Lie-algebras: Part 2

Transcription

1 Lie-algebras: Part 2 Thomas Gobet These are the notes of a one semester course on complex Lie algebras and their representation theory given at the TU Kaiserslautern during the wintersemester 2014, coming after a first part of one semester given by Gunter Malle. The aim is to give the classification of (semi)simple complex finitedimensional Lie algebras as well as their representation theory, assuming preliminary knowledge on complex Lie algebras (representation theory of sl 2 (C), nilpotent and solvable Lie algebras, root spaces of complex semisimple Lie algebras and root space decomposition). The first two chapters of the present script follow rather closely notes of a similar second part given by Gunter Malle a few years ago. If you find inaccuracies or typesetting mistakes, I would be very grateful if you take a few seconds to send me an gobet@mathematik.uni-kl.de. I thank Gunter Malle for his reading of the script and for his suggestions. Contents 1 Root systems Reflections in a euclidean vector space and root systems Bases of root systems The Weyl group of a root system Cartan-Matrices and Dynkin diagrams Classification of root systems Classification of Dynkin diagrams Root systems of classical Lie algebras Simplicity of sp 2n (C) Generators and relations Universal enveloping algebras Definition and first properties Poincaré-Birkhoff-Witt Theorem Tensor algebra and existence of universal enveloping algebras. 31 1

2 1 ROOT SYSTEMS 2 4 Highest weight theory Weights of a representation Coinduced modules and Verma modules Classification of simple highest weight modules Universal enveloping algebras are integral domains Classification of finite-dimensional simple modules The Weyl character formulas 51 1 Root systems Let E denote a euclidean vector space, that is, a finite-dimensional vector space over a totally ordered field k equipped with a scalar product For example one can take k = R. (, ) : E E k. 1.1 Reflections in a euclidean vector space and root systems Definition 1.1. For any 0 v E, the element s v End k (E) defined by is called a reflection along v. s v (x) = x 2 (x,v) (v,v) v One has s v (v) = v and s v (x) = x if x v. In particular, since E = kv v, one has s 2 v = id. Hence s v GL(E). We also say that s v is the reflection with reflecting hyperplane v. Notice that for any α k, one has s αv = s v. In particular, we can always choose v such that (v,v) = 2 and then simply write s v (x) = x (x,v)v. Notice that s v O(E), the orthogonal group of E. In other words, s v preserves the scalar product (, ). Indeed, for x,y E, we have ( (s v (x),s v (y)) = x 2 (x,v) ) (v,v) v,y 2(y,v) (v,v) v = (x,y)+4 (x,v)(y,v)(v,v) (v,v) 2 = (x,y). 2 (x,v)(y,v) (v,v) 2 (y,v)(x,v) (v,v)

3 1 ROOT SYSTEMS 3 Notation. For x,v E with v 0, we use the classical notation x,v for the scalar x,v := 2 (x,v) (v,v) k Notice that, is only linear in the first argument. Definition 1.2. Let R E. Then R is a root system if the following four conditions are satisfied: (R1) The set R is finite, generates E and 0 / E. (R2) For all α R, one has kα R = {±α}. (R3) For all α R, one has s α (R) R. (R4) For all α,β R, one has α,β Z. If R is a root system, then we call any α R a root. Lemma 1.3. Let R be a root system, α,β R with β ±α. Then 1. The product α,β β,α lies in {0,1,2,3}. 2. If α,β < 0, then α+β R. 3. If α,β > 0, then α β R. Proof. Proven in Part I. 1.2 Bases of root systems Definition 1.4. A subset R is a base of R if the following two conditions are satisfied: (B1) The set is a linear basis of E. (B2) Any β R can be written in the form β = α n α α with n α Z where all the n α have the same sign. Example 1.5. Let L = sl 3 (k). Then Φ = {±(e 11 e }{{ 22 ),±(e } 11 e 33 ),±(e }{{} 22 e 33 )} }{{} α α+β β is a root system; a possible basis is given by = {α,β}.

4 1 ROOT SYSTEMS 4 Lemma 1.6. Let R be a basis. Let α,β with α β. Then (α,β) 0 and α β / R. Proof. Assume that (α,β) > 0. By 1.3, 3), this implies that α β R, which contradicts the condition B2. Theorem 1.7 (Existence of bases). Any root system R has a basis. Proof. Let z E with (z,α) 0 for any α R. Notice that one can always find such a z E: if not, it would mean that for any z E, there exists an α R with (z,α) = 0. Now the kernel of (,α) is a hyperplane, hence this would mean that we can cover E by a finite number of hyperplanes, which is not possible. Set R + := {α R (z,α) > 0}, := {α R + β,γ R +,α β +γ}. We claim that is a basis of R. We first show the condition B2. Let β R. Then either β R +, or β R +. Hence it suffices to show that any β R + is a linear combination of elements from with nonnegative integral coefficients. Assume not. Then choose β R + which has not this form, with (z,β) minimal. Obviously β /, hence β = β 1 +β 2, β i R +. We have (z,β) = (z,β 1 ) +(z,β }{{} 2 ), }{{} >0 >0 hence 0 < (z,β i ) < (z,β), i = 1,2. If we could write both β 1,β 2 as linear combinations of with nonnegative integral coefficients, then the same would hold for β. Hence one of them is not such a linear combination, which is a contradiction to the choice of β with (z,β) minimal. We now show that B1 holds. Let α,β. We claim that (α,β) 0. Otherwise, one has γ := α β R by 1.3 3), hence α = β +γ R +, which implies γ / R + since α. It implies γ R +, hence β = γ + α is a linear combination of roots in R +, a contradiction to β. Now assume that α c αα = 0, with c α k. We put all the positive coefficients on the same side: x := c α α = ( c β )β α, c α>0 β, c β <0 and compute (x,x) = α,β c α ( c }{{} β )(α,β) }{{}}{{} >0 >0 0

5 1 ROOT SYSTEMS 5 from which it follows that (x,x) = 0, hence x = 0. Hence 0 = (x,z) = c α (α,z) c }{{} α = 0 α. α, c α>0 >0 Similarly we see that c β = 0 for all β so is linearly independent. Since R we deduce that generates E by (R1). Notice that bases of root systems are not unique in general. Definition 1.8. Let R be a basis. We write R + for the set of positive roots (with respect to ), that is, the roots which are nonnegative linear combinations of elements of. We set R := R + for the negative (with respect to ). The elements of are called the simple roots and a reflection along a simple root is a simple reflection. 1.3 The Weyl group of a root system Definition 1.9. Let R be a root system. The group W := s α α R generated by all the reflections along the roots is the Weyl group of R. Lemma The Weyl group of R is finite. Proof. It follows from (R3) that W permutes R. Hence we have a group homomorphism ϕ : W Aut(R). If w ker(ϕ), then g fixes pointwise any basis R which is also a basis of E by (B2), hence g = id E. It follows that ϕ is injective, hence W is finite since Aut(R) is finite. Lemma Let α. Then s α (R + \{α}) = R + \{α}. Proof. Let β R +, β α, β = γ c γγ with c γ 0. Since γ α, it follow from (R2) that there exists α γ such that c γ 0. But by (R3) we have s α (β) = β β,α α R. The coefficient c γ hence does not change when applying s α, hence by (B1) all the coefficients of the expansion of s α (β) R in must be positive, implying s α (β) R +. Fix a basis of R and set W 0 := s α α W. Lemma Let β R. There exists g W 0 and α such that β = g(α). Proof. First assume that β R +, β = γ c γγ with c γ 0. We define the height ht(β) of β as ht(β) = γ c γ Z >0.

6 1 ROOT SYSTEMS 6 The proof is by induction on ht(β). If ht(β) = 1 then β, in which case one can take g = id E, α = β. Hence assume that ht(β) 2. We claim that there is γ with (β,γ) > 0. Otherwise (β,γ) 0 for all γ which implies that (β,β) = γ c }{{} γ (β,γ) 0 (β,β) = 0 β = 0. }{{} 0 0 Let γ such that (β,γ) > 0. We have s γ (β) = β β,γ γ, hence by 1.11 we obtain ht(s γ (β)) = ht(β) β,γ < ht(β). By induction, there exist α, h W 0 such that s γ (β) = h(α), hence β = (s γ h)(α), s γ h W 0. Now if β R, then β R +, hence by the first part of the proof there is g W 0, α such that β = g(α). But it implies that β = (gs α )(α), gs α W 0, which concludes. Corollary The Weyl group W is generated, by the s α, α. In other words, we have W = W 0. Proof. Let β R. We show that s β W 0. Thanks to 1.12, there exists g W 0, α such that β = g(α). It implies that for any v E, s β (v) = v 2 (v,g(α)) ( ) (g(α),g(α)) g(α) = g g 1 (v) 2 (g 1 (v),α) α = (gs α g 1 )(v), (α,α) hence s β = gs α g 1 W 0. We now want to show that W acts transitively on the set of bases of R. To this end we need Lemma Let P,Q Z n n be matrices with nonnegative entries such that PQ = I n. Then P and Q are permutation matrices. Proof. Exercise (Hint: The set S of n-tuples of nonnegative integers is stable by multiplication by P). Theorem Let R be a root system with bases,. There exists w W such that w( ) =. Proof. Let R +, R be the sets of positive and negative roots with respect to and R +, R be the sets of positive and negative roots with respect to.

7 1 ROOT SYSTEMS 7 Notice that R + = R + = R 2. Write = {α 1,...,α n }, = {α 1,...,α n}. The proof is by induction on R + R. If R + R = 0 then R + = R +. It implies that any element from is an nonnegative integral combination of elements of, that is, α j = p ij α i and conversely α k = j q jkα j. Hence setting P = (p ij ), Q = (q jk ) we have PQ = I n. By 1.14, P and Q are permutation matrices, which implies =. Now assume that R + R = m > 0. Then R 0, otherwise one would have R +, hence R + R + which would imply R + = R +, a contradiction with R + R > 0. Let α R. By 1.11 we have s α (R + ) = (R + \{α}) { α}. In particular, α / s α (R + ) R. It implies that s α (R + ) R = m 1. It is clear that s α ( ) is again a basis of R with corresponding set of positive roots given by s α (R + ), hence by induction there exists y W such that y(s α ( )) =. Setting w = ys α we get the claim. 1.4 Cartan-Matrices and Dynkin diagrams Definition Let R be a root system with basis = {α 1,...,α l }. The matrix C := ( α i,α j ) ij is the Cartan matrix of R (with respect to ). Since s β (α i ),s β (α j ) = α i,α j for any i,j and β, one has thanks to Theorem 1.15 that the Cartan matrix does not depend on the choice of basis but only on the ordering of the simple roots. Example Type A 1, R = {±α}, = {α}, α,α = Type A 2, R + = {α,β,α+β}, = {α,β}, α,α = 2 = β,β. Using the tables given in Part I, one sees that it corresponds to an angle of θ = 2π between the roots. One has 3 α,β = 1 = β,α and s α (β) = α+β = s β (α). Hence ( ) 2 1 C = Type B 2, θ = 3π 4, α+β R, s α(β) = 2α+β R. One has C = ( )

8 1 ROOT SYSTEMS 8 4. Type G 2 ; one checks that C = ( ) The aim now is to represent root systems graphically. Let D = D(R) be the graph defined as follows: the vertices are the elements of. For α,β, we put d αβ = α,β β,α edges between α and β (recall that α,β β,α {0,1,2,3}). It is the Coxeter diagram of R. In case d αβ > 1, we must have(α,α) (β,β). We add an arrow from the longer to the shorter root. The resulting diagram is the Dynkin diagram D(R) of R. Example A 1 A 2 B 2 G 2 The root system Φ of the complex simple Lie algebra sl n (C) given by Φ = ±{α i + +α j 1 i,j n 1} = {α 1,...,α n 1 } with α i = e i e i+1, 1 i n 1, α i,α i+1 1, α i,α j = 0 for i j > 1 has corresponding Dynkin diagram... α 1 α 2 α n 1 Thanks to lemma 1.6, one has (α,β) 0 for any α,β with α β, implying α, β 0. The Cartan matrix can therefore be recovered from the Dynkin diagram. A natural question arises here: which information on R can we recover from D(R)? Definition Let R, R be root systems in euclidean spaces E and E. We say that (R,E) and (R,E ) are isomorphic is there exists a vector space isomorphism ϕ : E E such that ϕ(r) = R and α,β = ϕ(α),ϕ(β) for all α,β R. Notice that the second condition implies that α,β β,α = 4cos 2 (θ) = 4cos 2 (θ ), where θ (respectively θ ) is the angle between α and β (resp. ϕ(α) and ϕ(β)). It implies that cos(θ) = ±cos(θ ), but since α,β = ϕ(α),ϕ(β), the real numbers cos(θ) and cos(θ ) must have the same sign, whence θ = θ. Hence an isomorphism of root systems preserves the angles between the roots (but not necessarily the lengths of the roots).

9 1 ROOT SYSTEMS 9 Example Let R be a root system, c R. Then ϕ c : E E defined by v cv is an isomorphism from R to the root system R := {cα α R}. 2. For any α R, the reflection s α : E E is an automorphism of R. Proposition Let (R,E) and (R,E ) be root systems with D(R) = D(R ). Then (R,E) = (R,E ). Proof. Since D(R) = D(R ), there exist basis = {α 1,...,α n } of R, = {α 1,...,α n } of R such that α i,α j = α i,α j for all i,j = 1,...,n (see the remark before Definition 1.19). Define a linear map ϕ : E E, α i α i. It is an isomorphism of vector spaces and satisfies the second condition of Definition 1.19 for simple roots. We first show that ϕ(r) = R. For v E, α i we have ϕ(s αi (v)) = ϕ(v v,α i α i ) = ϕ(v) v,α i α i. Now for v = n j=1 a jα j we have the equality v,α i = j a j α j,α i = j a j α j,α i = ϕ(v),α i, Hence by replacing in the previous equality we get that ϕ(s αi (v)) = s α i (ϕ(v)). Since the s αi generate the Weyl group W we deduce that the image of the W-orbit of v under ϕ is contained in the W -orbit of ϕ(v). By Lemma 1.12 it implies that ϕ(r) R. By inverting the roles of R and R we can give the same argument and conclude that ϕ 1 (R ) R, hence ϕ(r) = R. It remains to show that α,β = ϕ(α),ϕ(β) for anyα,β R. We simply writes i ands i for the reflections alongα i andα i. By Lemma 1.12, there exists w W and α i such that w(α i ) = β. Since the s i generate W, we can write w as a product s i1 s i2 s ik, i j {1,...,n}. We set w = s i 1 s i 2 s i k. Then using the equality ϕ(s j (v)) = s j (ϕ(v)) for any j = 1,...,n and v E which we already showed above we get the equalities ϕ(α),ϕ(β) = ϕ(α),ϕ(w(α i )) = ϕ(α),w (ϕ(α i )) = w 1 (ϕ(α)),ϕ(α i ) = ϕ(w 1 (α)),ϕ(α i ). By the first part of the proof we have ϕ(w 1 (α)),ϕ(α i ) = w 1 (α),α i,

10 1 ROOT SYSTEMS 10 hence ϕ(α),ϕ(β) = w 1 (α),α i = α,w(αi ) = α,β.

11 2 CLASSIFICATION OF ROOT SYSTEMS 11 2 Classification of root systems 2.1 Classification of Dynkin diagrams The aim of this subsection is to prove the following Theorem: Theorem 2.1. Let R be an irreducible root system. Then D(R) is one of the following: A l, l l B l, l l C l, l l D l, l l E 6. l 1 E 7. E 8. F 4. G 2. The proof requires several steps. Definition 2.2. Let E be an euclidean vector space. A finite subset A = {v 1,...,v n } E is admissible if

12 2 CLASSIFICATION OF ROOT SYSTEMS The set A consists of linearly independent vectors, 2. For any i,j with i j, one has (v i,v j ) 0, and (v i,v i ) = 1, 3. One has 4(v i,v j ) 2 {0,1,2,3} for any v i v j. The associate graph Γ A has vertices v 1,...,v n with d ij := 4(v i,v j ) 2 edges between the v i, v j, i j. Example 2.3. Let R E be a root system with a basis; set { } α A := α R (α,α) is admissible. The graph Γ A is the Coxeter diagram of R. Notice that if A is admissible, then any subset A A is admissible. Lemma 2.4. Let A be admissible. The number N of pairs {v i,v j }, v i v j such that there exists at least one edge between v i and v j is at most A 1. Proof. Set v := n i=1 v i, Then v 0 since the v i are linearly independent. We compute (v,v) = i,j Hence we have (v i,v j ) = n i=1(v i,v i )+ i j (v i,v j ) = n+2 (v i,v j ) > 0, i<j Hence N < n = A. n > ( 2(v i,v j )) = }{{} i<j i<j 0 dij N, }{{} 1 iff joined Corollary 2.5. Let A be admissible. The graph Γ A contains no cycle. Proof. Assume that A contains a cycle with vertex set A A. Then A is again admissible. But A has at least A pairs of joined vertices, contradicting Lemma 2.4. Lemma 2.6. Let A be admissible. There are at most three edges starting at a given vertex v i of Γ A.

13 2 CLASSIFICATION OF ROOT SYSTEMS 13 Proof. Letv A with neighbours v 1,...,v k. Thanks to Corollary 2.5 one has (v i,v j ) = 0 for 1 i < j k. Consider the linear span U of {v,v 1,...,v k }. Extend {v 1,...,v k } to an orthonormal basis of U by adding v 0. One must then have (v 0,v) 0 since otherwise v would lie in the span of v 1,...,v k contradicting the linearly independence of the elements of A. We can assume (v,v 0 ) > 0. Thanks to the Gram-Schmidt process we then get that v = k i=0 (v,v i)v i, hence 1 = k i=0 (v,v i) 2. Now since (v,v 0 ) > 0 one must have that k (v,v i ) 2 < 1, Hence 4 > i=1 k (v,v i ) 2 = {edges starting at v}. i=1 Corollary 2.7. If Γ A is connected and possesses two vertices v 1,v 2 which are connected by three edges, then Γ A = v 1 v 2 Proof. Immediate consequence of Lemma 2.6 Lemma 2.8. Let Γ A having a subgraph of the form Γ sub =... v 1 v 2 v k, That is, with no multiple edges between the v i. Set A := (A\{v 1,...,v k }) {v}, where v := k i=1 v i. Then A is admissible and Γ A is obtained from Γ A by contracting Γ sub into a single vertex. Proof. The set A is clearly linearly independent. For 1 i < k we have that 4(v i,v i+1 ) 2 = 1 and since by definition of an admissible set we know that (v i,v i+1 ) 0, then 2(v i v i+1 ) = 1. Moreover, (v i,v j ) = 0 if i j > 1. We therefore get (v,v) = i,j (v i,v j ) = k i=1(v i,v i )+2 i<j (v i,v j ) = k (k 1) = 1. Now letw A\{v 1,...,v k }. Thenwis joined to at most onev i (Corollary 2.5). Hence(w,v) = 0 or(w,v) = (w,v i ) implying that4(w,v) 2 {0,1,2,3}. Hence A is admissible. The claim about the graph is now clear.

14 2 CLASSIFICATION OF ROOT SYSTEMS 14 Definition 2.9. Let A be admissible. A vertex of Γ A is called a branch vertex if it is joined to exactly three other vertices. Lemma Let A be admissible. Assume that Γ A is connected. Then Γ A has 1. At most one double edge 2. At most one branch vertex 3. No double edge if it has a branch vertex, and no branch vertex if it has a double edge. Proof. If Γ A has at least two double edges, then it must have a subgraph of the form Γ sub =... v 1 v 2 v k. Using Lemma 2.8 we can obtain an admissible subset A with graphγ A given by Γ A =, a contradiction to Lemma 2.6. This proves 1. The same argument can be given for 2. and v 1 v 2 v k be a subgraph of Γ A and set Lemma Let v := k i=1 iv i. Then (v,v) = k(k +1). 2 Proof. For 1 i < k one has 2(v i,v i+1 ) = 1 and (v i,v j ) = 0 for i j > 1. We compute (v,v) = k k 1 ij(v i,v j ) = i 2 +2 i(i+1)(v i,v i+1 ) i,j i=1 i=1 k k 1 k 1 = i 2 i(i+1) = k 2 k(k +1) i =. 2 i=1 i=1 i=1 Lemma Let A be admissible. Assume that Γ A has a double edge. Then Γ A is either... or. Proof. Thanks to Lemma 2.10, Γ A has the form v 1 v 2 v k w l w 2 w 1.

15 2 CLASSIFICATION OF ROOT SYSTEMS 15 We can without loss of generality assume that k l. Let v := k iv i, w := i=1 l jw j, j=1 then by Lemma 2.12 one has (v,v) = k(k+1), (w,w) = l(l+1). Furthermore, 2 2 4(v k,w l ) = 2, while (v i,w j ) = 0 otherwise. Hence (v,w) 2 = (kv k,lw l ) 2 = k2 l 2 2. (1) Since v,w are linearly independent, one has thanks to the Cauchy-Schwartz inequality that (v, w) < (v, v)(w, w), which implies k 2 l 2 2 < k(k +1)l(l+1). 4 It follows that 2kl < (k+1)(l+1), hence (k 1)(l 1) < 2. It forces l = 1 or l = k = 2 since we assumed k l. In the first case there is no restriction on k. Proposition If Γ A has a branch vertex, then Γ A is either of type D n (n 4) or of exceptional type E 6, E 7, E 8. Proof. Thanks to Lemma 2.10, the graph Γ A must have the form... v 1 v 2 v k z... w l w 2 w 1... x m x 2 x 1, and we can without loss of generality assume that k l m 1. Set v := k iv i, w := i=1 l iw i, x := i=1 m ix i. i=1 Then v,w and x are pairwise orthogonal. For a {v,w,x} set â := a a. In in the proof of Lemma 2.6, the linear span U of v,w,x,z has an orthonormal basis {ˆv,ŵ,ˆx,z 0 } where z 0 is such that (z,z 0 ) 0. Write z = (z,ˆv)ˆv +(z,ŵ)ŵ+(z,ˆx)ˆx+(z,ẑ 0 )ẑ 0,

16 2 CLASSIFICATION OF ROOT SYSTEMS 16 and deduce that 1 = (z,z) = a {v,w,x} (z,â) 2 +(z,z 0 ) 2 }{{} 0 a {v,w,x} (z,â) 2 < 1. (2) But (z,v) 2 = (z,kv k ) 2 = k2, similarly 4 (z,w)2 = l2, 4 (x,w)2 = m2. By Lemma we have that v 2 = k(k+1), w 2 = l(l+1) and x 2 = m(m+1). We can thus rewrite 2 as We deduce that 2k 2 1 > 4k(k +1) + 2l 2 4l(l+1) + 2m 2 4m(m+1) k = 2(k +1) + l 2(l+1) + m 2(m+1) = (k +1) (l+1) (m+1). 1 < 1 k l m+1, (3) 1 but since k l m 1 we have By (3) it k+1 l+1 m+1 2 implies that 1 < 3, hence that m = 1. By (3) again we then have that m+1 1 < implying that 2 > 1. We deduce that l {1,2}. 2 k+1 l+1 l+1 2 First assume that l = 2. Then > 1, hence k < 5, and k l = 2. k The case k = 2 yields a diagram of type E 6, k = 3 of type E 7 and k = 4 of type E 8. 1 Now assume that l = 1. It implies that > 0, hence there is no k+1 restriction on k 1. We obtain the Dynkin diagrams of type D 3+k. We have all the required tools to prove Theorem 2.1: Proof of 2.1. Let R be an irreducible root system with basis R and corresponding admissible A (see Example 2.3). Then Γ A is the Coxeter diagram of R. Then if we forget the arrows in the Dynkin diagrams, we know by Corollary 2.7, Lemma 2.12 and Proposition 2.13 that the diagram Γ A has the required form. Now if there is no multiple edge in Γ A, then the Coxeter and Dynkin diagrams are the same and we are done. If Γ A has a multiple edge, then for diagrams of type F 4 and G 2 the two possible orientations of the arrow give isomorphism diagrams, while for the remaining diagrams one obtains either type B or type C. At this point it is not clear that any diagram gives a root system. It is also not clear that any root system corresponds to a simple Lie algebra.

17 2 CLASSIFICATION OF ROOT SYSTEMS Root systems of classical Lie algebras The aim is to show that the infinite families A l, B l, C l and D l come from the so-called classical simple Lie algebras sl l+1, so 2l+1, sp 2l and so 2l. Lemma Let V be a k-vector space, S End(V) a finite dimensional subspace of End(V) such that s S, s is diagonalizable, s,t S, st = ts. Then V admits a so-called simultaneous decomposition into eigenspaces V = λ S V λ, where V λ = {v V xv = λ(x)v x S}. Proof. The proof is by induction on dims. Let s 1,...,s k be a basis of S. Since s 1 is diagonalizable, we have V = i V λi, where λ i (ks 1 ), λ i = λ i (s 1 ). Now let i > 1, v V λj. Then s 1 s i v = s i s 1 v = λ j s i v, hence s i v V λj. Therefore the linear span U = s 2,...,s k of s 2,...,s k preserves any eigenspace V λj of s 1, and by induction we have for any j that V λj = µ U V µ. We now define µ j S by setting for x = αs 1 +y S: And for v V µ we have µ j (x) = λ j (αs 1 )+µ(y), x(v) = αs 1 (v)+y(v) = αλ j (s 1 )+µ(y)v = µ j (αs 1 +y)v, Hence V µ = V µj. Hence V λj is the direct sum of the V µj for all the µ, and summing over all the λ j one gets the result.

18 2 CLASSIFICATION OF ROOT SYSTEMS 18 Let L be a classical Lie algebra, that is, a Lie algebra among sl l+1, so 2l+1, sp 2l or so 2l. Let H L be the subalgebra of diagonal matrices. The aim now is to show that these algebras are simple. We recall from Part I that a sub algebra H of a Lie algebra L is toral if it only consists of ad-semisimple elements (i.e., for any h H, the endomorphism ad : L L, x [h,x] is diagonalizable). If H L is toral, then H is abelian, hence by Lemma 2.14 there is a decomposition L = H α Φ H L α, where L α = {x L [h,x] = α(h) h H}. By convention, the set Φ only contains the α H such that L α 0. We would like to determine under which additional condition H is a Cartan subalgebra. Proposition Let L be a complex Lie algebra, H L a toral subalgebra. Assume that 0 / Φ. Then H is Cartan subalgebra. Proof. We know that H has only semisimple elements. It remains to show that H is maximal for this property. We know that a toral subalgebra is abelian. Hence let x H with [H, x] = 0. Let us write x = h x + α Φ n α i=1 c i α xi α, where for each α Φ, {x 1 α,...,xnα α } is a basis of L α. Then we have 0 = [h,x] = α Φ n α i=1 c i α α(h)xi α, h H. By linear independence of{x i α } α,i, if α(h) 0, then c i α = 0. Now since 0 / Φ, then for each α Φ we can find an h H such that α(h) 0. We conclude that c i α = 0, α,i. Hence x = h x H. We would like to find an additional condition which would imply semisimplicity. Proposition Let L be a complex Lie algebra with a toral subalgebra H such that L = H α ΦL α, with 0 / Φ. Assume that

19 2 CLASSIFICATION OF ROOT SYSTEMS h H α Φ with α(h) 0, 2. α Φ, dim C L α = 1, 3. If α Φ, L α = Cx α, y L such that [[x α,y],x α ] 0. Then L is semisimple. Proof. Recall that L is semisimple if and only if L has no nontrivial abelian ideal. Let A L be an abelian ideal. We then have that [H,A] [L,A] A. Since the ad-operation of H on A is diagonalizable and [H,A] A, it follows that the ad-operation of H on A must also be semisimple. Hence ( ) A = (A H) A L α. Now if A L α 0 for some α Φ, then by condition 2 we must have L α A, hence [L α,l] A. Now let y be as in 3 and set h := [x α,y] A. Since A is abelian and h,x α A, we then have that [h,x α ] = 0 contradicting 3. Hence A L α = 0 for each α Φ, hence A = A H. It follows that A H. Now let 0 h A. By 1 there exists α Φ such that α(h) 0. It implies that [h,x α ] = α(h)x α L α \{0}. But we also have that [h,x α ] A, hence A L α 0, a contradiction. Hence A = 0. We would like to find an additional condition implying simplicity of L. Recall from Part I that if L is complex semisimple. with root system Φ, then for any α Φ +, x α L α, 0 x α, y α L α, h α [L α,l α ] such that α Φ sl(α) := x α,y α,h α = sl 2 (C) with h α = [x α,y α ] H. It is called the sl 2 -triple associated to the positive root α Φ +. Moreover, for α,β Φ, we have that β,α := β(h α ) = (α,β) = κ(t α,t β ), where κ(, ) is the Killing form and t α the unique element of H such that for any h H, κ(t α,h) = α(h). Theorem Let L be a complex semisimple Lie algebra, with root system Φ. Then L is simple if and only if Φ is irreducible. Before proving the Theorem we need: Lemma Let Φ be a root system with decomposition Φ = Φ 1 Φ 2 such that (α,β) = 0 for any α Φ 1, β Φ 2. Then

20 2 CLASSIFICATION OF ROOT SYSTEMS α Φ 1, β Φ 2 α+β / Φ 1, 2. α,β Φ 1, α+β Φ α+β Φ 1. Proof. We begin by proving 1. By assumption, (α +β,α) = (α,α) 0. It implies that α + β / Φ 2. Similarly one has (α + β,β) = (β,β) 0, hence α+β / Φ 1. Let now assume that α+β Φ 2. Since β Φ 1, one then has A contradiction to 1. }{{} α = β +(α+β }{{}}{{} Φ 1 Φ 1 Φ 2 ), Proof of First assume that L is not simple. Let H L be a Cartan subalgebra. Let I L be a nontrivial ideal. Since H consists of semisimple elements and [H,I] I, the operation of H on I is semisimple. Hence one has ( ) I = H 1 L α I, α Φ where H 1 = H I. Since L is semisimple, we have that dim C L α = 1 for any α Φ, hence L α I {0,L α }. Set Φ 1 := {α Φ L α I = L α }. Similarly, I = H 2 ( α Φ 2 L α I ), where the symbol denotes the orthogonal with respect to the Killing form κ(, ). Thanks to the nondegeneracy of the Killing form one has that L = I I. Hence H = H 1 H 2, Φ = Φ 1 Φ 2. If Φ 2 =, then L α I for any α Φ, then I = H 2 H is an abelian ideal, hence must be trivial by semisimplicity of L. So we can assume that Φ 2. Let α i Φ i, i = 1,2. Then (α 1,α 2 ) = α 1,α 2 = α 1 (h α2 ) = 0 since α 1 (h α2 )x α1 = [h α2,x α1 ] I I = 0 (α 1,α 2 ) = 0 Φ is irreducible. Now let H L be a Cartan subalgebra with root space decomposition L = H α Φ L α. Assume that Φ is reducible, Φ = Φ 1 Φ 2, (α 1,α 2 ) = 0 for α 1 Φ 1, α 2 Φ 2 and Φ 1,Φ 2. Set I := e α,f α,h α α Φ 1, then using Lemma 2.18, part 2, one sees that I L. The aim it to show that I is an ideal, which will conclude. Let x L, a {e α,f α α Φ 1 }. Since

21 2 CLASSIFICATION OF ROOT SYSTEMS 21 the e α, f α generate I, it suffices to show that [x,a] I. We can without loss of generality assume that either x H, or x L β for some β Φ. If x H, then [x,e α ] = α(x)e α I, [x,f α ] = α(x)f α I, α Φ 1. If x L β, then [x,e α ] L α+β. If α + β / Φ, then L α+β = 0 I. Otherwise α+β I implies that β Φ 1 by Lemma 2.18, part 1. By Lemma 2.18, part 2, we then have that α +β Φ 1, hence L α+β I. The proof in case a = f α is similar. Hence I is a nontrivial ideal of L, implying that L is not simple Having Propositions 2.15, 2.16 and Theorem 2.17 in mind, the program to show the simplicity of the classical Lie algebras is the following: Consider the subalgebra H of L consisting of the diagonal matrices; explain why it is a toral subalgebra, consider the root space decomposition with respect to this toral subalgebra and check the condition of Proposition 2.15 three conditions of Proposition This will show that H is a Cartan subalgebra of L and that L is semisimple, Find explicitly a basis of Φ, Write down the Cartan matrix of Φ and deduce that Φ is irreducible. By Theorem 2.17, it implies that L is simple. We do it in the next subsection in case L = sp 2n (C). 2.3 Simplicity of sp 2n (C) ( ) M P First recall that a matrixa = lies in sp Q N 2n (C) if and only if P = P t, Q = Q t, N = M t. Choose as basis m ij := e ij e l+j,l+i, 1 i,j l, p ij := e i,l+j +e j,l+i, 1 i j l, q ji := p t ij = e l+j,i +e l+i,j, 1 i j l. Hence one has that dim C sp 2n (C) = l l(l+1) = 2l 2 + l. Define H := 2 m ii 1 i l. Let h = l i=1 a im ii = l i=1 a i(e ii e l+i,l+i ). A direct computation shows that [h,m ij ] = (a i a j )m ij, [h,p ij ] = (a i +a j )p ij, [h,q ji ] = (a i +a j )q ji, [h,h ] = 0, h H.

22 2 CLASSIFICATION OF ROOT SYSTEMS 22 Hence the operation of H on L is diagonalizable, and {m ij,p ij,q ji } is a basis of simultaneous eigenvectors. For i = 1,...,l, set ǫ i : H C defined by ǫ i (h) = a i. We find the following roots: Basis element m ij, (i j) p ij, (i < j) p ii q ji, (i < j) q ii Root ǫ i ǫ j ǫ i +ǫ j 2ǫ i (ǫ i +ǫ j ) 2ǫ i Hence we get Φ = {ǫ i ǫ j for i j,±(ǫ i +ǫ j ) for i < j,±2ǫ i }. In particular, Φ = l(l 1)+2 l(l+1) +2l. Since 0 / Φ, we have thanks to 2 Proposition 2.15 that H is a Cartan subalgebra. We now show that for any α Φ, [[x α,x α ],x }{{} α ] 0. We do it in case =:h α α = ǫ i ǫ j, the other cases being obtained by similar computations. We have that Hence h α = [x α,x α ] = [m ij,m ji ] = e ii e l+i,l+i (e jj e l+j,l+j ), [h α,x α ] = [h α,m ij ] = 2(e ij e l+j,l+i ) 0, i < j. We therefore get that the condition 3 in Proposition 2.16 is satisfied. Condition 1 is easy and condition 2 is clear. Hence sp 2n (C) is semisimple. Let us now write α i := ǫ i ǫ i+1 (1 i l 1), β = 2ǫ l. We claim that := {α 1,...,α l 1,β} is a basis of Φ. The linear independence is clear. Since dimh = dimh = l, must be a basis of H. It remains to show that any element of Φ is a linear combination of elements of with integral coefficients, which are either all nonnegative or either all nonpositive. We have ǫ i ǫ j = α i +α i+1 + +α j 1, 1 i < j l. ǫ i + ǫ j = ǫ i ǫ j + 2ǫ j = (α i + + α j 1 ) + 2(α j + + α l 1 ) + β, 1 i < j l. 2ǫ i = (ǫ i ǫ j )+(ǫ i +ǫ j ) = 2(α i + +α l 1 )+β, 1 i l. Direct computations show that the Cartan matrix of Φ with respect to is given by α i,α i±1 = 1, α i,α i = 2, α i,α j = 0 if i j > 1, α i,β = 0 if i l 1, α l 1,β = 1, β,α i = 0 if i l 1, β,α l 1 = 2, β,β = 2.

23 2 CLASSIFICATION OF ROOT SYSTEMS 23 We see that Φ is irreducible. We deduce using Theorem 2.17 that L is simple. 2.4 Generators and relations It remains to determine if an isomorphism of root systems (Φ,H) and(φ,h ) of two simple Lie algebras L, L implies an isomorphism L = L. Thanks to Proposition 1.21, it suffices to determine if two non isomorphic simple Lie algebras can have the same Dynkin diagram. An important tool towards this question is a definition of L by generators and relations. Such an approach is also useful to describe the simple Lie algebras of exceptional types. Recall that for a semisimple Lie algebra L, with Cartan subalgebra H L, root system Φ with basis = {α 1,...,α l }, then L contains subalgebras sl(α i ) = e αi,f αi,h αi L, 1 i l which are isomorphic to sl 2 (C). Are these elements generators for L? Example LetL = sl l+1 (C),e αi = e i,i+1,f αi = e i+1,i (1 i l). Then [e αi,f αi ] = e i,i e i+1,i+1 =: h i. We have that {h i } l i=1 is a basis of H, the subalgebra of diagonal matrices. If i+1 < j, then [e i,i+1,e i+1,j ] = e ij. Hence by induction, one obtains all the e ij for i < j from the e αi, f αi. Similarly one also obtains all the e ij for i > j. Hence we obtain all the e α, f α, α Φ and H. We therefore have that {e α1,...,e αl,f α1,...,f αl } generate L. This is a general fact: Proposition Let L be a complex semisimple Lie algebra, {α 1,,α l } a basis of the corresponding root system. Let {e i,f i,h i } be the standard basis of sl(α i ). Then L is generated as Lie algebra by {e 1,...,e l,f 1,...,f l }. Proof. Let L = e 1,...,e l,f 1,...,f l. We mimic the proof forsl l+1 (C) given in Example We first show that H L. Since h i := [e i,f i ] H L, it suffices to show that H = h 1,...,h l. But recall from Part I that H = H via the Killing form κ, t αi α i H, with [e i,f i ] a multiple of t αi. Hence {h 1,...,h l } is a linear basis of H, since {α 1,...,α l } is a linear basis of H. We now show that L β L for any β Φ. Recall that w W, 1 j l such thatβ = w(α j ), and thatw is a product ofs αi for variousi(lemma 1.12 and Corollary 1.13). Hence by induction it suffices to show that if β = s αi (γ) for γ Φ with L γ L, then L β L. We have that β = γ γ,α i α i.

24 2 CLASSIFICATION OF ROOT SYSTEMS 24 We claim that M := k Z L γ+kα i is an irreducible sl(α i )-module, which concludes, since L γ L. Indeed, we know that diml γ+kαi i. Let x L γ. Then [h i,x] = γ(h i )x and more generally for x L γ+kαi, one has [h i,x] = (γ(h i )+kα i (h i ))x = (γ(h i )+2k)x. Hence the eigenspaces ofh i onm are one-dimensional with eigenvalues either all odd or all even. Since sl(α i ) = sl ( C), it follows from the representation theory of sl 2 (C) that U is a finite direct sum of V d, for various d Z 0, where V d is the (unique up to isomorphism) simplesl 2 (C)-module of dimensiond+1. It follows from the representation theory of sl 2 (C) studied in Part I that U is isomorphic to a V d for some d Z 0. Hence L β L γ L. Now that we found a set of generators for a semisimple Lie algebra L, it is natural to try to find the relations one should give between these generators to get a presentation of L. Proposition Let = {α 1,...,α l } be a basis of the root system Φ of a semisimple complex Lie algebra L, sl(α i ) = e i,f i,h i. Denote by C = (c ij ) the Cartan matrix of Φ with respect to. Then the following relations hold: (S1) (S2) [h i,h j ] = 0 i,j, [h i,e j ] = c ji e j, [h i,f j ] = c ji f j, i,j, (S3) [e i,f i ] = h i i, [e i,f j ] = 0 for i j, (S4) (ade i ) 1 c ji (e j ) = 0, (adf i ) 1 c ji (e j ) = 0, i j. Proof. The relation (S1) is clear since H is a Cartan subalgebra, hence abelian. We have that [h i,e j ] = α j (h i )e j = α j,α i e j = c ji e j, and similarly if we replace e j by f j. Hence (S2) holds. The first relation in (S3) follows directly from the fact that {e i,f i,h i } is an sl 2 -triple. We have that [e i,f j ] L αi α j, but α i,α j α i α j / Φ, hence L αi α j = 0. Hence all the relations given in (S3) hold. Now assume that i j, set M := k Z L α j +kα i. Then M is an sl(α i )- module. Since α i α j / Φ, we have that L αj +kα i = 0 for k < 0. Hence the smallest eigenvalue of ad(h i ) on M is α j,α i = c ji 0. By the representation theory of sl 2 (C), it implies that c ji is the largest eigenvalue of adh i on M. Hence x := (ade i ) c ij (e j ) is an eigenvector of adh i with eigenvalue c ji, hence ad(x) = 0. This shows that the first relation of (S4) holds. The proof of the second one is similar (using the fact that is also a basis of Φ). Theorem 2.22 (Serre). Let C be a Cartan matrix of a (euclidean) root system R with basis = {α 1,...,α l }. Let L be the algebra generated by

25 2 CLASSIFICATION OF ROOT SYSTEMS 25 elements {e i,f i,h i, 1 i l} with the relations (S1)-(S4) from Proposition Then L is finite-dimensional, semisimple, with Cartan subalgebra H := h 1,...,h l, and Φ has Cartan matrix C. Sketch of the proof. The proof is in two steps. Firstly, let L be the Lie algebra with generators {e i,f i } 1 i l and relations S1-S3. Then L is infinite-dimensional. Set Ẽ := e i 1 i l, F := fi 1 i l, H := h i 1 i l. Then one can show that L = F H Ẽ. In the second step, setũ+ := (ade i ) 1 c ji (e j ) i,j,ũ := (adf i ) 1 c ji (f j ) i,j. These are ideals ofẽ and F respectively. Set E := Ẽ/Ũ+, F := F/Ũ. Then show that Ũ+, Ũ are ideals of L. It implies that L = E H F. One then shows that L is finite-dimensional with Cartan subalgebra H and Cartan matrix C. ( ) 2 0 Example LetΦbe of typea 1 A 1, C =. ThenΦis reducible. 0 2 One has that Ẽ is generated by e 1, e 2 with some relations (antisymmetry and Jacobi identity). This is the free Lie algebra on two generators e 1, e 2. Here c 12 = c 21 = 0, hence Ũ+ = ade 1 (e 2 ),ade 2 (e 1 ) = [e 1,e 2 ]. Hence Ẽ/Ũ+ is 2-dimensional, generated by the images of e 1 and e 2. The same holds for F/Ũ. Hence L = e 1 e 2 h 1,h 2 f 1,f 2. Corollary For each irreducible root system R, there exists a (up to isomorphism) unique simple complex Lie algebra with root system Φ = R. In particular, there exists a simple Lie algebra corresponding to the root systems of exceptional types G 2, F 4, E 6, E 7 and E 8.

26 3 UNIVERSAL ENVELOPING ALGEBRAS 26 3 Universal enveloping algebras 3.1 Definition and first properties In this section, k will be an arbitrary field, not necessarily of characteristic zero. The idea is to introduce an associative unital k-algebra U(L) associated to a Lie algebra L, such that L and U(L) have the same representation theory. One could then use tools from representation theory of associative algebras to understand the representation theory of L. Recall that any associative unital k-algebraacan be seen as a Lie algebra via the bracket [, ] : A A A, (a,a ) aa a a. We will denote by Lie(A) the algebra A viewed as a Lie algebra. Definition 3.1. Let L be an arbitrary Lie algebra over the field k. A universal enveloping algebra of L (or simply enveloping algebra of L) is a pair (U(L),ǫ) where U(L) is an associative, unital k-algebra and ǫ : L Lie(U(L)) a homomorphism of Lie algebras such that the following universal property holds: for any associative unital k-algebra A and any homomorphism of Lie algebras ϕ : L Lie(A), there exists a unique homomorphism ϕ : U(L) A of associative algebras such that ϕ = ϕ ǫ. L ǫ U(L) ϕ ϕ A Exercise. Show using the definition above that If L = 0, then U = k with the unique possible map ǫ is a universal enveloping algebra of L, If L = k, then U = k[x], with the map ǫ : α αx is a universal enveloping algebra of L. Notice that the associative algebra k[x] is infinite dimensional! In fact, except in case L = 0, a universal enveloping algebra will always be infinite dimensional. Lemma 3.2 (Unicity of the universal enveloping algebra). Assume that (U 1,ǫ 1 ) and (U 2,ǫ 2 ) are two enveloping algebras of L. There is a canonical isomorphism ψ : U 1 U 2.

27 3 UNIVERSAL ENVELOPING ALGEBRAS 27 We therefore say that a universal enveloping algebra is unique "up to unique isomorphism" (but there may be more than one isomorphism of associative algebras U 1 U 2 ; the "canonical" in Lemma 3.2 is here to emphasize that such an isomorphism should be compatible with the maps ǫ 1, ǫ 2 ). Hence "unique up to unique isomorphism" means "unique up to unique isomorphism satisfying some compatibility conditions". We may therefore simply call a universal enveloping algebra "the" universal enveloping algebra. Proof. Since ǫ 2 : L Lie(U 2 ) is a morphism of Lie algebras from L into an associative k-algebra viewed as Lie algebra, the universal property of the universal enveloping algebra applied with U 1 as universal algebra yields a homomorphism of associative algebras ǫ 2 : U 1 U 2 such that ǫ 2 ǫ 1 = ǫ 2. Similarly by reversing the roles of U 1 and U 2, one gets a homomorphism of associative algebras ǫ 1 : U 2 U 1 such that ǫ 1 ǫ 2 = ǫ 1. We then have that ǫ 1 ǫ 2 ǫ 1 = ǫ 1 ǫ 2 = ǫ 1, Hence f := ǫ 1 ǫ 2 : U 1 U 1 is a homomorphism of associative unital k- algebras such that f ǫ 1 = ǫ 1. But by the universal property of (U 1,ǫ 1 ), there should be a unique homomorphism with such a property, and f = id U1 satisfies it. Hence f = id U1. Similarly one shows that ǫ 2 ǫ 1 = id U2, hence ψ = ǫ 2 is the looked for isomorphism. Lemma 3.3. Let V be an abelian group, L a Lie algebra over k and ǫ : L U(L) a universal enveloping algebra for L. There is a bijection Structures of U(L) module on V Structures of representation of L over k on V, Where an U(L)-module V is made into a k-linear representation of L via α v = (α1 U(L) ) v, x v = ǫ(x) v, α k, x L, v V. Proof. A structure of V as U(L)-module is by definition a homomorphism of rings ϕ : U(L) End(V). The restriction of ϕ to k U(L) endows V with a structure of k-vector space. Hence ϕ can be seen as a morphism of associative unital k-algebras U(L) End k (V), and therefore also as a homomorphism of Lie algebras Lie(U(L)) Lie(End k (V)) = gl(v); we abuse notation and write this homomorphism again ϕ. The composition ϕ ǫ then yields a morphism of Lie algebras ϕ ǫ : L gl(v). This is by definition a k-linear representation of L.

28 3 UNIVERSAL ENVELOPING ALGEBRAS 28 To show that our assigment is bijective, we give the inverse map. A k- linear representation of V is a pair (V,ρ) where V is a k-vector space and ρ : L gl(v) = Lie(End k (V)) a homomorphism of Lie algebras. By the universal property of (U(L),ǫ), there exists a unique way of extending ρ to a homomorphism ρ : U(L) End k (V) of associative unital k-algebras. One then shows that the two maps are inverse of each other. We proved that if the universal enveloping algebra U(L) of L exists, then it is unique. We postpone the proof of existence to Subsection 3.3. Lemma 3.4. Let L be a Lie algebra with basis {X i } i I, where I is totally ordered, and let U(L) be the universal enveloping algebra of L. Then U(L) is generated as k-vector space by the monomials ǫ(x i1 )ǫ(x i2 ) ǫ(x im ), where m Z 0, i 1 i 2 i m. Proof. Let U := ǫ(x i1 )ǫ(x i2 ) ǫ(x im ) m Z 0,(i 1,...,i m ) I m k vector space. It is clear that U is a k-subalgebra of U(L). Moreover, Im(ǫ) U. Hence consider ϕ : L Lie(U ), x ǫ(x). By the universal property ofu(l), there is a unique homomorphism of associative unital k-algebras ϕ : U(L) U with ϕ = ϕ ǫ. Let ψ : U U be the inclusion. Then it is clear that ψ ϕ = ǫ. Hence we have that ψ ϕ ǫ = ψ ϕ = ǫ, hence by the universal property one must have ψ ϕ = id U(L). Hence ψ is surjective, but since it is also injective, one has that U = U. Now the products of the form ǫ(x i1 )ǫ(x i2 ) ǫ(x im ), m > 1, with i j > i j+1 for some 1 j < m can be written as linear combinations of order monomials using inductively the relation ǫ(x ij )ǫ(x ij+1 ) = ǫ([x ij,x ij+1 ])+ǫ(x ij+1 )ǫ(x ij ). 3.2 Poincaré-Birkhoff-Witt Theorem Theorem 3.5 (Poincaré-Birkhoff-Witt). Let L be a Lie algebra over a field k with universal enveloping algebra (U(L),ǫ), {X i } i I a basis of L, with I

29 3 UNIVERSAL ENVELOPING ALGEBRAS 29 totally ordered. The set A := m Z 0 {ǫ(x i1 )ǫ(x i2 ) ǫ(x im ) i 1 i 2 i m, (i j ) I m } is a k-basis of U(L). In particular, ǫ : L U(L) is injective. Proof. Thanks to Lemma 3.3, it suffices to show that the elements of A are linearly independent. For M = (i 1,...,i m ) I m, set l(m) = m. Let V be the vector space with basis m Z 0 {z M M = (i 1 i 2 i m ) I m }. In the sequel we write M for the set of ordered monomials in the elements of I. For i I, we write i M if i i 1 and we set im := (i,i 1,...,i m ). We claim that V can be equipped with a structure of L-module, such that X i z M = z im whenever i M. That is, there is a k-linear map L V V, (a,v) a v such that X i z M = z im whenever i M, and such that the induced map L gl(v), a {v a v} is a morphism of Lie algebras. Since {X i } i I is a k-basis of L and {z M } a k-basis of V, it suffices to give a formula for X i z M and check that it defines an L-module structure on V. Such a formula is defined inductively on the lexicographical order on M I, where M is ordered by l( ). That is, we fix (M,i) M I and assume that X j z N is already defined if l(n) < l(m), or if l(n) = l(m) and j < i. We furthermore assume that in that in case (N,j) < (M,i), then X j z N z P l(p) l(n)+1 k vector space. (4) To start the induction, we just define X i z = z (i) for all i I. We then set { zim if i M X i z M := X j (X i z N )+[X i,x j ] z N if M = jn,i > j. Notice that all the operations occurring in the second row are already defined sincel(n) = l(m) 1, andx i z N z P l(p) l(n)+1 = l(m) k vector space and j < i. First step. We show that condition 4 is still satisfied. This is clear for the first case since l(im) = l(m)+1 and for the second by induction. Second step. We show that such an inductive formula defines an operation of L on V, that is, we need to show that for any i,j I and N M, the following equality holds: [X i,x j ] z N = X i (X j z N ) X j (X i z N ).

30 3 UNIVERSAL ENVELOPING ALGEBRAS 30 The proof is by induction on l(n) and min(i,j). If i = j, then both sides are equal to zero. If j i, then by antisymmetry we can assume that i > j. If j N, then X j z N = z jn, hence we have X i (X j z N ) = X i z jn = by definition X j (X i z N )+[X i,x j ] z N, hence the formula holds. Now assume that j N. Write N = kk, with k < j < i. Hence we have z N = X k z K. We have to show that X i X j X k z K X j X i X k z K = [X i,x j ]X k z K, and we denote this equality simply by (ijk). We know that for any x,y L, the equality xy z K = yx z K +[x,y] z K holds since l(k) = l(n) 1. Therefore, the right hand side of (ijk) can be rewritten as [X i,x j ]X k z K = X k [X i,x j ] z K +[[X i,x j ],X k ] z K = X k X i X j z K X k X j X i z K +[[X i,x j ],X k ] z K. Notice that the equalities (jki) and (kij) hold by induction since k < j < i. By adding the right hand sides of (ijk), (jki) and (kij) we get X i X j X k z K X j X i X k z K +X j X k X i z K X k X j X i z K +X k X i X j z K X i X k X j z K, While by adding the right hand sides as rewritten above we get X k X i X j z K X k X j X i z K +[[X i,x j ],X k ] z K + X i X j X k z K X i X k X j z K +[[X j,x k ],X i ] z K + X j X k X i z K X j X i X k z K +[[X k,x i ],X j ] z K. The sum of the three terms with brackets vanishes thanks to the Jacobi identity. Hence the sum of the right hand sides of (ijk), (jki) and (kij) and the sum of the left hand sides are the same. Since (jki) and (kij) hold, it implies that (ijk) must also hold. Hence V is an L-module. Third step. For M = (i 1,i 2,...,i m ) M, we set X M := ǫ(x i1 )ǫ(x i2 ) ǫ(x im ).

31 3 UNIVERSAL ENVELOPING ALGEBRAS 31 We want to show that X M z = z M. The proof is by induction on l(m). If l(m) = 0, then X M = X = 1, and 1 z = z. Now assume that l(m) > 0, write M = in, i N. One has that X M z = ǫ(x i )X N z = ǫ(x i ) z N = X i z N = z in = z M. Fourth step. We conclude the proof by showing that the X M, M M are linearly independent. Assume that 0 = M M c MX M, where the C M are scalars, only finitely many of which are nonzero. Then one has that ( ) 0 = c M z M z = c M X M z = M z M V, M M M Mc M M hence c M = 0 for any M M since {z M } M M is a k-basis of V. 3.3 Tensor algebra and existence of universal enveloping algebras Definition 3.6. Let V be a vector space over a field k. The tensor algebra T(V) of V is the associative, unital k-algebra defined as vector space by T(V) = r Z 0 V r = k V (V V) (V V V)..., endowed with a product defined by (v 1 v 2 v n ) (w 1 w m ) = v 1 v n w 1 w m, and extended k-bilinearly. Lemma 3.7 (Universal property of the tensor algebra). Let V be a k-vector space and c : V T(V) the inclusion (as k-linear map). If A is an associative unital k-algebra and ϕ : V A a k-linear map, then there exists exactly one morphism of associative, unital k-algebras ˆϕ : T(V) A such that ϕ = ˆϕ c. V c T(V) ϕ ˆϕ A

32 3 UNIVERSAL ENVELOPING ALGEBRAS 32 Proof. Since the set V generates T(V) as associative unital k-algebra, there can be at most one such morphism. One the other hand, ˆϕ can be defined explicitly by setting ˆϕ(v 1 v n ) = ϕ(v 1 )ϕ(v 2 ) ϕ(v n ). Proposition 3.8. Let L be a Lie algebra over k. Consider the ideal I(L) of T(L) generated by the elements of the form (x y y x [x,y]), x,y L. Then the associative unital algebra U(L) := T(L)/I(L) together with the map ǫ : L c T(L) π U(L), where π is the canonical quotient map, is a universal enveloping algebra of L. Proof. One first has to show that ǫ = π c is a morphism of Lie algebras, since c is only a k-linear map. We have ǫ([x,y]) = π([x,y]) = π(x y y x) = π(x)π(y) π(y)π(x) = ǫ(x)ǫ(y) ǫ(y)ǫ(x) = [ǫ(x),ǫ(y)], Hence ǫ is a morphism of Lie algebras. It remains to show that U(L) satisfies the universal property of the universal enveloping algebra. Hence let ϕ : L Lie(A) be a morphism of Lie algebras, where A is an associative, unital k-algebra. In particular, ϕ is a k-linear map. By the universal property of the tensor algebra T(L), there exists a unique morphism of associative unital k-algebras ˆϕ : T(L) A such that ˆϕ c = ϕ. But ϕ is a morphism of Lie algebras, hence ˆϕ(x y y x [x,y]) = ϕ(x)ϕ(y) ϕ(y)ϕ(x) ϕ([x,y]) = 0, which implies that ˆϕ(I) = 0. Hence ˆϕ factors through U(L) via a map ϕ : U(L) A, that is, ˆϕ = ϕ π. T(L) π U(L) ˆϕ ϕ A It remains to show that ϕ is unique. Assume that we have another morphism of associative unital k-algebras ψ : U(L) A such that ψ π c = ϕ. Since π is surjective, it suffices to show that ψ π = ˆϕ. But we have ψ π c = ϕ π c = ˆϕ c, and since L generates T(L) as k-algebra, it follows that ψ π and ˆϕ agree on the set c(l) of generators of T(L), hence they must agree everywhere. We therefore have ψ π = ˆϕ, which concludes.

33 3 UNIVERSAL ENVELOPING ALGEBRAS 33

34 4 HIGHEST WEIGHT THEORY 34 4 Highest weight theory 4.1 Weights of a representation Definition 4.1. Let L be a complex semisimple Lie algebra with Cartan subalgebra H L. The elements of H are called weights (in German "Gewichte"). For V a representation of H and λ H a weight, we define the weight space V λ V as the H-submodule V λ := {v V hv = λ(v)h, h H}. If V λ 0, we say that λ is a weight of the representation V. We write P(V) := {λ H V λ 0} for the set of weights of V. Example 4.2. The Cartan subalgebra H acts on L via the adjunct operation. One has P(L) = R {0}, where R is the root system of L, and the various weight spaces are the L α as well as H (for the weight 0 H ). Definition 4.3. Fix a positive system R + R and define a partial order on H by µ λ λ = µ+ α R + n α α, n α Z 0 α R +. If V is a representation of L and if there exists λ P(V) such that µ P(V), µ λ, then λ is the highest weight of V and any 0 v V λ is a highest weight vector. An L-module generated by a highest weight vector of weight λ is called a highest weight module of weight λ. A weight vector v V such that L α v = 0 for any α R + is called a maximal vector. Since L α V µ V α+µ for any α R, µ H, it is clear that a highest weight vector must be maximal. The converse is false. Notice that the fact that a vector v V µ is maximal does not necessarily imply that the weight µ is maximal in P(V), but the converse is true (for the same reason as the one giving that a highest weight vector must be maximal). Can we always find a highest weight vector? Exercise. Let L be a semisimple complex Lie algebra with Cartan subalgebra H L. Let ǫ : L Lie(U(L)) be the canonical morphism. Then U(L) is an H-module via H L ǫ Lie(U(L)). Show that P(U(L)) =.