Lie groups and Lie algebras Wilfried Schmid Lecture notes by Tony Feng Spring 2012
Math 222
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Contents 1 Geometric preliminaries 1.1 Lie groups . . . . . . . . . . . 1.2 Lie algebras . . . . . . . . . . 1.3 Vector fields . . . . . . . . . . 1.4 Tangent and cotangent spaces 1.5 Pullbacks and pushforwards .
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7 7 9 10 11 13
2 The 2.1 2.2 2.3 2.4
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15 15 16 19 22
3 Lie 3.1 3.2 3.3 3.4
Lie algebra of a Lie group Invariant vector fields . . . . Integral curves . . . . . . . . The exponential map . . . . . The group law in terms of the
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algebras The Lie algebra gl(V ) . . . . . . . . . . The adjoint representation . . . . . . . . Poincar´e’s formula . . . . . . . . . . . . The Campbell-Baker-Hausdorff formula
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37 37 38 40 43 48
Universal Enveloping Algebra Construction of the universal enveloping algebra . . . . . . . . . . . Poincar´e-Birkhoff-Witt . . . . . . . . . . . . . . . . . . . . . . . . . . Proof of Poincar´e-Birkhoff-Witt . . . . . . . . . . . . . . . . . . . . .
51 51 52 54
4 Geometry of Lie groups 4.1 Vector bundles . . . . 4.2 Frobenius’ Theorem . 4.3 Foliations . . . . . . . 4.4 Lie subgroups . . . . . 4.5 Lie’s Theorems . . . . 5 The 5.1 5.2 5.3
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Math 222
CONTENTS
6 Representations of Lie groups 6.1 Haar Measure . . . . . . . . . 6.2 Representations . . . . . . . . 6.3 Schur orthogonality relations 6.4 The Peter-Weyl Theorem . .
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57 57 58 60 63
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69 69 70 72 76 78
8 Root systems 8.1 Structure of roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Weyl Chambers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83 83 89
7 Compact Lie groups 7.1 Compact tori . . . . . . . 7.2 Weight decomposition . . 7.3 The maximal torus . . . . 7.4 Trace forms . . . . . . . . 7.5 Representations of SU (2)
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9 Classificiation of compact Lie groups 97 9.1 Classification of semisimple Lie algebras . . . . . . . . . . . . . . . . 97 9.2 Simply-connected and adjoint form . . . . . . . . . . . . . . . . . . . 100 10 Representations of compact Lie groups 10.1 Theorem of the highest weight . . . . . 10.2 The Weyl character formula . . . . . . . 10.3 Proof of Weyl character formula . . . . 10.4 Proof of the Weyl integration formula. . 10.5 Borel-Weil . . . . . . . . . . . . . . . . .
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105 105 108 111 113 114
Disclaimer These are course notes that I wrote for Math 222: Lie Groups and Lie Algebras, which was taught by Wilfried Schmid at Harvard University in Spring 2012. There are, undoubtedly, errors, which are solely the fault of the scribe. I thank Eric Larson for pointing out some corrections on earlier versions of these notes.
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CONTENTS
6
Chapter 1
Geometric preliminaries 1.1
Lie groups
Definition 1.1.1. A Lie group, G, is a group endowed with the structure of a C ∞ manifold such that the inversion map s:G→G g 7→ g −1 and multiplication map m:G×G→G (g, h) 7→ gh are smooth. Remark 1.1.2. 1. It suffices to require that the single map (g, h) 7→ gh−1 is smooth. For example, the inversion map is smooth because it is the restriction of this map to e × G. 2. With moderate effort, one can show that requiring all structures to be C 2 is enough to imply smoothness. 3. Historically, Lie defined a Lie group not as above, but as (in modern times) the germ of a Lie group action. Let us now formally define the notion of “Lie group action.” Definition 1.1.3. Suppose G is a Lie group and M is a C ∞ manifold. An action of G on M is a map A : G × M → M such that A(g(A(h, m))) = A(gh, m). A more common notation for the action is g · m; in these terms, this condition is g · (h · m) = (gh) · m. 7
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Definition 1.1.4. Suppose G, H are Lie groups. A homomorphism from G to H is a C ∞ map F : G → H which is also a group homomorphism. Definition 1.1.5. Suppose G is a Lie group. A subgroup H of G in the algebraic sense is a Lie subgroup if H ⊂ G is a locally (with respect to H) closed submanifold of G. It is not obvious from this definition that a Lie subgroup is even a Lie group, so we will prove this presently. Proposition 1.1.6. A Lie subgroup is a Lie group when equipped with the induced C ∞ structure and the inclusion H → G is a Lie group homomorphism. Proof. First we clarify the meaning of “locally closed submanifold.” This means that for every h ∈ H, there exists an open neighborhood UH of h in H and an open neighborhood UG of h in G such that 1. UH = UG ∩ H. 2. UH ⊂ UG is a closed submanifold, e.g. looks locally like Rk ⊂ Rn . There is a subtlety here that we only have to consider a subset in H; this allows possible strange global behavior such as H actually “bunching up” at some point. Let N ⊂ M be a locally closed submanifold. A continuous function f on an open subset UN ⊂ N is C ∞ with respect to the induced C ∞ structure on N if and only if for all p in the domain of f , there exist open neighborhoods UN , UM of the domain of f such that f |UN is the restriction of a C ∞ function Fe : UM → R to UN . We just saw that H ⊂ G is a submanifold such that the inclusion H → G is a C ∞ map. It is a subgroup by definition. The map h 7→ h−1 is a C ∞ map to G because it is the restriction of the inversion map on G to H. That tells us that the inversion map is a C ∞ map to G taking values in H. By the definition of the induced C ∞ structure (namely, C ∞ maps to H are those which pull back C ∞ maps from H to C ∞ maps on the domain), this is a C ∞ map to H.
Example 1.1.7. The following are examples of Lie groups. 1. (Rn , +). 2. Giving Q the discrete topology, R×Q ⊂ R2 , + is a Lie subgroup of dimension 1 of (R2 , +). We recall that there are two topological restrictions on topological spaces that can be manifolds: Hausdorff, and second countable. Therefore, the same construction with R replacing Q would be a non-example. 3. (Rn /Zn , +): an n-dimensional torus. Smoothness, etc. follow because they are local properties. 8
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1.2. LIE ALGEBRAS
4. GLn (R), the group of invertible n × n matrices with real entries under matrix multiplication. This has a Lie group structure from its embedding as an open 2 subset of Rn by matrix entries. 5. GLn (C), with matrix multiplication. We can summarize Lie theory as the (shockingly successful) study of linearization, whereby one tries to study non-linear structure of a Lie group by the linear structure of its Lie algebra. Theorem 1.1.8. Let G be a Lie group, H ⊂ G a subgroup in the algebraic sense. If H is closed as a subset of G, then H has a unique structure of a Lie subgroup. Combining this result with example (4), we get lots of examples, including all the classical Lie groups: SO(n), SU(n), Sp(n, R), SO(p, q), SU(p, q), . . .. Exercise 1.1.9. (a) Show directly (i.e. not using the preceding theorem) that SO(n) has a natural Lie group structure, and (b) SO(n) ⊂ GL(n, R) is a Lie subgroup.
1.2
Lie algebras
Suppose K is a field. Definition 1.2.1. A Lie Algebra g over K is a vector space over K equipped with a bilinear map called the Lie bracket g×g→g (a, b) 7→ [a, b] such that 1. [a, a] = 0 for all a ∈ g. 2. [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0 for all a, b, c ∈ g. Remark 1.2.2. The definition tells us that the Lie bracket is skew-symmetric, so [a, b] = −[b, a]. Moreover, these conditions are equivalent if the characteristic of K is not 2. Definition 1.2.3. A homomorphism of Lie algebras F : g → g is a K-linear map which preserves the Lie bracket, i.e. [F a, F b] = F [a, b]. Definition 1.2.4. A Lie subalgebra of g is a K-linear subspace h ⊂ g such that [h, h] ⊂ h. 9
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Note that a Lie subalgebra is clearly a Lie algebra in its own right, and the inclusion h → g is a homomorphism of Lie algebras. Definition 1.2.5. An ideal in g is a K-linear subspace h ⊂ g such that [g, h] ⊂ h. If h ⊂ g is an ideal, then g/h has a unique Lie algebra structure so that the quotient map g → g/h is a homomorphism of Lie algebras. The Lie bracket on the quotient is [a, b] = [a, b] for lifts a, b of a, b. Example 1.2.6. 1. Suppose A is an associative algebra over K. Then we can turn A into a Lie algebra by defining [a, b] = ab − ba. The first condition is obvious, and the second can be checked by explicit computation. In particular, we may take A = End(V ) for some vector space V . 2. Suppose A is an associative algebra and D the vector space of derivations, i.e. linear maps d : A → A satisfying d(ab) = a(db) + (da)b for all a, b ∈ A. Then D is a Lie algebra under commutators, i.e. [d1 , d2 ] = d1 d2 − d2 d1 . In some sense, all Lie algebras arise in this way, though perhaps indirectly. Example 1.2.7. Let M be a C ∞ manifold, A = C ∞ (M ). Then D is the derivations of C ∞ (M ), i.e. the Lie algebra of vector fields.
1.3
Vector fields
Let M be a C ∞ manifold. By a “function” on M we mean either a real-valued function or complex-valued function unless otherwise specified. Recall that a vector field on M is a derivation of C ∞ (M ), which is the space of smooth functions on M . Lemma 1.3.1 (“Localization of vector fields.”). Suppose X is a vector field on M , p ∈ M , and f1 , f2 are functions on M such that f1 = f2 on some open neighborhood of p. Then Xf1 (p) = Xf2 (p). Proof. We use the existence of “cutoff functions.” Namely, suppose U0 ⊂ M is open and U1 ⊂ U0 is open such that U1 is compact and contained in U0 . Then there exists a function u ∈ C ∞ (M ) such that (i) f ≡ 1 on U1 , and (ii) Supp f ⊂ U0 . This follows from the existence of partitions of unity: just take an appropriate open cover (i.e. wherein U0 is the only open intersecting U1 ) and a partition of unity and consider the sum of those parts whose support intersects U1 . Apply this with U0 , U1 nested open neighborhoods containing p and f1 ≡ f2 on U0 . Then hf1 ≡ hf2 , since f1 ≡ f2 on the support of h. Then X(hf1 ) = X(hf2 ), so X(hfi )(p) = h(p)Xfi (p) + fi (p)Xh(p). Equating these for i = 1, 2, we conclude that Xf1 (p) = Xf2 (p).
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1.4. TANGENT AND COTANGENT SPACES
Corollary 1.3.2. Suppose X is a vector field on M and U ⊂ M is open. Then X can be restricted to U . Proof. If f ∈ C ∞ (U ), p ∈ U . Choose h ∈ C ∞ (M ) such that h ≡ 1 near p and Supp h ⊂ U . Define X|U f = X(hf )(p). This is well-defined by the lemma. Corollary 1.3.3. Vector fields can be patched, i.e. given open subset U1 , U2 ⊂ M and vector fields Xi on Ui for i = 1, 2 such that X1 |U1 ∩U2 = X2 |U1 ∩U2 , then there exists a unique vector field X on U1 ∪ U2 such that X|Ui = Xi . Proof. At any p ∈ Ui , define (Xf )(p) by Xi f |Ui (p). These two corollaries together tell us that vector fields constitute a sheaf. Definition 1.3.4. A coordinate neighborhood in M is a pair (U, x1 , . . . , xn ) with U ⊂ M open and x1 , . . . , xn ∈ C ∞ (M ) real-valued such that p 7→ (x1 (p), . . . , xn (p)) ∈ Rn establishes a diffeomorphism between U and its image in Rn (which must be open). Corollary 1.3.5. Suppose X is a vector field on M and (U, x1 , . . . , xn ) is a coordinate chart. Then X|U can be expressed uniquely as X|U =
u X
aj
j=1
∂ , ∂xj
aj ∈ C ∞ (M ).
Conversely, any such expression describes a unique vector field on M . Proof. Defining aj = Xxj . Suppose f ∈ C ∞ (U ) and p ∈ U . Then we may write f = f (p) · 1 +
X ∂f (p)(xi − xi (p)) + h, ∂dxi i
where h is a linear combination of products h1 h2 with h1 (p) = h2 (p) = 0 (this is just Taylor’s theorem). Note that X1 = 0 since X is a derivation. On a similar note, X(h1 h2 )(p) = h1 (p)Xh2 (p) + h2 (p)Xh1 (p) = 0. Therefore, X X ∂f ∂f (Xf )(p) = X (p)xi = ai (p). ∂xi ∂xi i
1.4
Tangent and cotangent spaces
Definition 1.4.1. A tangent vector at p is a linear map Xp : C ∞ (M ) → R or C such that Xp (f1 f2 ) = f1 (p)Xp f2 + f2 (p)Xp f1 . 11
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Corollary 1.4.2. If Xp is a tangent vector at p and f ∈ C ∞ (M ), then Xp f depends only on the values of f near p (i.e. in any neighborhood of p). This tells us that Xp f is well-defined for any C ∞ function f defined on some neighborhood of p; that is, tangent vectors are defined on germs of functions at p. Definition 1.4.3. The vector space (over R or C as appropriate) of tangent vectors at p is called the tangent space of M at p, and denoted Tp M . For p ∈ M , define Ip = {f ∈ C ∞ (M ) | f (p) = 0}, i.e. the ideal of functions vanishing at p. Given f ∈ C ∞ (M ), we may write f = f (p) · 1 + fp ,
fp ∈ Ip .
Denote by Ip2 the usual product of ideals. Suppose Xp ∈ Tp M and f ∈ Ip2 ; by the derivation property, Xp f = 0. Corollary 1.4.4. The map Xp 7→ (f 7→ Xp f ), where Xp ∈ Tp M , determines a linear inclusion Tp M → (Ip /Ip2 )∗ . Proof. We have a direct sum decomposition C ∞ (M ) = Ip ⊕ R (or C). Since Xp kills constants and Ip2 , it is uniquely determined by its action on (Ip /Ip2 ). Definition 1.4.5. Ip /Ip2 ' Tp∗ M is the cotangent space to M at p. In fact, the inclusion in the previous corollary is an isomorphism, which we will prove presently. Corollary 1.4.6. If (U ; x1 , . . . , xn ) is a coordinate neighborhood of p, then { ∂x∂ 1 |p , . . . , ∂x∂n |p } is a basis of Tp M . Proof. Completely analogous to Corollary 1.3.5. For f ∈ C ∞ (M ) and p ∈ M , define df |p ∈ T ∗ M as the image in T ∗ M ' Ip /Ip2 of f − f (p) · 1. If (U ; x1 , . . . , xn ) is a coordinate neighborhood of p, then {dx1 |p , . . . , dxn |p } is a basis of Tp∗ M . This is essentially Taylor’s theorem: we can write any function as a constant plus first order terms plus higher order terms about p, X ∂f X ∂f f − f (p) = (p)(xi − xi (p)) + . . . = (p)dxi . ∂xi ∂xi By counting dimensions, we conclude that Tp∗ M = (Tp M )∗ . Given a vector field X on M and p ∈ M , define Xp ∈ Tp M by Xp f = (Xf )(p). Therefore, any vector field X is completely determined by Xp ∈ Tp M for all p. Conversely, given an assignment p 7→ Xp ∈ Tp M , there exists a vector field X whose values are the Xp if and only if for every f ∈ C ∞ (M ), the function p 7→ Xp f is smooth. (If this is satisfied, it is clearly a derivation C ∞ (M ) → C ∞ (M ) since it is a derivation pointwise.) Since smoothness is a local property, we can check this criterion on coordinate neighborhoods. 12
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1.5
1.5. PULLBACKS AND PUSHFORWARDS
Pullbacks and pushforwards
Suppose F : M → N is a C ∞ map between manifolds. Then we can define the “pullback” F ∗ : C ∞ (N ) → C ∞ (M ) by F ∗ f = f ◦ F . 2 2 If p ∈ M , then F ∗ IN,f (p) ⊂ IM,p and F ∗ IN,f (p) ⊂ IM,p , so we get a linear ∗ ∗ ∗ map F : TF (p) N → Tp M . Dually, there is a map F∗ : Tp M → TF (p) N . Also, for any f ∈ C ∞ (N ), F ∗ sends df |F (p) ∈ TF∗ (p) N 7→ d(F ∗ f )|p ∈ Tp∗ M . In other words, d computes with pullbacks: F ∗ d = dF ∗ . This is clear from the interpretation of d as subtracting a constant from the function to put it in the ideal sheaf. We can pull back functions and differential forms under smooth maps, but a problem arises when we attempt to push vector fields forward. We will address this problem now. Definition 1.5.1. Let F : M → N be a smooth map. Two vector fields XM on M and XN on N are F -related if for every f ∈ C ∞ (N ), F ∗ (XN f ) = XM (F ∗ f ). In other words, the following diagram commutes. C ∞ (N )
XN
F∗
C ∞ (M )
XM
/ C ∞ (N )
F∗
/ C ∞ (M )
Yet another equivalent formulation is that F∗ XM |p = XN |F (p) for every p ∈ M . You can see the difficulty with pulling back in general: the left hand sides must agree for all p lying over a common point in N . Exercise 1.5.2. Check that if XM , XN and YM , YN are pairs of F -related vector fields, then [XM , YM ] is F -related to [XN , YN ]. Note that if F is a diffeomorphism, then for any vector field X on M there exists a unique vector field F∗ X on N which is F -related to X. After all, M and N are basically the same manifold. We can define F ∗ X by the formula above since we know that the pre-image of any point exists and is unique.
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Chapter 2
The Lie algebra of a Lie group 2.1
Invariant vector fields
Suppose G is a Lie group, g ∈ G. Define C ∞ maps `g : G → G
rg : G → G h 7→ hg −1 .
h 7→ gh
These are both diffeomorphisms by the definition of Lie group. Definition 2.1.1. The Lie algebra g of G is the Lie algebra of all left-invariant vector fields, i.e. vector fields X such that (`g∗ )X = X for all g ∈ G, with the Lie bracket being the commutator. Note that G-invariant means `g∗ Xh = Xgh for all g, h ∈ G. This is equivalent to `g∗ Xe = Xg for all g ∈ G. This gives an inclusion g → Te G sending X 7→ Xe (any left-invariant vector field is completely determined by its value at the identity). Exercise 2.1.2. Show that this map is an isomorphism: g ' Te G. Suppose that F : G → H is a homomorphism of Lie groups. Define F∗ : g → h as the composition F g ' Te G →∗ Te H ' h. Theorem 2.1.3. The map F∗ : g → h is a homomorphism of Lie algebras. Remark 2.1.4. If F1 , F2 are Lie group homomorphisms that can be composed, then F1∗ ◦ F2∗ = (F1 ◦ F2 )∗ . This follows from the analogous statement about pullbacks of functions. Therefore, “Lie algebra” is a covariant functor from the category of Lie groups (with Lie group homomorphisms) to Lie algebras. e ∈ h by the requirement that F∗ Xe = X ee ∈ Te H Proof. Suppose X, Y ∈ g. Define X e and similarly for Y , Y . 15
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e and Y, Ye are F -related. To see this, suppose g ∈ G. Then We claim that X, X eF (g) = `F (g)∗ X ee = `F (g)∗ F∗ Xe = (`F (g) ◦ F )∗ Xe X = (F ◦ `g )∗ Xe = F∗ `g∗ Xe = F∗ Xg , and similarly for Y, Ye . e Ye ] are F -related, which means that for all f ∈ C ∞ (H), Therefore, [X, Y ] and [X, e ) = X(F ∗ f ). F ∗ (Xf e Ye ] = X e Ye − Ye X e we see that Applying this to [X, ^ e Ye ] = [X, [X, Y ]. ^ By the same calculations as before, [X, Y ] and [X, Y ] are F -related.
2.2
Integral curves
Definition 2.2.1. Suppose M is a manifold, X a vector field on M . An integral curve to X is a pair (I, ϕ) where I ⊂ R is an open interval and ϕ : I → M is a smooth map satisfying d ϕ∗ |t0 = X|ϕ(t0 ) for all t0 ∈ I. dt Equivalently,
d dt
and X are ϕ-related.
Properties of integral curves. 1. If (I, ϕ) is an integral curve, then so is (I + a, ϕa ), where ϕa (t) = ϕ(t − a). 2. If (I, ϕ) is an integral curve, then (s−1 I, ϕ(st)) is an integral curve for sX, for s 6= 0. This is actually true for s = 0 if we interpret s−1 I = R. 3. Suppose (U ; x1 , . . . , xn ) is a coordinate neighborhood and X is a vector field on U with local expression X=
X
ai
∂ , ∂xi
ai ∈ C ∞ (U ).
Then (I, ϕ), where ϕ : I → M is given by t 7→ (x1 (t), . . . , xn (t)) is an integral i curve if and only if dx dt = ai ((xi (t))). The first and third facts follow directly from the chain rule. The third also follows from the chain rule: X dxi ∂ d ϕ∗ = . dt dt ∂xi 16
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2.2. INTEGRAL CURVES
Therefore, finding integral curves of vector fields amounts to solving an ODE. Basic facts on ODE. Theorem 2.2.2 (Existence of local solutions). If X is a vector field on M and p ∈ M , then there exists an integral curve (I, ϕ) with 0 ∈ I and ϕ(0) = p. Theorem 2.2.3 (Uniqueness). Suppose (I1 , ϕ1 ) and (I2 , ϕ2 ) are two integral curves for X, and ϕ1 (t0 ) = ϕ2 (t0 ) for some t0 ∈ I1 ∩ I2 . Then ϕ1 = ϕ2 on I1 ∩ I2 , and hence there exists a unique integral curve (I1 ∪ I2 , ϕ) with ϕ|Ui = ϕi . In particular, given an integral curve (I, ϕ) there exists a unique maximal exe ϕ) e such that I ⊂ Ie and ϕ| e I = ϕ with Ie maximal. tension, i.e. an integral curve (I, This is called a maximal integral curve. Theorem 2.2.4 (C ∞ dependence on initial conditions). If (I, ϕ) is an integral curve and p ∈ M , then there exists an open neighborhood U of p ∈ M and > 0, and a C ∞ map Φ : U × (−, ) → M such that (i) t 7→ Φ(·, t) is an integral curve for t ∈ (−, ). (ii) Φ(q, 0) = q for any q ∈ U . This says that we can choose the interval of an integral curve locally uniformly in p, and depending smoothly on p. Definition 2.2.5. A vector field X on M is complete if every maximal integral curve is defined on all of R. ∂ Example 2.2.6. 1. Let M = R2 − {(0, 0)} and X = ∂x . Then the integral curves are linearly parametrized by lines parallel to the x-axis. An integral curve on the x axis cannot be extended to all of R since there is a “hole” at (0, 0). d 2. The vector field dx on (−1, 1) is incomplete. Therefore, composing it with a diffeomorphism (−1, 1) ' R results in an incomplete vector field on R.
Proposition 2.2.7. Suppose G is a Lie group and X ∈ g. Then X is complete. Proof. Suppose ϕ : I → G is an integral curve and g ∈ G. Then t 7→ `g ◦ ϕ is also an integral curve, because X is G-invariant. In particular, if the integral curve through one point is infinitely extendable, then every integral curve is infinitely extendable and X is complete. The idea is simple: if the integral curve is incomplete, then it “runs out of steam” at some finite point. But since G is a Lie group and X is left-invariant, we can always translate it to keep it going a little longer. Suppose (I, ϕ) is a maximal integral curve with 0 ∈ I and ϕ(0) = e. Suppose that I has an endpoint t0 ; without loss of generality, suppose t0 > 0. Choose > 0 17
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CHAPTER 2. THE LIE ALGEBRA OF A LIE GROUP
such that || < t0 and choose t1 < t0 with |t1 − t0 | < 2 . Then t 7→ ϕ(t − t1 ) is welldefined for |t1 − t| < and takes the value e at t = t1 . Therefore, t 7→ ϕ(t1 )ϕ(t − t1 ) is an integral curve agreeing with ϕ at t = t1 and defined on an interval containing even t0 . Suppose X is a complete vector field on M . Define Φ : R × M → M by (i) Φ(0, p) = p for all p ∈ M . (ii) t 7→ Φ(t, p) is an integral curve for all p ∈ M . Theorem 2.2.8 (Global C ∞ dependence on parameters). Φ : R × M → M is a smooth map. Proof. By Theorem 2.2.4, Φ is smooth on an open neighborhood of {0} × M (as a function of both arguments), i.e. Φ is smooth near (0, p0 ). The map t 7→ Φ(t, Φ(s, p)) is an integral curve in t with value Φ(s, p) at t = 0, as is t 7→ Φ(s + t, p). By uniqueness of integral curves, we know that Φ(t, Φ(s, p)) = Φ(s + t, p) for all p ∈ M and all s, t ∈ R. For p ∈ M , let Ip = {t ∈ R | Φ is smooth near (t, p)}. Then Ip is open and 0 ∈ Ip . To prove the theorem, we just need to show that Ip = R for all p. Suppose this is not the case for some p0 ∈ M ; then the connected component of 0 in Ip has some finite endpoint t0 . Without loss of generality, assume that t0 > 0. Then we have by assumption, Φ is C ∞ (in both arguments) on a neighborhood of [0, t0 ) × {p0 }. Pick η satisfying 0 < η < 1. Then Φ(t0 , p) = Φ(ηt0 , Φ((1 − η)t0 , p)). If η is small enough, we can find an open neighborhood U of Φ(t0 , p0 ) such that p 7→ Φ(ηt0 , p) is C ∞ for p ∈ U . Making η smaller if necessary, we can arrange that Φ((1 − η)t0 , p0 ) ∈ U because Φ is continuous in the time variable. Moreover, for p near p0 , p 7→ Φ((1−η)t0 , p) is C ∞ by the definition of t0 . Therefore, the composition p 7→ Φ(ηt0 , Φ((1 − η)t0 , p)) = Φ(t0 , p) is C ∞ in p near p0 . We now show that this implies that (t, p) 7→ Φ(t, p) is smooth in a neighborhood of (t0 , p0 ), which furnishes the desired contradiction. To see this, observe that Φ(t, p) = Φ(t − t0 , Φ(t0 , p)). 18
Math 222
2.3. THE EXPONENTIAL MAP
By hypothesis, (t, q) 7→ Φ(t − t0 , q) is smooth near t = t0 and q = Φ(p0 , t0 ). Since p 7→ Φ(t0 , p) is smooth, Φ(t, p) is the composition of smooth functions (t, p) 7→ (t, Φ(t0 , p)) 7→ Φ(t − t0 , Φ(t0 , p)), hence is smooth.
2.3
The exponential map
Let G be a Lie group with Lie algebra g. We showed last time that any X ∈ g is complete. Fixing X ∈ g, we get by the previous proposition a smooth map ΦX : R × G → G satisfying ΦX (0, g) = g and t 7→ ΦX (t, g) is an integral curve for X. Therefore, we get a map Φ:R×G×g→G (t, g, X) 7→ ΦX (t, g). Proposition 2.3.1. Φ is C ∞ as a function of all variables. Before giving the proof, we set up the general framework. Consider the vector field X on a product of manifolds M × N . Say that X is tangential to the fibers of projection π2 : M × N → N if all the integral curves of X lie in the fibers. Equivalently, in local coordinates (U ; x1 , . . . , xk , xk+1 , . . . , xn ) such that x1 , . . . , xk are local coordinates on M and xk+1 , . . . , xn are local coordinates on N . The vector field X can be expressed locally as X=
n X i=1
ai
∂ . ∂xi
Then X is tangential to the fibers if and only if aj = 0 for j > k. In other words, for each m ∈ M and n ∈ N X(m,n) ∈ Tm M × {0} ⊂ Tm M ⊕ Tn N = T(m,n) M × N. We can view a vector field X of this type as a family of vector fields on M , parametrized by n ∈ N . Note that X is complete on M × N if and only if when considered as a family of vector fields on M parametrized by N , each member is complete as a vector field on M ; this is clear from the fact that the integral curve to any point of M × N stays in a signle fiber, which is isomorphic to M . Hence a family of complete vector fields with parameters in N determines a e : R × M × N → M × N by Theorem 2.2.8. Composing with the smooth map Φ projection M × N → M gives a smooth map Φ : R × M × N → M . 19
Math 222
CHAPTER 2. THE LIE ALGEBRA OF A LIE GROUP
Proof of Proposition 2.3.4. Let M = G and N = g in the above discussion. Properties of Φ. (i) Φ(0, g, X) = g. (ii) t 7→ Φ(t, g, X) is an integral curve for X. (iii) Φ(t, g, X) = gΦ(t, e, X). (iv) Φ(t, g, sX) = Φ(st, g, X). (v) Φ(s, e, X)Φ(t, e, X) = Φ(s + t, e, X). Proof. (i) and (ii) are the definition of Φ. (iii), (iv), and (v) follow from the observation that both sides of the respective equalities represent integral curves for X with the same initial conditions (we are using, of course, the fact that X is left-invariant). In particular, it follows from the above properties that Φ(t, g, X) = gΦ(1, e, tX). So Φ is completely determined by the map exp : g → G defined by exp X = Φ(1, e, X). This is the famous exponential map. Properties of the exponential map. (i) exp : g → G is smooth. (ii) t 7→ g exp(tX) is the integral curve to X with the initial point g at t = 0. (iii) exp(sX) exp(tX) = exp((s + t)X). Equivalently, exp(X + Y ) = exp(X) exp(Y ) if X, Y are linearly dependent. Proposition 2.3.2 (Universality of the exponential map.). Suppose F : G → H is a Lie group homomorphism. Then the following diagram commutes. F∗
g
exp
G
/h
F
20
exp
/H
Math 222
2.3. THE EXPONENTIAL MAP
Exercise 2.3.3. Prove this proposition. Proposition 2.3.4. The composition exp∗
g ' T0 g → Te G ' g is the identity. Exercise 2.3.5. Prove this proposition. Proposition 2.3.6. For g ∈ G and X ∈ g, f ∈ C ∞ (G), (Xf )(g) =
d f (g exp(tX))|t=0 . dt
Exercise 2.3.7. Prove this proposition. Corollary 2.3.8. The map exp : g → G is locally a diffeomorphism near 0 ∈ g. Proof. By Proposition 2.3.4, the derivative of the exponential map is the identity at the origin. The result then follows directly from the inverse function theorem. Any manifold is the countable union of its connected components. Note that connectedness and path-connectedness are equivalent notions for a manifold, since a manifold is locally path-connected. Let G be a Lie group and G0 the connected component containing the identity of G; we call this the identity component. Note that (G0 )2 = {g1 g2 | gi ∈ G0 }
and
(G0 )−1 = {g −1 | g ∈ G0 }
are connected and contain e (since we can go from a product to another product by a product of connecting paths), hence equal to G0 . This tells us that G0 is an open, closed subgroup. Also for g ∈ G, gG0 g −1 is a connected subgroup of G containing e. Therefore, the identity component is a normal subgroup. Now suppose U ⊂ G0 is a neighborhood of the identity. We can arrange that U = U −1 by replacing U with U ∩ U −1 if necessary. Then ∞ [ G1 = Uk k=1
is open (since U k is a union of translates of opens, hence open) and a subgroup of G0 . Therefore G1 is closed, because its complement is a union of its cosets. Since G0 was connected, it must be the case that G1 = G0 . We thus observe that if U is a neighborhood of the identity in a Lie group G, then U generates G0 in the algebraic sense. We have proved: Theorem 2.3.9. Let G be a Lie group with Lie algebra g. 21
Math 222
CHAPTER 2. THE LIE ALGEBRA OF A LIE GROUP
(i) exp(g) generates the identity component of G in the algebraic sense. (ii) Let F : G → H be a Lie group homomorphism. Then the restriction of F to G0 is completely determined by F∗ : g → h. Proof. (i) follows from the fact that exp is a local diffeomorphism from g to G, hence contains a neighborhood of e ∈ G. (ii) follows from the preceding observation and the universality of the exponential map, which tells us that knowledge of F∗ completely determines F restricted to the image of the exponential map. The general theme is that the Lie algebra recognizes everything about the Lie group, except in two cases 1. it doesn’t know about other connected components, and 2. it can’t distinguish covering spaces.
2.4
The group law in terms of the Lie algebra
If the Lie algebra knows “everything” about G, then it should know about the group operation. This is the subject of the Campbell-Baker-Hausdorff formula, which is our next topic. Let G be a Lie group and g its Lie algebra. We have seen that there exist neighborhoods V0 containing 0 in g and Ue containing e in G such that exp : V ' Ue . Let log : Ue → V0 denote the inverse. By shrinking V0 (and thus Ue ), we can arrange that V0 is ?-shaped about 0. Recall that this means that for any p ∈ V0 , the line between p and 0 lies in V0 . By further shrinking V0 if necessary, we may also assume that V0 = −V0 . By passing to a yet smaller neighborhood V , we assume that there exists a neighborhood U of e ∈ G such that log : U → g is smooth and well-defined, and (exp V )2 ⊂ U . To summarize, we have an open neighborhood V 3 0 ∈ g such that (i) V = −V (ii) V is ?-shaped about 0. (iii) There is a neighborhood U of e ∈ G such that log is well defined and smooth on U , and (exp V )2 ⊂ U . That allows us to define a C ∞ map M : V ×V → g sending M (X, Y ) = log(exp X exp Y ). This M is the group multiplication transferred to the Lie algebra via the exponential map. Note the following properties of this map: 1. M (X, 0) = X and M (0, Y ) = Y . 2. M (X, −X) = 0. 22
Math 222
2.4. THE GROUP LAW IN TERMS OF THE LIE ALGEBRA
3. M (−X, −Y ) = −M (Y, X). This follows from the fact that (exp X exp Y )−1 = (exp Y )−1 (exp X)−1 . 4. M (X, M (Y, Z)) = M (M (X, Y ), Z) if X, Y, Z is sufficiently small (for everything to lie in V ). This is associativity of the group law. Proposition 2.4.1. For X, Y ∈ g, ∂2 1 M (sX, tY )|s=t=0 = [X, Y ]. ∂s∂t 2 Proof. For g ∈ and f ∈ C ∞ (G), ∂ f (g exp(sX))|s=0 . ∂s ∂ Y f (g exp(sX))|s=0 . Now applying the Applying this to Y f , we have XY f (g) = ∂s same reasoning to Y f , we find that this is Xf (g) =
∂2 f (g exp(sX) exp(tY ))|s=t=0 . ∂s∂t Therefore, [X, Y ]f (e) = (XY − Y X)f (e) ∂2 (f (exp(sX) exp(tY )) − f (exp(tY ) exp(sX)))s=t=0 ∂s∂t ∂2 (f ◦ exp(M (sX, tY )) − f ◦ exp(M (tY, sX))) |s=t=0 = ∂s∂t ∂2 = (f ◦ exp(M (sX, tY )) − f ◦ exp(−M (−sX, −ty))) |s=t=0 . ∂s∂t By setting s and t to zero, we see that the Taylor expansion of M (sX, tY ) is =
M (sX, tY ) = sX + tY + stC + (terms of order ≥ 3) ∂2
where C = ∂s∂t M (sX, tY )|s=t=0 . We substitute the terms of order up to 2 into the expression above to obtain ∂2 [f ◦ exp(sX + tY + stC) + f ◦ exp(sX + tY − stC)]s=t=0 . ∂s∂t Replace f ◦ exp by ϕ, so ϕ is a C ∞ function defined on V . Since exp∗ is the identity map, the left hand side of the original equation is [X, Y ]ϕ(0). In other words, =
∂2 [ϕ(sX + tY + stC) − ϕ(sX + tY − stC)]s=t=0 ∂s∂t ∂2 =2 ϕ(sX + tY + stC)|s=t=0 ∂s∂t ∂2 =2 ϕ(M (sX, tY ))|s=t=0 . ∂s∂t
[X, Y ]ϕ(0) =
23
Math 222
CHAPTER 2. THE LIE ALGEBRA OF A LIE GROUP
Therefore, M (X, Y ) = X + Y + 12 [X, Y ] + . . .. Corollary 2.4.2. Multiplication on G determines the Lie bracket [·, ·] on g.
24
Chapter 3
Lie algebras 3.1
The Lie algebra gl(V )
Let V be a finite dimensional real or complex vector space, GL(V ) the group of invertible endomorphisms of V . This is open in End(V ). With the induced structure, GL(V ) becomes a Lie group of V , and a complex Lie group if V is a vector space over C. Definition 3.1.1. A complex Lie group is a group G endowed with the structure of a complex manifold such that multiplication and inversion are holomorphic maps. In this case, g is the Lie algebra of left-invariant holomorphic vector fields. This is the “holomorphic part” of the complexification of the Lie algebra of G considered as a real Lie group, essentially because of the Cauchy-Riemann equations. We will elaborate on this discussion later; we just want to indicate how the proceeding discussion applies to both real and complex vector spaces. Note also that in the complex case, exp, log, and M are holomorphic maps. Turn V into a normed vector space, i.e. pick a norm || · || : V → k satisfying (i) ||av|| = |a|||v|| for scalars a and vectors v. (ii) ||u + v|| ≤ ||u|| + ||v||. (iii) ||v|| ≥ 0 and equality holds only if v = 0. For T ∈ End(V ), define the “operator norm” ||T || = sup ||T v||. ||v||≤1
Note that ||S ◦ T || ≤ ||S|| ||T ||. Define Exp : End(V ) → End(V ) by ∞ X 1 k Exp(T ) = T . k! k=0
25
Math 222
CHAPTER 3. LIE ALGEBRAS
This converges uniformly on bounded subsets of End(V ). For formal reasons, Exp(S + T ) = Exp(S) Exp(T ) provided that S, T commute, because absolutely convergent series can be re-ordered. In particular, Exp(−T ) = Exp(T )−1 , so Exp lands in GL(V ). By the usual arguments, convergent power series can be differentiated term-by-term. So ∂ Exp(tT ) = Exp(tT ) · T. ∂t Therefore, t 7→ Exp(tT ) is tangential to the left-invariant (holomorphic) vector field whose value at e is T . Since this property also characterizes the exponential map, we must have Exp = exp. This shows: Proposition 3.1.2. The exponential map for GL(V ) (in both the real and complex cases) is given by ∞ X 1 k exp T = T . k! k=0
What about the logarithm? We have ∞
X 1 d log(1 − w) = − =− wk . dw 1−w k=0
Therefore, log z = −
∞ X (1 − z)k k=1
k
.
The map log : GL(V ) → End(V ) which is defined near 1V ∈ GL(V ) is given by log g = −
∞ X 1 (1 − g)k k k=1
for formal reasons concerning power series. We now compute the Lie bracket for gl(V ) := Lie(GL(V )). If 1 2 1 2 g = exp S exp T = 1 + S + S + . . . 1 + T + T + ... 2 2 1 1 = 1 + S + T + S 2 + T 2 + ST + . . . 2 2 where we omit all terms of order ≥ 3, because they are not relevant to the calculation. Then 1 M (S, T ) = log g = g − 1 − (1 − g)2 + . . . 2 1 2 1 2 1 2 = S + T + S + T + ST − S + T 2 − ST − T S 2 2 2 1 = S + T + [S, T ] + . . . 2 26
Math 222
3.2. THE ADJOINT REPRESENTATION
This discussion proves the following proposition. Proposition 3.1.3. The Lie bracket on End(V ), viewed as the Lie algebra of GL(V ), is [S, T ] = ST − T S. How should we view this? If we have an arbitrary Lie group, the Lie algebra can’t really detect things like covering spaces and different connected components. However, modulo these considerations, all Lie groups are captured as subgroups of GL(V ). In other words, every connected Lie group is a covering of some Lie group sitting inside a subgroup of GL(V ).
3.2
The adjoint representation
Suppose G is a Lie group with Lie algebra g. For g, h ∈ G, we have `g ◦ rg = ghg −1 , i.e. conjugation by g. So we get a group homomorphism G → Aut(G) sending g 7→ `g ◦ rg . Define Ad : G → End(g) by Ad (g) = (`g ◦ rg )∗ . Since (`g ◦ rg ) ◦ (`g−1 ◦ rg−1 ) is the identity, we have Ad (g −1 ) = (Ad )−1 . So Ad lands in Aut(g) ⊂ GL(g). We already know that GL(g) is a Lie group with Lie algebra End(g). Definition 3.2.1. We set ad := Ad∗ : g → End(g). Note that this is a Lie algebra homomorphism by the Jacobi identity. We cal Ad the adjoint representation and ad the infinitesimal adjoint representation. Proposition 3.2.2. (ad X)(Y ) = [X, Y ]. Proof. By definition of the exponential map, Y ∈ g ' Te G is the tangent vector at t = 0 to the curve t 7→ exp(tY ). Therefore, Adg(Y ) is the tangent vector at t = 0 to t 7→ g exp(tY )g −1 . Hence adX(Y ) is the tangent vector at s = 0 to the curve Ad(exp(sX))Y ). 27
Math 222
CHAPTER 3. LIE ALGEBRAS
Therefore, for any smooth f near e ∈ G, adX(Y ) ∈ g ' Te G, and (adX(Y ))f (e) = = = =
=
∂2 f (exp(sX) exp(tY ) exp(−sX))|s=t=0 ∂s∂t ∂2 f ◦ exp(M (sX, M (tY, −sX)))|s=t=0 ∂s∂t ∂2 1 f ◦ exp(sX + M (tY, −sX) + [sX, M (tY, −sX)] + . . .)|s=t=0 ∂s∂t 2 ∂2 1 f ◦ exp sX + tY − sX + [tY, −sX] ∂s∂t 2 1 + [sX, tY − sX + . . .] |s=t=0 2 2 ∂ f ◦ exp (tY + st[X, Y ] + . . .)s=t=0 . ∂s∂t
Letting ϕ = f ◦ exp, we have ϕ is smooth near 0 in g since exp∗ = 1 (and the inverse function theorem). In terms of ϕ, ∂2 ϕ(tY + st[X, Y ] + . . .)|s=t=0 ∂s∂t = ϕ([X, Y ]) = ([X, Y ]ϕ)(0).
((adX(Y ))ϕ)(0) =
Corollary 3.2.3. ad X for X ∈ g is a derivation of the Lie algebra g, i.e. ad X[Y, Z] = [ad X(Y ), Z] + [Y, ad X(Z)]. With adX = [X, ·], this condition is equivalent to the Jacobian identity. A derivation of the Lie algebra should be thought of as an “infinitesimal automorphism,” and ad X for some X ∈ g should be thought of as an “infinitesimal inner automorphism.”
3.3
Poincar´ e’s formula
Suppose X ∈ g; then g ' TX g. The exponential map induces exp |X
g ' TX g −−−∗−→ Texp X G. On the other hand, (`exp X )∗
g ' Te G −−−−−→ Texp X G. 28
´ FORMULA 3.3. POINCARE’S
Math 222
Since `exp(−X) = (`exp X )−1 , the map (`exp X )∗ is invertible and threading across the diagram gives a linear map g → g ∼ =
g
go
∼ =
exp∗ |X
/ TX g
Te G o
(`exp X )−1 ∗
/ Texp X G
Texp X G
One might ask, what is this map? Theorem 3.3.1 (Poincare’s formula for the differential of the exponential map). exp∗ |X = (`exp X )∗ ◦ Note that
1 − e− ad X . ad X
z 1 − e−z = 1 − + ..., z 2
an entire holomorphic function. So we can think of the expression convergent power series in the operator adX.
1−e−adX adX
as a
Remark 3.3.2. This formula is quite useful! By contrast, the exact coefficients of the Campbell-Baker-Hausdorff formula rarely comes up in practice. Proof. By the preceding discusson, there exists D(X) ∈ End(g) such that exp∗ |X = (`exp X )∗ ◦ D(X). ∼ =
g
exp∗ |X
/ TX g
(3.3.1)
/ Texp X G
D(X)
go
∼ =
Te G o
(`exp X )−1 ∗
Texp X G
The map D : g → End(g) is a smooth function of X. Exercise 3.3.3. Prove this. Define F (t) = tD(tX), which is then a smooth function of t with values in End(g). We claim that F 0 (t) = 1g − ad (X) ◦ F (t). Supposing this for the moment, define 1 − e−t ad X Fe(t) = := (1 − e−t ad X )(ad X)−1 . ad X 29
Math 222
CHAPTER 3. LIE ALGEBRAS
Then Fe0 (t) = e−tadX = 1 − Fe(t) ◦ ad(X) = 1 − adX ◦ Fe(t). Also, F (0) = Fe(0) = 0. Therefore, F = Fe, which is the desired conclusion. Now we prove the claim. First we give a more concrete interpretation of (3.3.1). Applying it to Y , we see that for f a smooth function near exp X in G, we have ∂ f (exp(X) exp(tY ))|t=0 = (D(X)Y )f (exp X). ∂t
(3.3.2)
To elaborate, let Y ∈ g. We are concerned with its image in Texp X G, which is a functional on C ∞ (M ); what is the value of its image at f ∈ C ∞ (M )? Along the map described by the right hand side of (3.3.1), it is just ((D(X)Y )f )(exp X), absorbing the (`exp X )∗ into the formula. Now consider the map described by the left hand side of (3.3.1). We know from the homework that ∂ f (exp(X + tY ))|t=0 = Y f (exp X), ∂t the reasoning being that t 7→ exp X exp(tY ) is an integral curve to Y at exp X. Since the ambiguity between left-invariant vector fields and elements of the Lie algebra will be confusing, for this proof only we use the notation r(Y ) for the left invariant vector field generated by Y ∈ g. In these terms, (3.3.2) says (r(Y )f )(g) =
d f (g exp uY )|u=0 . du
We used the notation r(Y ) because right translation commutes with left translation, so “infinitesimal right translation is a left-invariant vector field.” Recall that we wanted to prove that for F (t) = tD(tX), F 0 (t) = 1 − ad X ◦ F (t). So by the preceding equations, this has something to do with ∂2 f (exp(t(X + sY )))|s=0 ∂s∂t Let’s calculate it in two different ways. 1. By a homework problem, ∂ f (exp(t(X + sY ))) = (r(X + sY )f )(exp(t(X + sY ))). ∂t Now (3.3.2) tells us exactly how to differentiate this expression with respect to s. Doing so, we find that ∂ (∗)|s=0 = r(Y )f (exp tX) + r(D(tX)(tY ))(r(X)f )(exp(tX)). ∂s 30
´ FORMULA 3.3. POINCARE’S
Math 222 2. Reasoning as above, we first compute
∂ f (exp(t(X + sY )))|s=0 = r(tD(tX)Y )f (exp(tX)). ∂s Then differentiating with respect to t, we find that ∂ d (∗)|s=0 = r( (tD(tX))Y )f (exp tX) + r(X)r(tD(tX)Y )f (exp tX). ∂s dt Comparing the two expressions, we conclude that r(Y )f (exp tX) = r(
d (tD(tX)Y ))f (exp tX) + r([X, tD(tX)Y ])f (exp tX). dt
Since this holds for all f , we conclude that Y = Hence
d (tD(tX))Y + [X, tD(tX)Y ]. dt
d (tD(tX)) = 1g − adX ◦ tD(tX). dt That was a struggle... - Wilfried Schmid
We re-iterate Poincar´e’s formula for emphasis: exp∗ |X = (`exp X )∗ ◦
1 − e−adX . adX
Now let’s manipulate this identity. exp∗ |X = (rexp(−X) )∗ ◦ (rexp(X) )∗ ◦ (`exp(X) )∗ ◦
1 − e−adX adX
1 − e−adX adX 1 − e−adX = (rexp(−X) )∗ ◦ exp(ad X) ◦ (universality of exp) adX eadX − 1 = (rexp(−X) )∗ ◦ . adX = (rexp(−X) )∗ ◦ Ad exp X ◦
We have shown the right-translation analogue: Corollary 3.3.4. We have exp∗ |X = (rexp(−X) )∗ ◦ 31
ead X − 1 . ad X
Math 222
3.4
CHAPTER 3. LIE ALGEBRAS
The Campbell-Baker-Hausdorff formula
Now put a linear norm on g. Then there exists C > 0 such that ||[X, Y ]|| ≤ C||X|| ||Y || Choose an open neighborhood V of 0 in g such that (i) V = −V (ii) V is ?-shaped about 0. (iii) X, Y ∈ V =⇒ ||M (X, Y )|| ≤
2π C.
(This implies that ||adM (X, Y )|| < 2π.)
With this choice, for all X, Y ∈ V , hj (adM (X, Y )) is well-defined. Define z z e − 1 −1 h1 (z) = z = e −1 z −1 z 1 − e−z h2 (z) = = . 1 − e−z z As power series in z, h1 and h2 have radius of convergence 2π. Therefore, for ||X|| < 2π C , both h1 (adX) and h2 (ad(X)) are well-defined. So with this choice of V , hj (adM (X, Y )) is well-defined for all X, Y ∈ V . Since V is open, for X and Y in V there exists > 0 such that |t| < 1 + =⇒ tX, tY ∈ V. Define F (t) = M (tX, tY ). Then the map F : (−1 − , 1 + ) → g is smooth. Lemma 3.4.1. For X, Y ∈ V and |s|, |t| < 1 + we have the following identities: ∂ M (sX, tY ) = h1 (ad M (sX, tY ))X ∂s ∂ M (sX, tY ) = h2 (ad M (sX, tY ))Y ∂t Proof. Suppose f is a C ∞ function defined near exp sX exp tY . Consider ∂ ∂ f (exp sX exp tY ) = f (exp sX exp tY exp uY )|u=0 . ∂t ∂u Thinking of Y as a left-invariant vector field, this is equal to Y f (exp sX exp tY ). 32
Math 222
3.4. THE CAMPBELL-BAKER-HAUSDORFF FORMULA
Just by the chain rule, we have ∂ ∂ f (exp sX exp tY ) = exp∗ |M (sX,tY ) M (sX, tY ) f. ∂t ∂t | {z } ∈g'Te G
|
{z
}
∈Texp M G
Now applying the formula for the differential of the exponential map, we get
(`exp sX exp tY )∗ ◦
1 − e−adM (sX,tY ) ∂ M (sX, tY ) f. adM (sX, tY ) ∂t {z } | ∈g'Te G
Considering same as
1−e−adM (sX,tY ) ∂ adM (sX,tY ) ∂t M (sX, tY
) as a left-invariant vector field, this is the !
1 − e−adM (sX,tY ) ∂ M (sX, tY ) f (exp sX exp tY ). adM (sX, tY ) ∂t That tells us that Y =
1 − e−adM (sX,tY ) ∂ ∂ M (sX, tY ) = h2 (adM (sX, tY ))−1 M (sX, tY ). adM (sX, tY ) ∂t ∂t
as left-invariant vector fields. We conclude that ∂ M (sX, tY ) = h2 (adM (sX, tY ))Y. ∂t The other formula is obtained the same way. Actually, we have to identify g in that case with right-invariant vector fields, or consider the “anti-automorphism” g 7→ g −1 . In any case, there is no real difference between left and right. Corollary 3.4.2. With the notation above, F 0 (t) = h1 (ad F (t))X + h2 (ad F (t))Y. Proof. F 0 (t) =
∂ ∂t M (tX, tY
). Apply Lemma 3.4.1, using the chain rule.
So far we only know that F is a C ∞ function, but we can write down its Taylor series ∞ X F (t)“ = ” tk Mk (X, Y ). k=1
Differentiating and applying the corollary, ∞ X k=1
kt
k−1
Mk (X, Y ) = h1
∞ X
! `
t adM` (X, Y ) X + h2
`=1
∞ X
! `
t adM` (X, Y ) Y.
`=1
(3.4.1) 33
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Now, z z = 2 3 z −1 z + 2 + z6 + . . . 1 z = = 1 − + ..., z 1 + 2 + ... 2
h1 (z) =
and similarly h2 (z) =
ez
z z z = = 1 + + .... 2 −z z 1−e 2 z − 2 + ...
So h1 (0) = h2 (0) = 1 and h01 (0) = − 21 , h02 (0) = 21 . What is the first order term? On the right hand side of (3.4.1), we get M (tX, tY ) = t(X + Y ) + . . . Now we want to identify coefficients of tk . First considering only the linear term of h1 , we get 1 1 (k + 1)Mk+1 (X, Y ) = [X, Mk (X, Y )] − [Y, Mk (X, Y )] + . . . 2 2 Since the series inside the arguments of the hj have no constant term, the other contributions to tk come from linear combinations of terms adMj1 (X, Y )adMj2 (X, Y ) . . . adMjn (X, Y ) with 1 ≤ j1 , j2 , . . . , jn < k such that j1 + . . . + jn = k. Definition 3.4.3. A Lie monomial of degree k in variables X, Y is a well-formed expression in the terms of X, Y , and bracket. A Lie polynomial homogeneous of degree k in X and Y is a linear combination of linear monomials in X, Y of degree k. Example 3.4.4. The expression [X, [Y, X]] is a Lie monomial of degree 3. The expression [[X, Y ], [X, [X, Y ]]] is a Lie monomial of degree 5. Theorem 3.4.5 (Campbell-Baker-Hausdorff). There exist universal Lie polynomials Mk (X, Y ), homogeneous of degree k, with the following property: if G is a Lie group with Lie algebra g, then there exists a constant R(C) depending on C but not otherwise on G, g such that for X, Y ∈ g satisfying ||X||, ||Y || < R(C), the series M (X, Y ) =
∞ X
Mk (X, Y )
k=1
converges absolutely and uniformly on compact subsets of {||X||, ||Y || < R(C)} to the function M (X, Y ) satisfying exp M (X, Y ) = exp X exp Y. 34
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3.4. THE CAMPBELL-BAKER-HAUSDORFF FORMULA
Remark 3.4.6. Before embarking on the proof, we give some historical perspective. 1. The first proof (about 1885) was due to Friedrich Schur (not the usual Schur). This was the first proof standing up to modern levels of rigor. 2. Campbell treated the problem purely formally (not worrying about convergence). 3. The second proof was given in around 1890 by Poincar´e. 4. Baker came later, around 1902. 5. Hausdorff (1906) gave the modern formulation, in terms of Lie polynomials, with arguments essentially the same as ours. Proof of Theorem 3.4.5. At this point, the only issue left is that of convergence. The differential equation F 0 (t) = h1 (adF (t))X + h2 (adF (t))Y.
(3.4.2)
is a “holomorphic ordinary differential equation,” since we know that h1 (adF (t)) and h2 (adF (t)) are holomorphic on the relevant region. There is a general theorem about the existence of solutions in such a situation, precisely by looking at and bounding coefficients of power series. Set gC := C ⊗R g, a complex Lie algebra, and regard the ODE (3.4.2) as taking values in C with t as a complex parameter. Then (3.4.2) is a holomorphic ODE. By generalities on ODE, a solution certainly exists in a neighborhood of 0. Moreover, the point of going to the complex analytic context is that the radius of convergence of a complex-analytic function at a point is always the distance to the nearest singularity. So there exists r > 0, potentially depending on X and Y , such that the series X F (t) := tk Mk (X, Y ) has radius of convergence r. We need a nice estimate on r in terms of X and Y . If r = ∞, we are done. Otherwise, suppose that r < ∞. Recall that h1 , h2 have Taylor series with radius of convergence 2π, so by the existence and uniqueness of solutions of ODE, sup || ad F (t)|| ≥ 2π |t| 1, and thus X M (X, Y ) = F (1) = Mk (X, Y ).
36
Chapter 4
Geometry of Lie groups 4.1
Vector bundles
Let M be a C ∞ manifold of dimension n. Everything we say will apply equally well to the complex analytic context, replacing C ∞ by “holomorphic.” Definition 4.1.1. A rank k vector bundle over M with real (resp. complex) fibers consists of the following data: • A manifold E. • A surjective smooth map p : E → M . • The structure of a k-dim real (resp. complex) vector space on Ex := p−1 (x) for each x ∈ M , such that x has an open neighborhood U fitting into a commutative diagram ≈ / p−1 (U ) U × Rk p
≈ id
U
π
/U
which preserves the vector space structure (resp. replacing Rk by Ck ). This is called a local trivialization of the vector bundle. Definition 4.1.2. A section of E over U is a C ∞ map s : U → p−1 U such that p ◦ s = 1U . Sections can be added or multiplied by a scalar, so the sections over U comprise a vector space. Definition 4.1.3. A frame for E over U is a k-tuple of sections (s1 , . . . , sk ) which at each x ∈ U provide a basis for Ex . A local trivialization for the vector bundle over U is the same thing as a local frame over U . This is left as an exercise. 37
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Example 4.1.4. Let T M be the tangent bundle of M and T ∗ M be the cotangent bundle. These are vector bundles: if (U ; x1 , . . . , xn ) is a coordinate neighborhood, then { ∂x∂ 1 , . . . , ∂x∂n } is a local frame for T M and {dx1 , . . . , dxn } is a local frame for T ∗ M . Sections of the tangent bundle are vector fields; sections of the cotangent bundle are 1-forms. Definition 4.1.5. Let p : E → M be a vector bundle of rank k. A rank ` sub-bundle of E is a rank ` vector bundle pF : F → M together with a commutative diagram F
/E
pF
M
≈ id
p
/M
such that the inclusion F → E is an injective linear map on each fiber. Then each x ∈ M has an open neighborhood U with local frame s1 , . . . , sk for E such that s1 , . . . , s` is a local frame for F .
4.2
Frobenius’ Theorem
Consider a rank k sub-bundle E ⊂ T M . For each x ∈ M , Ex is a k-dimensional subspace of Tx M . Definition 4.2.1. An integral submanifold for E is a k-dimensional submanifold S ⊂ M such that for each x ∈ S, Tx S = Ex ⊂ Tx M . Definition 4.2.2. A tangent sub-bundle E ⊂ T M is involutive if for any two vector fields X, Y on an open U ⊂ M taking values in E, the commutator [X, Y ] also takes values in E. Remark 4.2.3. The more common terminology for tangent sub-bundle is “distribution.” That is, a vector field is a section of the tangent bundle. It takes values in E if its value over x ∈ M lies in Ex ⊂ Tx M . The bundle E is involutive if for any two such vector fields taking values in E, so does their Lie bracket. Definition 4.2.4. The tangent sub-bundle E is integrable if each point in M has a coordinate neighborhood (U ; x1 , . . . , xn ) such that ∂x∂ 1 , . . . , ∂x∂ k constitute a local frame for E. Note that in this situation, 1. The set {xk+1 = ck+1 , . . . , xn = cn } where the ci are constant, are the integral submanifolds for E. 2. E is involutive, because k k k k X X X X ∂ ∂ ∂bi ∂ ∂ai ∂ aj , bj = aj − bj . ∂xj ∂xj ∂xj ∂xi ∂xj ∂xi j=1
j=1
i,j=1
38
i,j=1
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4.2. FROBENIUS’ THEOREM
Theorem 4.2.5 (Frobenius). A tangent sub-bundle is integrable if and only if it is involuble. As remarked above, it is obvious that integrable implies involuble; the substance of the theorem is the other direction. Example 4.2.6. Suppose E is a rank one sub-bundle. Given any point m ∈ M , there exists an open neighborhood U and a vector field X on U which is a frame for E on U . The values of X must be non-zero at every point of U (by definition of local frame). Then any two vector fields Y1 , Y2 defined near m and taking values in E can be expressed as Yi = fi X. [Y1 , Y2 ] = (f1 Xf2 − f2 Xf1 )X So in the rank one case, involutivity is automatic. According to the theorem, E should be integrable. Let’s try to see this. By the theory of ODEs, after shrinking U if necessary, we can construct a map Φ : (−, ) × U → M such that for all u ∈ U , t 7→ Φ(t, u) is an integral curve of X and Φ(0, u) = u. After shrinking U more if necessary, we can introduce coordinates x1 , . . . , xn on U such that xj (m) = 0. Then after making a linear change of coordinates, we can arrange that ∂ Xm = . ∂x1 m Define a C ∞ map from a neighborhood V of 0 ∈ Rn to M by (y1 , . . . , yn ) 7→ Φ(y1 ; 0, y2 , . . . , yn ). It is easy to see that the differential of this map at 0 is invertible at m. By the inverse function theorem, this map can be inverted locally near 0, which tells us that near 0, we can regard y1 , . . . , yn as coordinates. But in terms of these coordinates, we have locally ∂y∂ 1 = X since Φ defines an integral curve for X. Let’s recap at a conceptual level what we did here. By a linear change of coordinates, we arranged coordinates so that X = ∂x∂ 1 at a single point. Then using the theory of integral curves, we showed that this must hold locally. We summarize our work in the theorem below. Theorem 4.2.7 (“Local structure of a non-zero vector field”). Let M be a manifold, m ∈ M , and X a vector field defined near m such that Xm 6= 0. Then there exist local coordinates x1 , . . . , xn defined near m such that X = ∂x∂ 1 locally. This proves the rank one case of Frobenius’ theorem. The general case proceeds by induction on the rank, and the induction step is a clever reduction to the rank one case. (See e.g. Warner.) 39
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CHAPTER 4. GEOMETRY OF LIE GROUPS
Example 4.2.8. Now consider the rank 2 case on a neighborhood of 0 in R3 . There exists a local frame which, up to linear change of coordinates, can be written as follows. ∂ ∂ + a(x) ... ∂x1 ∂x3 ∂ ∂ X2 = + b(x) ... ∂x2 ∂x3
X1 =
where a(0) = b(0) = 0. Then [X1 , X2 ] =
∂b ∂a − ∂x1 ∂x2
∂ . ∂x3
∂a ∂b − ∂x is the obstruction to involutivity, so generically a rankd 2 tangent The term ∂x 1 2 3 sub-bundle of T R is not involutive, hence not integral.
Example 4.2.9. Let G be a Lie group with Lie algebra g and h ⊂ g a Lie subalgebra. Define a tangent sub-bundle Eh ⊂ T G as follows: (Eh )g = `g∗ h ⊂ Tg G ⊂ g = Te H. Is this a tangent sub-bundle, i.e. is it spanned locally by a frame? Yes; this is true even globally. Let Y1 , . . . , Yk be a basis of h ⊂ g, and view them as left-invariant vector fields; then (Y1 , . . . , Yk ) is a global frame for Eh . Then [Yh , Yj ] ∈ h =⇒ Eh is involutive. (It suffices to check involutivity on the level of frames, since any vector fields can be written as a C ∞ (M ) linear combination of these.)
4.3
Foliations
Proposition 4.3.1. Let M be a connected manifold, E ⊂ T M an integrable rank k tangent subbundle. Given any point m ∈ M , there exists a unique maximal connected integral submanifold which contains m. Let (U ; x1 , . . . , xn ) be a coordinate neighborhood such that ∂x∂ 1 , . . . , ∂x∂ k is a frame for E on U . If S ∩ U 6= 0, then by definition S ∩ U must be contained in a countable union (because S is second-countable) of submanifolds of the form {x ∈ U | xk+1 = ck+1 , . . . , xn = cn } By maximality, it must contain all of the slices that it meets. The issue to be established is second-countability. We establish some (non-standard) terminology. Definition 4.3.2. A coordinate neighborhood (U ; x1 , . . . , xn ) is adapted in E if 40
Math 222
4.3. FOLIATIONS
1. the image of U in Rn is (−1, 1)n 2.
∂ ∂x1
. . . , ∂x∂ k is a frame for E|U .
e; x 3. there exists a coordinate neighborhood (U e1 , . . . , x en ) such that (a) U is com∂ ∂ e , (b) x pact and contained in U ej |U = xj , and (c) ∂f is a frame for E x1 , . . . , ∂ x f k e. over U Note that M , being second-countable, can be covered by a countable number of adapted coordinate neighborhoods U1 , U2 , . . .. Definition 4.3.3. A slice of an adapted coordinate neighborhood (U ; x1 , . . . , xn ) is a set of the form {(x1 , . . . , xn ) | xk+1 = ck+1 , . . . , xn = cn }. So “slice” means maximal connected integral submanifold of E|U . The slices in U are parametrized by (xk+1 , . . . , xn ) ∈ (−1, 1)n−k . We shall call a continuous (not necessarily C 1 ) path γ : [0, 1] → M tangential to E if it satisfies the following condition: (*) Let U be an adapted coordinate neighborhood of γ(t0 ) ∈ U , t0 ∈ [0, 1]. Then for t near t0 , γ(t) lies in the same slice as γ(t0 ). This is equivalent to the weaker condition (**) Given any t0 ∈ [0, 1] there exists an adapted coordinate neighborhood U of γ(t0 ) such that for t near t0 , γ(t) lies in the same slice as γ(t0 ) because locally, integral submanifolds are unique. There is just no way to jump from one slice to another. Lemma 4.3.4. Let Uα , Uβ be adapted coordinate neighborhoods and S a slice of Uα . Then S ∩ Uβ is contained in a finite union of slices in Uβ . Proof. Suppose not, i.e. suppose that there exists a sequence {pk } in S ∩ Uβ such that for k 6= `, pk and p` lie in different slices of Uβ . But S is compact, so the sequence {pk } converges to some p∞ ∈ S. eα and S is contained in a slice of U eα , Recall that Uα has compact closure in U e e e say S. Furthermore, p∞ ∈ Uβ ⊂ Uβ so p∞ lies in a slice of Uβ , say Sβ . But integral submanifolds for E are locally unique, in the sense that Se ∩ Seβ e of p∞ that is also contained in contains an open neighborhood (with respect to S) e Sβ . So the limiting pk cannot all be in distinct slices. 41
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To prove the existence of a maximal connected integral submanifold passing through a given point m ∈ M , choose U0 , U1 , . . . with m ∈ U0 . Let γ be a path tangential to E, with γ(0) = m. By compactness of [0, 1], there exists a partition 0 < t0 < . . . , < tN = 1 of [0, 1] such that for 0 < j ≤ N , γ([ti , ti+1 ]) ⊂ U`j , so in particular γ(1) ∈ U`N . By the lemma, there is only a finite number of possible slices for U`N on which γ(1) can lie. The possibilities for these slices is completely determined by the datum of U0 = Uα1 = . . . = UαN . So to prove existence of a maximal connected integral submanifold S, consider all finite chains Uα0 = U0 , Uα1 , . . . , UαN such that Uαj−1 ∩ Uαj 6= ∅. Then take successive unions of slices, starting with the slice of U0 on which M lies, then dictated by intersections of Uαj−1 and Uαj . The union will be a union of a countable number of slices of the U` , connected by construction. To summarze, we have proved: Theorem 4.3.5. Let M be a manifold, E ⊂ T M an integrable tangent sub-bundle, m ∈ M . Then there exists a unique maximal connected integral submanifold passing through m. Theorem 4.3.6. Let G be a Lie group with Lie algebra g and h ⊂ g a Lie subalgebra. Then there exists a unique connected Lie subgroup H ⊂ G whose Lie algebra, considered as as subalgebra of g, coincides with h. Proof. Define Eh ⊂ T G as before, (Eh )g = `g∗ h. By the previous theorem, there exists a unique maximal integral submanifold H which contains e. Suppose h ∈ H; we need to check that hH ⊂ H. Then `h−1 H is also an integral submanifold because Eh is invariant under left translation. Therefore, `h−1 H is an integral submanifold containing e and is therefore contained in H by maximality. So H ⊂ G is a subgroup. This shows that multiplication H × H → G is C ∞ and takes values in H. By the discussion of adapted coordinate neighborhoods, given h ∈ H there exist open neighborhoods U of h in G and UH of h in H such that UH → U is a closed embedding with UH connected. Therefore, any C ∞ map from N to G which takes values in H is smooth as a map N → H. So H ⊂ G is a connected Lie subgroup. The Lie algebra of H, as a subspace of g ' Te G is by construction Te H = (Eh )e = h. To establish uniqueness, recall that h _
g
exp
exp
/ H _ /G
is a commutative diagram. Let U be an open neighborhood of 0 ∈ G. Then exp(U ∩ h) contains an open neighborhood of e in H, and we saw earlier than any neighborhood of e in a connected Lie group generates the whole group in the algebraic sense. Since this discussion applies equally well to any potential other 42
Math 222
4.4. LIE SUBGROUPS
connected Lie subgroup H 0 with Lie algebra h, we have the desired uniqueness of H. As a corollary, we see that H can also be described as the smallest subgroup of G, in the algebraic sense, containing exp(U ∩ h). In particular, this gives a bijection between connected subgroups of G and Lie subalgebras of g.
4.4
Lie subgroups
Let M be a manifold. We discuss various notions of “submanifolds” of M . Immersion. S a manifold equipped with a map ϕ : S → M such that ϕ∗ is locally 1-1 for every s ∈ S. It follows from the inversion function theorem that ϕ is locally injective. This is a very weak notion. For instance, consider the map R → R2 sending t 7→ (cos t, sin t). This is an immersion, but it is from from being injective. Injective Immersion. S a manifold equipped with a globally injective immersion ϕ : S → M . This is the notion of submanifold that we have been using. One can get pathological examples where the topology on S is incompatible with the subspace topology from M . For instance, the inclusion R×Q ⊂ R2 or the “figure 8” immersion. Embedding. This is an injective immersion where the intrinsic topology of S must agree with the induced (subspace) topology. For example, the inclusion of an open interval in R2 . Closed embedding. An embedding in which ϕ(S) ⊂ M is closed. However, for submanifolds which are Lie groups, many of these notions collapse. Theorem 4.4.1. For a Lie subgroup H ⊂ G, the following are equivalent. 1. H is closed as a subset of G. 2. The intrinsic and induced topologies agree. 3. ι : H ,→ G is a closed embedding. Example 4.4.2. R × Q ,→ R2 is a Lie subgroup which is not a closed embedding. The submanifold (0, 1) → (0, 1) × (0, 1) looking like a σ with a tail at the boundary of the square is closed but does not have the induced topology. The open interval in R2 has the induced topology but is not closed. 43
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Proof. By definition, the first two conditions together are the third. So we must show that in the setting of a Lie subgroup H ⊂ G, (1) ⇐⇒ (2). Both (1) and (2) are local (with respect to G) properties. So it suffices to prove (1’) ⇐⇒ (2’): (1’) For any open neighborhood U of g ∈ G, H ∩ U is closed in G ∩ U (2’) For any open neighborhood U of g ∈ G, the inclusion U ∩ H ⊂ H is a homeomorphism onto the image. We know that any open neighborhood of g ∈ G contains an adapted (to Eh ) coordinate neighborhood. So we may assume that U ' (−1, 1)n is an adapted coordinate neighborhood and P the set of slices, P ' (−1, 1)n−k and U ' (−1, 1)k × P (these are all topological isomorphisms, i.e. homeomorphisms). Let PH ⊂ P denote the set of slices that are contained in U ∩ H. We know that PH ⊂ P is at most countable. For the induced topology, we have the homeomorphism U ∩ H ' (−1, 1)k × PH . On the other hand, for the intrinsic topology we have the homeomorphism U ∩ H ' (−1, 1)k × PH when PH is given the discrete topology. This reduces the problem to showing that for any adapted U , PH ⊂ P is closed if and only if PH is discrete. Suppose PH ⊂ P is not closed; then there exists an adapted U such that PH is not closed with this U . Then there exists a sequence pn ∈ PH converging to some p ∈ P where p ∈ / PH . The pn , p ∈ P correspond to hn , h with hn ' 0 × pn and h ' 0 × p. For N large and all n ≥ N , h−1 N hn will lie in a particular adapted coordinate neighborhood of −1 the identity. Now, h−1 / H, so h−1 N hn ∈ H but hN h ∈ N hn lie in slices of Ue ∩ H which accumulate at the slice of e. Since by construction the hn are “infinitely far apart” in H, this tells us that PH is not discrete. Now suppose that there exists U such that PH is not discrete. By left translation, we may arrange that U is an adapted neighborhood of the identity and the slice of e in U is not isolated in PH . Translating again, we see that no slice is isolated. If PH were closed, then for p ∈ PH the set PH − {p} would be dense and open in PH . So \ (PH \ {p}) p∈PH
is a countable intersection of open dense subsets, which is empty. This contradicts the Baire category theorem if PH were a complete metric space in its induced topology. Hence PH is not closed. Lemma 4.4.3. Suppose that H ⊂ G is a closed Lie group. Then G/H, equipped with the quotient topology, is Hausdorff. The fact that H is closed implies that the identity coset in G/H is closed, so every point of G/H is closed, i.e. G/H has the “T1” axiom. It is a general fact that for topological groups, this implies “T2.” However, we will go through the details. 44
Math 222
4.4. LIE SUBGROUPS
Proof. Suppose g ∈ G and g ∈ / H. Then g has an open neighborhood which does not intersect H. We can expression this open neighborhood as U1 g, where U1 is an open neighborhood of e. Now we may choose neighborhoods U2 , U3 of e such that U22 ⊂ U1 (because multiplication is continuous) and U2 = U2−1 (by replacing U2 with U2 ∩ U2−1 if necessary) and U3 ⊂ U2 and gU3 g −1 ⊂ U2 . If gU3 H ∩ U3 H = ∅, then these two opens descend to open neighborhoods of gH and eH in G/H which are disjoint, as had to be shown. If not, there exists some x ∈ gU3 H ∩ U3 = gU3 g −1 gH ∩ U3 ⊂ U2 gH ∩ U2 . This gives an element h ∈ H lying in U2−1 U2 g ⊂ U1 g, which is a contradiction. Theorem 4.4.4. Keeping the notation above, let p : G → G/H denote the quotient map. There exists a unique structure of a C ∞ manifold on G/H which is compatible with the quotient topological structure, such that for a continuous function f defined on an open subset U ⊂ G/H, f ∈ C ∞ (U ) if and only if p∗ f ∈ C ∞ (p−1 U ). Furthermore, this structure has the following properties. • p : G → G/H and the action map G × G/H → G/H are smooth. • p∗ induces a linear isomorphism g/h ' TeH (G/H). • Each point in G/H has an open neighborhood U over which p : p−1 (U ) → U has a C ∞ section. Proof. Uniqueness is obvious: to know the notion of C ∞ fnction on open subsets is to know the manifold structure. Assuming that G/H equipped with this structure is a manifold, p : G → G/H is C ∞ because the pullbacks of C ∞ functions are C ∞ . m The fact that G × G → G is C ∞ is C ∞ , since the pullback of a C ∞ function on G/H is a function on G × G/H whose pullback to G × G is right H-invariant and smooth. We shall see, in the proof of existence of the smooth structure, that p∗ is surjective and dim(G/H) = dim G−dim H. Now, TeH G/H ' g/ ker p∗ . The fact that C ∞ functions on G/H are right H-invariant implies that h ⊂ ker p∗ : Te G → TeH (G/H). By dimension reasons, h = ker p∗ . To complete the proof, we must construct the manifold structure and local sections near the identity coset (because left translation by any g ∈ is a diffeomorphism) which relates local sections at the identity coset to sections near gH. To establish the manifold structure, we need to show that locally G/H, with the presumed smooth structure, is diffeomorphic to an open set in Rn . Again, it suffices to do this for an open neighborhood of e. Note that this is an entirely local question. As before, let Eh denote the left-invariant tangent sub-bundle of T G whose fiber at e is h. Let U be an adapted coordinate neighborhood of e. We may suppose that the coordinates are centered at e, i.e. xj (e) = 0 for all j. In the following, we will shrink U and and tacitly scale the coordinates so that the shrunk U is 45
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an adapted coordinate neighborhood (recall that adapted coordinates have as their image (−1, 1)n ). As before, we have U ' P × S0 where P = (−1, 1)n−k and S0 is the slice through e ' 0, isomorphic to (−1, 1)k . Note that each slice is contained in a coset gH (by left invariance of the tangent sub-bundle h, integral manifolds are preserved). After shrinking U , we can arrange that U ∩ H = S0 . Shrinking further if necessary, we claim that we may arrange that no two distinct slices in U lie in the same coset gH. If this were not true, there would exist sequences p0n and p00n lying in distinct slices in P both converging to 0 and lying in the same coset. Let gn0 and gn00 be the group elements to (p0n , 0) and (p00n , 0). Then gn0 → e, and gn00 → e, but gn0 H = gn00 H. Then (gn00 )−1 gn0 → e and (gn00 )−1 gn0 H = H, so (gn00 )−1 gn0 ∈ S0 for n 0. So we get convergence (gn00 )−1 gn0 → e in H, which implies that gn00 → gn0 in H, but this is impossible because they are “far apart” in H (the different slices having the discrete topology). It must then be the case that p−1 (P ) = U ' P × H. That implies p(U ) ' P ' (−1, 1)n−k and hence p(U ) is isomorphic to an open subset of Rn−k , the identification preserving the smooth structure. The functions on p(U ) are the right H-invariant functions on P × H, i.e. functions on P . The section s : (−1, 1)n−k ' P → U sending x 7→ (x, 0) is smooth by construction. Then p : G → G/H “looks like” the projection U ' (−1, 1)n−k × (−1, 1)k → (−1, 1)n−k . Theorem 4.4.5. Let G be a Lie group, H ⊂ G a subgroup in the algebraic sense such that H as a subset of G is closed. Then H has a natural, unique structure of a Lie subgroup. Proof. Let g be the Lie algebra of G. Define h = {X ∈ g | exp tX ∈ H for all t}. Lemma 4.4.6. h ⊂ g is a linear subspace. Proof. From the defninion it is immediate that h is closed under scalar multiplicaiton, so it suffices to show that h is closed under addition. Define M : V × V → g as before, i.e. multiplication on G pulled back to g near 0 via exp. Suppose X, Y ∈ h and t ∈ R. For n 0, t t X, Y ∈ V. n n Then t t t 1 M X, Y = (X + Y ) + O( 2 ). n n n n Then exp(t(X + Y )) is the limit as n → ∞ of n t t t t exp nM X, Y = exp M X, Y n n n n n t t ≈ exp X exp Y ∈H n n 46
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The fact that H ⊂ G is closed implies that exp(t(X + Y )) ∈ H for all t. Let h0 ⊂ g be a linear complement, i.e. g ' h ⊕ h0 . Lemma 4.4.7. If U is a sufficiently small open neighborhood of 0 ∈ g, then exp(U ∩ h) = exp(U ) ∩ H. Proof. Note that exp h ⊂ H by definition. Therefore, exp(V ∩h) ⊂ H ∩exp V for any open neighborhood V of 0 in g. If the lemma were false, we could find a sequence hn ∈ H converging to e such that hn ∈ / exp h. Let q ⊂ g be a linear complement to h. Then the map q⊕h→G (X, Y ) → exp X exp Y is a local diffeomorphism of a neighborhood of 0 in q ⊕ h into a neighborhood of e in G. Eventually, the hn lie in the image of this map, so their pre-images have non-zero q-component. This says that there exist sequences {Xn } ⊂ q and {Yn } ⊂ h such that exp Xn exp Yn = hn , Xn → 0, Yn → 0, and Xn 6= 0 for all n. fn where ||X en || = 1. Introduce a linear norm || · || on g and write Xn = tn X e e Passing to a subsequence if necessary, we may assume that Xn → X ∈ q. We must e = 1. Let t ∈ R be fixed. We know that tn → 0, so we may necessarily have ||X|| choose kn ∈ Z such that t − tn < kn tn ≤ t. Then kn tn → t, so en ) → exp(tX). e exp(kn Xn ) = exp(kn tn X But on the other hand, exp(kn Xn ) = (exp Xn )kn , so e exp(kn Xn ) = (exp Xn )kn → exp tX. e ∈ H. Since this holds for all t, it must be the case that X e ∈ h. Hence So exp(tX) e e X ∈ h ∩ q = {0}. But X is a unit vector, so this is impossible. Near the origin, exp is a diffeomorphism. By the preceding lemma, e has an open neighborhood Ue such that H ∩ Ue is a closed, embedded submanifold of Ue . By translation, every h ∈ H has an open neighborhood Uh with the same property. Since H is closed, for any g ∈ G − H there exists an neighborhood Ug of g such that Ug ∩ H = ∅. Therefore, H ,→ G is a closed embedding of a submanifold. Hence the multiplication map H × H → G and the inversion map H → G are restrictions to H × H (resp. H) of smooth maps from G × G (resp. G), hence are smooth maps. Moreover, they take values in H, hence they are C ∞ as maps into H (precisely because this is a smooth embedding; we gave counterexamples in the other cases). 47
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4.5
CHAPTER 4. GEOMETRY OF LIE GROUPS
Lie’s Theorems
Theorem 4.5.1. Suppose Φ : G → H is a homomorphism of Lie groups such that on the level of Lie algebras, Φ∗ : g → h is injective. Then (i) ker Φ ⊂ G is discrete. (ii) If Φ is onto, then Φ : G → H is a covering map. Proof. ker Φ is a closed, normal Lie subgroup (e.g. by the theorem we just proved). By a homework problem, G/ ker Φ has a natural Lie group structure. The map Φ : G → H factors through G/ ker Φ → H by the definition of the C ∞ structure. Therefore, Φ∗ : g → h factors through g/Lie(ker Φ) → h. Since Φ∗ is injective, ker Φ must have dimension 0, which implies that ker Φ has the discrete topology. This establishes (i). Note that we have implicitly used the Baire category theorem; it is an exercise to give a proof not using this theorem. Φ factors through G/ ker Φ → H. If Φ is onto, then this is an isomorphism of Lie groups. Also, G/ ker Φ has the same Lie algebras as G, so Φ∗ : g → h is a linear isomorphism. Locally near the identity, Φ is a diffeomorphism, so there exist open neighborhoods UG,0 of eG ∈ G and UH,0 of eH ∈ H such that UG,0 ∩ ker Φ = {e} and Φ restricts to a diffeomorphism UG,0 → UH,0 . Now let UG be an open neighborhood of e such that UG = UG−1 and UG2 ⊂ UG,0 . Let UH = Φ(UG ). Then Φ : UG → UH is a diffeomorphism. We claim that for g ∈ ker Φ with g 6= 0, then UG g ∩ UG = ∅ (otherwise, we would have a non-identity element g ∈ UG−1 UG ⊂ UG,0 , which is impossible by construction). By translation, UG g1 ∩ UG g2 = ∅ whenever g1 , g2 are distinct elements of ker Φ. This implies that [ Φ−1 (UH ) = UG g, g∈ker Φ
i.e. the pre-image of UH is a disjoint union of neighborhoods of the form UG g such that Φ : UG g → UH is a diffeomorphism for any g ∈ ker Φ. By left translation, we obtain an analogous description for suitably small neighborhoods of any given g ∈ G and UΦ(g) of Φ(g) ∈ H, hence Φ is a covering map. Corollary 4.5.2. Suppose Φ : G → H is a Lie group homomorphism such that Φ∗ : g → h is an isomorphism. If G, H are connected then Φ is a covering map. In particular, if H is also simply connected, Φ is an isomorphism. Proof. The point is to show surjectivity. By the universality of exp, the image of Φ contains an open neighborhoods of the identity in H. Since H is connected, this open neighborhood generates H, so Im Φ = H. By the previous theorem, Φ : G → H is a covering map. If H is simply connected, then obviously Φ must be a homeomorphism. 48
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Corollary 4.5.3. Let G, H be Lie groups with Lie algebras g, h and ϕ : g → h a Lie algebra homomorphism. Suppose G and H are connected and G is simply connected. Then there exists a unique Lie group homomorphism Φ : G → H such that Φ∗ = ϕ. Proof. Consider the graph of ϕ as a subalgebra of g ⊕ h: {(X, ϕ(X)) | X ∈ g} ⊂ g ⊕ h Then we know that there exists a closed Lie subgroup Γ ⊂ G × H with this Lie algebra. Let p1 , p2 be the two projection maps G × H → G and G × H → H. Well, p1∗ restrict to the graph of ϕ is 1-1 and onto, so p1 : Γ → G is a covering map. Since G is simply connected, this is an isomorphism. So p−1 1 : G → Γ is a well-defined Lie −1 group homomorphism. Define Φ = p2 ◦ p1 ; then Φ∗ = ϕ by construction. Corollary 4.5.4. Let G, H be Lie groups and Φ : G → H a homomorphism of groups in the algebraic sense. If Φ is continuous, it is C ∞ , hence a Lie group homomorphism. Proof. Let Γ be the graph of Φ, Γ = {(g, Φ(g)) | g ∈ G} ⊂ G × H. This is a subgroup in the algebraic sense because Φ is a homomorphism, and a closed subset since Φ is continuous. Hence this is a Lie subgroup of G × H. By construction, p1 |Γ : Γ → G is 1-1 and onto, hence an isomorphism. By definition of Γ, Φ = p2 ◦ p−1 1 . Proposition 4.5.5. A discrete normal subgroup of a connected Lie group G is central (lies in the center). Proof. Suppose h ∈ H is not the identity. The map g 7→ ghg −1 is continuous, so given an open Uh of h, there exists an open neighborhood U of e ∈ G such that g ∈ U =⇒ ghg −1 ∈ Uh . We can do this for some Uh such that Uh ∩ H = h. By normality, ghg −1 ∈ H ∩ UH , so ghg −1 = h. That tells us that an open neighborhood of e ∈ G commutes with h, but such a neighborhood generates G, so all of G commutes with h. A more concise way to say this: the map g 7→ ghg −1 sends G to a discrete set, hence is constant. Note that subgroups of this type arise as kernels of Lie group homomorphisms Φ : G → H such that Φ∗ is 1-1 and G is connected. In particular, this applies to Lie group shomomorphisms Φ : G → H that are covering maps, with G connected. e the universal cover. Suppose G is a connected Lie group and G e such that the Theorem 4.5.6. There exists a natural structure of a Lie group on G e covering map G → G is a Lie group homomorphism. 49
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Exercise 4.5.7. Prove this theorem. Exercise 4.5.8. Let g be a finite dimensional Lie algebra over R. Does there exist a Lie group G with Lie algebra g? An approach to this problem is through Ado’s theorem: Theorem 4.5.9 (Ado). Any such g can be realized as a Lie subalgebra of End(RN ) for some N . Combined with our work above, this tells us the answer is yes.
50
Chapter 5
The Universal Enveloping Algebra 5.1
Construction of the universal enveloping algebra
Let K be an arbitrary field, g a Lie algebra over K. We define O
g=
∞ M
g⊗k .
k=0
This is a graded associative algebra over K with unit, freely generated by g⊗1 = g. N These conditions characterize g as an algebra. Similarly, we define ∞ M S(g) = S k g, k=0
where S k g is the k th symmetric power of g. This is a commutative algebra over K with unit, freely generatd by S 1 g = g, which uniquely characterizes S(g). Now define O U (g) = g/({XY − Y X − [X, Y ] : X, Y ∈ g}). This is the tensor algebra modded out by the two sided ideal generated by relations of the form XY − Y X − [X, Y ]. The composition g = g⊗1 → U (g) is a Lie algebra homomorphism when U (g) is considered as a Lie algebra via commutators. We have constructed U (g) to have the following properties: 51
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• it is an associative algebra over K with unit (*), • it is filtered (**), • it is equipped with a linear map i : g → U1 (g) which is a Lie algebra homomorphism via commutators, generated by i(g). L (**) Let Uk (g) denote the image of `≤k g⊗` . Then we have a filtration U0 (g) ⊂ U1 (g) ⊂ . . . ⊂ Uk (g) . . . ⊂∞ k=0 Uk (g) = U (g) satisfying Uk (g)U` (g) ⊂ Uk+` (g). (*) Note that O
g=K ⊗g . | {z· 1} ⊕ g | {z } L g⊗0
∞ k=1
g ⊗k
The second summand is called the augmentation ideal. Note that the ideal we are quotienting out by lies in the augmentation ideal, and hence never harms the constants. Hence U0 (g) = K · 1. Definition 5.1.1. The pair (U (g), i) is called the universal enveloping algebra of g.
5.2
Poincar´ e-Birkhoff-Witt
Question. Is i : g → U1 (g) injective? Well, there isn’t anything that is transparently in the kernel, but it also isn’t necessarily obvious that the relations can’t somehow kill something at degree 1. In fact, it is injective, as we shall eventually prove. Observe that for each k, “Uk (g) is commutative modulo Uk−1 (g).” In other words, the commutator of two elements in Uk (g) differ by an element of Uk−1 (g). This is because the elements of i(g) in U2 (g) commute modulo i(g). For instance, in U3 (g) the elements XY Z, Y ZX, XZY , etc. all commute up to elements in U2 (g) (note that this is a different meaning of “commute” from what is sometimes used elsewhere). Consider the associated graded algebra to U (g): gr(U (g)) :=
∞ M
Uk (g)/Uk−1 (g).
k=0
This is a graded associative algebra over K with unit, generated by i(g). It is commutative by the remarks above. Hence by the universality of S(g), the map linear map i : g → U1 (g) induces a unique algebra homomorphism S(g) → gr(U (g)) of commutative algebras with unit. Moreover, this map is surjective by construction, since g generates gr(U (g)). 52
´ 5.2. POINCARE-BIRKHOFF-WITT
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Theorem 5.2.1 (Poincar´e-Birkhoff-Witt). This morphism of algebras is an isomorphism: S(g) ' gr(U (g)). Corollary 5.2.2. i : g → U1 (g)/U0 (g) is an isomorphism, hence i : g → U (g) is injective. This tells us how to think of U (g): it is the universal associative algebra which contains g as a Lie sub-algebra with the Lie bracket of commutators. Before embarking on the proof, we observe some more consequences. Suppose dim g < ∞, and let X1 , . . . , Xn be a basis of g over K. Corollary 5.2.3. {X1j1 . . . Xnjn | 0 ≤ j1 , . . . , jn < ∞} is a basis for U (g). Proof. The subset of elements above with j1 + . . . + jn = k obviously comprise a basis for S k (g), hence they push forward to a basis for Uk /Uk−1 . What if g is infinite-dimensional? Similar considerations apply, but one has to be more careful about stating the result. Corollary 5.2.4. Let h ⊂ g be a Lie subalgebra. Then U (h) is a subalgebra of U (g). Proof. Choose the basis of g to contain a basis for h as a subset, and then use the form of solutions in Corollary 5.2.3. Corollary 5.2.5. Suppose h1 , h2 are Lie subalgebras of g such that g = h1 ⊕ h2 as vector spaces. Then U (g) ' U (h1 ) ⊗K U (h2 ) as left U (h1 )-modules and right U (h2 )-modules. Proof. This is clear by considering bases: hoose the basis of g so that the first things are a basis of h1 and the second things are a basis of h2 . Any basis element in the form of Corollary 5.2.3 may be expressed as a tensor of two such elements for h1 and h2 . Now suppose that K has characteristic 0. Define sek : g⊗k → Uk (g) 1 X Y1 ⊗ . . . ⊗ Yk 7→ Yσ(1) . . . Yσ(k) . k! σ∈Sk
We initially define this for basis elements only, and then extend by linearity to the whole tensor product. Note that sek factors through S(g), hence induces sk : S k (g) → Uk (g). This map sk is called the “symmetrization map.” Note that this definition doesn’t depend on PBW and, in fact, even without PBW it is clear that Im Sk spans Uk (g)/Uk−1 (g). What PBW tells us is that this is not only a surjection, but an isomorphism. 53
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Corollary 5.2.6. The symmetrization map s = isomorphism S(g) ' U (g).
5.3
L∞
k=0 sk
defines a vector space
Proof of Poincar´ e-Birkhoff-Witt
We shall prove PBW in the following situation: g is the Lie algebra of a Lie group, so in particular K = R or C, and dim g < ∞. Granted Ado’s theorem, this furnishes a proof for arbitrary real or complex Lie algebras. We know that the map from the symmetric algebra to the universal enveloping algebra is surjective; we want it to be injective. To do this, we want to show that the universal enveloping algebra is as big as we expect: that there are no unexpected relations. We achieve this by finding a space (the space of functions on G) on which it acts sufficiently nontrivially. We first set up some preliminaries. Let M by a C ∞ manifold of dimension n. To an open set U ⊂ M , define D(U ) to be the algebra of linear differential operators on U with C ∞ coefficients of finite order. Let Dk (U ) be the space of differential operators of order ≤ k. Let (U ; x1 , . . . , xn ) be a coordinate neighborhood. Then k X X ∂ j1 ∂ j2 ∂ jn ∞ Dk (U ) = aj1 ...jn j1 j2 . . . jn | aj1 ...jn ∈ C (M ) . ∂x1 ∂x2 ∂xn `=0 0≤j1 ,...,jn 0, and in particular Eα spans the line in gα pairing non-degenerately with Eα . Recale both Eα and E−α by the same positive factor so that B(Eα , E−α ) =
2 . (α, α)
while preserving E−α = −E α . We have the relations eα. [Eα , E−α ] = B(Eα , E−α )H [Hα , Eα ] = hα, Hα iEα since Hα ∈ tR . We want to compute this quantity hα, Hα i, so pair with E−α . hα, Hα iB(Eα , E−α ) = B([Hα , Eα ], E−α ) = B([Eα , E−α ], Hα ) Inserting the normalizations we made above, we conclude that 2 = B(Hα , Hα ) (α, α) α Hα 1 h , i = ||α|| · ||Hα ||. ||α|| ||Hα || 2
hα, Hα i
Hα By construction, ||H is a unit vector in (α⊥ )⊥ (see above for a discussion of what α || this means). Therefore, we must have
h
α Hα , i = ±1. ||α|| ||Hα || 84
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8.1. STRUCTURE OF ROOTS
From the equation we see that the sign must be +1, so 1 1 = ||α||||Hα ||. 2 Therefore, we have hα, Hα i =
1 (||α||||Hα ||)2 = 2. 2
Summary. With our normalization, we have the relations [Hα , Eα ] = 2Eα [Hα , E−α ] = −2E−α [Eα , E−α ] = Hα and H α = −Hα , E α = −E−α . Define ϕα : sl(2, C) → g by
1 0 7→ Hα 0 −1 0 1 7→ Eα 0 0 0 0 7→ E−α 1 0
Then by the data in the conclusion, ϕ is a Lie algebra homomorphism. By construction, for X ∈ sl(2, R) we have ϕ(X) = ϕ(X), where the complex conjugation is respect to su(2) and gR , respectively. Let sα = ϕα (sl(2, C)). We shall see later that the dim gα = 1. Therefore, sα = gα ⊕ g−α ⊕ [gα , g−α ]. This shows that everything depends only on α, not on the particular choice of Eα , etc. Now, SU (2) is simply connected with center {±1}. Therefore, ϕα lifts to a Lie group homomorphism ϕ fα : SU (2) → G Let Sα ⊂ G be the image. The kernel is a discrete central subgroup, hence a subgroup of {±1}. Therefore, it must be the case that Sα ' SU (2) or SU (2)/{±1} Theorem 8.1.2. Let α, β ∈ Φ. Then 1. dim gα = 1, and 85
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2. [gα , gβ ] = gα+β provided that α + β ∈ Φ. 3. The only roots proportional to α are ±α. 4. There exist integers p, q ∈ Z such that {k ∈ Z | β + kα ∈ Φ ∪ {0}} = {k ∈ Z | p ≤ k ≤ q}. (α,β) Moreover, p + q = −2 (α,α) .
Proof. Let S be the set of integers k such that β + kα ∈ Φ ∪ {0}. Define ( M CHα 0 = β + kα for some k ∈ S, g(S) = gβ+kα ⊕ . 0 otherwise k∈S We claim that [sα , g(S)] ⊂ g(S). This is clear because bracketing any summand above with Eα lands in another thing of this form, i.e. [Eα , gβ+kα ] ⊂ gβ+(k+1)α . Via ϕα , we get a representation of sl(2, C) on g(S). What are the weights for this representation? hkα, Hα i + hβ, Hα i = 2k +
(β, α) (β, α) hα, Hα i = 2k + 2 . (α, α) (α, α)
because Hα ∈ (α⊥ )⊥ . These must be integers by our classification of finite-dimensional su2 -representations, so we must have 2
(β, α) ∈ Z. (α, α)
0 1 maps 0 0 the weight spaces corresponding to to strictly negative weights injectively to the subsequent weight space. Then
We know that in any finite dimensional representation of sl(2, C),
[Eα , g−α ] ⊂ CHα . This is injective, hence dim g−α = 1. The conclusion is that the weight spaces are one dimensional with all weights having the same parity. Therefore, g(S) is irreducible under the action of sα . 0 1 In an irreducible finite-dimensional representation of sl(2, C), maps the 0 0 n-weight space isomorphically to the n + 2 weight space unless one of two spaces is zero. Therefore, [Eα , gβ ] ' gβ+α if β + α is a root (β is a root by assumption). This proves the second assertion. 86
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Now suppose α, β = cα for c ∈ R are roots and β 6= ±α. Interchanging α and β or negating α if necessary, we may assume that 0 < c < 1. From earlier, we know that (β, α) 1 2 ∈ Z =⇒ c = . (α, α) 2 Now apply part (2) with α/2 in place of α and β: [gα/2 , gα/2 ] = gα . Since these are one-dimensional spaces, [CEα/2 , CEα/2 ] = gα but the left hand side of the equation is obviously 0. This establishes (3). The set of weights of a finite dimensional representation of sl(2, C) is symmetric about the origin, and is an uninterrupted string of integers of the same parity. Therefore, S = {p ≤ k ≤ q} for some p, q ∈ Z. So it must be the case that (β, α) (β, α) 2p + 2 = − 2q + 2 (α, α) (α, α) so −2
(β, α) = p + q. (α, α)
By definition of S, it contains 0, so p ≤ 0 ≤ q. Corollary 8.1.3. Suppose α, β are roots and α 6= ±β. 1. If (α, β) > 0 then β − α ∈ Φ. 2. If (α, β) < 0 then β + α ∈ Φ. 3. If (α, β) = 0 then p + q = 0, so either α + β and α − β are both roots, or neither is a root. 4. If α ± β ∈ / Φ, then (α, β) = 0. Suppose α, β are roots and α 6= ±β. If α ± β ∈ / Φ, then one says that α, β are “strongly orthogonal.” In this situation, [sα , sβ ] = 0, i.e. sα and sβ commute. In fact, this is an if and only if condition. Suppose now that (π, V ) is an irreducible, finite-dimensional representation of G. Suppose α ∈ Φ and λ is a weight of π. Define S = {k ∈ Z | λ + kα is a weight.} 87
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Corollary 8.1.4. In the notation above, S = {k ∈ Z | p ≤ k ≤ q} for some p + q ∈ Z, and p + q = −2
(λ, α) . (α, α)
M
V λ+kα .
Proof. Let V (S) =
k∈S
Then sα V (S) ⊂ V , and the rest of the proof goes through as before. Corollary 8.1.5. In the notation above, 1. If (α, λ) > 0 so then λ − α ∈ Φ. 2. If (α, λ) < 0 then λ + α ∈ Φ. 3. If (α, λ) = 0 then p + q = 0, so either α + λ and α − λ are both roots, or neither is a root. In SU (2), consider 0 1 . −1 0
We saw that this normalizes the diagonal maximal torus in SU (2) and acts on it by sending −1 α 0 α 0 . → 7 0 α 0 α−1 Now, t normalizes sα and T normalizes Sα , and ker{eα : T → C∗ } centralizes Sα . Define
0 1 nα := ϕ . −1 0 −1 α 0 This normalizes T and inverts . On the level of Lie algebras, Ad nα is 0 α the identity on {X ∈ t | hα, Xi = 0} and sends Hα to −Hα . Dually, on it∗R , Ad nα is (orthogonal) reflection about α⊥ . Definition 8.1.6. There exists an element sα ∈ W , with W viewed as a group acting on it∗R , which is reflection about α (namely, sα = Ad nα ). 88
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8.2
8.2. WEYL CHAMBERS
Weyl Chambers
For each root α, we constructed a subgroup Sα ⊂ G with complexified Lie algebra sα = gα ⊕ g−α ⊕ [gα , g−α ], so Sα ' SU (2) or SU (2)/{±1}. Sα is normalized by T and commutes with ker{eα : T → C∗ }. We constructed sα ∈ W , sα = Ad nα the image of
0 1 −1 0
∈ SU (2), such that sα
is the reflection about α⊥ : for µ ∈ it∗R , sα µ = µ − 2
(µ, α) α. (α, α)
(µ,α) We saw that 2 (α,α) α ∈ Z, and has significance concerning the roots of the form µ + kα. The point is that SU (2) commutes with ker eα . So sα fixes the kernel of eα and acts as −1 on the other piece of t (or inversion on the other piece of T ).
Definition 8.2.1. An element H ∈ itR , respectively µ ∈ it∗R , is regular if hα, Hi 6= 0 (resp. (α, µ) 6= 0), for all α ∈ Φ. The picture is that the regular set in itR (resp. it∗R ) is the complement of a finite number of hyperplanes, which slice up the vector space into wedges.
Definition 8.2.2. An open Weyl chamber in itR (resp. it∗R ) is a connected component of the regular set. A Weyl chamber is the closure of an open Weyl chamber. 89
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We can think of itR ' it∗R under the inner product, sending the regular sets to each other. Therefore, the Weyl chambers in itR correspond bijectively to those in it∗R via this isomorphism. Choose a regular H0 ∈ itR and define Φ+ = {α ∈ Φ | hα, H0 i > 0}. Lemma 8.2.3. Keeping the notation above, (i) Φ = Φ+ t Φ− . (ii) For α, β ∈ Φ+ , if α + β ∈ Φ then α + β ∈ Φ+ . (iii) Φ+ depends only on the Weyl Chamber in which H0 lies, not on the particular choice of H0 . (iv) The Weyl chamber C + in which H0 lies is characterized by int(C + ) = {H ∈ itR | hα, Hi > 0 for all α ∈ Φ+ }. C + = {H ∈ itR | hα, Hi ≥ 0 for all α ∈ Φ+ }. Proof. (i) and (ii) are obvious. For (iii), observe that if we can connected H0 to some other regular H by a path in this regular set, then by the intermediate value theorem hα, H0 i > 0 ⇐⇒ hα, H1 i > 0. Therefore, int(C + ) ⊂ {H ∈ itR | hα, Hi > 0 for all α ∈ Φ+ }. If the two sets were not equal, there would exist H1 ∈ / int(C + ) such that hα, H1 i > 0 for all α ∈ Φ+ . By convexity, the entire line connecting H1 and H0 is in the regular set, furnishing a contradiction. This proves (iv). An obvious observation that we used in the proof is: Corollary 8.2.4. The open Weyl chambers are convex. Theorem 8.2.5. W is generated, as a group, by the sα for α ∈ Φ. Moreover, it acts simply transitively on the set of Weyl chambers. (Recall that a simply transitive action is one in which no element but the identity fixes anything, i.e. the stabilizers are all trivial). 90
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Proof. The Weyl group W acts on Φ, hence on the regular set, hence also on the Weyl chambers. Also, sα ∈ W for every α ∈ Φ. Suppose C + and C1 are two Weyl chambers with H0 ∈ int(C + ), and choose H1 ∈ int(C1 ). Moving H1 slightly, we can arrange that the straight line segment ` from H0 to H1 does not cross any intersection of two or more root hyperplanes (think about the codimension).
Let N denote the number of hyperplanes α⊥ crossed by `, and let N (C + , C1 ) denote the minimum possible N over all possible choices of H0 , H1 . Then N (C + , C1 ) measures the “distance” between C + and C1 ; in particular, N (C + , C1 ) = 0 if and only if C + = C1 . Now choose H0 and H1 such that N = N (C1+ , C1 ). Replace H1 by a “new” H1 which is very close to one root hyperplane α⊥ but not comparably close to any other (traverse the line ` from H0 to H1 , and choose the new H1 to be just after the last crossing of a hyperplane).
Then sα H1 is regular (just the reflection of H1 across across the very close hyperplane α⊥ ) and lies in a different Weyl chamber, namely sα C1 . By construction, 91
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N (C + , sα C1 ) < N (C + , C1 ). Repeating this argument, we find α1 , . . . , αn ∈ Φ such that N (C + , sα1 . . . sαn C1 ) = 0 i.e. C + = sα1 . . . sαn C1 . This shows that the subgroup of W generated by the sα for α ∈ Φ acts transitively on the set of Weyl chambers. To complete the proof, we must show that if w ∈ W and wC + = C + , then e 0 ∈ int(C + ). By convexity, w = e. Supposing that w fixes C + , choose H |W |
1 X ke w H0 = H0 ∈ int(C + ) |W | k=1
and wH0 = H0 . Choose n ∈ NG (T ) such that w = Ad n. Let S be the closure of {exp(itH0 ) | t ∈ R} ⊂ T (since H0 ∈ itR ). Then S is connected, compact, abelian, and contained in T , hence it is a torus. Also Ad n(H0 ) = H0 , therefore n ∈ ZG (S), which is connected. Since T ⊃ S the complexified Lie algebra z of ZG (S) contains t, hence has a weight decomposition M t⊕( gα ) B ⊂ Φ. α∈B
By construction, for each α ∈ B we must have eα = 1 on S, hence hα, H0 i = 0 (since ZG (S) = Zg (H0 ), but there are no such α since H0 is in the interior of a Weyl chamber, so B = ∅. Therefore z = t, hence ZG (S)0 = T , and by connectedness ZG (S) = T , so n ∈ T . Therefore, w = id.
With the original choices of H0 , Φ+ , int(C + ) = {H ∈ itR | hα, Hi > 0∀α ∈ Φ+ }. Φ+ is a “system of positive roots” and C + is called the “dominant Weyl Chamber.” Definition 8.2.6. α ∈ Φ+ is a simple root if α cannot be expressed as α = β + γ with β, γ ∈ Φ+ . Of course, this notion depends on the choice of dominant Weyl chamber C + which was used to define Φ+ . Let Ψ ⊂ Φ+ denote the set of simple roots. Theorem 8.2.7. Keeping the notation above, (a) For α, β ∈ Ψ such that α 6= β, then (α, β) ≤ 0, and if equality holds then α and β are strongly orthogonal (α ± β are both not roots). (b) Ψ is a vector space basis of the R-linear span of Φ. 92
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(c) If any α ∈ Φ is expressed as a linear combination of simple roots, then the coefficients are integers, either all nonnegative (in which case α ∈ Φ+ ) or all nonpositive (in which case α ∈ Φ− ). (d) Every α ∈ Φ is W -conjugate to a simple root. (e) W is generated by sα with α ∈ Ψ. Proof. (a) Suppose α, β ∈ Ψ are distinct simple roots with (α, β) > 0. From the discussion last time, we know that β − α ∈ Φ. Interchange the roles of α and β if necessary so that α − β = γ ∈ Φ+ . Then α = β + γ, which is a contradiction. In particular, this argument shows that α − β ∈ / Φ. Therefore, if (α, β) = 0 then they must be strongly orthogonal (again, look at the corollary from last time: for orthogonal roots, either both their sum and difference are roots or neither is). (b),(c) Now suppose that α ∈ Φ+ . If α ∈ / Ψ, then there exist positive roots 0 α , α00 ∈ Φ+ such that α = α0 + α00 . Observe that hα, H0 i = hα0 , H0 i + hα00 , H0 i with all terms positive. Since there are only finitely many positive roots, Φ is a discrete set, and this process must eventually terminate, in which we have expressed α as a integer combination of simple roots. Therefore, Ψ spans the R-linear span of Φ. We also get (c) after showing linear independence of the simple roots. Suppose that Ψ is not linearly dependent over R. Say r X
cj αj = 0
j=1
where Ψ = {α1 , . . . , αr } and the cj real and not all zero. Renumbering the simple roots and multiplying the identity by −1 if necessary, we may assume that c1 , . . . , ck > 0 and k ≥ 1, while ck+1 , . . . , cr ≤ 0. Then X X cj αj = −cj αj , j≤k
j>k
so taking the inner product gives something non-negative: X X 0 ≤ || cj αj ||2 = cj (−ci )(αj , αi ) j≤k
j≤k
but cj > 0, −ci ≥ 0, and (αj , αi ) ≤ 0. That forces X X cj αj = 0 =⇒ h cj , H0 i = 0 j≤k
j≤k
which is a contradiction since the αj were positive roots and cj ≥ 0 in this range. 93
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By (a)-(c), we have now characterized C + = {H ∈ itR | hα, Hi ≥ 0 for all α ∈ Ψ}. Note that itR is the direct sum of the central part and the linear span of the duals of the roots t = z ⊕ (t ∩ [g, g]), and C + is a “partial quadrant” in the latter summand. In particular, every face of C + (i.e. any codimension 1 component of ∂C + is contained in exactly one of the hyperplanes α⊥ for α ∈ Ψ). Conversely, every one of these hyperplanes contains a face of C + since the simple roots impose independent conditions.
Let α be a root. Then the hyperplane α⊥ contains a face of some Weyl chamber C1 . We know that C1 = wC + for some w ∈ W since W acts simply transitively on the set of Weyl chambers. So (wα)⊥ contains a face of C + , so (wα)⊥ = β ⊥ for β ∈ Ψ. By independence, wα = ±β. But sα α = −α, so β = wα or β = wsα α. This establishes (d). To prove (e), recall the proof of the generation of W by the sα , α ∈ Φ. Given any two Weyl chambers C1 , C2 we defined N (C1 , C2 ) with N (C1 , C2 ) ≥ 0 and N (C1 , C2 ) = 0 ⇐⇒ C1 = C2 . We had seen that there exists α ∈ Φ such that α⊥ is a face of C1 and N (sα C1 , C2 ) < N (C1 , C2 ), 94
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Apply this with C1 = C + : there exists a simple root such that N (sα C + , C2 ) < N (C + , C2 ). W acts on the chambers, so N (sα C + , C2 ) = N (C + , sα C2 ). Repeating this argument, we eventually arrive at C + = sαn . . . sα1 C2 with αj ∈ Ψ, which establishes 5.
We have almost completed the ingredients for the classification of the compact Lie groups. As a consequence of the theorem, we see that knowledge of Ψ as an abstract set, along with the data ||α||, α ∈ Ψ, and the inner products (α, β) with α, β ∈ Ψ, completely determine Φ as a subset of the inner product space spanned by Φ over R. Now suppose α, β ∈ Φ such that α 6= ±β. Switching the roles of α and β if necessary, we may suppose that ||α|| ≥ ||β||. Theorem 8.2.8. In this situation, if α and β are not orthogonal then ||α||2 = k||β||2 and (cos ∠(α, β))2 =
k = 1, 2, 3 (α, β)2 k = . 2 2 ||α|| ||β|| 4
Proof. We know that 2
(α, β) (α, β) ,2 ∈ Z. ||α||2 ||β||2
Taking the product, 4
(α, β)2 ∈Z ||α||2 ||β||2 2
(α,β) k but α and β are neither proportional nor perpendicular, so ||α|| 2 ||β||2 = 4 where k = 1, 2, or 3. Since ||α|| ≥ ||β||, the first term in the first equation is the smaller
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integer, hence (α, β) 2 ||α||2 = 1. Straightforward manipuulations of this imply what we want. Corollary 8.2.9. Suppose α, β are distinct simple roots and (α, β) 6= 0. Then ◦ 120 k = 1 ∠(α, β) = 135◦ k = 2 . ◦ 150 k = 3 with ||α||2 = k||β||2 , or vice versa. This potentially allows us to reconstruct the root system from the knowledge of a couple of roots.
96
Chapter 9
Classificiation of compact Lie groups 9.1
Classification of semisimple Lie algebras
Definition 9.1.1. A Lie algebra g over a field k is simple if it contains no proper ideal and has dimension at greater than 1. Note that this rules out abelian Lie algebras, since every subalgebra of a Lie algebra is an ideal. Definition 9.1.2. A Lie algebra g over a field k is semisimple if it is a direct sum of commuting simple ideals. Definition 9.1.3. A Lie group G is simple (resp. semisimple) if its Lie algebra gR is simple (resp. semisimple) as a Lie algebra over R. Proposition 9.1.4. Suppose G is a connected, compact, simple Lie group. Then g is simple over C. This says that gR is “absolutely simple” (simple when tensored up with the algebraic closure). Proof. Suppose not; then g contains a simple ideal g1 . Then g1 6= g1 because gR is simple. We claim that g = g1 ⊕ g1 . Indeed, if g1 ⊕ g1 isn’t the whole space, then its intersection with gR is a proper ideal (because the Galois group of C/R interchanges the two factors). Note also that [g1 , g1 ] = 0. Define J : g → g by ( iX X ∈ g1 JX = . −iX X ∈ g1 Then J = J, so J : gR → gR 97
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with J 2 = −1. Suppose u, v ∈ gR . Then u = u1 + u ¯1 , v = v1 + v¯1
u1 , v1 ∈ g1
so [Ju, v] = [J(u1 + u ¯1 ), v1 + v¯1 ] = [iu1 − i¯ u1 , v1 + v¯1 ] = i[u1 , v1 ] − i[¯ u1 , v¯1 ] Observe that this coincides with J[u, v] = J[u1 + u ¯1 , v1 + v¯1 ] = J([u1 , v1 ] + [¯ u1 , v¯1 ]) = i[u1 , v1 ] − i[¯ u1 , v¯1 ]. So J : gR → gR with J 2 = −1 and [Ju, v] = J[u, v] = [u, Jv]. Turn gR into a Lie algebra over C by defining iX = JX. What are we doing? Turning a finite dimensional real vector space into a complex vector space amounts to giving a linear map that squares to −1. We have just proved that the complex structure is compatible with the Lie bracket. We had seen earlier that this implies that G is a complex Lie group. By assumption, it is compact and non-abelian, which is impossible (see homework). On the homework, we saw that if G is a connected, compact Lie group then G is semisimple if and only if ZG is finite. If so, then there exist connected compact simple Lie groups G1 , . . . , Gn with a finite covering map G1 × . . . × Gn → G. So, to understand connected, compact, semisimple Lie groups, it suffices to understand connected, compact, simple Lie groups and their centers. As usual, let G be a connected compact Lie group and T a torus. Observation: G is semisimple if and only if the R-linear span of Φ is it∗R , which is the case if and only if Ψ is an R-basis of it∗R . Definition 9.1.5. Φ is irreducible if Φ cannot be expressed as a disjoint union Φ = Φ1 t Φ2 with Φi = −Φi non-empty, and Φ1 ⊥ Φ2 . Definition 9.1.6. Ψ is irreducible if Ψ cannot be expressed as Ψ = Ψ1 t Ψ2 with Ψi non-empty and Ψ1 ⊥ Ψ2 . Another observation: if G is semisimple, then G is simple if and only if Φ is irreducible, which is the case if and only if Ψ is irreducible. This is again straightforward, and its proof is left as an exercise. Now suppose that G is connected, compact, and simple. The Dynkin diagram of G is the graph constructed as follows: 98
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• The vertices are the simple roots. • α1 , α2 ∈ Ψ are connected by k edges if (α1 , α2 ) 6= 0 and ||α1 || = k||α2 ||2 or vice versa, and are not connected if (α1 , α2 ) = 0 (equivalently strongly orthogonal). If k 6= 1, the edge is directed from the longer root to the shorter. By the observation, the Dynkin diagram must be connected, since the connected components correspond to reducible pieces. Theorem 9.1.7. The only possible Dynkin diagrams of connected, compact, simple Lie groups are:
Each of these does correspond to a connected, compact, simple Lie group, which is unique up to covering. Proof sketch. The first part follows from some elementary analysis using inner products, etc. For instance, one quickly realizes that cycles are impossible, and then only one pair of nodes can have multiple edges. An is SU (n + 1), Bn = SO(2n + 1), Cn = Sp(2n) ∩ U (4n), and Dn = SO(2n). These are the classical groups. 99
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Why the uniqueness? From the Dynkin diagram we can reconstruct the roots and root system (see remarks above). For simple α, take Eα , E−α , and Hα as above, and normalize the scaling. Additional roots are obtained by bracketing these for simple roots. Compact semisimple Lie groups are adjoing groups (the image under the adjoint representation). The last question concerns the centers; we will address this next time.
9.2
Simply-connected and adjoint form
The homework, together with the discussion last time, implies that to “understand” all connected, compact Lie groups, it suffices to “understand” the coverings of simple, connected, compact Lie groups. Suppose G is connected, compact, and semisimple, T ⊂ G a maximal torus and L, Λ as usual. We define GAd := G/ZG which is isomorphic to the image of G under the adjoint representation (something commutes with a neighborhood of 0 if and only if it commutes with everything, since a neighborhood of 0 generates everything). Obviously, ZG ⊂ T is discrete, hence finite. Similarly, we have TAd = T /ZG , etc. Identifying tR /L ' T , this maps to TAd ' tR /LAd with kernel LAd /L, so ZG ' LAd /L. Then LAd = {H ∈ tR | exp H ∈ ZG } = {H ∈ tR | exp H ∈ ker Ad } = {H ∈ tR | eα exp H = 1 for all α} = {H ∈ tR | hα, Hi ∈ 2πiZ for all α ∈ Φ} = {H ∈ tR | hα, Hi ∈ 2πiZ for all α ∈ Ψ} We have seen that (α, µ) ∈ Z for all α ∈ Φ} (α, α) = {µ ∈ itR | hµ, Hα i ∈ Z for all α ∈ Φ} (∗)
Λ ⊂ {µ ∈ itR | 2
= {µ ∈ itR | hµ, Hα i ∈ Z for all α ∈ Ψ} := Λsc The inclusion (∗) is actually an equality for G simply connected, and depends on knowing that every µ ∈ Λ is the weight of some finite dimensional irreucible representation. We’ll get to this later. 100
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If Λsc were the weight lattice of some covering group of G, then the corresponding unit lattice would be Lsc = the Z-linear span of 2πiHα for α ∈ Ψ. Theorem 9.2.1. With the notation above, 1. π1 (G) is finite. 2. Let Gsc be the universal covering. Then its unit lattice is Lsc and its weight lattice is Λsc . In particular, π1 (GAd ) ' ZGsc ' LAd /Lsc . Sketch. This is only a sketch; a more detailed argument will be posted online. Define Greg := {g ∈ G | dim ZG (g) = rank G}, where rank G is the dimension of the maximal torus. Then Greg is a conjugation invariant. Then G − Greg = {g ∈ G | dim ZG (g) > rank G} = {g ∈ G | dim ker(1 − Ad g) > rank G} since ker(1 − Ad g) = ZG (g). This implies that G − Greg is a real analytic subvariety of G, i.e. cut out by real analytic equations. By invariance under conjugation, and the fact that every element of G is conjugate to an element of T , G − Greg = {gtg −1 | t ∈ T − Treg , g ∈ G, “g ∈ G/T 00 } where Treg = Greg ∩ T and “g ∈ G/T 00 means that we need only take coset reprentatives of G/T (representatives of the same coset will give the same gtg −1 ). The centralizer of t ∈ T contains T , and is bigger if and only if eα (t) = 1 for some α. T − Treg = {t ∈ T | eα (t) = 1 for some α ∈ Φ}. Then G − Greg =
[
{gtg −1 | eα (t) = 1, “g ∈ G/T Sα00 }
α
where Sα is defined as in our proof of the structure theorem for roots, since any such t will commute with sα as well, and T normalizes Sα . The condition eα (t) = 1 imposes a codimension-one condition, and T Sα has dimension dim T + 2 since Sα has dimension 3 but intersects T in a one-dimensional piece. So dim(G − Greg ) ≤ (dim T − 1) + dim G − (dim T + 2) ≤ dim G − 3. In fact, this is an equality. What does cutting out a codimension 3 thing mean? Cutting out codimension 1 disconnects. Cutting out codimension 2 doesn’t disconnect, but changes the fundamental group. Codimension 3 doesn’t change the fundamental group, since homotopies occur on a surface, and we can homotope things to be transverse to the codimension 3 things. 101
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Therefore, the inclusion Greg ,→ G induces an isomorphism on π1 . Consider the map G/T × Treg → Greg (gT, t) 7→ gtg −1 . Facts: 1. This is a covering map with covering group W , which acts on G/T by right translation and on Treg by conjugation. 2. Treg is connected. Now, π1 (G/T ) ' π1 (G/T × {pt}) ⊂ π1 (G/T × Treg ). Since the map is a covering map, the induced map on fundamental groups is injective, so π1 (G/T × Treg ) → π1 (Greg ) ' π1 (G) is injective. Let g(s) represent a loop in G/T , i.e. g(1)T = g(0)T and s 7→ g(s) is continuous. Then the image loops in π1 (G) is s 7→ g(s)t0 g(s)−1
t0 ∈ Treg .
Note that we couldn’t choose t0 = e because t0 must be regular. However, this can be contracted in G by letting t0 → e. The conclusion is that π1 (G/T ) = 0. The map G → G/T is a fibration with fiber bundle T , so π1 (T ) → π1 (G) → π1 (G/T ) is exact. Since the last term is 0, π1 (T ) → π1 (G) is surjective. But T = tR /L, so we get a surjection L π1 (G). That tells us that π1 (G) is finitely generated and abelian. What we’ve seen is that if we pass to a covering, the unit lattice shrinks and the weight lattice grows. But the inclusion (∗) that we haven’t proven yet shows that the weight lattice is “bounded above.” 102
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This already implies π1 (G) is finite because Λ is “bounded above.” (We know that π1 (G) is finitely generated and abelian.) Consider the map SU (2) → G, with image Sα . “L of SU (2)00
/L
/ π1 (G)
π1 (SU (2))
3 We know that SU (2) ' S is simply connected, and the path which represents the 2πi 0 generator in the unit lattice of SU (2) is 0 −2πi
2πis e 0 s 7→ →G 0 e−2πis the composite map being s 7→ exp(2πisHα ). Conclusion. Under the surjection L → π1 (G), the 2πiHα all go to e. So π1 (G) is a quotient of L/Lsc , in particular finite. To complete the proof, we must show that this is an isomorphism when G = Gsc , the universal cover of G. This will be posted online. We now establish the inclusion (*) from earlier. Suppose µ ∈ Λ, i.e. µ lifts to a character eµ : T → C∗ . Consider {f ∈ C(G) | f (gt) = eµ (t)f (g) for all t ∈ T }. This is non-zero because locally the fibration G → G/T is a product, i.e. there are local sections. So we can take a function on G/T with compact support contained within the appropriate open set, and extend to the pre-image in G by multiplying by eµ on the T factor. By Peter-Weyl, the L2 closure of this space is M [ Vi ⊗ (Vi∗ )µ b i∈G
b hence µ is a weight of V ∗ . so (Vi∗ )µ cannot be zero for all i ∈ G, i
103
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104
Chapter 10
Representations of compact Lie groups 10.1
Theorem of the highest weight
Now suppose that G is connected and compact (not necessarily semisimple, so we could have a center of positive dimension). Let T ⊂ G be the maximal torus, C + a dominant Weyl chamber. Thus we have Φ+ and Ψ. Let (π, V ) be an irreducible, finite-dimensional representation of G. Theorem 10.1.1 (“Theorem of the highest weight”). The following conditions on a weight λ of π are equivalent. (1) λ + α, for α ∈ Φ+ , is not a weight. (2) There exists a non-zero vλ ∈ V λ such that Eα vλ = 0 for all α ∈ Φ+ . (3) Any weight of π can be expressed as λ − α1 − α2 − . . . − αN with α1 , . . . , αN ∈ Φ+ (repetitions allowed). A weight with these properties exists and is unique. It is called the “highest weight” of π, and it has the following properties. 1. dim V λ = 1. 2. (λ, α) ≥ 0 for every α ∈ Φ+ , i.e. λ is dominant. 3. λ determines π up to isomorphism. 105
Math 222 CHAPTER 10. REPRESENTATIONS OF COMPACT LIE GROUPS Remark 10.1.2. We shall see later that every dominant λ ∈ Λ is the highest weight of some irreducible representation. Therefore, b ' {dominant λ ∈ Λ}. G Proof. (1) =⇒ (2). This is because Eα V λ ⊂ V λ+α (in fact, we get the result for any such vλ ). (2) =⇒ (3). Define L+ = linear span of t and gα , α ∈ Φ+ M M− = g−α . α∈Φ+
Both are Lie subalgebras of g, so g = L+ ⊕ M − (a vector space direct sum). By PBW, U (g) = U (M− )U (L+ ). Suppose vλ ∈ V λ is non-zero. By irreducibility, V = U (g)vλ = U (M− )U (L+ )vλ . Now, ( Cvλ tvλ = 0
λ 6= 0 λ=0
so if Eα vλ = 0 then for all α ∈ Φ+ , U (g)vλ = U (M− ) U (L+ )vλ = U (M− )vλ . | {z } Cvλ
This is the linear span of E−α1 , E−α2 , . . . , E−αN vλ for all possible N -tuples of positive roots. This proves (3). Also, U (M− )vλ = U (M− )(M− ⊕C)vλ . The summand coming from U (M− )M− has weight strictly less than λ, so dim V λ = 1. (3) =⇒ (1). Suppose λ + α is a weight for some α ∈ Φ+ . By (3), λ + α = λ − α1 − . . . − αN , so therefore, α + α1 + . . . + αN = 0. Apply this to some regular H0 ∈ C + : hα + α1 + . . . + αN , H0 i = 0, 106
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which is a contradiction. For uniqueness, suppose for the sake of contradiction that λ1 and λ2 both satisfy properties (1)-(3). Then αi ∈ Φ+
λ1 = λ2 − α1 . . . − αN λ2 = λ1 − β1 − . . . − βM
β j ∈ Φ+
Equating these, we find that α1 + . . . + αN + β1 + . . . + βM = 0 which contradicts positivity by evaluating on any H0 in the dominant Weyl chamber. To prove existence, suppose that µ0 is a weight of π. If µ0 satisfies (1), set λ = µ0 . If not, we may choose a positive α ∈ Φ+ such that µ1 = µ0 + α1 is a weight. If µ1 satisfies (1), set λ = µ1 ; if not, continue. The process must stop because V is finite dimensional, so there are only finitely many weights. Eventually, we will produce a λ = µk which satisfies (1). Let λ be as in (1)-(3). For α ∈ Φ+ , λ + α is not a weight. Hence (λ, α) ≥ 0 (by the p + q thing), which verifies dominance. The last thing to verify is that λ determines π up to isomorphism. As before, define subalgebras of g L+ = linear span of t and gα , α ∈ Φ+ M− =
M
g−α .
α∈Φ+
So g = L+ ⊕ M− . Define the U (g) module Mλ , where Mλ = U (g)/left ideal generated by gα , H − hλ, Hi1
α ∈ Φ+ , H ∈ t.
This is called the “Verma module of highest weight λ.” Let mλ be the image of 1 ∈ Mλ under the quotient. Note that mλ 6= 0. Why? We had expressed U (g) = U (M− )U (L+ ). We are essentially dividing out by U (L+ ): the relations that lie in the augmentation ideal (the gα ) obviously can’t kill 1, since they raise degrees; the other relations take the constants to the Lie algebra of the torus, which doesn’t intersect gα for α ∈ Φ+ , hence also can’t kill 1. By definition, t acts on mλ by λ and the gα for α ∈ Φ+ kill mλ , so U (L+ )mλ = Cmλ . 107
Math 222 CHAPTER 10. REPRESENTATIONS OF COMPACT LIE GROUPS Then we claim that U (M− ) ' Mλ X 7→ Xmλ as vector spaces. This is because Mλ is obviously generated by mλ as a g module (this is true even before quotienting), and U (g) = U (M− )U (L+ ), so Mλ ' U (g)mλ ' U (M− )U (L+ )mλ ' U (M− )mλ . We next claim that t acts finitely, i.e. every m ∈ Mλ lies in a finite dimensional, t-stable subspace. Indeed, for X ∈ U (M− ), H ∈ t, HXmλ = (HX − XH + XH)mλ = ad H(X)mλ + XHmλ = ad H(X)mλ + hλ, HiXmλ . So H preserves the degree (by which we mean the filtered components of the filtration). This also implies that t acts semisimply (since it acts semisimply on the gα ), with eigenvalues of the form λ − α1 − . . . − αN , αi ∈ Φ+ (the λ coming from the second term in the equation above, and the other stuff from the first term.) In particular, the weight λ occurs with multiplicity 1. Let N ⊂ Mλ be a proper U (g) submodule. Then t acts on N , with weights as above except λ cannot be a weight for N since mλ generates all of Mλ . So Mλ contains a unique maximal, proper, U (g) submodule (i.e. not Mλ , but conceivably zero), namely the linear span of all proper submodules. Therefore, Mλ has a unique irreducible, simple U (g) quotient. If V is an irreducible representation with highest weight λ, then note that the ideal defining Mλ annihilates V λ by (2), so we get a surjective homomorphism of U (g) modules Mλ → V sending mλ 7→ vλ . Hence V is this unique irreducible, simple quotient.
10.2
The Weyl character formula
Let G be a compact, connected Lie group, and T, Φ+ , Ψ be as usual. Define ρ=
1 X α ∈ it∗R . 2 + α∈Φ
2ρ is a sum of roots, hence 2ρ ∈ ΛAd , i.e. the Z-linear span of Ψ. Lemma 10.2.1. We have 1. For w ∈ W , wρ − ρ ∈ ΛAd . 108
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10.2. THE WEYL CHARACTER FORMULA
(ρ,α) 2. 2 (α,α) = 1 for every α ∈ Ψ.
Proof. Choose α ∈ Ψ and let β ∈ Φ+ , β 6= α. Then sα β = β − 2
(β, α) α ∈ Φ. (α, α)
We know that when β is expressed as a linear combination of simple roots, all coefficients are non-negative integers. Therefore, the same must be true of sα β because some coefficient in the expression for sα β is positive. That says sα (Φ+ − {α}) = Φ+ − {α}. (Obviously, sα α = −α.) So 1 sα ρ = sα 2 =
1 2
X β∈Φ+ ,β6=α
X
β−
β∈Φ,β6=α
=ρ−
1 β+ α 2
α 2
α α − = ρ − α. 2 2
So we’ve found that sα ρ = ρ − α. On the other hand, we know that sα ρ = ρ − 2 so we can conclude that 2
(ρ, α) α (α, α)
(ρ, α) = 1 for all α ∈ Ψ. (α, α)
This also proves that sα ρ ∈ ρ + ΛAd and sα generate W , establishing (1). The lemma implies that ρ ∈ Λsc , since Λsc = {µ ∈ it∗R | 2
(µ, α) ∈ Z for all α ∈ Ψ}. (α, α)
In the following, we could take one of two courses: (a) either use the fact that Λsc is the weight lattice of the universal covering of G, or (b) go to a 2-fold cover of T so that e2ρ : T → C∗ has a square root, namely eρ from the twofold cover. Then keep track of computations to see that they make sense. 109
Math 222 CHAPTER 10. REPRESENTATIONS OF COMPACT LIE GROUPS We will do the second. Define the formal expression Y Y ∆= (eα/2 − e−α/2 ) = eρ (1 − e−α ) . |{z} α∈Φ+
defined up to ±1 on T α∈Φ+
|
{z
}
well-defined on T
Weyl integration formula. Normalize the Haar measure on G, T to have total measure 1, so the “quotient measure” on G/T also has this property (the quotient measure satisfies the property that the measure of a set in the quotient is the measure of its pre-image). Then for f ∈ C(G), Z
1 f dg = #W G
Z Z T
|∆(t)|2 f (gtg −1 ) dg dt.
G/T
Corollary 10.2.2. If f is conjugation invariant, then Z Z 1 f dg = |∆(t)|2 f (t) dt. #W G T Weyl character formula. Let (π, V ) be an irreducible representation for G of highest weight λ and χλ its character. Then P χλ |T =
w∈W
(w)ew(λ+ρ) . ∆
Moreover, each dominant λ ∈ Λ is the highest weight of some finite dimensional irreducible (π, V ). Definition 10.2.3. For w ∈ W , we define (w) = det{w : tR → tR } = ±1. The value is ±1 because the Weyl group is finite dimensional, hence the determinant is a root of unity; but this is a map of real vector spaces. Remark 10.2.4. Some observations: 1. We have w(λ + ρ) = wλ + wρ = wλ + wρ − ρ + ρ. Obviously wλ ∈ Λ, and Lemma 10.2.1 tells us that wρ − ρ ∈ Λ, so we deduce that w(λ + ρ) ∈ Λ + ρ. w(λ+ρ) has an ambiguity in sign, but it is the same ambiguity as Therefore, Q e ρ −α ), so the quotient is well-defined. ∆=e α∈Φ+ (1 − e 2. Observe that χλ |T is a class function and T meets all conjugacy classes, so this completely determines χ.
110
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10.3. PROOF OF WEYL CHARACTER FORMULA
3. A priori, the expression on the right hand side of the Weyl character formula might have “poles.” However, this can only happen if e−α = 1, so the function is well-defined on Treg = {t ∈ T | eα (t) 6= 1∀α ∈ Φ}, which is open and dense in T .
10.3
Proof of Weyl character formula
b let χi be the character. We know that {χi | i ∈ G} b is an orthonormal For i ∈ G, 2 basis of the space of L conjugation-invariant functions. Define ϕi = χi |T · ∆. By the Weyl integration formula, Z T
( 0 ϕi ϕj dt = #W χi χj dg = #W G Z
i 6= j . i=j
Also, if f is an L2 conjugation invariant function and ϕ = f |T · ∆, then Z b =⇒ f = 0. ϕϕi dt = 0 for all i ∈ G T
b and let λ ∈ Λ be the highest weight. We know that Fix i ∈ G (λ, α) > 0 for all α ∈ Φ+ (ρ, α) > 0 for all α ∈ Φ+ so (λ + ρ, α) > 0 for all α ∈ Φ+ , i.e. λ + ρ lies in the interior of the dominant Weyl chamber in it∗R . In particular, the terms w(λ + ρ) for w ∈ W are distinct because W acts freely on the interiors of the Weyl chambers. We obviously have the identity Y X ϕi = dim V µ eµ eρ (1 − e−α ). α∈Φ+
µ∈Λ
The highest weight appears in the left term, with multiplicity one. So we can write this as ϕi = eλ+ρ + (an integral linear combination of terms) eµ+ρ where µ = λ− sum of positive roots. In particular, the eλ+ρ term certainly doesn’t get canceled. 111
Math 222 CHAPTER 10. REPRESENTATIONS OF COMPACT LIE GROUPS Note that χi |T is W -invariant, since W acts on the torus by conjugation. We had seen that for any simple root α ∈ Ψ, sα (Φ+ − {α}) = Φ+ − {α} and sα (α) = −α (i.e. sα preserves all the positive roots except α, which it negates). Therefore Y sα ∆ = sα eβ/2 − e−β/2 = −∆ β∈Φ+
because exactly one factor changes sign. This implies that ∆ is “W -alternating” in the sense that w∆ = (W )∆ for any w ∈ W . Then ϕi is W -alternating, so X combination of other terms ϕi = (w)ew(λ+ρ) + integral linear . with no cancellation w∈W
By orthogonality, Z
Z
2
|ϕi | dt = #W +
|other terms|
T
but we also know that
Z
|ϕi |2 dt = #W.
T
Therefore, the other terms must vanish. This proves the Weyl character formula: if χ is the character of the irreducible representation with highest weight λ, then P (w)ew(λ+ρ) χ|Treg = Q w∈W α/2 . − e−α/2 ) α∈Φ+ (e This can also be written as χ|Treg
P w(λ+ρ)−ρ w∈W (w)e = Q , −α ) α∈Φ+ (1 − e
which shows it to be well-defined. Applying this to the trivial representation, where the highest weight is λ = 0, we obtain what is called the Weyl denominator formula. Corollary 10.3.1 (Weyl denominator formula). Y X (eα/2 − e−α/2 ) = (w)ewρ . α∈Φ+
w∈W
Remark 10.3.2. The Weyl character formula also implies the following statement which we previously proved by algebraic means: the highest weight of an irreducible representation π determines π up to isomorphism. We had stated this proposition earlier. Proposition 10.3.3. If λ ∈ Λ is dominant, i.e. (λ, α) ≥ 0 for each α ∈ Φ+ , then λ is the highest weight of some irreducible representation. 112
Math 222
10.4. PROOF OF THE WEYL INTEGRATION FORMULA.
b Define Proof. Let λi be the highest weight of πi , λ 6= λi for all i ∈ G. X ϕ= (w)ew(λ+ρ) w∈W
There’s no cancellation here because the w(λ + ρ) are all distinct (as mentioned at ϕ the beginning of the lecture). Also, ϕ is W -alternating, hence W is W -invariant. Then there exists a unique W -invariant (conjugation-invariant, since the group of deck transformations for the covering in the section below is precisely W ) function ϕ f ∈ C(Greg ) such that f |Treg = ∆ . Since G − Greg has measure 0, we can regard f 2 as an element of L (G). By the Weyl integration formula, Z Z 1 f χi dg = f ∆χi ∆ dt #W T G Z X 1 (w)ew(λ+ρ) (v)e−v(λi +ρ) dt = #W w,v∈W
= 0 by orthogonality.
10.4
Proof of the Weyl integration formula.
We know that the map F
G/T × Treg → Greg (gT, t) 7→ gtg −1 is a covering map with group of deck transformations equal to W = NG (T )/T . Therefore, Z Z Z 1 f (g) dg = F ∗ f dt dg #W G/T T G Z Z 1 | det F∗ |f (gtg −1 ) dt dg = #W G/T T Left multiplication by g gives a map TgT G/T ← TeT G/T ' g/t '
M
gα
α∈Φ
and Tt T ' t (this is the tangent space to T at t). So M T(gT,t) (G/T × Treg ) ' gα ⊕ t ' g ' Tgtg−1 Greg . α∈Φ
113
Math 222 CHAPTER 10. REPRESENTATIONS OF COMPACT LIE GROUPS These identifications we made are a little sketchy, but the measure is left invariant, etc. so things are OK. L Suppose X ∈ α∈Φ gα ∩ gR and let f ∈ C ∞ (G). We want to compute F∗ ; let’s trace through the identifications made above. F (g exp(sX)T, t) = g exp(sX)t exp(−sX)g −1 = g exp(sX) exp(−s Ad tX)tg −1 . So ∂ f (g exp(sX) exp(−s Ad tX)tg −1 )|s=0 ∂s = `(− Ad g(1 − Ad t)X) f (gtg −1 ) | {z }
F∗ (X, 0)f (gtg −1 ) =
left infinitesimal translation by ...
Recall that if f ∈ C ∞ (G) and X ∈ gR , then r(X)f (g) =
∂ f (g exp(sX))|s=0 . ∂s
and `(X)f (g) =
∂ f (exp(−sX)g)|s=0 . ∂s
For Y ∈ tR , F (gT, exp(sY )t) = g exp(sY )tg −1 . Hence F∗ (0, Y )f (gtg −1 ) =
∂ f (g exp(sY )tg −1 )|s=0 = `(− Ad g(Y ))f (gtg −1 ). ∂s
That means that ( L X ∈ α∈Φ gα −1 Ad g ◦ F∗ : Y ∈t
7→ (1 − Ad t)X 7→ Y
so det(Ad g −1 ◦ F∗ (t)) =
Y
(1 − eα (t))
α∈Φ
=
Y
(1 − eα (t))(1 − e−α (t))
α∈Φ+
= |∆(t)|2 . 114
Math 222
10.5
10.5. BOREL-WEIL
Borel-Weil
Suppose (π, V ) is irreducible of highest weight λ. The theorem of the highest weight implies that V N+ (= space of N+ -invariant vectors) = V λ . L where N+ = α∈Φ+ gα (for groups, we look at the elements that map to the identity. For Lie algebras, we look at things that get mapped to 0). This is an equality of vector spaces with an action of T (or t) since T normalizes N+ . Also let M N− = g−α . α∈Φ+
Dually we have (V ∗ )N− = (V ∗ )−λ . By Peter-Weyl, V = {f ∈
M [ N Vi ⊗ (Vi∗ − )−λ } b i∈G
i.e. all functions f on G on which T acts via right translation by e−λ . What this says is: n o r(X)f =0 for all X∈N V ' f ∈ C ∞ (G) | f (gt)=e−λ (t)f (g)∀t∈T− . Facts: G/T can be turned into a complex manifold such that (a) G acts on G/T holomorphically. (b) The holomorphic tangent space to G/T at eT is isomorphic to M N+ ⊂ gα ' Tet G/T α∈Φ
(Note that N+ is in the complexified Lie algebra.) (c) If λ ∈ Λ, there exists a unique (up to isomorphism) holomorphic line bundle LX → G/T such that (i) the action of G on G/T lifts to Lλ and (ii) T acts on the fiber of Lλ at eT by eλ . So G acts on the space of holomorphic sections of Lλ . Theorem 10.5.1 (Borel-Weil). For λ ∈ Λ dominant, H 0 (G/T, O(Lλ )) is the space of irreducible representations of highest weight λ and H p (G/T, O(Lλ )) = 0 for p > 0. Remark 10.5.2. The Borel-Weil-Bott theorem gives a description of H p (G/T, O(Lλ )) for all λ ∈ Λ and integers p. 115