVARIANT THEORY
-

81

. Then H is the stabilizer of V 1, and G=H is quasia ne. Further, there is an open G-equivariant embedding G=H , X , where X is an a ne G-variety. Take Y := X \ (G=H ), and consider the ideal I(Y ) O(X ). There is a non-zero G-eigenvector f in I(Y ). Thus G=H = Xf .

82

IVAN V. ARZHANTSEV
K

0 -1 a t0 n в 0 t-1 : t K ; N = T 1 0 T ; Un = 0 -1 : = 1; a ta B = 0 t-1 : t Kв ; a K ; SL(2). First we classify algebraic Lie subalgebras g in sl(2) (and thus connected subgroups). If g = x , then either x is nilpotent (then it is unique up to conjugation, and we get U = U1 ) or semisimple (it is unique up to conjugation and proportionality, and we get T ). Assume that g = x; y . If x is semisimple, 0 then we may set x = 1 -1 and y = aE12 + bE21 . Here [x; y] = 2aE12 - 0 2bE21 g, hence ab = 0 and we obtain B . Finally, if x and y are nilpotent, then G0 is unipotent, but a maximal unipotent subgroup in SL(2) is one-dimensional. In order to describe non-connected subgroups, we need to calculate the normalizers: NSL(2) U = B , NSL(2) T = N , NSL(2) B = B . 1.9.27. Denote Uk = e1 ; : : : ; ek and A(i) the operator induced by A Pk1 ;:::;ks on Uki =Uki-1 . Consider H = {A Pk1 ;:::;ks : A(i) = id; i = 1; : : : ; s}. Clearly, H is a closed normal unipotent subgroup in Pk1 ;:::;ks , and 1.9.26. T =
P
k1 ;:::;ks

;

=H GL(k1 ) =

в

GL(k

2 - k1 ) в · · · в

GL(ks

- ks-1

)

is reductive. Hence H = Ru (P
R(P
k1 ;:::;ks

k1 ;:::;ks

). Similar arguments show that
Z (GL(k
s - ks-1

) = (Z (GL(k1 ))

в ··· в

))) H:

1.9.28. See Exercise 1.3.30 (b). 1.9.29. If p : G G=Ru (G) and H G=Ru (G) is a normal unipotent subgroup, then, by Lemma 1.9.1, p-1 (H ) is a normal unipotent subgroup of G. 1.9.30. Take G1 = Ga and G2 = Gm . 1.9.31. If G1 and G2 are reductive, then G1 вG2 satis es condition (5) of Theorem 1.9.16. For H G, one has Ru (H ) G. Finally, if H G, then any representation of G=H may be considered as a representation of G, thus G=H satis es Theorem 1.9.16 (4). 1.9.32. (Ru (F )) is a normal unipotent subgroup of G. Thus Ru (F ) Ker, a contradiction with Lemma 1.6.11. 1.9.33. Consider p : G G=R(G). Since G=R(G) is semisimple, so is its tangent algebra, and r(g) Ker(dp) = Lie(R(G)). Now we shall show that if R(G) is a solvable algebraic group, then Lie(R(G)) is a solvable Lie algebra. Indeed, R(G) contains a non-trivial closed normal commutative subgroup A R(G); a = Lie(A) is a commutative ideal of Lie(R(G)). Consider R(G)=A with Lie(R(G)=A) = Lie(R(G))=a and apply induction. 1.9.34. Study the linear span of an orbit, Exercise 1.3.27. 1.9.35. Semisimple classical groups are : SL(n), O(n) (n = 2), SO(n) (n = 2) and Sp(2n) (calculate the center using Exercise 1.9.34 and the Schur Lemma). Simple classical groups are SL(n), O(n) (n = 2; 4), SO(n) (n = 2; 4) and Sp(2n). In order to prove it, one should study ideals in the corresponding Lie algebra. 1.9.36. One may assume that H and N are connected. Applying Proposition 1.2.6 to H N we get that H N is closed. Note that the group H в N acts transitively on H в N , (h; n) (h1 ; n1 ) = (hh1 ; n1 n-1 ), on H N , (h; n) h1 n1 = hh1 n1 n-1 , and the morphism : H в N H N , (h; n) = hn is (H в N )-equivariant. Theorem 3.0.33 implies that di erential of at (e; e) is surjective. But Lie(H в N ) = Lie(H ) Lie(N ) (a direct sum of vector spaces) and the di erential of the restriction of on H в {e} (resp. on {e} в N ) de nes the embedding Lie(H ) Lie(G) (resp. Lie(N ) Lie(G)). 1.9.37. If x g is nilpotent, then G(x) G, a contradiction. Thus all elements of g are semisimple. It is su cient to prove that G0 is commutative. If it is not the case, then g is not commutative (otherwise all elements of g may be diagonalized simultaneously and G0 is contained in D(n)). Take any x g \ z(g) and let t be a maximal commutative algebraic Lie subalgebra in g containing x. Then t is the tangent algebra of a subtorus T G0 . The Ad(T )-module g is a direct sum

ALGEBRAIC GROUPS AND INVARIANT THEORY

83
t;

of one-dimensional submodules. This shows that g=t g ; where g := {x g : [y ; x] = (y )x} for all y

where is a nite set of non-zero linear functions on t. It follows Identity that [g ; g ] g + . Since + s for all ; = big s, the operator ad(x) is nilpotent for any x g . But this op trivially on t, thus it is non-zero. Then it can not be an image element of g, a contradiction.

from the Jacobi and su ciently erator acts nonof a semisimple

84

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2.1.9. K[x1 ; x1 x2 ; x1 x2 ; x1 x3 ; : : : ] or xi xj : i 2j . 2 2 12 2.1.10. Let p : O(X ) O(X )G be G-invariant pro jection. Prove that any simple Gsubmodule of non-zero type in O(X ) lies in Ker p. 2.1.11. By Exercise 2.1.10, it is su cient to check that the formula 1 f g·f |G| gG de nes a G-invariant pro jection O(X ) O(X )G . 2.1.12. Clearly, O(Kn )GL(n) = K and f f ((0; : : : ; 0)) is a G-invariant map. 2.1.13. For n = 2 it is su cient to prove that {xk ; (x1 + x2 )k ; : : : ; (x1 + kx2 )k } 1 is a basis in the space of homogeneous polynomials of degree k. Here one may use the Vandermonde determinant. For n > 2 use induction. 2.1.14. Let f be a homogeneous invariant of degree k. There are linear forms l1 ; : : : ; lN k k with f = l1 + · · · + lN . Then N 1 f= ( (g · li )k ): |G| gG i=1

Any gG (g · li )k is a symmetric polynomial in g · li , and it may be expressed in elementary symmetric polynomials in g · li . But these elementary symmetric polynomials are homogeneous invariants of degree |G|. 2.1.15. (a) The invariant bilinear symmetric form q de nes a quadratic invariant on Kn : F (x) := q(x; x). Consider the standard orthogonal basis e1 ; : : : ; en , set S = e1 , and prove that O(Kn )SO(n) = K[F ]. (b) The action Sp(2n) : K2n is transitive on the set of non-zero vectors, so O(K2n )Sp(2n) = K. (c) For s < n the action has open orbit and O(X )SL(n) = K. If s = n, then de ne F (v1 ; : : : ; vn ) := det(v1 ; : : : ; vn ), set S = (e1 ; : : : ; en ) , and prove that O(X )SL(n) = K[F ]. n 2.1.16. (a) xn ; xn-1 x2 ; : : : ; x1 x2 -1 ; xn ; 11 2 n ; xn ; x x . (b) x1 2 1 2 2.1.17. x1 x3 ; x2 x3 ; x2 x3 ; x2 x4 ; x1 x2 x2 ; x1 x3 x4 3 14 2 4 Prove that there exists a generating set consisting of monomials, and generating monomials correspond to non-decomposable non-negative integer solutions of the equation 3a1 + a2 - a3 - 2a4 = 0.

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2.2.11. Put X = SL(2) and G = U (2) with the action by right translations. Here Spec(O(X )G ) K2 , but (X ) = K2 \ {0} (cf. Proposition 1.4.6 and Exam= ple 1.4.8). 2.2.12. All semisimple elements in -1 (y) are conjugate to the same diagonal matrix, thus form one GL(n)-orbit O . By Corollary 1.7.12, the centralizer of a semisimple element in -1 (y) has the largest dimension among centralizers of elements in -1 (y), thus the orbit O has the smallest dimension. 2.2.13. Since for any f O(X ) all coe cients of the polynomial Ff (T ) = (T - g · f ) are G-invariants, the extension O(X )G O(X ) is integral. Moreover, any Gorbit is closed in X . 2.2.14. Consider Kв : K3 , t · (x1 ; x2 ; x3 ) = (tx1 ; tx2 ; t-1 x3 ). 2.2.15. Follow the proof of Theorem 2.2.7. 2.2.16. Let 1 : X Y1 and 2 : X Y2 be categorical quotients. Applying De nition 2.2.5 to the G-invariant morphisms 2 : X Y2 and 1 : X Y1 , we get morphisms 1 and 2 : XA A
AA 1 AA AA ~}} 1 2/ / Y2 Y1 Y1
1 } } } }} 2 gG

}

The uniqueness implies that 2 1 = idY1 . Similarly, 1 2 = idY2 . 2.2.17. Assume that y Y := X==G, y (X ). Take Z := Y \ {y}. Then : X Z = is a G-invariant morphism, and there is : Y Y \ {y} making the diagram commutative: X x GG
Y Let i : Y \ {y} Y be the embedding. Then uniqueness implies that i = idY . But (i )(y) = y, a contradiction. m The condition O(Kn )T = K is equivalent to the existence of a non-constant a1 : : : xan with a + · · · + a = 0. monomial x1 11 nn n It is su cient to prove that : O(Y ) O(X )G is an isomorphism. Since is dominant, the homomorphism is injective. Take any f O(X )G . We know, that f is constant on a generic ber of . Corollary 3.0.29 guarantees that there is O(Y ) with ( ) = f . For s = n, consider : X Sym(n), (v1 ; : : : ; vn ) = (q(vi ; vj )). This morphism is surjective, because any symmetric matrix S has a form AT A for some A Mat(n в n). Moreover, for a non-degenerate S the preimage -1 (S ) consists of a unique G-orbit: AT A = B T B A = (AT )-1 B T B , (AT )-1 B T O(n). Now apply Igusa's criterion. For s < n, use Lemma 2.1.5 and surjectivity of . Consider X = G as an a ne H -variety with the right H -action. All orbits of the action have the same dimension, thus are closed, and any ber of the quotient morphism : G G==H is an H -coset. Since is the categorical quotient, there is a (bijective) morphism : G==H G=H , which is an isomorphism. If zi Gxi Gxi+1 , then (xi ) = (zi ) = (xi+1 ). It is su cient to prove that any two points on Pn are equivalent. Renumbering, one may assume that there is a one-parameter subgroup : Kв T m such that ; 1 ; i ; 2 and ; 1 < ; 2 . For any x Pn denote by xj the point [y1 : ::: : yn+1 ] with yi = xi for ; i = ; j and yi = 0 otherwise. Check that for any x; y Pn x x1 (x1 + y2 ) y2 y:
x {xx
xxx xx

GG GG GG G# i / Y \ {y } / Y:

2.2.18. 2.2.19.

2.2.20.

2.2.21.

2.2.22. 2.2.23.

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2.3.17. Consider the diagonal action SL(2) : K2 K2 , and set Z = e1 e1 , X = SL(2)Z . Here O(X )SL(2) = K, because any orbit in Z contains zero in its closure. But X does not have open SL(2)-orbit, so K(X )SL(2) = K. 2.3.18. Use Proposition 2.3.6. 2.3.19. Assume that G-orbits in X are of the same dimension. Then (G1), (G2) and (G4) follow from Theorem 2.2.2, and (G3) follows from Theorem 3.0.30. In fact, one may remark that (U ) = (GU ) and deduces (G3) from Theorem 2.2.2 (2). 2.3.20. Follows from Exercise 2.3.19. 2.3.21. If QO(X )G = K(X )G , then elements of O(X )G separate G-orbits on the open subset U of Corollary 2.3.14. By Lemma 2.3.13, the intersection of a generic ber of with U is dense in this ber. Conversely, if W X is an open subset consisting of G-orbits of maximal dimension, then generators of O(X )G separate orbits on W -1 (V ), and thus generate K(X )G (Proposition 2.3.16). 2.3.22. Let f1 ; : : : ; fm be a generating set of K(X )G . There is an open subset W X=G such that all fi are regular on -1 (W ). Then f1 ; : : : ; fm O(-1 (W ))G = (O(W )) (K(X=G)): For a categorical quotient, consider Kв : Kn , t · (x1 ; : : : ; xn ) = (tx1 ; : : : ; txn ). 2.3.23. If : K2 \ {0} Y is the geometric quotient, then Y is a curve. Since any curve is either a ne or pro jective, the points y1 = (Ge1 ) and y2 = (Ge2 ) lies in a common a ne chart U Y . There is a function f O(U ) with f (y1 ) = f (y2 ). Then (f ) is a rational invariant on K2 \ {0} that separates Ge1 and Ge2 . But в в K(K2 \ {0})K = K(K2 )K = K(x1 x2 ), a contradiction.

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References
[Bo91] A. Borel, Linear Algebraic Groups. Graduate Texts in Math. 126, Springer-Verlag, 1991. [Ha99] J. Hausen, Geometrische Invariantentheorie. Universit Konstanz, 1999. at [Hu72] J.E. Humphreys, Introduction to Lie Algebras and Representation Theory. Second Printing, Revised. Graduated Texts in Math. 9, Springer-Verlag, 1972. [Hu75] J.E. Humphreys, Linear Algebraic Groups. Graduated Texts in Math. 21, Springer-Verlag, 1975. [Kr84] H. Kraft, Geometrische Methoden in der Invariantentheorie. Aspekte der Mathematik D1, Vieweg-Verlag, Braunschweig, 1984. [OV90] A.L. Onishchik and E.B. Vinberg, Lie Groups and Algebraic Groups. Springer-Verlag Berlin Heidelberg, 1990. [PV94] V.L. Popov and E.B. Vinberg, Invariant Theory. In: Encyclopaedia of Math. Sciences 55, (A.N. Parshin, I.R. Shafarevich Eds.), Algebraic Geometry IV, Springer-Verlag, 1994. [Sp81] T.A. Springer, Linear Algebraic Groups. Progress in Math. 9, Birkh auser, 1998. [TY05] P. Tauvel and R.W.T. Yu, Lie Algebras and Algebaic Groups. Springer Monographs in Math., Springer-Verlag, 2005.

Department of Higher Algebra, Faculty of Mechanics and Mathematics, Moscow State University, Leninskie Gory, GSP-1, Moscow, 119991, Russia Mathematical Institute, Tubingen University, Auf der Morgenstelle 10, 72076 Tubingen, Germany E-mail address : arjantse@mccme.ru