Classification of Finite-Dimensional Simple Lie Algebras over the Complex Numbers

Transcription

1 Classification of Finite-Dimensional Simple Lie Algebras over the Complex Numbers by Shawn Baland A thesis submitted to the Faculty of the University of Colorado in partial fulfillment of the requirements for graduation with departmental honors Department of Mathematics November 12, 2007 Updated December 6, 2007 Defense Committee Prof. Arlan Ramsay, Mathematics (Thesis Advisor) Prof. Richard Clelland, Mathematics Prof. James Meiss, Applied Mathematics

2 ii Baland, Shawn Classification of Finite-Dimensional Simple Lie Algebras over the Complex Numbers Thesis directed by Prof. Arlan Ramsay Lie algebras were first introduced by S. Lie in 1880 as algebraic structures used to study local problems in Lie groups. Specifically, the tangent space of a Lie group at the identity has the natural structure of a Lie algebra. Aside from their many applications, Lie algebras have attracted interest in their own right. This paper deals primarily with the classification of finite-dimensional simple Lie algebras over the complex numbers, first achieved through the independent work of E. Cartan and W. Killing during the decade Contemporary methods of studying Lie algebras and their representations were later developed by H. Weyl. The classification theorem begins by studying the decomposition of a semisimple Lie algebra into a direct sum of subspaces with respect to a carefully chosen subalgebra called the Cartan subalgebra. One then forms a Euclidean space from the nonzero linear functionals associated with these subspaces via an inner product called the Killing form. The geometry of this Euclidean space allows us to encode vital information about a Lie algebra s structure into a graph called the Dynkin diagram. It is then shown that there are only a limited number of possible connected Dynkin diagrams, and that these diagrams are in one-to-one correspondence with the finite-dimensional simple Lie algebras over the complex numbers. The paper s structure closely follows that of [Car05]. It is assumed that the reader has a thorough knowledge of linear algebra, although this is the only major prerequisite.

3 Dedication To my mother, for supporting me when I was less ambitious.

4 iv Acknowledgments I would first like to thank Professor Peter Elliott for encouraging me to go beyond what is usually expected of an undergraduate student. I would also like to thank Professor Richard Green for suggesting this project, and for pointing me toward many valuable resources. I owe the greatest thanks to Professor Arlan Ramsay, my thesis advisor, for spending a great deal of time answering all of my questions, and for asking many questions that I had not considered. Thanks also to Mike Noyes, my reader, for improving this paper s clarity and accuracy. Finally, I would like to thank my wife, Lyndsie, for her support and understanding throughout my undergraduate career.

5 v Contents Chapter 1 Basic concepts Lie algebras, subalgebras, and ideals Lie homomorphisms and quotient algebras Nilpotent and solvable Lie algebras Representations of solvable and nilpotent Lie algebras Basic representation theory A transition to finite-dimensional complex Lie algebras Representations of solvable Lie algebras Representations of nilpotent Lie algebras Cartan subalgebras Existence of Cartan subalgebras Derivations and automorphisms Concepts from algebraic geometry Conjugacy of Cartan subalgebras The Cartan decomposition Root spaces The Killing form The Cartan decomposition of a semisimple Lie algebra

6 vi 5 Root systems and the Weyl group Positive systems and fundamental systems of roots The Weyl group The Cartan matrix and the Dynkin diagram The Cartan matrix The Dynkin diagram Classification of Dynkin diagrams Classification of Cartan matrices The existence and uniqueness theorems Properties of structure constants The uniqueness theorem Some generators and relations in a simple Lie algebra The Lie algebras g(a) and g(a) The existence theorem Bibliography 133

7 vii Figures Figure 6.1 Cartan matrices and Dynkin diagrams for l = 1 and Dynkin diagrams and quadratic forms for l = 1 and Standard list of connected Dynkin diagrams

8 Chapter 1 Basic concepts 1.1 Lie algebras, subalgebras, and ideals Let k be a field. A k-algebra is a k-vector space A equipped with a bilinear operation A A A called multiplication. The dimension of A is the dimension of A as a k-vector space. A Lie algebra is a k-algebra g with multiplication (x, y) [x, y] satisfying the following conditions: (i) [x, x] = 0 for all x g. (ii) [[x, y], z] + [[y, z], x] + [[z, x], y] = 0 for all x, y, z g. Condition (ii) is called the Jacobi identity. Proposition 1.1. [y, x] = [x, y] for all x, y g, i.e., multiplication in a Lie algebra is anticommutative. Proof. If x, y g, then It follows that [y, x] = [x, y]. 0 = [x + y, x + y] = [x, x] + [x, y] + [y, x] + [y, y] = 0 + [x, y] + [y, x] + 0.

9 2 Notice that anticommutativity implies [x, x] = [x, x] for all x g, yielding 2[x, x] = 0. Hence if char k 2, then anticommutativity is equivalent to condition (i) for a Lie algebra. Example 1.2. An associative algebra is a k-algebra A satisfying the associative law, that is, (xy)z = x(yz) for all x, y, z A. Lie algebras are easily obtained from associative algebras in the following way: Given any associative algebra A, we define the commutator of x, y A by [x, y] = xy yx. The commutator is clearly bilinear. Moreover, [x, x] = xx xx = 0 and [[x, y], z] + [[y, z], x] + [[z, x], y] = (xy yx)z z(xy yx) + (yz zy)x x(yz zy) + (zx xz)y y(zx xz) = 0 for all x, y, z A. Thus A forms a Lie algebra via the commutator product. We call this algebra the Lie algebra of A and denote it by [A]. If a and b are subspaces of a Lie algebra g, we define [a, b] to be the subspace of g generated by all elements of the form [x, y] with x a and y b. Proposition 1.3. [a, b] = [b, a] for all subspaces a, b of g. Proof. Any element in [a, b] is of the form λ 1 [x 1, y 1 ] + + λ r [x r, y r ] where x i a, y i b, λ i k for i = 1,..., r. Now [x i, y i ] = [y i, x i ] [b, a] by Proposition 1.1. It follows that [a, b] [b, a]. By a symmetric argument, we also have [b, a] [a, b], and so [a, b] = [b, a]. A subalgebra of a Lie algebra g is a subspace h g such that [h, h] h. Hence h forms a Lie algebra by restricting the multiplication on g to the subspace h.

10 3 Proposition 1.4. If h and k are subalgebras of g, then h k is a subalgebra of g. Proof. We know that h k is a subspace of g from linear algebra. Now [h k, h k] is generated by elements of the form [x, y] with x, y h k. We have x, y h and x, y k so that [x, y] [h, h] [k, k] h k. It follows that [h k, h k] h k, and so h k is a subalgebra of g. An ideal of g is a subspace a g such that [a, g] a. This condition is equivalent to [g, a] a by Proposition 1.3, and so every ideal of a Lie algebra is two-sided. Note also that [a, a] a since a g. Thus ideals are also subalgebras. Proposition 1.5. If a and b are ideals of g, then a b, a + b, and [a, b] are ideals of g. Proof. We know that these sets are subspaces of g from linear algebra. For the first part, notice that [a b, g] is generated by elements of the form [x, y] with x a b, y g. We have x a and x b so that [x, y] [a, g] [b, g] a b. It follows that [a b, g] a b, and so a b is an ideal of g. For the second part, notice that [a + b, g] is generated by elements of the form [x+y, z] with x a, y b, z g. We have [x+y, z] = [x, z]+[y, z] [a, g]+[b, g] a+b. It follows that [a + b, g] a + b, and so a + b is an ideal of g. For the third part, notice that [[a, b], g] is generated by elements of the form [[x, y], z] with x a, y b, z g. We have [[x, y], z] = [[x, z], y] [[y, z], x] by the Jacobi identity. Now [x, z] a and [y, z] b since a and b are ideals. Thus [[x, y], z] [a, b]. It follows that [[a, b], g] [a, b], and so [a, b] is an ideal of g. 1.2 Lie homomorphisms and quotient algebras As with Lie subalgebras and ideals, the concepts of Lie homomorphisms and quotient algebras are analogous to those of associative algebras. Let g, h be Lie algebras over a field k. A map φ : g h is a homomorphism of Lie algebras if φ is k-linear

11 4 and φ([x, y]) = [φ(x), φ(y)] for all x, y g. An isomorphism of Lie algebras is a bijective homomorphism of Lie algebras. If there exists an isomorphism φ : g h, then we say g and h are isomorphic and write g = h. Proposition 1.6. If g is a Lie algebra and a is an ideal of g, then the quotient space g/a forms a Lie algebra via the multiplication [x + a, y + a] = [x, y] + a for all x, y g. Proof. To prove this multiplication is well defined, suppose x + a = x + a and y + a = y + a. We must show that [x, y] + a = [x, y ] + a. Now x + a = x + a implies x = x + a 1 for some a 1 a. Likewise, y + a = y + a implies y = y + a 2 for some a 2 a. Thus [x, y] + a = [x + a 1, y + a 2 ] + a = [x, y ] + [x, a 2 ] + [a 1, y ] + [a 1, a 2 ] + a. The last three summands are in a since a is an ideal, and so [x, y] + a = [x, y ] + a. Next, we have [x + a, x + a] = [x, x] + a = 0 + a = a for all x a. Finally, the Jacobi identity follows from the Jacobi identity in g. Proposition 1.7. Let φ : g h be a homomorphism of Lie algebras. Then im φ is a subalgebra of h, ker φ is an ideal of g, and g/ ker φ = im φ. Proof. We know that im φ is a subspace of h from linear algebra. If x, y g, then [φ(x), φ(y)] = φ([x, y]) im φ. This shows that [im φ, im φ] im φ, and so im φ is a subalgebra of h. We also know that ker φ is a subspace of g. If x ker φ and y g, then φ([x, y]) = [φ(x), φ(y)] = [0, φ(y)] = 0,

12 showing that [x, y] ker φ. It follows that [ker φ, g] ker φ, and so ker φ is an ideal of g. To prove the last statement, consider the linear map 5 This map is well defined since g/ ker φ im φ x + ker φ φ(x). x + ker φ = y + ker φ x y ker φ φ(x y) = 0 φ(x) = φ(y). The leftward implication proves that the map is injective. If φ(x) im φ, then x + ker φ maps to φ(x), showing that the map is surjective. Finally, we prove that the map is a homomorphism of Lie algebras. If x, y g, then [x + ker φ, y + ker φ] = [x, y] + ker φ φ([x, y]) = [φ(x), φ(y)]. Hence the map respects Lie multiplication, and so it is an isomorphism of Lie algebras. Lemma 1.8. If h is a subalgebra of g and a is an ideal of g, then h + a is a subalgebra of g. Proof. We know that h + a is a subspace of g from linear algebra. Now [h + a, h + a] is generated by elements of the form [x + y, x + y ] with x, x h and y, y a. We have [x + y, x + y ] = [x, x ] + [x, y ] + [y, x ] + [y, y ] [h, h] + ([h, a] + [a, h] + [a, a]) h + a. It follows that [h + a, h + a] h + a, and so h + a is a subalgebra of g. Proposition 1.9. Let h be a subalgebra of g and a an ideal of g. Then a is an ideal of h + a, h a is an ideal of h, and (h + a)/a = h/(h a).

13 6 Proof. We know that h+a and h a are subalgebras of g by Lemma 1.8 and Proposition 1.4, respectively. We have [a, h + a] [a, g] a, and so a is an ideal of h + a. Also, [h a, h] [h, h] and [h a, h] [a, h] so that [h a, h] [h, h] [a, h] h a. Thus h a is an ideal of h. Notice that (h + a)/a forms a Lie algebra by Proposition 1.6. Consider the linear map φ : h (h + a)/a defined by φ(x) = x + a. We have φ([x, y]) = [x, y] + a = [x + a, y + a] = [φ(x), φ(y)], and so φ is a homomorphism of Lie algebras. It is surjective since each element of (h + a)/a can be written as x + a with x h. Finally, we have ker φ = {x h φ(x) = 0} = {x h x + a = a} = {x h x a} = h a. Thus h/(h a) = (h + a)/a by Proposition Nilpotent and solvable Lie algebras The key to studying a Lie algebra s structure is knowing about its ideals. The existence of the right kind of ideals will be useful in the sequel and is introduced in this section. Let g be a Lie algebra. We consider the subspaces of g defined recursively by g 1 = g, g n+1 = [g n, g] for n 1. Proposition g n is an ideal of g and g = g 1 g 2 g 3. Proof. We prove the first statement by induction on n. If n = 1, then g 1 = g is an ideal since [g, g] g. Now suppose the statement is true for n = r, that is, g r is an ideal. Then g r+1 = [g r, g] is an ideal by Proposition 1.5, completing the induction. To prove

14 7 the second statement, let n 1. Then g n+1 = [g n, g] g n since g n is an ideal. Given Proposition 1.10, we call the chain of ideals g = g 1 g 2 g 3 the lower central series of g. We say that g is nilpotent if g n = 0 for some n 1. If g 2 = 0, then we say g is abelian. This is equivalent to the condition [g, g] = 0, i.e., [x, y] = 0 for all x, y g. We now consider the subspaces of g defined recursively by g (0) = g, g (n+1) = [ g (n), g (n)] for n 0. Proposition g (n) is an ideal of g and g = g (0) g (1) g (2). Proof. We prove the first statement by induction on n. If n = 0, then g (0) = g is clearly an ideal. Now suppose the statement is true for n = r, that is, g (r) is an ideal. Then g (r+1) = [ g (r), g (r)] is an ideal by Proposition 1.5, completing the induction. To prove the second statement, let n 0. Then g (n+1) = [ g (n), g (n)] [ ] g (n), g g (n) since g (n) is an ideal. Given Proposition 1.11, we call the chain of ideals g = g (0) g (1) g (2) the derived series of g. We say that g is solvable if g (n) = 0 for some n 0. Proposition (i) [g m, g n ] g m+n for all m, n 1. (ii) g (n) g 2n for all n 0.

15 8 (iii) Every nilpotent Lie algebra is solvable. Proof. For part (i), we fix m and use induction on n. If n = 1, then [g m, g] = g m+1. Now suppose the statement is true for n = r. By definition, [ g m, g r+1] = [g m, [g r, g]]. This subspace is generated by elements of the form [x, [y, z]] with x g m, y g r, z g. We have [x, [y, z]] = [[z, x], y] + [[x, y], z] (by the Jacobi identity) [[g, g m ], g r ] + [[g m, g r ], g] [ g m+1, g r] + [ g m+r, g ] g m+r+1 (by the inductive hypothesis). It follows that [g m, g n ] g m+n for all n. For part (ii), we again use induction on n. For n = 0, we have g (0) = g = g 20. Now suppose the statement is true for n = r. We have g (r+1) = [ g (r), g (r)] [ g 2r, g 2r ] g 2 2 r = g 2r+1 by part (i), completing the induction. For part (iii), if g is nilpotent, then g n = 0 for some n 1. Thus g (n) g 2n g n = 0 by part (ii), and so g is solvable. Proposition If g is solvable, then every subalgebra and every quotient algebra of g is solvable. Proof. We first show that if φ : g h is a surjective homomorphism of Lie algebras, then φ ( g (n)) = h (n) for all n 0. We proceed by induction on n. For n = 0, we have φ ( g (0)) = φ(g) = h = h (0). Now suppose the statement is true for n = r. We have ( φ g (r+1)) ([ = φ g (r), g (r)]) [ ( = φ g (r)) (, φ g (r))] [ = h (r), h (r)] = h (r+1), completing the proof of the claim.

16 9 To prove the proposition, choose n such that g (n) = 0. If h is a subalgebra of g, then h (n) g (n) = 0, and so h is solvable. Finally, if a is an ideal of g, then (g/a) (n) is the image of g (n) under the canonical homomorphism g g/a by the preceding result. We have g (n) = 0, yielding (g/a) (n) = 0, and so g/a is solvable. Proposition If a is an ideal of g such that a and g/a are solvable, then g is solvable. Proof. We first show that g (m+n) = ( g (m)) (n) for all m, n 0. We fix m and use induction on n. For n = 0, we have g (m+0) = g (m) = ( g (m)) (0). Now suppose the statement is true for n = r. We have g (m+r+1) = [ g (m+r), g (m+r)] [ ( = g (m)) (r) (, g (m)) ] (r) ( = g (m)) (r+1), completing the proof of the claim. To prove the proposition, notice that since g/a is solvable, we have (g/a) (m) = 0 for some m 0, that is, g (m) a. And since a is solvable, we have a (n) = 0 for some n 0. By the preceding result, this yields g (m+n) = ( g (m)) (n) a (n) = 0, and so g is solvable. Proposition Every finite-dimensional Lie algebra g contains a unique maximal solvable ideal r. Proof. We first show that if a and b are solvable ideals of g, then a+b is a solvable ideal of g. Notice that a + b is an ideal by Proposition 1.5. Because a is solvable, we know that a/(a b) is solvable by Proposition Now (a + b)/b = a/(a b) by Proposition 1.9. Thus (a + b)/b is solvable. Since b is also solvable, we see that a + b is solvable by Proposition 1.14.

17 10 To prove the proposition, we choose a solvable ideal r of g having maximal dimension. Let a be any solvable ideal of g. Then a+r is a solvable ideal by the preceding result. We have a + r = r since dim r is maximal, and so a r. Thus r contains every solvable ideal of g. Given Proposition 1.15, we call the maximal solvable ideal r the solvable radical of g. A Lie algebra g is semisimple if r = 0, that is, if g contains no nonzero solvable ideal. A Lie algebra g is simple if it contains no proper ideals, i.e., no ideals other than 0 and g. Suppose g is 1-dimensional over k. Then for any nonzero x g, {x} forms a basis of g. Since [x, x] = 0, we see that g 2 = 0, and so g is abelian. This is true for any Lie algebra of dimension 1. Thus any two 1-dimensional Lie algebras over k are isomorphic. Because the dimension of any subspace is either 0 or 1, we see that g contains no proper subspaces. Hence g contains no proper ideals, and so g is simple. We call the 1-dimensional Lie algebra the trivial simple Lie algebra. Proposition Every non-trivial simple Lie algebra is semisimple. Proof. Suppose g is simple but not semisimple. Then r 0, and since r is an ideal, we must have r = g. Because r is solvable, g must be solvable, and so g (n) = 0 for some n 0. Now g (1) is an ideal. Hence g (1) = g or g (1) = 0. But g (1) cannot equal g, because this would imply g (n) = g for all n 0, contradicting the solvability of g. Thus g (1) = 0, i.e., [g, g] = 0. If a is any subspace of g, then [a, g] [g, g] = 0 a. It follows that every subspace of g is an ideal, and so g cannot have any proper subspaces. Hence dim g = 1, and so g is the trivial simple Lie algebra.

18 Chapter 2 Representations of solvable and nilpotent Lie algebras 2.1 Basic representation theory Recall from Example 1.2 that any associative algebra A gives rise to a Lie algebra [A] via the commutator product. We consider the case where A = M n (k), the ring of n n matrices over k. One easily checks that this is an associative k-algebra. We define the Lie algebra gl n (k) by gl n (k) = [M n (k)]. A representation of a Lie algebra g is a homomorphism of Lie algebras ρ : g gl n (k) for some n 0. In this case, ρ is called a representation of degree n. Two representations ρ, ρ of degree n are equivalent if there exists an invertible matrix A M n (k) such that ρ (x) = A 1 ρ(x)a for all x g. A left g-module is a k-vector space V equipped with a left action g V V (x, v) xv satisfying the following conditions: (i) (x, v) xv is linear in x and v. (ii) [x, y]v = x(yv) y(xv) for all x, y g, v V.

19 12 Note that we can define right g-modules in a similar way, although we will not have occasion to use them in this paper. Thus from now on, we will assume that all g-modules are left g-modules. There is a one-to-one correspondence between representations of degree n and n-dimensional g-modules. Suppose ρ is a representation of g of degree n. Let V be an n-dimensional vector space and choose a basis e 1,..., e n of V. For each x g, we define the action of x on an element of V by x j λ j e j = ij λ j ρ ij (x)e i. Thus for each x g, the matrix ρ(x) acts as a linear transformation on V with respect to the basis e 1,..., e n. For all x, y g, v V, we have ρ([x, y])v = [ρ(x), ρ(y)]v = (ρ(x)ρ(y) ρ(y)ρ(x))v = ρ(x)ρ(y)v ρ(y)ρ(x)v, showing that this action transforms V into a g-module. Conversely, suppose V is an n-dimensional g-module and let e 1,..., e n be a basis of V. For each x g, we write xe j = i ρ ij (x)e i where ρ ij (x) k. This gives rise to the matrix ρ(x) = (ρ ij (x)). We have [x, y] e j = x(ye j ) y(xe j ) ( ) ( ) = x ρ kj (y)e k y ρ kj (x)e k k k = k = k = i = i ρ kj (y)xe k ρ kj (x)ye k k ( ) ρ kj (y) ρ ik (x)e i ( ) ρ kj (x) ρ ik (y)e i i k i ( ) (ρ ik (x)ρ kj (y) ρ ik (y)ρ kj (x)) k (ρ(x)ρ(y) ρ(y)ρ(x)) ij e i. e i

20 13 It follows that ρ([x, y]) = ρ(x)ρ(y) ρ(y)ρ(x) = [ρ(x), ρ(y)], and so ρ is a representation of g. Now let f 1,..., f n be another basis of V and ρ the representation obtained from this basis. I claim that ρ is equivalent to ρ. We know there exists an invertible n n matrix A such that f j = i A ije i. We have xf j = A kj xe k = ( ) A kj ρ ik (x)e i = k k i i We also have xf j = ρ kj (x)f k = ( ) ρ kj (x) A ik e i = k k i i ( ) ρ ik (x)a kj e i. k ( ) A ik ρ kj (x) e i. It follows that ρ(x)a = Aρ (x) so that ρ (x) = A 1 ρ(x)a for all x g. Thus ρ is equivalent to ρ as desired. Let V be a g-module, a a subspace of g, and U a subspace of V. We define a U to be the subspace of V spanned by all elements of the form xu with x a and u U. A submodule of V is a subspace U of V such that g U U. Notice that both 0 and V are submodules of V. A proper submodule of V is a submodule distinct from 0 and V. A g-module V is irreducible if it contains no proper submodules. We say that V is completely reducible if it is a direct sum of irreducible submodules. Finally, V is said to be indecomposable if it cannot be written as a sum of two proper submodules. We see that every irreducible g-module is indecomposable, although the converse need not hold. Example 2.1. Let V be a g-module and W a submodule of V. Then the quotient space V/W forms a g-module via the action k (x, v + W ) xv + W. To prove this action is well defined, let v + W = v + W. We must show that xv + W = xv + W.

21 14 Now v + W = v + W implies v = v + w for some w W. Thus xv + W = x(v + w) + W = xv + xw + W = xv + W since xw W. The module properties in V/W follow easily from those in V. Example 2.2. The vector space g can be made into a g-module via the action (x, y) [x, y]. This is certainly linear in x and y. By the Jacobi identity, we also have [[x, y], z] = [x, [y, z]] [y, [x, z]], showing that g is indeed a g-module. We call this module the adjoint module and denote the action of x on the vector y by ad x y. By the module properties, we have ad [x, y] = ad x ad y ad y ad x. Notice that a submodule a of g is defined by the property [g, a] a. Hence the submodules of the adjoint module g are precisely the ideals of g. 2.2 A transition to finite-dimensional complex Lie algebras With the exception of Propositions 1.15 and 1.16, every definition and result in this paper has been applicable to Lie algebras with arbitrary base fields and dimensions. We wish now to draw a line in the sand. From this point forward, we shall assume that every Lie algebra g is finite-dimensional and that the base field k is the field C of complex numbers. These types of Lie algebras have particularly nice representations, as we shall witness in the following sections. 2.3 Representations of solvable Lie algebras This section will produce some very useful results about solvable Lie algebras.

22 As one might expect, a 1-dimensional representation of g is a homomorphism of Lie algebras ρ : g [C]. 15 Lemma 2.3. A linear map ρ : g C is a 1-dimensional representation of g if and only if ρ vanishes on g 2. Proof. Suppose ρ is a 1-dimensional representation of g and let x, y g. Then ρ([x, y]) = [ρ(x), ρ(y)] = ρ(x)ρ(y) ρ(y)ρ(x) = 0, and so ρ vanishes on g 2. Conversely, if ρ vanishes on g 2, then ρ([x, y]) = 0 = ρ(x)ρ(y) ρ(y)ρ(x) = [ρ(x), ρ(y)] so that ρ is a homomorphism from g into [C]. Theorem 2.4 (Lie s theorem). Let g be a solvable Lie algebra and V a finite-dimensional irreducible g-module. Then dim V = 1. Proof. We proceed by induction on dim g. If dim g = 1, then g = Cx for some x g. Since C is algebraically closed, we can find an eigenvector v of x in V. Then xv Cv, and so Cv is a g-submodule of V. Because V is irreducible, this forces V = Cv and dim V = 1. Now suppose dim g > 1 and that the theorem is true for all solvable Lie algebras of dimension dim g 1. Notice that since g is solvable, we have g 2 g, or else we would have g (n) = g for all n 0. Let a be a subspace of g containing g 2 such that dim a = dim g 1. Then [a, g] [g, g] = g 2 a, and so a is an ideal of g. We may regard V as an a-module. Let W be an irreducible a-submodule of V. By induction, we have dim W = 1. Then yw = λ(y)w for all y a, where λ is the 1-dimensional representation induced by W. Define the set U = {u V yu = λ(y)u for all y a}.

23 We have 0 W U V. I claim that U is a g-submodule of V. To prove this, let x g, y a, u U. Then 16 y(xu) = x(yu) [x, y]u = x(λ(y)u) λ([x, y])u = λ(y)xu λ([x, y])u. We shall show that λ([x, y]) = 0. Having proved this, it follows that xu U, and so U is a g-submodule of V. Since V is irreducible, this forces U = V, and so yv = λ(y)v for all y a, v V. Now because a has codimension 1, we have g = a Cz for some z g\a. Let v be an eigenvector of z in V. Then Cv is invariant under the actions of both a and z, and so Cv is a g-submodule of V. Because V is irreducible, this forces V = Cv and dim V = 1. To prove that λ([x, y]) = 0 for all x g, y a, it suffices to consider the element x = z such that g = a Cz since λ vanishes on a 2 by Lemma 2.3. Let u be any nonzero vector in U. We define the vectors v 0 = u, v 1 = zu, v 2 = z(zu),... Let V 0 = 0 and V i = v 0, v 1,..., v i 1 for i > 0. We have V 0 V 1 V 2. I claim that yv i = λ(y)v i (mod V i ) for all y a, i 0. We proceed by induction on i. For i = 0, we have yv 0 = yu = λ(y)u = λ(y)v 0. Now suppose i > 0 and that the statement is true for i 1. We have yv i = y(zv i 1 ) = z(yv i 1 ) [z, y]v i 1. By induction, yv i 1 = λ(y)v i 1 + v and [z, y]v i 1 = λ([z, y])v i 1 + v for some v, v V i 1. Hence yv i = z(λ(y)v i 1 + v ) (λ([z, y])v i 1 + v ) = λ(y)v i + zv λ([z, y])v i 1 v λ(y)v i (mod V i ),

24 17 completing the proof of the claim. Now because V is finite-dimensional, there exists a minimal r > 0 such that V r = V r+1 = V r+2 =. By construction, V r is invariant under the action of z, and given the above result, V r is also invariant under the action of a. Hence V r is a g-submodule of V, and since V is irreducible, we have V = V r. Then by the above result, any element y a acts on V via an upper-triangular matrix with λ(y) along the main diagonal. This is certainly true for the element [z, y], and so tr V ([z, y]) = rλ([z, y]). But we also have tr V ([z, y]) = tr V (zy yz) = tr V (zy) tr V (yz) = 0. Hence rλ([z, y]) = 0, and since C is an integral domain, this yields λ([z, y]) = 0. Corollary 2.5. Let g be a solvable Lie algebra and V a finite-dimensional g-module. Then a basis can be chosen for V with respect to which we obtain a matrix representation ρ of g of the form ρ(x) = 0 for all x g. Thus elements of g are represented by upper-triangular matrices. Proof. It suffices to show there exists a chain of submodules 0 = V 0 V 1 V n = V such that dim V i = i for i = 0,..., n. We proceed by induction on dim V. The statement is trivial for dim V = 1. So suppose dim V = n > 1 and that the statement is true for all g-modules of dimension n 1. Choose an irreducible g-submodule W of V. By Theorem 2.4, we have dim W = 1. Then V/W is a g-module of dimension n 1, and

25 18 by the inductive hypothesis, we may assume there exists a chain of submodules 0 = V 0 V 1 V n 1 = V/W with the desired properties. For i = 1,..., n, let V i denote the preimage of V i 1 in V under the canonical homomorphism. We claim that the chain 0 = V 0 V 1 V n = V has the desired properties for V. Let v 1,..., v i 1 be a basis of V i 1 where v j V j \V j 1 for j = 1,..., i 1. Consider the elements w, v 1,..., v i 1 V i where 0 w W and v j is in the preimage of v j. We shall prove these elements forms a basis of V i. Let φ be the canonical homomorphism V V/W. If λ 0 w + i 1 j=1 λ jv j = 0 where λ j C, then i 1 i 1 i 1 0 = φ(0) = φ λ 0 w + λ j v j = λ 0 φ(w) + λ j φ(v j ) = λ j v j. j=1 j=1 j=1 Because v 1,..., v i 1 are linearly independent, we must have λ 1 = = λ i 1 = 0. It follows that λ 0 w = 0, and since w 0, this yields λ 0 = 0. Thus w, v 1,..., v i 1 are linearly independent, and because dim V i = dim V i /W + dim W = (i 1) + 1 = i, we see that w, v 1,..., v i 1 is indeed a basis of V i. Finally, we show that V i is a submodule of V. Let x g, v V i. Then φ(v) V i 1, and by induction, we have ρ(x)v+w V i 1. Hence φ(ρ(x)v) V i 1, and so ρ(x)v V i. Corollary 2.6. Every solvable Lie algebra g has a chain of ideals 0 = a 0 a 1 a n = g where dim a i = i. Proof. We apply the proof of Corollary 2.5 to the adjoint module g. The statement follows immediately since each submodule of g is an ideal by Example 2.2.

26 Representations of nilpotent Lie algebras Recall from Proposition 1.12 that every nilpotent Lie algebra is solvable. In this section, we will apply our results about solvable Lie algebras to nilpotent Lie algebras in order to gain some very useful representations. We first introduce some standard results from linear algebra. Let V be a finitedimensional vector space and let T : V V be a linear transformation with eigenvalues λ 1,..., λ r. The generalized eigenspace of V with respect to λ i is the set V i of all v V annihilated by some power of T λ i 1. Let χ(t) = (t λ 1 ) m 1 (t λ 2 ) m2 (t λ r ) mr be the characteristic polynomial of T. It follows from the Jordan canonical form (see [DF04, 12.3]) that (i) V = V 1 V r. (ii) Each V i is invariant under the action of T. (iii) dim V i = m i, and the characteristic polynomial of T Vi is (t λ i ) m i. The relation between generalized eigenspaces and representations of nilpotent Lie algebras will be exhibited in the next theorem. A useful proposition is needed first. Proposition 2.7. Let g be a Lie algebra and V a g-module. For each y g, define the linear map ρ(y) : V V by ρ(y)v = yv. Let v V, x, y g, α, β C. Then (ρ(y) (α + β)1) n xv = n i=0 ( ) n ((ad y β1) i x ) ( (ρ(y) α1) n i v ). i Proof. We proceed by induction on n, the case where n = 0 being obvious. Suppose the statement is true for n = r. For each i, let x i = (ad y β1) i x g. Then (ρ(y) (α + β)1) r+1 xv = (ρ(y) (α + β)1) r i=0 ( ) r ρ(x i )(ρ(y) α1) r i v. i

27 20 Now (ρ(y) (α + β)1)ρ(x i ) = ρ([y, x i ]) + ρ(x i )ρ(y) (α + β)ρ(x i ) = ρ((ad y β1)x i ) + ρ(x i )(ρ(y) α1) = ρ(x i+i ) + ρ(x i )(ρ(y) α1). It follows that (ρ(y) (α + β)1) r+1 xv r ( ) r r = ρ(x i+1 )(ρ(y) α1) r i v + i i=0 i=0 r+1 ( ) r r+1 ( r = ρ(x i )(ρ(y) α1) r (i 1) v + i 1 i i=0 i=0 ( ( ) r since = 0 and 1 r+1 ( ) r + 1 ((ad = y β1) i x ) ( ρ(y) α1) r+1 i v ). i=0 i (The last line follows from the identity ( r+1 i ) ( = r ( i 1) + r i).) This completes the induction. ( ) r ρ(x i )(ρ(y) α1) r i+1 v i ) ρ(x i )(ρ(y) α1) r+1 i v ( ) ) r = 0 r + 1 Theorem 2.8. Let g be a nilpotent Lie algebra and V a finite-dimensional g-module. For y g, let ρ(y) be the linear map described in Proposition 2.7. Then the generalized eigenspaces of V associated with ρ(y) are all submodules of V. Proof. Let V i be a generalized eigenspace of V. If x g and v V i, then (ρ(y) λ i 1) n xv = n i=0 ( ) n ((ad y) i x ) ( (ρ(y) λ i 1) n i v ) i by Proposition 2.7 with α = λ i, β = 0. Since g is nilpotent, we have (ad y) i x = 0 if i is sufficiently large. And since v V i, we have (ρ(y) λ i 1) n i v = 0 if n i is sufficiently large. Hence (ρ(y) λ i 1) n xv = 0 if n is sufficiently large. It follows that xv V i, and so V i is a submodule of V. Corollary 2.9. Let g be a nilpotent Lie algebra and V a finite-dimensional indecomposable g-module. Then a basis can be chosen for V with respect to which we obtain a

28 21 representation ρ of g of the form λ(x) ρ(x) = 0 λ(x) for all x g. Proof. By Corollary 2.5, we may choose a basis of V with respect to which ρ(x) is uppertriangular. The generalized eigenspaces of V with respect to ρ(x) are all submodules of V by Theorem 2.8, and V is their direct sum. Since V is indecomposable, only one of the generalized eigenspaces can be nonzero. Thus ρ(x) has only one eigenvalue. Let λ(x) be this eigenvalue. Then all of the diagonal entries of the upper-triangular matrix ρ(x) will be equal to λ(x). One easily checks that the map x λ(x) is a 1-dimensional representation of g. Theorem 2.8 showed us that for any element y g, we obtain a decomposition of V into a direct sum of g-submodules. We wish now to obtain a more general decomposition of V, i.e., one that does not rely on the choice of y. Let g be a nilpotent Lie algebra and V a finite-dimensional g-module. For any 1-dimensional representation λ of g, we define the set V λ = {v V ( x g)( N(x) 1) (ρ(x) λ(x)1) N(x) v = 0}. Theorem V = λ V λ, and each V λ is a submodule of V. Proof. We first decompose V into a direct sum of indecomposable submodules. Then by Corollary 2.9, each such submodule will induce a 1-dimensional representation λ of g. Let W λ be the direct sum of all indecomposable submodules giving rise to λ. We have V = λ W λ.

29 I shall prove that W λ = V λ, and thus W λ is independent of the way we obtained our indecomposable components. Suppose W λ V λ. Observe that W λ V λ by Corollary It follows from a standard result in linear algebra that W λ + W µ V λ = W λ + V λ W µ. µ λ µ λ We have W λ + ( µ λ W µ = V, and thus V λ = W λ + V λ ) µ λ W µ. Since V λ W λ, there must exist a nonzero v V λ µ λ W µ. We write v = µ λ w µ where w µ W µ. Now the vector space g over the infinite field C cannot be expressed as the union of finitely-many proper subspaces. One easily checks that for each µ λ, the set of x g satisfying µ(x) = λ(x) is a proper subspace. Thus there exists an x g such that µ(x) λ(x) for all µ λ. Now since v V λ, there exists an N λ 1 such that (ρ(x) λ(x)1) N λ v = 0. And since w µ W µ, there exists an N µ 1 such that (ρ(x) µ(x)1) Nµ w µ = 0. Hence (ρ(x) µ(x)1) Nµ v = 0. µ λ Because µ(x) λ(x) for all µ λ, we see that the polynomials (t λ(x)) N λ, (t µ(x)) Nµ µ λ are coprime. Thus there exist polynomials a(t), b(t) C[t] such that a(t)(t λ(x)) N λ + b(t) µ λ(t µ(x)) Nµ = 1. It follows that a(ρ(x))(ρ(x) λ(x)1) N λ v + b(ρ(x)) µ λ(ρ(x) µ(x)1) Nµ v = v. The left-hand side of this equation is zero by the above results, and thus v = 0. This is a contradiction, so we must have W λ = V λ. It follows that V = λ V λ, and each V λ is a submodule of V.

30 23 If λ is a 1-dimensional representation of g and V λ 0, then we call λ a weight of V, and we call V λ the weight space of λ. Given Theorem 2.10, we call the decomposition λ V λ the weight space decomposition of V. Since each V λ is the direct sum of the indecomposable components giving rise to λ, it follows from Corollary 2.9 that a basis can be chosen for V λ with respect to which a representation ρ of g on V λ has the form λ(x) ρ(x) = 0 λ(x) for all x g. This fact will prove extremely useful throughout the rest of this paper. The final theorem of this chapter illustrates the connection between nilpotent Lie algebras and nilpotent linear transformations. Theorem 2.11 (Engel s theorem). A Lie algebra g is nilpotent if and only if ad x : g g is nilpotent for all x g. Proof. Suppose g is nilpotent. Then g n = 0 for some n. Let y g. We have ad x y g 2, (ad x) 2 y g 3,..., (ad x) i 1 y g i,... Thus (ad x) n 1 y g n = 0. It follows that (ad x) n 1 = 0, and so ad x is a nilpotent linear map. Now suppose ad x is nilpotent for all x g. We wish to show that g is nilpotent. Suppose g is not nilpotent. Choosing a maximal nilpotent subalgebra h of g, we may regard g as an h-module. Then h is an h-submodule of g. Let m be an h-submodule of g properly containing h such that m/h is irreducible. By Theorem 2.4, we have dim m/h = 1. Then m/h = C(x + h) for some x m\h, and m = h Cx.

31 24 The 1-dimensional representation induced by m/h must be the zero map. To see why, suppose there existed a y h such that ad y (x + h) = λx + h with λ 0. Then ad y x = λx + z for some z h. Thus (ad y) n x = λ n x + (ad y) n z 0 for all n since the second term is in h. Hence ad y would fail to be nilpotent, contrary to our assumption. Because the 1-dimensional representation induced by m/h is the zero map, we have [h, m] h. It follows that [m, m] = [h Cx, h Cx] [h, h] + [h, Cx] + [Cx, h] + [Cx, Cx] = [h, h] + [h, Cx] h. Thus m is a subalgebra of g and h is an ideal of m. We shall prove that for each i 1, there exists a positive integer e(i) such that m e(i) h i. We proceed by induction on i. The statement is true for i = 1 since m 2 h. Now suppose the statement is true for i = r. Then m e(r)+1 = [ ] m e(r), h Cx [ ] [h r, h] + m e(r), Cx = h r+1 + ad x m e(r). I claim that m e(r)+j h r+1 + (ad x) j m e(r) for all j 1. We proceed by induction on j, the case for j = 1 being proved. Assuming the statement is true for j = s, we have [ ] m e(r)+s+1 h r+1 + (ad x) s m e(r), m [ ] h r+1 + (ad x) s m e(r), h Cx h r+1 + (ad x) s+1 m e(r) since h r+1 is an ideal of m and (ad x) s m e(r) (ad x) s h r h r. Thus we have shown m e(r)+j h r+1 + (ad x) j m e(r) for all j. Since ad x is nilpotent, (ad x) j = 0 if j is sufficiently large. For such j, we have m e(r)+j h r+1. Setting e(r + 1) = e(r) + j, we obtain m e(r+1) h r+1, completing the induction. Now because h is nilpotent, h i = 0 if i is sufficiently large. For such i, we have m e(i) = 0, showing that m is nilpotent. This contradicts the maximality of h, and so g must be nilpotent.

32 25 Note: The traditional statement of Theorem 2.11 applies to arbitrary finitedimensional g-modules and linear maps. (See [Ser92, V.2].) The proof, however, involves embedding g into gl n (C) for some n. This requires a result known as Ado s theorem, the proof of which lies beyond the scope of this paper. (For a proof, refer to [Jac71, VI.2].) Corollary A Lie algebra g is nilpotent if and only if g has a basis with respect to which the adjoint representation of g has the form 0 ρ(x) = 0 0 for all x g. Proof. Suppose g is nilpotent. Then g has a chain of ideals g g 2 g 3 g r = 0 for some r. We make this chain finer by choosing a series of subspaces between consecutive ideals, each having codimension 1 in its predecessor. Let a be a subspace such that g i a g i+1. Then [a, g] [ g i, g ] = g i+1 a, showing that a is an ideal. Hence we obtain a chain of ideals g = a n a n 1 a 1 a 0 = 0 with dim a k = k and [g, a k ] a k 1. Choosing a basis of g adapted to this chain of ideals, the representation ρ(x) will be of strictly upper-triangular form for all x g. Conversely, suppose a basis can be chosen for g with respect to which ρ(x) is strictly upper-triangular for all x g. Then ρ(x) is clearly nilpotent, whence ad x is nilpotent for all x g. Thus g is nilpotent by Theorem 2.11.

33 Chapter 3 Cartan subalgebras If g is a Lie algebra and h is a subalgebra of g, then g forms an h-module via the adjoint action. If g is semisimple and h is chosen carefully, then the action of h induces a decomposition of g that tells us essentially all we need to know about the structure of g. In this chapter, we focus on finding the subalgebra h yielding such a decomposition. 3.1 Existence of Cartan subalgebras Let g be a Lie algebra and h a subalgebra of g. We define the normalizer of h to be the set N(h) = {x g [h, x] h for all h h}. Proposition 3.1. N(h) is a subalgebra of g, h is an ideal of N(h), and N(h) is the largest subalgebra of g containing h as an ideal. Proof. For the first statement, let x, y N(h), h h. Then [h, [x, y]] = [[y, h], x] + [[h, x], y] h by the Jacobi identity, and so N(h) is a subalgebra of g. The second statement is obvious from the definition of N(h). For the third statement, notice that if h is an ideal of m, then [h, m] h so that m N(h). A subalgebra h of g is a Cartan subalgebra if h is nilpotent and N(h) = h. We wish to show that every Lie algebra contains a Cartan subalgebra.

34 Let x g and consider the map ad x : g g. Define g 0,x to be the generalized eigenspace of ad x with eigenvalue 0, that is, 27 g 0,x = {y g (ad x) n y = 0 for some n 1}. We call g 0,x the null component of g with respect to x. An element x g is regular if dim g 0,x is as small as possible. Clearly, any Lie algebra will contain regular elements. Theorem 3.2. If x is a regular element of g, then the null component g 0,x is a Cartan subalgebra of g. Proof. Let h = g 0,x. We first show that h is a subalgebra of g. Let y, z h. Then (ad x) n [y, z] = n [ (ad x) i y, (ad x) n i z ] i=0 by Proposition 2.7 with V = g and α = β = 0. Since y h, we have (ad x) i y = 0 if i is sufficiently large. And since z h, we have (ad x) n i z = 0 if n i is sufficiently large. Hence (ad x) n [y, z] = 0 if n is sufficiently large. This shows that [y, z] h, and so h is a subalgebra of g. We next show that h is nilpotent. Set dim h = l and let b 1,..., b l be a basis of h. Let y = λ 1 b i + +λ l b l h where λ 1,..., λ l C. Since h is a subalgebra, the linear map ad y : g g restricts to ad y : h h. This induces the linear map ad y : g/h g/h. Let χ(t) be the characteristic polynomial of ad y on g, χ 1 (t) the characteristic polynomial of ad y on h, and χ 2 (t) the characteristic polynomial of ad y on g/h. We have χ(t) = χ 1 (t)χ 2 (t). Now χ(t) = det(t1 ad y) and y depends linearly on λ 1,..., λ l. Hence the coefficients of χ(t) are polynomial functions of λ 1,..., λ l. The same applies to χ 1 (t) and χ 2 (t). Let χ 2 (t) = d 0 + d 1 t + d 2 t 2 +

35 28 where d 0, d 1, d 2,... are polynomial functions of λ 1,..., λ l. Note that d 0 is not the zero polynomial, for in the special case y = x, the eigenvalues of ad y on g/h are nonzero and d 0 is their product. Let χ 1 (t) = t m ( c 0 + c 1 t + c 2 t 2 + ) where c 0, c 1, c 2,... are polynomial functions of λ 1,..., λ l and c 0 is not the zero polynomial. Then m l = deg χ 1 (t). It follows that χ(t) = t m (c 0 d 0 + a sum involving positive powers of t). Now c 0 d 0 is not the zero polynomial, and thus we may choose elements λ 1,..., λ l in C such that c 0 d 0 is nonzero. For such an element y we have dim g 0,y = m. Since x is a regular element with dim g 0,x = l, we have l m. But we also have m l, and so m = l. Thus t l divides χ 1 (t), and since dim χ 1 (t) = l, we have χ 1 (t) = t l. It follows by the Cayley-Hamilton theorem that for any y h, (ad y) l : h h is the zero map. Thus ad y is nilpotent for all y h, and so h is nilpotent by Theorem Finally, we show that N(h) = h. Clearly h N(h). So let z N(h). Then [x, z] h, and hence (ad x) n [x, z] = 0 for some n 1. But this means (ad x) n+1 z = 0, and so z h. It follows that N(h) h, and thus N(h) = h. 3.2 Derivations and automorphisms A derivation of a Lie algebra g is a linear map D : g g satisfying D[x, y] = [Dx, y] + [x, Dy] for all x, y g. Proposition 3.3. ad x is a derivation for all x g. Proof. If y, z g, then ad x [y, z] = [x, [y, z]] = [[x, y], z] + [y, [x, z]] = [ad x y, z] + [y, ad x z] by the Jacobi identity. Thus ad x is a derivation of g.

36 An automorphism of g is an isomorphism φ : g g. The automorphisms of g form a group Aut(g) under composition of maps. 29 Proposition 3.4. If D is a nilpotent derivation of g, then exp(d) is an automorphism of g. Proof. Since D is nilpotent, we have D n = 0 for some n. It is easy to check that the Leibniz rule holds for the linear map D, that is, D r [x, y] = r i=0 ( ) r [D i x, D r i y ] for all x, y g. i It follows that n 1 [exp(d)x, exp(d)y] = = = = i=0 2n 2 k=0 2n 2 k=0 2n 2 k=0 D i x i! l=0 n 1, j=0 D j y j! k [ D l ] x, Dk l y l! (k l)! 1 k! k l=0 D k [x, y] k! = exp(d)[x, y]. ( ) k [ D l x, D y] k l l (by the Leibniz rule) Hence exp(d) : g g is a homomorphism. Likewise, exp( D) : g g is a homomorphism, and we see that exp(d) exp( D) = 1. Thus exp(d) is an automorphism. An inner automorphism of g is an automorphism of the form exp(ad x) for x g with ad x nilpotent. The inner automorphism group Inn(g) is the subgroup of Aut(g) generated by all inner automorphisms. Proposition 3.5. Inn(g) is a normal subgroup of Aut(g). Proof. Let φ Aut(g). Every element of Inn(g) is of the form exp(ad x 1 ) exp(ad x 2 ) exp(ad x r )

37 30 where x 1,..., x r g and ad x 1,..., ad x r are nilpotent. We have exp(ad x 1 ) exp(ad x 2 ) exp(ad x r ) = exp(ad x 1 + ad x ad x r ) = exp(ad (x 1 + x x r )). Let x = x 1 + x x r. For all y g, we have φ(ad x)φ 1 y = φ[x, φ 1 y] = [φx, y] = ad φx y. Thus φ(ad x)φ 1 = ad φx. By linearity, we have φ(exp(ad x))φ 1 = exp(ad φx). Hence φ(exp(adx 1 ) exp(adx 2 ) exp(adx r ))φ 1 = exp(adφx 1 ) exp(adφx 2 ) exp(adφx r ). Since φ is an automorphism, we see that ad φx 1,..., ad φx r are nilpotent, and so φ(exp(ad x 1 ) exp(ad x 2 ) exp(ad x r ))φ 1 Inn(g). Two subalgebras h, k of g are conjugate in g if there exists a φ Inn(g) such that φ(h) = k. We shall prove that any two Cartan subalgebras of g are conjugate in g. We first require some concepts from algebraic geometry. 3.3 Concepts from algebraic geometry If h is a nilpotent subalgebra of a Lie algebra g, then g forms an h-module via the restriction of the adjoint action. This action gives rise to the weight space decomposition g = λ g λ as in Theorem 2.10 where g λ = {x g ( h h)( n 1) (ad h λ1) n x = 0}. It follows from Corollary 2.12 that h g 0. Suppose our nilpotent subalgebra h is equal to g 0. Then there exist nonzero 1-dimensional representations λ 1,..., λ r of h such that g = h g λ1 g λr. Thus if x g, then x = x 0 + x x r where x 0 h and x i g λi for i = 1,..., r.

38 31 Proposition 3.6. ad x i is nilpotent for all i 0. Proof. Let µ : h C be a weight of the h-module g and let y g µ. We have n (ad h µ(h)1 λ i (h)1) n [ [x i, y] = (ad h λi (h)1) j x i, (ad h µ(h)1) n j y ] j=0 for all h h by Proposition 2.7. Since x i g λi, we have (ad h λ i (h)1) j x i = 0 if j is sufficiently large. And since y g µ, we have (adh µ(h)1) n j y = 0 if n j is sufficiently large. Hence (ad h µ(h)1 λ i (h)1) n [x i, y] = 0 if n is sufficiently large. This shows that [x i, y] g λi +µ, and thus ad x i g µ g λi +µ. Now because λ i 0 and there are only finitely-many µ : h C such that g µ 0, we see that (ad x i ) N = 0 if N is sufficiently large. Thus ad x i is nilpotent. f : g g by It follows immediately that exp(ad x i ) Inn(g) for i 0. We define the map f(x) = exp(ad x 1 ) exp(ad x 2 ) exp(ad x r ) x 0. Let {b ij } be a basis of g where i = 1,..., r and, for a fixed i, the elements b ij form a basis of g λi with respect to which elements of h are represented by upper-triangular matrices as in Corollary 2.5. (Here λ 0 = 0.) Proposition 3.7. f : g g is a polynomial function, that is, ( ) f λij b ij = µ ij b ij where each µ ij is a polynomial in the λ kl. Proof. Each map ad x i : g g is clearly linear. We also have N (ad x i ) k exp(ad x i ) = k! k=0 for some N 1 since ad x i is nilpotent. Thus exp(ad x i ) : g g is a polynomial function. The map f is a composition of the linear map x x 0 with the polynomial functions exp(ad x i ) and is thus a polynomial function.

39 32 We now write µ ij = f ij (λ kl ). We define the Jacobian matrix J(f) = ( f ij / λ kl ) and the Jacobian determinant det J(f) of f. Clearly det J(f) is an element of the polynomial ring C[λ kl ]. Proposition 3.8. det J(f) is not the zero polynomial. Proof. We shall show that det J(f) 0 when evaluated at a carefully chosen element of h. Let h h and consider ( f ij / λ kl ) h. Suppose k 0. We have Now suppose k = 0. We have f(h + tb kl ) f(h) ( f/ λ kl ) h = lim t t exp(ad tb kl )h h = lim t t h + t[b kl, h] + h = lim t t Thus J(f) is a block matrix of the form = [b kl, h] = [h, b kl ] λ k (h)b kl (mod b k1,..., b kl 1 ). f(h + tb 0l ) f(h) ( f/ λ 0l ) h = lim t t h + tb 0l h = lim = b 0l. t t J 0 0 J 1 0 J r

40 33 where J 0 is the identity matrix and J i is of the form λ i (h) 0 λ i (h) for i = 1,..., r. It follows that (det J(f)) h = ± r i=1 λ i(h) d i where d i = dim g λi. Now the vector space h over the infinite field C cannot be expressed as the union of finitelymany proper subspaces. For each i, the set of h h satisfying λ i (h) = 0 is a proper subspace of h since the linear maps λ i are all nonzero. Thus there exists an h h such that λ i (h) 0 for all i. For such an h, we have (det J(f)) h 0, showing that det J(f) is not the zero polynomial. Proposition 3.9. The polynomial functions f ij are algebraically independent. Proof. Suppose there exists a nonzero polynomial F (x ij ) C(x ij ) such that F (f ij ) = 0. We choose an F with minimal total degree in the variables x ij. For each index k, l, we have λ kl F (f ij ) = 0 so that i,j F f ij f ij λ kl = 0. Let v be the vector ( F/ f ij ). Then vj(f) = (0,..., 0). Since det J(f) = 0, we must have v = (0,..., 0), and hence F/ f ij = 0 for all f ij. Notice that F/ x ij is a polynomial in C[x ij ] of smaller total degree than F. By our choice of F, F/ x ij must be the zero polynomial, that is, F does not involve the indeterminate x ij. This is true for all x ij, and so F must be a constant polynomial. Since F (f ij ) = 0, this constant must be zero. This contradicts the fact that F is not the zero polynomial, and so the f ij must be algebraically independent.

41 Let B = C[f ij ] be the polynomial ring in the f ij and A = C[λ ij ] the polynomial ring in the λ ij. Consider the homomorphism θ : B A uniquely determined by 34 θ(f ij ) = f ij (λ kl ) A. Proposition The homomorphism θ : B A is injective. Proof. Let F ker θ, that is, θ(f ) = 0. Then F (f ij ) = 0 as a function of the λ kl. Since the f ij are algebraically independent, this implies F = 0. Thus ker θ = 0, and so θ is injective. It follows that θ is an embedding of B into A. The following is a standard result from algebraic geometry. Proposition Let A and B be integral domains such that B A, A, B have common identity element 1, and A is finitely-generated over B. If p is a nonzero element of A, then there exists a nonzero element q of B such that any homomorphism φ : B C with φ(q) 0 extends to a homomorphism ψ : A C with ψ(p) 0. Proof. We refer the reader to [Car05, Proposition 3.10]. We wish to apply this result to our earlier situation. Let d = dim g. We have the polynomial function f : C d C d (λ ij ) (f ij (λ kl )). Let V = C d. For each polynomial p C[x ij ], we define the set V p = {v V p(v) 0}. Corollary For each nonzero polynomial p C[x ij ], there exists a nonzero polynomial q C[x ij ] such that V q f(v p ).