Lie groups and Lie algebras. Wilfried Schmid Lecture notes by Tony Feng

Transcription

1 Lie groups and Lie algebras Wilfried Schmid Lecture notes by Tony Feng Spring 2012

2 Math 222 2

3 Contents 1 Geometric preliminaries Lie groups Lie algebras Vector fields Tangent and cotangent spaces Pullbacks and pushforwards The Lie algebra of a Lie group Invariant vector fields Integral curves The exponential map The group law in terms of the Lie algebra Lie algebras The Lie algebra gl(v ) The adjoint representation Poincaré s formula The Campbell-Baker-Hausdorff formula Geometry of Lie groups Vector bundles Frobenius Theorem Foliations Lie subgroups Lie s Theorems The Universal Enveloping Algebra Construction of the universal enveloping algebra Poincaré-Birkhoff-Witt Proof of Poincaré-Birkhoff-Witt

4 Math 222 CONTENTS 6 Representations of Lie groups Haar Measure Representations Schur orthogonality relations The Peter-Weyl Theorem Compact Lie groups Compact tori Weight decomposition The maximal torus Trace forms Representations of SU(2) Root systems Structure of roots Weyl Chambers Classificiation of compact Lie groups Classification of semisimple Lie algebras Simply-connected and adjoint form Representations of compact Lie groups Theorem of the highest weight The Weyl character formula Proof of Weyl character formula Proof of the Weyl integration formula Borel-Weil

5 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 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. 5

6 Math 222 CONTENTS 6

7 Chapter 1 Geometric preliminaries 1.1 Lie groups Definition A Lie group, G, is a group endowed with the structure of a C manifold such that the inversion map and multiplication map are smooth. s : G G g g 1 m : G G G (g, h) gh Remark It suffices to require that the single map (g, h) 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 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

8 Math 222 CHAPTER 1. GEOMETRIC PRELIMINARIES Definition 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 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 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 U H of h in H and an open neighborhood U G of h in G such that 1. U H = U G H. 2. U H U G is a closed submanifold, e.g. looks locally like R k R n. 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 U N 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 U N, U M of the domain of f such that f UN is the restriction of a C function F : U M R to U N. 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 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 The following are examples of Lie groups. 1. (R n, +). 2. Giving Q the discrete topology, R Q R 2, + is a Lie subgroup of dimension 1 of (R 2, +). 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. (R n /Z n, +): an n-dimensional torus. Smoothness, etc. follow because they are local properties. 8

9 Math LIE ALGEBRAS 4. GL n (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 subset of R n2 by matrix entries. 5. GL n (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 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 (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 A Lie Algebra g over K is a vector space over K equipped with a bilinear map called the Lie bracket such that 1. [a, a] = 0 for all a g. g g g (a, b) [a, b] 2. [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0 for all a, b, c g. Remark 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 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 A Lie subalgebra of g is a K-linear subspace h g such that [h, h] h. 9

10 Math 222 CHAPTER 1. GEOMETRIC PRELIMINARIES 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 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 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. [d 1, d 2 ] = d 1 d 2 d 2 d 1. In some sense, all Lie algebras arise in this way, though perhaps indirectly. Example 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 ( Localization of vector fields. ). Suppose X is a vector field on M, p M, and f 1, f 2 are functions on M such that f 1 = f 2 on some open neighborhood of p. Then Xf 1 (p) = Xf 2 (p). Proof. We use the existence of cutoff functions. Namely, suppose U 0 M is open and U 1 U 0 is open such that U 1 is compact and contained in U 0. Then there exists a function u C (M) such that (i) f 1 on U 1, and (ii) Supp f U 0. This follows from the existence of partitions of unity: just take an appropriate open cover (i.e. wherein U 0 is the only open intersecting U 1 ) and a partition of unity and consider the sum of those parts whose support intersects U 1. Apply this with U 0, U 1 nested open neighborhoods containing p and f 1 f 2 on U 0. Then hf 1 hf 2, since f 1 f 2 on the support of h. Then X(hf 1 ) = X(hf 2 ), so X(hf i )(p) = h(p)xf i (p) + f i (p)xh(p). Equating these for i = 1, 2, we conclude that Xf 1 (p) = Xf 2 (p). 10

11 Math TANGENT AND COTANGENT SPACES Corollary 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 Vector fields can be patched, i.e. given open subset U 1, U 2 M and vector fields X i on U i for i = 1, 2 such that X 1 U1 U 2 = X 2 U1 U 2, then there exists a unique vector field X on U 1 U 2 such that X Ui = X i. Proof. At any p U i, define (Xf)(p) by X i f Ui (p). These two corollaries together tell us that vector fields constitute a sheaf. Definition A coordinate neighborhood in M is a pair (U, x 1,..., x n ) with U M open and x 1,..., x n C (M) real-valued such that p (x 1 (p),..., x n (p)) R n establishes a diffeomorphism between U and its image in R n (which must be open). Corollary Suppose X is a vector field on M and (U, x 1,..., x n ) is a coordinate chart. Then X U can be expressed uniquely as X U = u j=1 a j, x j a j C (M). Conversely, any such expression describes a unique vector field on M. Proof. Defining a j = Xx j. Suppose f C (U) and p U. Then we may write f = f(p) 1 + i f dx i (p)(x i x i (p)) + h, where h is a linear combination of products h 1 h 2 with h 1 (p) = h 2 (p) = 0 (this is just Taylor s theorem). Note that X1 = 0 since X is a derivation. On a similar note, X(h 1 h 2 )(p) = h 1 (p)xh 2 (p) + h 2 (p)xh 1 (p) = 0. Therefore, (Xf)(p) = i X ( ) f (p)x i x i ( f = ai x i ) (p). 1.4 Tangent and cotangent spaces Definition A tangent vector at p is a linear map X p : C (M) R or C such that X p (f 1 f 2 ) = f 1 (p)x p f 2 + f 2 (p)x p f 1. 11

12 Math 222 CHAPTER 1. GEOMETRIC PRELIMINARIES Corollary If X p is a tangent vector at p and f C (M), then X p f depends only on the values of f near p (i.e. in any neighborhood of p). This tells us that X p 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 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 T p M. For p M, define I p = {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 + f p, f p I p. Denote by Ip 2 the usual product of ideals. Suppose X p T p M and f Ip 2 ; by the derivation property, X p f = 0. Corollary The map X p (f X p f), where X p T p M, determines a linear inclusion T p M (I p /I 2 p ). Proof. We have a direct sum decomposition C (M) = I p R (or C). Since X p kills constants and I 2 p, it is uniquely determined by its action on (I p /I 2 p ). Definition I p /I 2 p T p 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 If (U; x 1,..., x n ) is a coordinate neighborhood of p, then { x 1 p,..., x n p } is a basis of T p M. Proof. Completely analogous to Corollary For f C (M) and p M, define df p T M as the image in T M I p /Ip 2 of f f(p) 1. If (U; x 1,..., x n ) is a coordinate neighborhood of p, then {dx 1 p,..., dx n 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, f f(p) = f (p)(x i x i (p)) +... = f (p)dx i. x i x i By counting dimensions, we conclude that Tp M = (T p M). Given a vector field X on M and p M, define X p T p M by X p f = (Xf)(p). Therefore, any vector field X is completely determined by X p T p M for all p. Conversely, given an assignment p X p T p M, there exists a vector field X whose values are the X p if and only if for every f C (M), the function p X p 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

13 Math PULLBACKS AND PUSHFORWARDS 1.5 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. If p M, then F I N,f(p) I M,p and F IN,f(p) 2 I M,p 2, so we get a linear map F : TF (p) N T p M. Dually, there is a map F : T p M T F (p) N. Also, for any f C (N), F sends df F (p) TF (p) N 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 Let F : M N be a smooth map. Two vector fields X M on M and X N on N are F -related if for every f C (N), F (X N f) = X M (F f). In other words, the following diagram commutes. C (N) X N C (N) F C (M) X M F C (M) Yet another equivalent formulation is that F X M p = X N 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 Check that if X M, X N and Y M, Y N are pairs of F -related vector fields, then [X M, Y M ] is F -related to [X N, Y N ]. 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. 13

14 Math 222 CHAPTER 1. GEOMETRIC PRELIMINARIES 14

15 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 l g : G G r g : G G h gh h hg 1. These are both diffeomorphisms by the definition of Lie group. Definition The Lie algebra g of G is the Lie algebra of all left-invariant vector fields, i.e. vector fields X such that (l g )X = X for all g G, with the Lie bracket being the commutator. Note that G-invariant means l g X h = X gh for all g, h G. This is equivalent to l g X e = X g for all g G. This gives an inclusion g T e G sending X X e (any left-invariant vector field is completely determined by its value at the identity). Exercise Show that this map is an isomorphism: g T e G. Suppose that F : G H is a homomorphism of Lie groups. Define F : g h as the composition g T e G F T e H h. Theorem The map F : g h is a homomorphism of Lie algebras. Remark If F 1, F 2 are Lie group homomorphisms that can be composed, then F 1 F 2 = (F 1 F 2 ). 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. Proof. Suppose X, Y g. Define X h by the requirement that F X e = X e T e H and similarly for Y, Ỹ. 15

16 Math 222 CHAPTER 2. THE LIE ALGEBRA OF A LIE GROUP We claim that X, X and Y, Ỹ are F -related. To see this, suppose g G. Then X F (g) = l F (g) Xe = l F (g) F X e = (l F (g) F ) X e = (F l g ) X e = F l g X e = F X g, and similarly for Y, Ỹ. Therefore, [X, Y ] and [ X, Ỹ ] are F -related, which means that for all f C (H), F ( Xf) = X(F f). Applying this to [ X, Ỹ ] = XỸ Ỹ X we see that [ X, Ỹ ] = [X, Y ]. By the same calculations as before, [X, Y ] and [X, Y ] are F -related. 2.2 Integral curves Definition 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 ϕ d dt t 0 = X ϕ(t0 ) for all t 0 I. d Equivalently, dt and X are ϕ-related. Properties of integral curves. 1. If (I, ϕ) is an integral curve, then so is (I + a, ϕ a ), where ϕ a (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 0. This is actually true for s = 0 if we interpret s 1 I = R. 3. Suppose (U; x 1,..., x n ) is a coordinate neighborhood and X is a vector field on U with local expression X = a i, x i a i C (U). Then (I, ϕ), where ϕ : I M is given by t (x 1 (t),..., x n (t)) is an integral curve if and only if dx i dt = a i ((x i (t))). The first and third facts follow directly from the chain rule. The third also follows from the chain rule: d ϕ dt = dx i. dt x i 16

17 Math INTEGRAL CURVES Therefore, finding integral curves of vector fields amounts to solving an ODE. Basic facts on ODE. Theorem (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 (Uniqueness). Suppose (I 1, ϕ 1 ) and (I 2, ϕ 2 ) are two integral curves for X, and ϕ 1 (t 0 ) = ϕ 2 (t 0 ) for some t 0 I 1 I 2. Then ϕ 1 = ϕ 2 on I 1 I 2, and hence there exists a unique integral curve (I 1 I 2, ϕ) with ϕ Ui = ϕ i. In particular, given an integral curve (I, ϕ) there exists a unique maximal extension, i.e. an integral curve (Ĩ, ϕ) such that I Ĩ and ϕ I = ϕ with Ĩ maximal. This is called a maximal integral curve. Theorem (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 ɛ > 0, and a C map Φ : U ( ɛ, ɛ) M such that (i) t Φ(, t) is an integral curve for t ( ɛ, ɛ). (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 A vector field X on M is complete if every maximal integral curve is defined on all of R. Example Let M = R 2 {(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). 2. The vector field d dx on ( 1, 1) is incomplete. Therefore, composing it with a diffeomorphism ( 1, 1) R results in an incomplete vector field on R. Proposition 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 l 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 t 0 ; without loss of generality, suppose t 0 > 0. Choose ɛ > 0 17

18 Math 222 CHAPTER 2. THE LIE ALGEBRA OF A LIE GROUP such that ɛ < t 0 and choose t 1 < t 0 with t 1 t 0 < ɛ 2. Then t ϕ(t t 1) is welldefined for t 1 t < ɛ and takes the value e at t = t 1. Therefore, t ϕ(t 1 )ϕ(t t 1 ) is an integral curve agreeing with ϕ at t = t 1 and defined on an interval containing even t 0. 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 Φ(t, p) is an integral curve for all p M. Theorem (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, p 0 ). The map t Φ(t, Φ(s, p)) is an integral curve in t with value Φ(s, p) at t = 0, as is t 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 I p = {t R Φ is smooth near (t, p)}. Then I p is open and 0 I p. To prove the theorem, we just need to show that I p = R for all p. Suppose this is not the case for some p 0 M; then the connected component of 0 in I p has some finite endpoint t 0. Without loss of generality, assume that t 0 > 0. Then we have by assumption, Φ is C (in both arguments) on a neighborhood of [0, t 0 ) {p 0 }. Pick η satisfying 0 < η < 1. Then Φ(t 0, p) = Φ(ηt 0, Φ((1 η)t 0, p)). If η is small enough, we can find an open neighborhood U of Φ(t 0, p 0 ) such that p Φ(ηt 0, p) is C for p U. Making η smaller if necessary, we can arrange that Φ((1 η)t 0, p 0 ) U because Φ is continuous in the time variable. Moreover, for p near p 0, p Φ((1 η)t 0, p) is C by the definition of t 0. Therefore, the composition p Φ(ηt 0, Φ((1 η)t 0, p)) = Φ(t 0, p) is C in p near p 0. We now show that this implies that (t, p) Φ(t, p) is smooth in a neighborhood of (t 0, p 0 ), which furnishes the desired contradiction. To see this, observe that Φ(t, p) = Φ(t t 0, Φ(t 0, p)). 18

19 Math THE EXPONENTIAL MAP By hypothesis, (t, q) Φ(t t 0, q) is smooth near t = t 0 and q = Φ(p 0, t 0 ). Since p Φ(t 0, p) is smooth, Φ(t, p) is the composition of smooth functions hence is smooth. (t, p) (t, Φ(t 0, p)) Φ(t t 0, Φ(t 0, p)), 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 Φ X : R G G satisfying Φ X (0, g) = g and t Φ X (t, g) is an integral curve for X. Therefore, we get a map Φ : R G g G (t, g, X) Φ X (t, g). Proposition 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 π 2 : M N N if all the integral curves of X lie in the fibers. Equivalently, in local coordinates (U; x 1,..., x k, x k+1,..., x n ) such that x 1,..., x k are local coordinates on M and x k+1,..., x n are local coordinates on N. The vector field X can be expressed locally as X = n i=1 a i. x i Then X is tangential to the fibers if and only if a j = 0 for j > k. In other words, for each m M and n N X (m,n) T m M {0} T m M T n 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 smooth map Φ : R M N M N by Theorem Composing with the projection M N M gives a smooth map Φ : R M N M. 19

20 Math 222 CHAPTER 2. THE LIE ALGEBRA OF A LIE GROUP Proof of Proposition Let M = G and N = g in the above discussion. Properties of Φ. (i) Φ(0, g, X) = g. (ii) t 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 This is the famous exponential map. Properties of the exponential map. (i) exp : g G is smooth. exp X = Φ(1, e, X). (ii) t 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 (Universality of the exponential map.). Suppose F : G H is a Lie group homomorphism. Then the following diagram commutes. g F h exp G F H exp 20

21 Math THE EXPONENTIAL MAP Exercise Prove this proposition. Proposition The composition is the identity. Exercise Prove this proposition. g T 0 g exp T e G g Proposition For g G and X g, f C (G), Exercise Prove this proposition. (Xf)(g) = d dt f(g exp(tx)) t=0. Corollary 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 G 0 the connected component containing the identity of G; we call this the identity component. Note that (G 0 ) 2 = {g 1 g 2 g i G 0 } and (G 0 ) 1 = {g 1 g G 0 } are connected and contain e (since we can go from a product to another product by a product of connecting paths), hence equal to G 0. This tells us that G 0 is an open, closed subgroup. Also for g G, gg 0 g 1 is a connected subgroup of G containing e. Therefore, the identity component is a normal subgroup. Now suppose U G 0 is a neighborhood of the identity. We can arrange that U = U 1 by replacing U with U U 1 if necessary. Then G 1 = U k k=1 is open (since U k is a union of translates of opens, hence open) and a subgroup of G 0. Therefore G 1 is closed, because its complement is a union of its cosets. Since G 0 was connected, it must be the case that G 1 = G 0. We thus observe that if U is a neighborhood of the identity in a Lie group G, then U generates G 0 in the algebraic sense. We have proved: Theorem Let G be a Lie group with Lie algebra g. 21

22 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 G 0 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 V 0 containing 0 in g and U e containing e in G such that exp : V U e. Let log : U e V 0 denote the inverse. By shrinking V 0 (and thus U e ), we can arrange that V 0 is -shaped about 0. Recall that this means that for any p V 0, the line between p and 0 lies in V 0. By further shrinking V 0 if necessary, we may also assume that V 0 = V 0. 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 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

23 Math 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) 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 For X, Y g, 2 s t M(sX, ty ) s=t=0 = 1 [X, Y ]. 2 Proof. For g and f C (G), Xf(g) = s f(g exp(sx)) s=0. Applying this to Y f, we have XY f(g) = s Y f(g exp(sx)) s=0. Now applying the same reasoning to Y f, we find that this is Therefore, [X, Y ]f(e) = (XY Y X)f(e) 2 s t f(g exp(sx) exp(ty )) s=t=0. = 2 s t (f(exp(sx) exp(ty )) f(exp(ty ) exp(sx))) s=t=0 = 2 s t (f exp(m(sx, ty )) f exp(m(ty, sx))) s=t=0 = 2 s t (f exp(m(sx, ty )) f exp( M( sx, ty))) s=t=0. 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) where C = 2 s t M(sX, ty ) s=t=0. We substitute the terms of order up to 2 into the expression above to obtain = 2 s t [f exp(sx + ty + stc) + f exp(sx + ty stc)] s=t=0. 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, [X, Y ]ϕ(0) = 2 s t [ϕ(sx + ty + stc) ϕ(sx + ty stc)] s=t=0 = 2 2 s t ϕ(sx + ty + stc) s=t=0 = 2 2 s t ϕ(m(sx, ty )) s=t=0. 23

24 Math 222 CHAPTER 2. THE LIE ALGEBRA OF A LIE GROUP Therefore, M(X, Y ) = X + Y [X, Y ] Corollary Multiplication on G determines the Lie bracket [, ] on g. 24

25 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 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 Note that S T S T. Define Exp : End(V ) End(V ) by T = sup T v. v 1 Exp(T ) = 25 k=0 1 k! T k.

26 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 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 The exponential map for GL(V ) (in both the real and complex cases) is given by 1 exp T = k! T k. What about the logarithm? We have Therefore, k=0 d 1 log(1 w) = dw 1 w = w k. log z = (1 z) k. k The map log : GL(V ) End(V ) which is defined near 1 V GL(V ) is given by 1 log g = (1 g)k k for formal reasons concerning power series. We now compute the Lie bracket for gl(v ) := Lie(GL(V )). If ( g = exp S exp T = 1 + S + 1 ) ( 2 S T + 1 ) 2 T ( = 1 + S + T S2 + 1 ) 2 T 2 + ST +... k=1 k=1 where we omit all terms of order 3, because they are not relevant to the calculation. Then M(S, T ) = log g = g (1 g) = S + T S T 2 + ST 1 2 = S + T [S, T ] +... k=0 ( S 2 + T 2 ST T S ) 26

27 Math THE ADJOINT REPRESENTATION This discussion proves the following proposition. Proposition 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 l g r g = ghg 1, i.e. conjugation by g. So we get a group homomorphism G Aut(G) sending g l g r g. Define Ad : G End(g) by Ad (g) = (l g r g ). Since (l g r g ) (l g 1 r g 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 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 (ad X)(Y ) = [X, Y ]. Proof. By definition of the exponential map, Y g T e G is the tangent vector at t = 0 to the curve t exp(ty ). Therefore, Adg(Y ) is the tangent vector at t = 0 to t g exp(ty )g 1. Hence adx(y ) is the tangent vector at s = 0 to the curve Ad(exp(sX))Y ). 27

28 Math 222 CHAPTER 3. LIE ALGEBRAS Therefore, for any smooth f near e G, adx(y ) g T e G, and (adx(y ))f(e) = 2 s t f(exp(sx) exp(ty ) exp( sx)) s=t=0 = 2 s t f exp(m(sx, M(tY, sx))) s=t=0 = 2 s t f exp(sx + M(tY, sx) [sx, M(tY, sx)] +...) s=t=0 = 2 s t f exp ( sx + ty sx + 1 [ty, sx] [sx, ty sx +...]) s=t=0 = 2 s t f exp (ty + st[x, Y ] +...) s=t=0. 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, ((adx(y ))ϕ)(0) = 2 s t ϕ(ty + st[x, Y ] +...) s=t=0 = ϕ([x, Y ]) = ([X, Y ]ϕ)(0). Corollary 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 Poincaré s formula Suppose X g; then g T X g. The exponential map induces On the other hand, g T X g exp X T exp X G. g T e G (l exp X) T exp X G. 28

29 Math POINCARÉ S FORMULA Since l exp( X) = (l exp X ) 1, the map (l exp X ) is invertible and threading across the diagram gives a linear map g g g = T X g exp X T exp X G g = T e G (l exp X ) 1 T exp X G One might ask, what is this map? Theorem (Poincare s formula for the differential of the exponential map). Note that exp X = (l exp X ) 1 e z z 1 e ad X ad X. = 1 z , an entire holomorphic function. So we can think of the expression 1 e adx adx convergent power series in the operator adx. as a Remark 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 = (l exp X ) D(X). (3.3.1) g = T X g exp X T exp X G g D(X) = T e G (l exp X ) 1 T exp X G The map D : g End(g) is a smooth function of X. Exercise Prove this. Define F (t) = td(tx), which is then a smooth function of t with values in End(g). We claim that F (t) = 1 g ad (X) F (t). Supposing this for the moment, define F (t) = 1 e t ad X ad X := (1 e t ad X )(ad X) 1. 29

30 Math 222 CHAPTER 3. LIE ALGEBRAS Then F (t) = e tadx = 1 F (t) ad(x) = 1 adx F (t). Also, F (0) = F (0) = 0. Therefore, F = F, 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 t f(exp(x) exp(ty )) t=0 = (D(X)Y )f(exp X). (3.3.2) To elaborate, let Y g. We are concerned with its image in T exp 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 (l 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 t f(exp(x + ty )) t=0 = Y f(exp X), the reasoning being that t 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 du f(g exp uy ) u=0. 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 (t) = 1 ad X F (t). So by the preceding equations, this has something to do with Let s calculate it in two different ways. 1. By a homework problem, 2 s t f(exp(t(x + sy ))) s=0 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 ( ) s=0 = r(y )f(exp tx) + r(d(tx)(ty ))(r(x)f)(exp(tx)). 30

31 Math POINCARÉ S FORMULA 2. Reasoning as above, we first compute s f(exp(t(x + sy ))) s=0 = r(td(tx)y )f(exp(tx)). Then differentiating with respect to t, we find that s ( ) s=0 = r( d (td(tx))y )f(exp tx) + r(x)r(td(tx)y )f(exp tx). 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 = d (td(tx))y + [X, td(tx)y ]. dt Hence d dt (td(tx)) = 1 g adx td(tx). That was a struggle... - Wilfried Schmid We re-iterate Poincaré s formula for emphasis: Now let s manipulate this identity. exp X = (l exp X ) 1 e adx adx. exp X = (r exp( X) ) (r exp(x) ) (l exp(x) ) 1 e adx adx = (r exp( X) ) Ad exp X 1 e adx adx = (r exp( X) ) exp(ad X) 1 e adx adx = (r exp( X) ) eadx 1 adx. We have shown the right-translation analogue: Corollary We have exp X = (r exp( X) ) ead X 1 ad X. 31 (universality of exp)

32 Math 222 CHAPTER 3. LIE ALGEBRAS 3.4 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, h j (adm(x, Y )) is well-defined. Define h 1 (z) = h 2 (z) = z e z 1 = z 1 e z = ( e z ) 1 1 z ( ) 1 e z 1. z As power series in z, h 1 and h 2 have radius of convergence 2π. Therefore, for X < 2π C, both h 1(adX) and h 2 (ad(x)) are well-defined. So with this choice of V, h j (adm(x, Y )) is well-defined for all X, Y V. Since V is open, for X and Y in V there exists ɛ > 0 such that t < 1 + ɛ = tx, ty V. Define F (t) = M(tX, ty ). Then the map F : ( 1 ɛ, 1 + ɛ) g is smooth. Lemma For X, Y V and s, t < 1 + ɛ we have the following identities: s M(sX, ty ) = h 1(ad M(sX, ty ))X t M(sX, ty ) = h 2(ad M(sX, ty ))Y Proof. Suppose f is a C function defined near exp sx exp ty. Consider f(exp sx exp ty ) = t u f(exp sx exp ty exp uy ) u=0. Thinking of Y as a left-invariant vector field, this is equal to Y f(exp sx exp ty ). 32

33 Math THE CAMPBELL-BAKER-HAUSDORFF FORMULA Just by the chain rule, we have t f(exp sx exp ty ) = exp M(sX,tY ) M(sX, ty ) f. } t {{} g T eg }{{} T exp M G Now applying the formula for the differential of the exponential map, we get Considering 1 e adm(sx,ty ) adm(sx,ty ) ((l exp sx exp ty ) 1 e adm(sx,ty ) ) M(sX, ty ) f. adm(sx, ty ) t }{{} g T eg t same as ( 1 e adm(sx,ty ) That tells us that adm(sx, ty ) Y = 1 e adm(sx,ty ) adm(sx, ty ) as left-invariant vector fields. We conclude that M(sX, ty ) as a left-invariant vector field, this is the ) M(sX, ty ) f(exp sx exp ty ). t t M(sX, ty ) = h 2(adM(sX, ty )) 1 M(sX, ty ). t t M(sX, ty ) = h 2(adM(sX, ty ))Y. 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 g 1. In any case, there is no real difference between left and right. Corollary With the notation above, F (t) = h 1 (ad F (t))x + h 2 (ad F (t))y. Proof. F (t) = tm(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 F (t) = t k M k (X, Y ). k=1 Differentiating and applying the corollary, ( ) ( ) kt k 1 M k (X, Y ) = h 1 t l adm l (X, Y ) X + h 2 t l adm l (X, Y ) Y. k=1 l=1 33 l=1 (3.4.1)

34 Math 222 CHAPTER 3. LIE ALGEBRAS Now, h 1 (z) = = z e z 1 = z z + z2 2 + z z = 1 z , and similarly h 2 (z) = z 1 e z = z z z = 1 + z So h 1 (0) = h 2 (0) = 1 and h 1 (0) = 1 2, h 2 (0) = 1 2. 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 t k. First considering only the linear term of h 1, we get (k + 1)M k+1 (X, Y ) = 1 2 [X, M k(x, Y )] 1 2 [Y, M k(x, Y )] +... Since the series inside the arguments of the h j have no constant term, the other contributions to t k come from linear combinations of terms adm j1 (X, Y )adm j2 (X, Y )... adm jn (X, Y ) with 1 j 1, j 2,..., j n < k such that j j n = k. Definition 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 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 (Campbell-Baker-Hausdorff). There exist universal Lie polynomials M k (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 ) = M k (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

35 Math THE CAMPBELL-BAKER-HAUSDORFF FORMULA Remark 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 Poincaré. 4. Baker came later, around Hausdorff (1906) gave the modern formulation, in terms of Lie polynomials, with arguments essentially the same as ours. Proof of Theorem At this point, the only issue left is that of convergence. The differential equation F (t) = h 1 (adf (t))x + h 2 (adf (t))y. (3.4.2) is a holomorphic ordinary differential equation, since we know that h 1 (adf (t)) and h 2 (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 g C := 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 F (t) := t k M k (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 h 1, h 2 have Taylor series with radius of convergence 2π, so by the existence and uniqueness of solutions of ODE, sup ad F (t) 2π t <r since the differential equation (3.4.2) is sensible as long as ad F (t) < 2π. By our choice of C, sup F (t) 2π t <r C. So there exists r 1 with 0 < r 1 < r such that sup t r 1 F (t) = π C 35

36 Math 222 CHAPTER 3. LIE ALGEBRAS since the sup of F (t) on disks depends continuously on the radius of the disk, and must assume any value less than 2π C. Because h 1, h 2 have Taylor series with radius of convergence 2π, D = max{ a k π k, b k π k } < where h 1 (t) = a k t k and h 2 = b k t k. Then for t < r 1, So Hence F (t) π CD ad F (t) π = F (t) D( X + Y ). t So r > r 1 C and D), we have r > 1, and thus 0 ( F r t ) dr Dr 1 ( X + Y ) for t r 1. t sup F (t) = π t r 1 C Dr 1( X + Y ). 1 X + Y. For X and Y sufficiently small (depending only on M(X, Y ) = F (1) = M k (X, Y ). 36

37 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 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 E x := p 1 (x) for each x M, such that x has an open neighborhood U fitting into a commutative diagram p 1 (U) U R k U p id which preserves the vector space structure (resp. replacing R k by C k ). This is called a local trivialization of the vector bundle. Definition A section of E over U is a C map s : U p 1 U such that p s = 1 U. Sections can be added or multiplied by a scalar, so the sections over U comprise a vector space. Definition A frame for E over U is a k-tuple of sections (s 1,..., s k ) which at each x U provide a basis for E x. 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 U π

38 Math 222 CHAPTER 4. GEOMETRY OF LIE GROUPS Example Let T M be the tangent bundle of M and T M be the cotangent bundle. These are vector bundles: if (U; x 1,..., x n ) is a coordinate neighborhood, then { x 1,..., x n } is a local frame for T M and {dx 1,..., dx n } is a local frame for T M. Sections of the tangent bundle are vector fields; sections of the cotangent bundle are 1-forms. Definition Let p : E M be a vector bundle of rank k. A rank l sub-bundle of E is a rank l vector bundle p F : F M together with a commutative diagram F E M p F id 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 s 1,..., s k for E such that s 1,..., s l is a local frame for F. 4.2 Frobenius Theorem Consider a rank k sub-bundle E T M. For each x M, E x is a k-dimensional subspace of T x M. Definition An integral submanifold for E is a k-dimensional submanifold S M such that for each x S, T x S = E x T x M. Definition 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 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 E x T x M. The bundle E is involutive if for any two such vector fields taking values in E, so does their Lie bracket. Definition The tangent sub-bundle E is integrable if each point in M has a coordinate neighborhood (U; x 1,..., x n ) such that x 1,..., x k constitute a local frame for E. Note that in this situation, 1. The set {x k+1 = c k+1,..., x n = c n } where the c i are constant, are the integral submanifolds for E. 2. E is involutive, because k k a j, x j j=1 j=1 b j = x j M k i,j=1 p b i a j x j x i k i,j=1 a i b j. x j x i 38

39 Math FROBENIUS THEOREM Theorem (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 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 Y 1, Y 2 defined near m and taking values in E can be expressed as Y i = f i X. [Y 1, Y 2 ] = (f 1 Xf 2 f 2 Xf 1 )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 : ( ɛ, ɛ) U M such that for all u U, t Φ(t, u) is an integral curve of X and Φ(0, u) = u. After shrinking U more if necessary, we can introduce coordinates x 1,..., x n on U such that x j (m) = 0. Then after making a linear change of coordinates, we can arrange that X m = m. x 1 Define a C map from a neighborhood V of 0 R n to M by (y 1,..., y n ) Φ(y 1 ; 0, y 2,..., y n ). 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 y 1,..., y n 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 = 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 ( 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 X m 0. Then there exist local coordinates x 1,..., x n defined near m such that X = 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 x 1 x 1