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Representation Theory Spring 2013, taught by Joe Harris. Contents 1 Basic Definitions 2 2 Examples of Lie Groups 2 3 Isogenies of Lie Groups 3 4 Maps Between Lie Groups 4 5 Lie Algebras 6 6 The Exponential Map 7 7 Classes of Lie Algebras 9 8 Low-Dimensional Simple Lie Algebras 13 8.1 sl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2 8.2 Plethysm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 8.3 sl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3 9 Simple Lie Algebras 20 10 Analysis of sl 23 n 11 Geometric Plethysm 25 12 The Symplectic Groups Sp 27 2n 13 Orthogonal Lie Algebras 30 13.1 Clifford Algebras and Spin Representations . . . . . . . . . . . . . . . . . . . . . . . 34 14 Classification Theorem 35 15 Construction of g and G 40 2 2 1 16 Complex Lie Groups Associated to Simple Lie Algebras 41 16.1 Forms of sl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 n+1 16.2 Forms of sp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 2n 16.3 Forms of so . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 m 17 Representation Rings and Characters 43 18 The Weyl Character Formula 46 19 Real Forms 48 1 Basic Definitions A C∞ manifold is a set X with the structure of a topological space and an atlas: an open cover {U } and homeomorphisms ϕ : U −→∼ ∆ ∼= Rn such that ϕ ◦ϕ−1 is C∞ where defined. α α α α β We can talk about C∞ functions on X and C∞ maps X → Y via working with atlases. Recall that a group is a set X with the structure of e ∈ X, m : X ×X → X, and i : X → X satisfying appropriate axioms. A Lie group is a C∞ manifold G with a group structure such that m and i are C∞. We say that a map G → H is a Lie group homomorphism if it is a group homomorphism which is also a C∞ map. A Lie subgroup H of G is an H ⊆ G which is a subgroup and a closed submanifold. As a nonexample, consider a line through the origin in R2 having an irrational slope. The map R2 → R2/Z2 takes this line to a dense subset of the torus, called an immersed submanifold of the torus. A representation of G on a vector space V is a map G → GL(V), the set of invertible linear maps on V. 2 Examples of Lie Groups • GL (R) ⊆ Rn2 is a Lie group because of the explicit descriptions of matrix multiplication and n inversion. This makes GL(V) a Lie group for any finite dimensional vector space V. • SL (R) is a submanifold of GL (R) because det : M (R) → R) has nonvanishing differential. n n n (It’senoughtocheckattheidentity.)WecanalsolookatSL(V),thesetofϕpreservingavol- ume form. That is, for some nonzero α ∈ ΛnV∗, we have α(ϕ(v ),...,ϕ(v )) = α(v ,...,v ). 1 n 1 n • A flag V in V is a nested sequence of subspaces 0 (cid:40) V (cid:40) ··· (cid:40) V (cid:40) V (2.1) 1 k 2 Define B(V) to be the set of automorphisms of V preserving V; that is, ϕ such that ϕ(V ) ⊆ V i i for every i (in this case, ϕ(V ) necessarily equals V .) i i A full flag is a V where V has dimension i. In an appropriate basis for V, B(V) consists of i upper triangular matrices. More generally, B(V) consists of block upper triangular matrices in an appropriate basis. • Let N(V) be the subgroup of B(V) consisting of those ϕ such that (ϕ−I)(V ) ⊆ V . That i i−1 is, ϕ is the identity on successive quotients. • Let Q be a symmetric bilinear form on V. Q : V × V → R is an element of Sym2V∗. Assume that Q is nondegenerate. Define O(V,Q) to be the set of ϕ preserving Q; that is, Q(ϕ(v),ϕ(w)) = Q(v,w). In matrix form, express Q as a symmetric n × n matrix M, so Q(v,w) = vTMw. Then O(V,Q) = {A ∈ GL (R) : ATMA = M}. (2.2) n If M = I, then O(V,Q) is called O(n). More generally, given Q, we can find a basis such that M is diagonal with diagonal entries 1 and −1. We say that M has signature (k,n−k), and O(V,Q) is called O . k,n−k SO(V,Q) is the further subgroup of determinant-1 transformations. • Take Q to be a skew-symmetric nondegenerate bilinear form, so Q ∈ Λ2V∗. Q can be specified by a skew-symmetric M, and under an appropriate basis, we can write (cid:18) (cid:19) O I M = (2.3) −I 0 The group of matrices preserving Q is called Sp . 2n • If V is 7-dimensional and α is a general skew-symmetric trilinear form (α ∈ Λ3V∗), then the group of automorphisms of V preserving α is nontrivial and called G . 2 3 Isogenies of Lie Groups Proposition 3.1. If G is a connected Lie group and U is an open subset containing e, then U generates G. Proof. ByreplacingU byU∩U−1,wecanwithoutlossofgeneralityassumeU = U−1.LetH bethe subgroup generated by U. Suppose H (cid:54)= G. Then there exists g ∈ ∂H = H\H. Then gU ∩H (cid:54)= ∅; choose h in the intersection. If h = gu for u ∈ U ⊆ H, then g = hu−1 ∈ H, a contradiction. Let H be a Lie group and Γ ⊆ Z(H) discrete. Set G = H/Γ. Claim. G has the unique structure of a Lie group such that the quotient map H → G is a map of Lie groups. (In fact, the map is a covering map.) On the other hand, now start with G a Lie group and let H be a connected topological space such that H → G is a covering map. Clearly H inherits the structure of a manifold. Choose an element e(cid:48) ∈ H mapping to e. Then: 3 Theorem 3.2. There exists a unique group structure on H such that H is a Lie group and H → G is a map of Lie groups. Proof. First consider the case where H is the universal cover of G. Consider the group law on G as a map G×G → G. We can uniquely lift H ×H mH H (3.1) G×G mG G We get a unique map H ×H → H mapping (e(cid:48),e(cid:48)) to e(cid:48). Inversion is similar. Toseethatm isassociative,considera(bc)((ab)c)−1.ThisisacontinuousmapH×H×H → H H lying in the fiber of e. So this map is a constant, identically e(cid:48). Proposition 3.3. If G is a connected Lie group and Γ ⊆ G is discrete and normal, then Γ ⊆ Z(G). Proof. For h ∈ Γ, the map G → Γ by g (cid:55)→ ghg−1 is continuous, so must be constant. The value at e is h, so ghg−1 = g for every g ∈ G. We say that connected groups G,H are isogenous if there exists a Lie group map G (cid:16) H with discrete kernel (that is, a covering space map). The isogeny classes are the equivalence classes of the equivalence relation generated by isogenies. In each isogeny class, there exists a unique initial member,whichwe’llcallH,thesimplyconnectedform:theuniversalcoveringspaceofanymember. If Z(H) is discrete, then there also exists a final object H/Z(H), called the adjoint form. Our basic approach is to describe all representations of the simply connected form, and then say which ones descend to a given member of the isogeny class. 4 Maps Between Lie Groups Here is a problem: given G,H, describe all Lie group maps G → H. (The case H = GL(V) corresponds to representation theory.) We’ll assume that G is simply connected. In fact, ρ : G → H is determined not just by its values on any open neighborhood of e, but even dρ : T G → T H. In e e e other words we have an inclusion Hom (G,H) (cid:44)→ Hom(T G,T H). (4.1) Lie e e We would like to know which linear maps T G → T H actually arise as differentials of Lie group e e maps. If ρ is a Lie group map, then ρ(gh) = ρ(g)ρ(h). In other words, the following diagram commutes for every g: 4 ρ G H mg mρ(g) (4.2) ρ G H where m is multiplication by g. To focus our attention on G,H near the identity, consider g instead ψ which is conjugation by g. We also have the commutative diagram g ρ G H (4.3) ψg ψρ(g) ρ G H But now ψ (e) = e so we can consider (dψ ) : T G → T G. A get a map G → GL(T G) by g g e e e e g (cid:55)→ (dψ ) , called the adjoint representation Ad, a distinguished representation of G. Taking the g e differential of Ad, we get a map ad : T G → End(T G). By taking transpose, we can also view ad e e as a map T G×T G → T G by (X,Y) (cid:55)→ ad(X)(Y) = [X,Y]. e e e The point is that if ρ is any Lie group map, then (dρ) respects the binary operation ad on T G. e e That is, dρ×dρ T G×T G T H ×T H e e e e (4.4) [ , ] [ , ] dρ T G T H e e In fact, if G is simply connected, then any linear map ϕ : T G → T H is the differential of a Lie e e group map ρ : G → H if and only if [ϕ(X),ϕ(Y)] = ϕ([X,Y]). Fact. We can describe the map [ , ] explicitly. Start with G = GL(V). Then T G = End(V). Observe that ψ : G → G by h (cid:55)→ ghg−1 e G extends naturally to End(V), and this extension is the differential of ψ , since ψ is linear. We have g g dψ (X) = gXg−1. g To describe ad, we start with a given X ∈ End(V) and choose an arc γ : (−(cid:15),(cid:15)) → G with γ(0) = e and γ(cid:48)(0) = X. Then [X,Y] = ad(X)(Y) (4.5) d (cid:12) = (cid:12) γ Yγ−1 (4.6) dt(cid:12)t=0 t t = γ(cid:48)Yγ−1+γ Y(γ−1)(cid:48) (4.7) 0 0 0 0 = γ(cid:48)Yγ−1−γ Y(γ−1γ(cid:48)γ−1) (4.8) 0 0 0 0 0 0 = XY −YX. (4.9) This formula also applies to any subgroup of GL(V). This has two consequences: 5 • [X,Y] = −[Y,X]; that is, the bracket is skew-symmetric. • The Jacobi identity holds: for X,Y,Z ∈ T G, e [[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y] = 0. (4.10) 5 Lie Algebras We define a Lie algebra g to be a vector space with a skew-symmetric bilinear map g × g → g satisfying the Jacobi identity. So one of our above claims is that for G simply connected and H any Lie group, if g = T G and e h = T H, then e Hom (G,H) = Hom (g,h). (5.1) Liegrp Liealg We will split this assertion into the following two theorems: Theorem 5.1. If G is connected, any Lie group map ρ : G → H is determined by its differential dρ : g → h. Theorem 5.2. If G is simply connected, a linear map ϕ : g → h is the differential of a Lie group map ρ : G → H if and only if ϕ is a Lie algebra map. Whycouldn’twejustdefine[X,Y] = XY −YX?TheproblemisthatXY isnotalwaysdefined: given G (cid:44)→ GL(V) and X,Y ∈ T G viewed as endomorphisms of V, XY depends on the embedding e of G into GL(V), and is not canonical. In fact, it may not even lie in the image of T G. e We may sometimes write End(V) as gl(V) if we are interested in the Lie algebra structure. ALiealgebramorphismfromgtohisalinearmapϕ : g → hsuchthatϕ([X,Y]) = [ϕ(X),ϕ(Y)]. A representation of g on V is a Lie algebra morphism g → gl(V). This can be viewed as an action of g on V such that for every X,Y ∈ g, X(Y(v))−Y(X(v)) = [X,Y](v). Observe thatgiven agroupGactingon vectorspacesV andW,weget anactionofGonV ⊗W by g(v ⊗w) = g(v)⊗g(w). In other words, given ρ : G → GL(V) and ρ : G → GL(W), we V W obtain a map ρ = ρ ⊗ρ . V⊗W V W How do the corresponding Lie algebra maps relate to this? Suppose ϕ : g → gl(V) and ϕ : V W g → gl(W)aregiven.Letγ beanarcinGwithγ = eandγ(cid:48) = X.Wehaveγ (v⊗w) = γ (v)⊗γ (w). 0 0 t t t Differentiating at t = 0, X(v⊗w) = X(v)⊗w+v⊗X(w). (5.2) In general, given two Lie algebra representations, the tensor product representation will be defined by the above. Here are some other examples: • If G acts on V, then G correspondingly acts on Sym2V characterized by g(v2) = g(v)2. The corresponding representation on g is given by X(v2) = 2vX(v). 6 • Given ϕ : V → W, recall that the transpose map ϕT : W∗ → V∗ is the map described by composition: [ϕT(λ)](v) = λ(ϕ(v)). Now if G acts on V, then G acts on V∗ by ρ (g) = ρ (g−1)T. The corresponding represen- V∗ V tation of g is given by ϕ (X) = −ϕ (X)T. V∗ V Remark. • The definition of a Lie algebra doesn’t require g to be finite dimensional, but we’ll only deal with finite dimensional ones. • The definition also doesn’t specify the base field. We’ll assume that the base field is either R or C, and will use R by default. • In fact, every Lie algebra can be embedded in gl (R) for some n. This implies that every Lie n algebra arises from a Lie group. (cid:94) • But not every Lie group is a subgroup of GL (R). An example is the universal cover SL (R) n 2 of SL (R). Another example is given by 2      1 a b (cid:44) 1 0 b     0 1 c 0 1 0 : b ∈ Z . (5.3)  0 0 1   0 0 1  6 The Exponential Map Suppose G is any Lie group, and X ∈ g = T G. We can define a vector field V = V on G by e X V(g) = (dm ) (X). (We may write α instead of dα for a map α on manifolds. If the domain is a g e ∗ subset of R, then α(cid:48) may be used.) Now recall from differential geometry the following Theorem 6.1. If M is any manifold, p ∈ M, and V is a vector field on M, then there exists a unique germ of an arc γ : (−(cid:15),(cid:15)) → M such that γ(0) = p and γ(cid:48)(t) = v(γ(t)). Applythistheoremtotheabovesituation.GivenX ∈ g = T GandV thecorrespondingvector e X field, let γ : (−(cid:15),(cid:15)) → G be the corresponding integral curve with γ(0) = e. Claim. γ is a group homomorphism where defined. That is, γ(s+t) = γ(s)γ(t). Proof. Fix s, and set α(t) = γ(s+t) and β(t) = γ(s)γ(t). Then α(cid:48)(t) = V(γ(s+t)) while β(cid:48)(t) = (dm ) V(γ(t)). Since V is a left invariant vector field, α(t) and β(t) are both integral curves γ(s) γ(t) for V starting at γ(s), so they must be the same arc. Now using the claim, we can extend γ to R. In summary, for each X ∈ g, we get a Lie group morphism ϕ : R → G such that ϕ(cid:48) (0) = X. X X These properties uniquely determine ϕ . Such a map is called a 1-parameter subgroup. X By uniqueness, we get the following naturality: if ρ : G → H is any Lie group map and X ∈ g, then ρ◦ϕ = ϕ . (6.1) X dρ(X) 7 Remark. • ϕ does not need to be injective. However, if X (cid:54)= 0, the kernel will be a nontrivial X closed subgroup of R, therefore either {0} or rZ for some r > 0. In these cases, the image of ϕ will be isomorphic to R or S1. • If ϕ is injective, the image may be a closed subgroup or simply an immersed subgroup. X • We have ϕ (t) = ϕ (λt). λX X We now define the exponential map exp : g → G by exp(X) = ϕ (1). So exp restricted to each X line through 0 ∈ g is a 1-parameter subgroup. Fact. The map exp is the unique C∞ map g → G such that the differential at 0 is the identity and the restriction to lines is a 1-parameter subgroup. By this characterization of exp, given a Lie group map ρ : G → H, we have ρ G H exp exp (6.2) dρ g h We’ll now show that exp is C∞. Explicitly, if G = GL(V) and g = gl(V), then X2 X3 exp(X) = 1+X + + +··· (6.3) 2 6 This is also true for any subgroup of GL(V), and therefore any Lie group whose Lie algebra embeds into gl(V) for some V. On a small neighborhood of e ∈ G, we can define an inverse to exp, called log. In the case of GL(V), (g−I)2 (g−I)3 log(g) = (g−I)− + −··· (6.4) 2 3 Given X,Y ∈ g sufficiently close to 0, we will define X(cid:63)Y = log(exp(X)exp(Y)). We then have (cid:18)(cid:18) X2 (cid:19)(cid:18) Y2 (cid:19)(cid:19) X (cid:63)Y = log 1+X + +··· 1+Y + +··· (6.5) 2 2 (cid:18) (cid:18)X2 Y2(cid:19) (cid:19) = log 1+(X +Y)+ +XY + +··· (6.6) 2 2 (cid:18) (cid:18)X2 Y2(cid:19) (cid:19) 1 (cid:18) (cid:18)X2 Y2(cid:19)(cid:19) = X +Y + +XY + +··· − X +Y + +XY + +··· (6.7) 2 2 2 2 2 (cid:18)X2 Y2 (X +Y)2(cid:19) = X +Y + +XY + − +··· (6.8) 2 2 2 [X,Y] = X +Y + +··· (6.9) 2 The final formula of Campbell-Hausdorff-Dynkin-Baker-etc. is 8 [X,Y] 1 X (cid:63)Y = X +Y + + ([X,[X,Y]]−[Y,[X,Y]])+··· (6.10) 2 12 The important fact is that this expression involves only addition and Lie brackets. Corollary 6.2. Given h a vector subspace of g, the subgroup H of G generated by exp(h) is an immersed subgroup with tangent space T H = h if and only if h is a Lie subalgebra of G. e We get a bijection between connected immersed subgroups of G and Lie subalgebras of g. We conclude by proving Theorems 5.1 and 5.2. Theorem5.1followsfromthenaturalityofexp:asexpislocallyadiffeomorphism,dρdetermines ρ on a neighborhood of the identity in G. Proof of Theorem 5.2. Given ϕ : g → h, let j ⊆ g × h be the graph. This is a Lie subalgebra if and only if ϕ is a Lie algebra map. In this case, there exists J ⊆ G×H an immersed subgroup. Composing with the projection G × H → G gives a covering map J → G. G simply connected implies J = G, so J is the graph of a Lie group map G → H. 7 Classes of Lie Algebras Assume G is connected, and let g be its Lie algebra. g is said to be abelian if [X,Y] = 0 for every X,Y ∈ g. More generally, for any g, we define the center Z(g) = {X ∈ g : [X,Y] = 0∀Y ∈ g}. (7.1) Connected subgroups H ⊆ G correspond to subalgebras h of g. Now suppose H is normal. Then gHg−1 = H for every G ∈ G. Choose γ : I → G with γ(0) = e and γ(cid:48)(0) = X ∈ g. Then γ Hγ−1 = H for every t, so ad(X) = Ad(γ ) takes h to h. t t t We say that h is an ideal if [g,h] ⊆ h. Then H is normal if and only if h is an ideal. Remark. If h ⊆ g is a subalgebra, then [ , ] descends to g/h (meaning there exists a Lie algebra structure on the vector space g/h such that g → g/h is a Lie algebra map) if and only if h is an ideal. We say that g is simple if g has no proper nontrivial ideals. Given a Lie algebra g, we define two sequences of ideals in g: 1. The lower central series: start with D g = g, and inductively define D g = [g,D g]. (In 0 i i−1 particular, D g = [g,g]. We get a sequence of subalgebras 1 D g ⊇ D g ⊇ D g ⊇ ··· (7.2) 0 1 2 2. The derived series: start with D0g, and this time Dig = [Di−1g,Di−1g]. The Jacobi relation can be used to show that the Dig are in fact ideals. 9 We say that g is nilpotent if D g = 0 for some k, and that g is solvable if Dkg = 0 for some k. k We say that g is semisimple if g has no nonzero solvable ideals. (We’ll see that this is equivalent to being a direct sum of simple algebras.) AkeyexampleofanilpotentLiealgebraistheLiealgebranofstrictlyuppertriangularmatrices in gl . This is the Lie algebra of the subgroup of unipotent upper triangular matrices. We will prove n that any nilpotent Lie algebra can be realized as a subalgebra of n (for some n). Another example of a solvable Lie algebra is b, the space of all upper triangular matrices in gl . n In fact, every solvable Lie algebra is a subalgebra of b (for some n). Observe that g is solvable if and only if there exists a sequence of subalgebras 0 ⊆ g ⊆ g ⊆ ··· ⊆ g ⊆ g (7.3) 1 2 k such that g is an ideal in g and g /g is abelian. As a consequence, if h ⊆ g is an ideal, i i+1 i+1 i then g is solvable if and only if h and g/h are solvable. Remark. Ifa,b ⊆ garesolvableideals,thena+bisalsosolvable.Thisisbecause(a+b)/b = a/(a∩b), which is solvable. Corollary 7.1. There exists a maximal solvable ideal in g. This is called the radical of g and denoted Rad(g). We now have an exact sequence 0 → Rad(g) → g → g/Rad(g) → 0 (7.4) with g/Rad(g) semisimple. We will denote g/Rad(g) by g . ss Theorem 7.2 (Engel). If ϕ : g → gl(V) is a representation of g such that ∀X ∈ g, ϕ(X) is nilpotent, then there exists 0 (cid:54)= v ∈ V such that X(v) = 0 for every X ∈ g. Observe that in this case, there exists a basis v ,...,v of V such that ϕ(g) ⊆ n. 1 n Remark. If X ∈ gl(V) is nilpotent, then there is a sequence of spaces V ⊇ V ⊇ V ⊇ ··· ⊇ V ⊇ 0 (7.5) 1 2 k suchthatX mapsV intoV .Thenad(X)viewedasanendomorphismofgl(V)isalsonilpotent. i i+1 Proof of Engel’s theorem. Induct on dimg. Lemma 7.3. g contains an ideal h of codimension 1. Proof. Lethbeanymaximalpropersubalgebraofg.Weclaimthathisanidealandthatcodim(h ⊆ g) = 1. Look at the adjoint action ad(h) ⊆ gl(g). Since ad(h) carries h into itself, ad(h) acts on g/h. Every element acts nilpotently. For if Xk = 0 on V, then ad(X)2k−1Y = 0 for every Y ∈ g. By the inductive hypothesis, there exists Y ∈ g/h nonzero such that [h,Y] ⊆ h. By maximality, we must have h+Y = g, so h has codimension 1. 10

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