Lie groups, Lie algebras, and their representations

Transcription

1 Lie groups, Lie algebras, and their representations Gwyn Bellamy gbellamy September 16, 2016 Abstract These are the lecture notes for the 5M reading course Lie groups, Lie algebras, and their representations at the University of Glasgow, autumn 2015.

2 Contents 1 Introduction i 2 Manifolds - a refresher 2 3 Lie groups and Lie algebras 11 4 The exponential map 20 5 The classical Lie groups and their Lie algebras 30 6 Representation theory 35 7 The structure of Lie algebras 40 8 Complete reducibility 48 9 Cartan subalgebras and Dynkin diagrams The classification of simple, complex Lie algebras Weyl s character formula Appendix: Quotient vector spaces 76

3 1 Introduction Lie groups and Lie algebras, together called Lie theory, originated in the study of natural symmetries of solutions of differential equations. However, unlike say the finite collection of symmetries of the hexagon, these symmetries occurred in continuous families, just as the rotational symmetries of the plane form a continuous family isomorphic to the unit circle. The theory as we know it today began with the ground breaking work of the Norwegian mathematician Sophus Lie, who introduced the notion of continuous transformation groups and showed the crucial role that Lie algebras play in their classification and representation theory. Lie s ideas played a central role in Felix Klein s grand Erlangen program to classify all possible geometries using group theory. Today Lie theory plays an important role in almost every branch of pure and applied mathematics, is used to describe much of modern physics, in particular classical and quantum mechanics, and is an active area of research. You might be familiar with the idea that abstract group theory really began with Galois work on algebraic solutions of polynomial equations; in particular, the generic quintic. But, in a sense, the idea of groups of transformations had been around for a long time already. As mentioned above, it was already present in the study of solutions of differential equations coming from physics (for instance in Newton s work). The key point is that these spaces of solutions are often stable under the action of a large group of symmetries, or transformations. This means that applying a given transformation to one solution of the differential equation gives us another solution. Hence one can quickly and easily generate new solutions from old one, buy one, get one free. As a toy example, consider a black hole centred at (0, 0) in R 2. Then, any particle on R 2 {(0, 0)}, whose position at time t is given by (x(t), y(t)), will be attracted to the origin, the strength of the attraction increasing as it nears (0, 0). This attraction can be encoded by the system of differential equations dx dt = x (x 2 + y 2 ), dy 3/2 dt = In polar coordinates (r(t), θ(t)), this becomes y. (1) (x 2 + y 2 ) 3/2 dr dt = 1 r. (2) 2 i

4 The circle S 1 = {0 ϑ < 2π} (this is a group, ϑ 1 ϑ 2 = (ϑ 1 +ϑ 2 ) mod 2π) acts on R 2 by rotations: ( ) ( ) ( ) cos ϑ sin ϑ x x cos ϑ y sin ϑ ϑ (x, y) = =. sin ϑ cos ϑ y x sin ϑ + y cos ϑ Then it is clear from (2) that if (x(t), y(t)) is a solution to (1), then so too is ϑ (x(t), y(t)) = (x(t) cos ϑ y(t) sin ϑ, x(t) sin ϑ + y(t) cos ϑ). Thus, the group S 1 acts on the space of solutions to this system of differential equations. The content of these lecture notes is based to a large extent on the material in the books [5] and [8]. Other sources that treat the material in these notes are [1], [2], [4], [9] and [7]. 1

5 2 Manifolds - a refresher In this course we will consider smooth real manifolds and complex analytic manifolds. 2.1 Topological terminology Every manifold is in particular a topological space, therefore we will occasionally need some basic topological terms. We begin by recalling that a topological space is a pair (X, T ) where X is a set and T is a collection of subsets of X, the open subsets of X. The collection of open subsets T must include X and the emptyset, be closed under arbitrary unions and finite intersections. Recall that a topological space M is disconnected if there exist two non-empty open and closed subsets U, V M such that U V = ; otherwise M is connected. It is path connected if, for any two points x, y X, there is a path from x to y (every path connected space is connected, but the converse is not true). We say that X is simply connected if it is path connected and the fundamental group π 1 (X) of X is trivial i.e. every closed loop in X is homotopic to the trivial loop. The space X is compact if every open covering admits a finite subcover. The space X is said to be Hausdorff if for each pair of distinct points x, y X there exists open sets x U, y V with U V =. For each l R + and x R n, we denote by B l (x), resp. B l (x), the set of all points y R n such that x y l, resp. x y < l. We will always consider R n as a topological space, equipped with the Euclidean topology i.e. a base for this topology is given by the open balls B l (x). Recall that the Heine-Borel Theorem states: Theorem 2.1. Let M be a subset of R n. Then, M is compact if and only if it is closed and contained inside a closed ball B l (0) for some l 0. Remark 2.2. If M is some compact subspace of R n and {X i } i N a Cauchy sequence in M, then the limit of this sequence exists in M. This fact is a useful way of showing that certain subspaces of R n are not compact. 2.2 Manifolds Recall that a real n-dimensional manifold is a Hausdorff topological space M with an atlas A(M) = {φ i : U i V i i I}, where {U i i I} is an open cover of M, each chart φ i : U i V i is a homeomorphism onto V i, an open subset of R n, and the composite maps φ i φ 1 j : V i,j V j,i 2

6 are smooth morphisms, where V i,j = φ j (U i U j ) R n and V j,i = φ i (U i U j ) R n. Example 2.3. The circle S 1 = {(x 1, x 2 ) x 2 1+x 2 2 = 1} R 2 is a manifold. We can use stereographic projection to construct charts on S 1. Let N = (0, 1) and S = (0, 1) be the north and south poles respectively. N P = (x 1, x 2 ) Q = (z, 0) S A line through N and P meets the x 1 -axis in a unique point. This defines a map S 1 {N} R. To get an explicit formula for this, the line through N and P is described by {t(x 1, x 2 ) + (1 t)(0, 1) t R}. This implies that z = x 1 1 x 2. Therefore we define U 1 = S 1 {N}, V 1 = R and φ 1 : U 1 V 1, φ 1 (x 1, x 2 ) = x 1 1 x 2. Similarly, if we perform stereographic projection from the south pole, then we get a chart where U 2 = S 1 {S} and V 2 = R. φ 2 : U 2 V 2, φ 2 (x 1, x 2 ) = x 1 x 2 + 1, For this to define a manifold, we need to check that the transition map φ 2 φ 1 1 is smooth. Since U 1 U 1 = S 1 {N, S}, φ 1 (U 1 U 2 ) = φ 2 (U 1 U 2 ) = R {0}. Hence φ 2 φ 1 1 : R {0} R {0}. A direct calculation shows that φ 2 φ1 1 (z) = 1. This is clearly smooth. z A continuous function f : M R is said to be smooth if f φ 1 i : V i R is a smooth, i.e. infinitely differentiable, function for all charts φ i. The space of all smooth functions on M is denoted C (M). Example 2.4. The real projective space RP n of all lines through 0 in R n+1 is a manifold. The set of points in RP n are written [x 0 : x 1 : : x n ], where (x 0, x 1,..., x n ) R n+1 {0} and [x 0 : x 1 : : x n ] represents the line through 0 and (x 0, x 1,..., x n ). Thus, for each λ R, [x 0 : x 1 : : x n ] = [λx 0 : λx 1 : : λx n ]. 3

7 The n + 1 open sets U i = {[x 0 : x 1 : : x n ] RP n x i 0} cover RP n. The maps φ i : U i R n, [x 0 : x 1 : : x n ] ( x 0 x i,..., x i,..., xn x i ) define an atlas on RP n. Thus, it is a manifold. Exercise 2.5. Check that the maps φ i are well-defined i.e. only depends on the line through (x 0, x 1,..., x n ). Prove that RP n is a manifold by explicitly describing the maps φ j φ 1 i U j ) R n φ j (U i U j ) R n and checking that they are indeed smooth map. The following theorem is extremely useful for producing explicit examples of manifolds. : φ i (U i Theorem 2.6. Let m < n and f : R n R m, f = (f 1 (x 1,..., x n ),..., f m (x 1,..., x n )), a smooth map. Assume that u is in the image of f. Then f 1 (u) is a smooth submanifold of R n, of dimension n m if and only if the differential ( fi d v f = x j ) i=1,...,m j=1,...,n x=v : R n R m is a surjective linear map for all v f 1 (u). The proof of the above theorem is based on the inverse function theorem, which implies that for each v in the closed set f 1 (u), there is some l > 0 such that B l (v) f 1 (u) B l (v) for some (n m)-dimensional ball B l (v). This implies that f 1 (u) is a submanifold of R n of dimension (n m). A proof of this theorem can be found in [10, Corollary 1.29]. Exercise 2.7. By considering the derivatives of f = x x 2 n 1, show that the (n 1)-sphere S n 1 R n is a manifold. Exercise 2.8. Consider the smooth functions f 1 = x x 3 2 x 1 x 2 x 3 1, f 2 = x 2 1x 2 x 3 and f 3 = (2x 2 x 3 3)e x 1 from R 3 to R. For which of these functions is the differential d v f i always surjective for all v f 1 i (0)? For those that are not, the closed subset f 1 i (0) is not a submanifold of R 3. Remark 2.9. Since every point in a manifold admits an open neighborhood homeomorphic to an open subset of R n, for some n, they are especially well-behaved topological spaces. We mention, in particular, that a connected manifold is path connected and always admits a universal cover. 2.3 Tangent spaces The tangent space at a point of manifold is extremely important, playing a key role when trying to compare manifolds, study functions on the manifold etc. Intuitively, I hope it is clear what the tangent space at a point should be - it s simply all vectors tangent to the manifold at that point. Unfortunately, to make this mathematically precise, some of the intuition is lost in the 4

8 technicalities. Remarkably, there are several equivalent definitions of the tangent space T m M to M at the point m M. In this course we will see three of these definitions, and show that they are equivalent. The first definition, which we will take as the definition is in terms of point derivations. Definition Fix a point m M. A point derivation at m is a map ν : C (M) R such that the following two properties are satisfied 1. ν is linear i.e. ν(αf + βg) = αν(f) + βν(g) for α, β R and f, g C (M). 2. It satisfies the derivation rule: ν(fg) = ν(f)g(m) + f(m)ν(g), f, g C (M). One can easily check that if ν, µ are point derivations at m and α R, then ν + µ and αν are point derivations at m. Thus, the set of point derivations forms a vector space. The tangent space to M at m is defined to be the vector space T m M of all point derivations at m. To get some intuition for this notion, lets first consider the case where the manifold M is some just R n. For a = (a 1,..., a n ) R n, we define the point derivation v a at u by I claim that a v a defines an isomorphism R n v a (f) = a 1 f x 1 x=u + + a n f x n x=u. T u R n. One can easily check that v a is a point derivation. Also, v a (x i ) = a i, which implies that the map is injective. Therefore the only thing to check is that every point derivation at u can be written as v a for some a. Let ν be an arbitrary point derivation and set a i = ν(x i ) R. Then if a = (a 1,..., a n ), we need to show that ν v a = 0. It is certainly zero on each of the coordinate functions x i. Locally, every smooth function has a Taylor expansion f(x) = k 1,...,k n 0 k f x k 1 1 x k (x 1 1 u 1 ) k1 (x n u n ) kn. 2 x kn x=u n Using the derivation rule for point derivations it is easy to see that ν(f) v a (f) = 0 too. Thus, we can think of x 1,..., x n as being a basis of T u R n for any u R n. In particular, it is clear that dim T u R n = n for all u R n. Remark The tangent space T m M at m is a locally property of M i.e. it only sees what happens locally on M near m. This statement can be made precise as follows. Let U M be 5

9 an open neighborhood of m. Then, the fact that any smooth function f C (M) restricts to a smooth function on U i.e. f U C (U) means that we can define a canonical map T m U T m M, ν ν by ν(f) := ν(f U ). This map is an isomorphism. In order to prove this, one needs to use the existence of partitions of unity, which in this case imply that every function g C (U) can be extended to a smooth function on M i.e. for each g C (U) there exists f C (M) such that f U = g. Since a manifold locally looks like R n, and the tangent space at m only sees what happens around m, it is not surprising that: Proposition If M is an n-dimensional manifold, then dim T m M = n for all m M. Proof. To prove the proposition, we will show that a chart ϕ : U R n around m defines an isomorphism of vector spaces ϕ : T m M T ϕ(m) R n. The definition of ϕ is very simple. Given ν T m M a point derivation and f C (R n ) a function, we define ϕ (ν)(f) := ν(f ϕ). We leave it to the reader to check that ϕ (ν) T ϕ(m) R n is a point derivation. To show that it is an isomorphism, it suffices to note that we also have a map ϕ 1 : Im ϕ U and hence we can define a map (ϕ 1 ) : T ϕ(m) R n T m. Unpacking the definitions, one sees that (ϕ 1 ) ϕ = id and ϕ (ϕ 1 ) = Id, as required. For instance, (ϕ (ϕ 1 ) )(ν)(f) = ϕ ((ϕ 1 ) (ν))(f) = (ϕ 1 ) (ν)(f ϕ) = ν(f ϕ ϕ 1 ) = ν(f). The second definition of tangent space uses the notion of embedded curves. A curve on M is a smooth morphism γ : ( ɛ, ɛ) M, where ɛ R >0 { }. We say that γ is a curve through m M if γ(0) = m. Given a curve γ through m, we can construct a point derivation γ at m by the simple rule γ(f) = d dt (f γ) t=0. The key point here is that f γ is just a function ( ɛ, ɛ) R, which we can easily differentiate. Let s consider the case where M = R n. If ρ : ( ɛ, ɛ) R n is a curve in R n then we can differentiate 6

10 m γ Figure 1: A curve through m and its point derivation. it and get a vector ρ (0) T ρ(0) R n = R n. Concretely, if ρ(t) = (ρ 1 (t),..., ρ n (t)), then ρ (0) is the point derivation ρ (0) = n i=1 ( ) dρi dt t=0 (3) x i at ρ(0). For instance, consider ρ : R R 3, ρ(t) = (t 2, 3t, 2 sin t), then ρ (0) = 3 x x 1. Now the question becomes: when do two curves through m define the same point derivation? Well we see from the definition that if γ 1 γ 2 if and only if d dt (f γ 1) = d t=0 dt (f γ 2), t=0 f C (M), then γ 1 γ 2 γ 1 = γ 2 T m M. Denote by [γ] the class of curves through m that are equivalent to γ. The claim is that, as a set at least, the tangent space T m M at m can be identified with the equivalence classes of curves through m, under the above equivalence relation. By construction, there is an injective map from the set of equivalence classes to T m M. So we just need to show that it is surjective i.e. Lemma For any ν T m M, there exists a curve γ through m such that γ = ν. Proof. Recall from the proof of Proposition 2.12 that, give a chart ϕ : U R n around m, we constructed an isomorphism ϕ : T m M T ϕ(m) R n, where (ϕ ν)(f) = ν(f ϕ). Let s assume that we can find a curve γ : ( ɛ, ɛ) R n through ϕ(m) such that γ = ϕ ν. Then let µ = ϕ 1 γ. We 7

11 have µ(f) = d dt (f µ) t=0 = d dt (f ϕ 1 ) γ t=0 = γ(f ϕ 1 ) = (ϕ ν)(f ϕ 1 ) = ν(f ϕ 1 ϕ) = ν(f). Thus, it suffices to assume that M = R n and m = (m 1,..., m n ) R n. In this case we have seen that ν = a i + + a n x 1 x n for some a i R. Let γ be the curve γ(t) = (a 1 t + m 1,..., a n t + m n ). Then γ(0) = m and equation (3) shows that γ = ν. Putting all the tangent spaces into a single family, we get Definition The tangent bundle of M is the set T M = {(m, v) m M, v T m M}. The tangent bundle T M is itself a manifold and comes equipped with smooth morphisms i : M T M, i(m) = (m, 0) and π : T M M, π(m, v) = m. Exercise If M = f 1 (0) R n, where f : R n R m, then the tangent space to M at m is the subspace Ker(d m f : R n R m ) of T m R n = R n. Describe the tangent space to S n 1 = f 1 (0) in R n at (1, 0,..., 0), where f = x x 2 n 1. The differential of a function f C (M), at a point m, is a linear map d m f : T m M R. In terms of the first definition of a tangent space, d m f([γ]) = (f γ) (0). In terms of the second definition of a tangent space, if X i T φi (m)r n, then f φ 1 i : V i R. Differentiating this function gives d φi (m)(f φ 1 i ) : R n R and we define (d m f)([x i ]) = d φi (m)(f φ 1 i )(X i ). Of course, one must check that both these definitions are actually well-define i.e. independent of the choice of representative of the equivalence class. Let f : M N be a smooth map between manifolds M and N. Then, for each m M, f defines a linear map d m f : T m M T f(m) M between tangents spaces, given by (d m f)([γ]) = [f γ]. Since we get one such map for all points in m and they vary continuously over M, we actually get 8

12 a smooth map df : T M T N, (df)(m, [γ]) = (f(m), d m f([γ])). The following fact, which is a consequence of the inverse function theorem, will be useful to us later. Let f : M N be a smooth map between manifolds M and N such that the differential d m f is an isomorphism at every point m M. If N is simply connected and M connected then f is an isomorphism. 2.4 Vector fields A vector field is a continuous family of vectors in the tangent bundle i.e. it is a rule that assigns to each m M a vector field X m T m M such that the family {X m } m M varies smoothly on M. The notion of vector field will be crucial later in relating a Lie group to its Lie algebra. Definition A vector field on M is a smooth morphism X : M T M such that π X = id M. The space of all vector fields on M is denoted Vect(M). The key point of defining a vector field is that one can differentiate functions along vector fields. Let X be a vector field on M and f : M R a smooth function. We define X(f)(m) = (f γ) (0) for some (any) choice of curve through m such that [γ] = X m. Lemma The vector field X defines a map C (M) C (M) satisfying the product rule X(fg) = X(f)g + fx(g), f, g C (M). (4) Proof. Let f, g be smooth maps and m M. Then, X(fg)(m) = ((fg) γ) (0) = ((f γ)(g γ)) (0) = (f γ)(0)(g γ) (0) + (f γ) (0)(g γ)(0) f(m)x(g)(m) + X(f)(m)g(m). Hence X(fg) = X(f)g + fx(g). A linear map C (M) C (M) satisfying equation (4) is called a derivation. Thus, every vector field defines a derivation. The converse is also true - every derivation defines a unique vector field on M (we won t need this fact though). An equivalent definition of the action of a vector field is that X(f) is the function on M whose value at m equals (d m f)(x m ). Once should think of vector fields, or derivations, as being continuous families of point derivations. 9

13 Exercise Let M, N and P be manifolds and f : M N, g : N P smooth maps. Show that the linear map d m (g f) : T m M T g(f(m)) P equals (d f(m) g) (d m f). Hint: Using the first definition of the tangent space, this is virtually a tautology. Exercise Since S 2 R 3, we can define the function f : S 2 R by saying that it is the restriction of 2x 1 x 2 2 +x 1 x 3. Recall the description of the tangent space T (1,0,0) S 2 given in exercise Describe d (1,0,0) f : T (1,0,0) S 2 R. Similarly, let g : R 3 R 2 be the function f = (f 1 (x 1, x 2, x 3 ), f 2 (x 1, x 2, x 3 )), where f 1 (x 1, x 2, x 3 ) = x 3 2x 1 x 3 sin x 1 and f 2 (x 1, x 2, x 3 ) = e x 3 x 2 cos x 2. What is is the linear map d (0,π,1) g : R 3 R 2? 2.5 Integral curves Let X be a vector field on a manifold M and fix m M. An integral curve (with respect to X) γ : J M through m is a curve such that γ(0) = m and (d x γ)(1) = X γ(x) for all x J. Then γ is a solution to the equation d dt γ(t) t=x = X γ(x) for all x J. By choosing a chart containing m, the problem of finding an integral curve through m is easily seen to be equivalent to solving a system of linear, first order differential equations. Therefore the fundamental theorem of ordinary differential equations says that there exists some ɛ > 0 and an integral curve γ : ( ɛ, ɛ) M. Moreover, γ is unique. One can try to make the open set J R as large as possible. There is a unique largest open set on which γ exists; if J is this maximal set then γ : J M is called the maximal integral curve for X through m. Definition A vector field X on M is said to be complete if, for all m M, the maximal integral curve through m with respect to X is defined on the whole of R. Exercise Let X be the vector field x 1 x 1 (2x 1 + 1) γ for X through (a, b) R 2. Is X complete? x 2 on R 2. Construct an integral curve 2.6 Complex analytic manifolds All the above definitions and results hold for complex analytic manifolds, where M is said to be a complex analytic manifold if the atlas consists of charts into open subsets of C n such that the transition maps φ i φ 1 j are biholomorphic. All maps between complex analytic manifolds are holomorphic e.g. holomorphic vector fields or C hol (M), the spaces of holomorphic functions on M. 10

14 3 Lie groups and Lie algebras In this section we introduce the stars of the show, Lie groups and Lie algebras. 3.1 Lie groups Let (G, m, e) be a group, where m : G G G is the multiplication map and e G the identity element. Definition 3.1. The group (G, m, e) is said to be a Lie group if G is a manifold such that both the multiplication map m, and inversion g g 1, are smooth maps G G G, and G G respectively. We drop the notation m and simply write gh for m(g, h) if g, h G. Example 3.2. Let S 1 C be the unit circle. Multiplication on C restricts to S 1 S 1 S 1, making it a Lie group. It is the group of rotations of the real plane. Example 3.3. The set of invertible n n matrices GL(n, R) is an open subset of R n2 and hence a manifold. Matrix multiplication and taking inverse are smooth maps. Therefore GL(n, R) is a Lie group. Example 3.4. Every finite group can be considered as a zero-dimensional Lie group. Example 3.5. Let SO(3) denote the set of all 3 by 3 real matrices A such that AA T = 1 and det(a) = 1. It is clear that this set is a group under the usual multiplication of matrices. 3.2 Morphisms of Lie groups A morphism φ : G H between Lie groups is a group homomorphism which is also a smooth map between manifolds. Exercise 3.6. For g G, let L g : G G be the map L g (h) = gh of left multiplication. 1. Using the fact that i g : G G G, i g (h) = (g, h), is a smooth map, show that L g : G G is a smooth map. 2. Using part (1), show that if U G is an open subset and g G, then gu is also open in G. 3. Similarly, show that if C G is a closed subset then gc is also closed in G. 11

15 Hints: In part 1. use the fact that the composite of two smooth maps is smooth. In parts 2. and 3. recall that a map f : X Y between topological spaces is continuous if and only if f 1 (U) is open, resp. f 1 (C) is closed, in X for all U Y open, resp. C Y closed. Assume that the Lie group G is connected. Then, as the following proposition shows, one can tell a great deal about the group by considering the various neighborhoods of the identity. Proposition 3.7. Let G be a connected Lie group and U an open neighborhood of the identity. Then the elements of U generate G. Proof. Recall that G connected implies that the only non-empty closed and open subset of G is G itself. Since the map g g 1 is smooth, U 1 = {g 1 g U} is open in G. Thus, U U 1 is also open. It is non-empty because e U U 1. Replacing U by this intersection we may assume that g 1 U if and only if g U. By exercise 3.6, if g U then gu is open in U. Hence U U = g U gu is open in G. This implies by induction that H = {g 1 g k g i U} is an open subset of G. But it s easy to check that H is also a subgroup of G. Therefore, to show that H = G, it suffices to show that H is closed in G. Let C = G H. Since H is open in G, C is closed. We assume that C. Notice that if g C, then gh C and hence H g 1 C, a closed subset of G. Thus, H is contained in the intersection C = g C g 1 C. The arbitrary intersection of closed sets is closed, thus C is closed. Hence it suffices to show that H = C. If f C H then, in particular, f (f 1 ) 1 (G H) i.e. there exists g G H such that f = (f 1 ) 1 g = fg. But this implies that g = e belongs to G H; a contradiction. In particular, if f : G H is a morphism of Lie groups with G connected then Proposition 3.7 implies that f is uniquely defined by what it does on neighborhoods of the identity. Taking smaller and smaller neighborhoods of e, one eventually arrives at the tangent space of G at e and the map d e f : T e G T e H. Remarkably this linear map captures all the information about the original morphism f, Theorem 3.8. Let G and H be Lie groups, with G connected. Then a morphism f : G H is uniquely defined by the linear map d e f : T e G T e H. What this really means is that if f, g : G H are morphisms of Lie groups then f = g if and only if d e f = d e g. The proof of Theorem 3.8 is given in section 4; see Corollary Exercise 3.9. Let G = R (=: R {0}), where the multiplication comes from the usual multiplication on R. Show that the map φ : G G, φ(x) = x n is a homomorphism of Lie groups. What is T e G? Describe d e φ : T e G T e G. 12

16 Naturally, one can ask, as a converse to Theorem 3.8, which linear maps T e G T e H extend to a homomorphism of groups G H? Surprisingly, there is a precise answer to this question. But before it can be given we will need to introduce the notion of Lie algebras and describe the Lie algebra that is associated to each Lie group. 3.3 The adjoint action In this section we will define the Lie algebra of a Lie group. The idea is that geometric objects are inherently non-linear e.g. the manifold M R 3 defined by the non-linear equation x 5 +y 5 z 7 = 1. The same applies to Lie groups. But humans don t seem to cope very well with non-linear objects. Therefore Lie algebras are introduced as linear approximations to Lie groups. The truly remarkable thing that makes Lie theory so successful is that this linear approximation captures a great deal (frankly unjustifiable) of information about the Lie group. We begin with automorphisms. An automorphism φ of a Lie group is an invertible homomorphism G G. Exercise Show that the set Aut(G) of all automorphisms of G forms a group. An easy way to cook up a lot of automorphisms of G is to make G act on itself by conjugation. Namely, for each g G, define Ad(g) Aut(G) by Ad(g)(h) = ghg 1. This defines a map G Aut(G), g Ad(g), called the adjoint action. Exercise Check that Ad is a group homomorphism. If φ Aut(G) belongs to the image of Ad, we say that φ is an inner automorphism of G. We may also consider Ad as a smooth map G G G, Ad(g, h) = ghg 1. The key reason for introducing the adjoint action is that it fixes the identity i.e. Ad(g)(e) = e for all g G. If we return to the setting of Theorem 3.8, then the homomorphism f : G H also sends e G to e H. Moreover, the diagram G f H (5) Ad(g) G f H Ad(f(g)) is commutative; that is, Ad(f(g))(f(u)) = f(ad(g)(u)) for all u G. 13

17 So we can being our linear approximation process by differentiating diagram (5) at the identity, to get a commutative diagram of linear maps T e G d ef T e H (6) d e Ad(g) T e G d ef T e H d e Ad(f(g)) However, if we want to check that d e f really is the differential of some homomorphism f : G H then (6) doesn t really help because the right vertical arrow is d e Ad(f(g)) and we would need to know the value f(g) for g e i.e. we still need to see what f is doing away from the identity element. To overcome this problem, we will use the different interpretation of Ad as a map G G G, (g, h) ghg 1. Just as in diagram 5, we get a commutative diagram G G f f H H (7) Ad G f H. Ad The temptation now is just to differentiate this diagram at (e, e). But this is not quite the right thing to do. Instead, we differentiate each entry of Ad, resp. of f f, separately to get a bilinear map. You may not have see the definition of bilinear before. As a remainder, Definition Let k be a field and U, V and W k-vector spaces. A map b : U V W is said to be bilinear if both b(u, ) : V W and b(, v) : U W are linear maps, u U, v V, i.e. b(αu 1 + βu 2, γv 1 + δv 2 ) = αγb(u 1, v 1 ) + αδb(u 1, v 2 ) + βγb(u 2, v 1 ) + βδb(u 2, v 2 ), for all u 1, u 2 U, v 1, v 2 V, α, β, γ, δ k. Remark The bidifferential: if M, N and K are manifolds and f : M N K is a smooth map, the the bidifferential b (m,n) f of f at (m, n) M N is a bilinear map T m M T n N T f(m,n) K. Informally, one first fixes m M to get a smooth map f m : N K. Differentiating at n, we get a linear map d n f m : T n N T f(m,n) K. Then we fix w T n N and define f w : M T K by f (m) = d n f m (w). Differentiating again we get d m f w : T m M T f(m,n) K and hence b (m,n) f : T m M T n N T f(m,n) K given by (v, w) (d m f w)(v). If one differentiates along M first and then along N then we get the same bilinear map. This follows from the fact that 14 2 f x i x j = 2 f x j x i

18 for a smooth map f : R n R. Thus, we get a commutative diagram of bilinear maps T e G T e G b (e,e)f f T e H T e H (8) b (e,e) Ad T e G d ef T e H. b (e,e) Ad Notice that only the biderivative of f at the identity appears in the above diagram. We have no need to know what f does away from the identity. To make the notation less cumbersome we fix g = T e G and h = T e H. Then [, ] G := b (e,e) Ad : g g g is a bilinear map. This is the Lie bracket on the vector space g. Thus, we have shown Proposition Let f : G H be a morphism of Lie groups. Then, the linear map d e f : g h preserves brackets i.e. (d e f)([x, Y ] G ) = [(d e f)(x), (d e f)(y )] H, X, Y g. (9) This leads us to the following key result, which is one of the main motivations in the definition of Lie algebras. Theorem Let G and H be Lie groups, with G simply connected. Then a linear map g h is the differential of a homomorphism G H if and only if it preserves the bracket, as in (9). The proof of Theorem 3.15 will be given in section 4; see Theorem Example As an example to keep some grasp on reality, we ll consider the case G = GL(V ), where V is some n-dimensional real vector space (so V R n ). Then, for matrices A, B GL(V ), Ad(A)(B) = ABA 1 is really just naive matrix conjugation. For G = GL(V ), the tangent space T e GL(V ) equal End(V ), the space of all linear maps V V (after fixing a basis of V, End(V ) is just the space of all n by n matrices over R so that End(V ) R n2 ). Y End(V ), then differentiating Ad with respect to B in the direction of Y gives d (A,1) Ad(Y ) = lim ɛ 0 A(1 + ɛy )A 1 A(1)A 1 ɛ = AY A 1. If A GL(V ) and Thus, d (A,1) Ad is just the usual conjugation action of A on End(V ). Next, for each Y End(V ), we want to differential the map A d (A,1) Ad(Y ) = AY A 1 at 1 GL(V ). This will give a linear 15

19 map ad(y ) : End(V ) End(V ). If X, Y End(V ), then (1 + ɛx) 1 = 1 ɛx + ɛ 2 X 2. So, (1 + ɛx)y (1 ɛx + ɛ 2 X 2 ) Y ad(y )(X) = lim ɛ 0 ɛ ɛxy ɛy X + ɛ 2 = lim = XY Y X. ɛ 0 ɛ Thus, b (1,1) Ad(X, Y ) = [X, Y ] = XY Y X, the usual commutator of matrices. As explained in remark 3.13, [, ] G is defined by differentiating the map Ad Y : G g given by Ad Y (g) = (d e Ad(g))(Y ). Swapping arguments, we may also consider the map Ad(g) : g g again given by Y (d e Ad(g))(Y ). Exercise Show that Ad(g) Ad(h) = Ad(gh) for all g, h G. Conclude that Ad(g) is an invertible linear map and hence defines a group homomorphism Ad : G GL(g). The map Ad : G GL(g) is a morphism of Lie groups. Its differential at the identity is denoted ad : g End(g). Applying Proposition 3.14 to this situation gives the following lemma, which will be useful later. Lemma The map ad preserves brackets i.e. ad([x, Y ] G ) = [ad(x), ad(y )] E, where [A, B] E := AB BA is the bracket on End(g). Remark One can check that ad(y )(X) = ad(x)(y ) = [X, Y ] G for all X, Y g. 3.4 Lie algebras We define here the second protagonist in the story - the Lie algebra. We ve essential already seen above that for each Lie group G, the space g is an example of a Lie algebra. Definition Let k be a field and g a k-vector space. Then g is said to be a Lie algebra if there exists a bilinear map [, ] : g g g, called the Lie bracket, such that 1. The Lie bracket is anti-symmetric meaning that [X, Y ] = [Y, X], X, Y g. 2. The Jacobi identity is satisfied: [X, [Y, Z]] + [Z, [X, Y ]] + [Y, [Z, X]] = 0, X, Y, Z g. 16

20 Exercise Assume that char k 2. Show that the first axiom in the definition of a Lie algebra is equivalent to the condition [X, X] = 0 for all X g. Example Let M be a manifold. Then the space Vect(M) of vector fields on M is a Lie algebra via the rule [X, Y ] := X Y Y X, where X Y is the vector field that acts on C (M) by first applying Y and then applying X. Example Let V be a k-vector space and gl(v ) = End(V ) the space of all linear maps V V. Then, as we have already seen, gl(v ) is a Lie algebra, the general linear Lie algebra, with bracket [F, G] = F G G F. If V is n-dimensional, then we may identify gl(v ) with gl(n, k), the Lie algebra of n n matrices, where the bracket of two matrices A and B is just the commutator [A, B] = AB BA. The Lie algebra gl(n, k) contains many interesting Lie subalgebras such as n(n, k) the Lie algebra of all strictly upper triangular matrices or b(n, k) the Lie algebra of upper triangular matrices. Exercise Prove that gl(n, k) is a Lie algebra, and that b(n, k) and n(n, k) are Lie subalgebras of gl(n, k). 3.5 The Lie algebra of a Lie group Recall that we defined in section 3.3 a bracket [, ] on the tangent space g at the identity of a Lie group G. As expected, Proposition The pair (g, [, ]) is a Lie algebra. Proof. The bracket is bilinear by construction. Therefore we need to check that it is anti-symmetric and satisfies the Jacobi identity. By exercise 3.21, to check that the first axiom holds it suffices to show that [X, X] = 0 for all X g. Recall from the first definition of tangent spaces that an element X g can be written γ (0) for some γ : ( ɛ, ɛ) G. Let Y = ρ (0) be another element in g. We can express the bracket of X and Y in terms of γ and ρ. First, for each t ( ɛ, ɛ) and g G, Ad(γ(t))(g) = γ(t)gγ(t) 1. Then, taking g = ρ(s) and differentiating ρ to get Y, (d e Ad(γ(t)))(Y ) = ( ) d ds γ(t)ρ(s)γ(t) 1 s=0. Thus, ad(x)(y ) = [( ) ] d ds γ(t)ρ(s)γ(t) 1 s=0 t=0. 17

21 In particular, [ d [X, X] = ad(x)(x) = dt [ ( ) d d = dt ds γ(s) [ ] d = dt X t=0 = 0. ( ) ] d ds γ(t)γ(s)γ(t) 1 s=0 t=0 s=0 ] t=0 We have implicitly used the fact that γ(t)γ(s) = γ(s)γ(t) for all s, t. This is proved in Lemma 4.6 below. Using the anti-symmetric property of the bracket, the Jacobi identity is equivalent to the identity [X, [Y, Z]] [Y, [X, Z]] = [[X, Y ], Z] for all X, Y, Z g. Recall that ad(x)(y ) = [X, Y ]. Therefore the above identity can be written [ad(x) ad(y ) ad(y ) ad(x)](z) = ad([x, Y ])(Z), which would follow from the identity ad(x) ad(y ) ad(y ) ad(x) = ad([x, Y ]) in End(g). But this is exactly the statement of Lemma 3.18 that ad preserves brackets. A Lie algebra h is said to be abelian if [X, Y ] = 0 for all X, Y h. Exercise Let H be an abelian Lie group. Show that its Lie algebra h is abelian. 3.6 Ideals and quotients Let g be a Lie algebra. A subspace h of g is called a subalgebra if the bracket on g restricts to a bilinear map [, ] : h h h. This makes h into a Lie algebra. A subalgebra l is said to be an ideal if [l, g] l. If l is an ideal of g then the quotient vector space g/l is itself a Lie algebra, with bracket [X + l, Y + l] := [X, Y ] + l. Exercise Let l g be an ideal. Check that the bracket on g/l is well-defined and that g/l is indeed a Lie algebra. Exercise Let l be a Lie subalgebra of g such that [g, g] l. Show that l is an ideal and that the quotient g/l is abelian. Exercise Show that n(2, k) is a one-dimensional ideal in b(2, k). More generally, show that n(n, k) is an ideal in b(n, k). What are the dimensions of b(n, k) and n(n, k)? Is the quotient b(n, k)/n(n, k) abelian? 18

22 3.7 Lie algebras of small dimension Let g be a one-dimensional Lie algebra. Then g = k{x} for any 0 X g. What is the bracket on g? The fact that the bracket must be anti-symmetric implies that [X, X] = 0. Therefore the bracket is zero and g is unique up to isomorphism. When dim g = 2, let X 1, X 2 be some basis of g. If the bracket on g is not zero, then the only non-zero bracket can be [X 1, X 2 ] = [X 2, X 1 ] and hence [g, g] is a one dimensional subspace of g. Let Y span this subspace. Let X be any element not in [g, g] so that X, Y is also a basis of g. Then [X, Y ] must be a non-zero element in [g, g], hence it is a multiple of Y. By rescaling X, we may assume that [X, Y ] = Y. This uniquely defines the bracket on g (and one can easily check that the bracket does indeed make g into a Lie algebra). Thus, up to isomorphism there are only two Lie algebras of dimension two. In dimension three there are many more examples, but it is possible to completely classify them. The most important three dimensional Lie algebra is sl(2, C), the subalgebra of gl(2, C) consisting of matrices of trace zero. Exercise Let E = ( ), F = ( ), H = ( ), Show that {E, F, H} is a basis of sl(2, C). Calculate [H, E], [H, F ] and [E, F ]. 19

23 4 The exponential map In this section, we define the exponential map. This allows us to go back from the Lie algebra of a Lie group to the group itself. 4.1 Left-invariant vector fields First we note that there is a natural action of G on C (G), the space of all smooth functions on G. Namely, given f C (G) and g, h G, define the new function g f by (g f)(h) = f(g 1 h). (10) Now let X Vect(G) be a vector field on G. Since X is uniquely defined by its action on C (G), we can define an action of G on Vect(G) by (g X)(f) = g [X(g 1 f)], f C (G). (11) It might seem a bit strange that I ve just decided to put in the ( ) 1 into equation (10), but this is necessary to ensure that both C (G) and Vect(G) become left G-modules. The vector field X is said to be left invariant if g X = X for all g G. We denote by Vect L (G) Vect(G) the subspace of all left invariant vector fields. Exercise 4.1. Show that Vect L (G) is a Lie subalgebra of Vect(G). Lemma 4.2. Let G be a Lie group. The map X X e defines an isomorphism of vector spaces Vect L (G) g. Proof. Let X Vect L (G). We will show that X is unique defined by its value X e at e. Let g, h G and f C (G). First we consider the differential d h (g f) : T h G R. Since (g f)(h) = f(g 1 h), we have g f = f L g 1, where L g : G G is defined by L g (h) = gh. Thus, d h (g f) = d g 1 hf d h L g 1. Then, [(g X)(f)](h) = [g (X(g 1 f))](h) = [X(g 1 f)](g 1 h) = [d g 1 h(g 1 f)](x g 1 h) = (d h f) (d g 1 hl g )(X g 1 h). Since X is assumed to be left invariant, this implies that (d h f)(x h ) = (d h f) (d g 1 hl g )(X g 1 h) 20

24 for all f C (G). Hence X h = (d g 1 hl g )(X g 1 h), g, h G. (12) In particular, taking h = g shows that X g = (d e L g )(X e ) i.e. X is uniquely defined by X e. Thus, the map X X e is injective. Conversely, for each X g, define the vector field X on G by X g = (d e L g )(X). This means that X(f)(h) = (d h f)(d e L h (X)) for all f C (G). belongs to Vect L (G). It is left as an exercise to check that X Exercise 4.3. Check that the vector field X defined in the proof of Lemma 4.2 is left invariant. Remark 4.4. One can actually show that the isomorphism X X e of Lemma 4.2 is an isomorphism of Lie algebras. Example 4.5. Let G = (R, ). Then T G = R R and Vect(G) = {f x f C (R )}. If α R and f C (G), then the group R acts on C (R ) by (α f)(p) = f(α 1 p) e.g. α x n = α n x n since This implies that C (R ) G = R. Similarly, (α x n )(β) = x n (α 1 β) = (α 1 β) n. ( ( α f )) ( (x n ) = α f ) (α n x n ) x x = (α f)nα n x n 1 ( = α(α f) ) (x n ). x Thus, α (f ) (x ) x = α(α f) n ; for instance α x x = α n+1 x n. This implies that x VectL (R ) = R{x }. x 4.2 One parameter subgroups By Lemma 4.2 each element X g defines a unique left invariant vector field ν(x) such that ν(x) e = X. Then, associated to ν(x) is an integral curve ϕ X : J G through e. Lemma 4.6. The integral curve ϕ X is defined on the whole of R. Moreover, it is a homomorphism of Lie groups R G i.e. ϕ(s + t) = ϕ(s)ϕ(t) for all s, t R. Proof. Choose s, t J such that s + t also belongs to J. Then we claim that ϕ(s + t) = ϕ(s)ϕ(t). Fix s and let t vary in a some small open set J 0 J containing 0 such that s + t still belongs to J. 21

25 Then α(t) = ϕ(s + t) and β(t) = ϕ(s)ϕ(t) are curves in G such that α(0) = β(0). Differentiating, α(t) = ϕ(s + t) = ν(x) ϕ(s+t) = ν(x) α(t). To calculate β(t), we first write β(t) = (L ϕ(s) ϕ)(t). Then, β(t) = (d ϕ(t) L ϕ(s) )(ϕ (t)) = (d ϕ(t) L ϕ(s) ) ( ν(x) ϕ(t) ) = ν(x) ϕ(s) ϕ(t) = ν(x) β(t), where we have used (12) in the last equality but one. Thus, they are both integral curves for ν(x) through the point ϕ(s). By uniqueness of integral curves, they are equal. We can use the equality ϕ(s + t) = ϕ(s)ϕ(t) to extend ϕ to the whole of R: for each s, t J such that s + t is not in J, set ϕ(s + t) = ϕ(s)ϕ(t). The uniqueness property of integral curves shows that this is well-defined. The homomorphism ϕ X : R G is called the one-parameter subgroup associated to X g. The uniqueness of integral curves implies that it is the the unique homomorphism γ : R G such that γ (0) = X. Definition 4.7. The exponential map exp : g G is defined by exp(x) = ϕ X (1). The uniqueness of ϕ X implies that ϕ sx (t) = ϕ X (st) for all s, t R (since the derivatives with respect to t of ϕ sx (t) and ϕ X (st) at 0 are equal). Hence (d 0 exp)(x) = lim ɛ 0 exp(ɛx) exp(0) ɛ = lim ɛ 0 ϕ ɛx (1) ϕ 0 (1) ɛ = lim ɛ 0 ϕ X (ɛ) ϕ X (0) ɛ = ϕ X(0) = X, where we have used the fact that ϕ 0 (t) = ϕ X (0) = e for all t R. Thus, the derivative d 0 exp of exp at 0 g is just the identity map. In the case of G = GL(n, R) and hence g = gl(n, R), the exponential map can be explicitly written down, it is just the usual exponential map exp(x) = i=0 X i i!. 22

26 The same formula applies for any closed subgroup G of GL(n, R). This function behaves much the same way as the exponential function e x : R R. However, it is not true that exp(x + Y ) equals exp(x) exp(y ) in general - see subsection 4.6 for more. 4.3 Cartan s Theorem So far the only significant example of a Lie group we have is GL(n, R). Cartan s Theorem allows us to cook up lots of new Lie groups. Our proof of Cartan s Theorem is based on the proof in [?]. The proof of the theorem involves some basic analysis, which we give as a separate lemma. Let S n 1 denote the unit sphere in R n. For each x R n {0}, let [x] = x x Sn 1 i.e. [x] is the point on the unit sphere in R n that also lies on the line through 0 and x. Lemma 4.8. Let x n, x R n {0} with lim n x n = 0. Then, lim n [x n ] = [x] if and only if there exist positive integers a n such that lim n a n x n = x. Proof. If such integers exist then clearly lim n [x n ] = [x] since [a n x n ] = [x n ] for all n. Conversely, assume that lim n [x n ] = [x]. To say that lim n a n x n = x means that the distance a n x n x between a n x n and x becomes arbitrarily small as n. For each n choose a positive integer a n such that a n x x n < 1. If v, w Rn and α, β R then the triangle inequality implies that αv w βv w + α β v. In our situation, this implies that a n x n x x n x x n x + a n x x n x n Now lim n x n = 0 implies that lim n x n = 0 and x n x x n x = 1 x x n x n x x also tends to zero as n tends to infinity since lim n [x n ] = [x]. x n x x n = 1 x [x n ] [x] x + x n. Theorem 4.9 (Cartan s Theorem). Let G be a Lie group and H a closed subgroup. Then H is a submanifold of G. Hence H is a Lie subgroup of G. Proof. We begin by noting that it suffices to construct some open neighborhood U of e in G such 23

27 that H U is a closed submanifold of U. Assuming this, then for each h H, hu is an open neighborhood of h in G such that H hu = h(h U) is a closed submanifold. We fix a norm on the vector space g. The key to the proof of Cartan s Theorem is to choose carefully a candidate subspace W g for the Lie algebra of H. The space W is defined to be all w g such that either w = 0 or there exists a sequence {w n } in g such that w n 0, exp(w n ) H, w n 0 and [w n ] [w]. We will show: 1. exp(w ) H. 2. W is a subspace of g. 3. There is an open neighborhood U of 0 in g and a diffeomorphism φ : U φ(u) G with φ(0) = e such that φ(u W ) = φ(u) H. Assume (1)-(3) hold. Then, since W is a subspace of g, U W is clearly a submanifold of U. Therefore, part (3) implies that φ(u W ) is a submanifold of φ(u). Hence φ(u) H is a submanifold of φ(u) as required. Proof of part (1): Let w W {0}, with {w n } as in the definition of W. By Lemma 4.8, there exist positive integers a n such that a n w n w. Since exp(a n w n ) = exp(w n ) an H, and H is closed in G, the limit exp(w) = lim n exp(w n ) an belongs to H. Part (2): Since [w] = [tw] for all t R, tw W if w W. Thus, it suffices to show that if v, w W then v +w W. We can assume without loss of generality that v, w, v +w are non-zero. Recall that exp is a diffeomorphism in a neighborhood of 0. Therefore, for sufficiently small t, there exists a smooth curve t u(t) in g such that exp(tv) exp(tw) = exp(u(t)), (13) and u(0) = 0. Equation (13) implies that exp(u(t)) H and ( ) 1 lim nu u(ɛ) = lim n n ɛ 0 ɛ = u (0) = v + w. Then, since u ( 1 n) u(0) = 0, exp ( u ( 1 n)) belongs to H for all n, and [u ( 1 n) ] [v + w], we conclude that v + w W. Part (3): Let V be a complement to W in g. We define φ : g = V W G, (v, w) exp(v) exp(w). 24

28 The differential d 0 φ is just the identity map on g. Therefore there is some neighborhood U of 0 in g such that φ is a diffeomorphism from U onto φ(u). Clearly, φ(u W ) φ(u) φ(w ) H φ(u). Therefore, we need to show that H φ(u) is contained in φ(u W ). Assume that this is not the case. Then, in every open neighbourhood U n of 0 in g there exist (v n, w n ) V W such that φ(v n + w n ) H but v n 0. In particular, exp(v n ) H. We take U 1 U 2 such that n U n = {0}, so that {v n } is a sequence converging to 0. Since the unit sphere S in V is compact, there exists some v V {0} and subsequence {v n} of {v n } such that [v n] [v] (this is the result from metric spaces saying that every sequence in a compact space has a convergent subsequence). But this implies that v W ; a contradiction. Remark We ve actually shown in the proof of Theorem 4.9 that there is some neighborhood U of e in G such that U H = exp(h) U for any closed subgroup H of G. Example As a consequence, the following are Lie groups B(n, R) = {All upper triangular, invertible matrices.}, N(n, R) = {A B(n, R) Diagonal entries of A all equal 1.}, T (n, R) = {All diagonal matrices.} since they are all subgroups of GL(n, R) defined as the zeros of some polynomial equations. Exercise Using the criterion of example 2.6, show directly that B(n, R), N(n, R) and T (n, R) are submanifolds of GL(n, R). The analogue of Cartan s Theorem holds for closed subgroups of GL(n, C); they are complex Lie subgroups. 4.4 Simply connected Lie groups We recall that a path connected topological space X is simply connected if the fundamental group π 1 (X) of X is trivial i.e. every closed loop in X is homotopic to the trivial loop. In general it is not true that a Lie group is uniquely defined by its Lie algebra i.e. it is possible to find non-isomorphic Lie groups whose Lie algebras are isomorphic. However, if we demand that our Lie group be simply connected, then: Theorem 4.13 (S. Lie). Let g be a finite dimensional, real Lie algebra. Then, there exists a unique simply connected Lie group G whose Lie algebra is g. Moreover, if G is any other connected Lie group with Lie algebra g, then G is a quotient of G. 25

29 Recall that a covering map f : M N is a map such that every n N is contained in some open neighborhood U with f 1 (U) a disjoint union of open sets, each mapping homeomorphically onto U. Before we prove the theorem, we require some preparatory results. A covering map always satisfies the path lifting property: if γ : [0, 1] N is a path with n = γ(0) and m is a lift of n, then there is a unique path γ : [0, 1] M lifting γ such that γ(0) = m (we say that γ is a lift of γ if f γ = γ). Using the path lifting property one can easily show that Lemma Assume that M is simply connected and let g : Z N be a smooth morphism from a simply connected manifold Z sending z to n. Then there exists a unique morphism g : Z M, sending z to m such that f g = g. An easy, but important consequence of Lemma 4.14 is that every Lie group admits a simply connected cover. Proposition Let G be a Lie group, H a connected manifold, and ϕ : H G a covering map. Choose e, an element lying over the identity of G. Then there is a unique Lie group structure on H such that e is the identity and ϕ is a map of Lie groups; and the kernel of ϕ is in the centre of H. Proof. We first assume that H is a simply connected manifold. Define the map α : H H G by α(h 1, h 2 ) = ϕ(h 1 )ϕ(h 2 ) 1. By Lemma 4.14, there exists a unique map α : H H H such that α (e, e ) = e and ϕ α = α. Then we define h 1 := α (e, h) and h 1 h 2 = α (h 1, h 1 2 ) for h, h 1, h 2 H. This defines smooth morphisms H H and H H H resp. and one can use the uniqueness of the lift α to show that this makes H into a group. The proof of the following proposition follows easily from the proof of Cartan s Theorem, Theorem 4.9. Proposition Let G be a Lie group with Lie algebra g and h a Lie subalgebra of g. Then the subgroup H generated by exp(h) is a closed, Lie subgroup of G, whose Lie algebra is h. We also note: Lemma Let G be a connected group and φ : G H a morphism of Lie groups. d e φ : g h is an isomorphism, then d g φ : T g G T φ(g) H is an isomorphism for all g G. If Proof. This is simply the case of rewriting the map φ in a clever way. Recall that we have L g : G G, L g (u) = gu. Fix g G and define ϕ : G H by ϕ(u) = φ(g)φ(u). We can rewrite 26

30 this as ϕ = L φ(g) φ. This implies that d e ϕ = d φ(g) L φ(g) d e φ is an isomorphism. On the other hand, ϕ also equals u φ(gu), which we can write as ϕ = φ L g. Therefore, d e ϕ = d g φ d e L g. Since both d e ϕ and d e L g are invertible linear maps, this implies that d g φ is too. Proof of Theorem The key to Theorem 4.13 is Ado s Theorem, Theorem 7.27, which says that there exists some n 0 such that g is a Lie subalgebra of gl(n, R). Therefore Proposition 4.16 implies that there is some connected, closed Lie subgroup G GL(n, R) with Lie G = g. Let ϕ : G G be the universal cover of G. Proposition 4.15 says that we can endow G with the structure of a Lie group such that ϕ is a quotient of Lie groups. Moreover, d e ϕ : g g is the identity. Thus, it suffices to show that if G is another simply connected Lie group with Lie algebra g then G G. To show this, we consider the product G G. It s Lie algebra is g g and the diagonal copy g of g in g g is a Lie subalgebra. Therefore Proposition 4.16 implies that there is some connected, closed Lie subgroup K G G such that Lie K = g. The maps φ 1 : K G G G and φ 2 : K G G G are homomorphisms of Lie groups whose differential at the identity is the identity map on g. Thus, we have maps φ i between Lie groups whose differential is an isomorphism on Lie algebras. Hence, Lemma 4.17 implies that d k φ i is an isomorphism for all k K. Now, we have maps φ i : K G, G between connected manifolds, where G and G are simply connected. As mentioned at the end of section 2.3, this implies that each φ i is an isomorphism. Hence we may identify K G G. 4.5 The proof of Theorems 3.15 and 3.8 Finally, we have the theorem that motivated the definition of a Lie algebra in the first place. It was stated as Theorem Theorem Let G and H be Lie groups with G simply connected, and let g and h be their Lie algebras. A linear map µ : g h is the differential of a morphism φ : G H if and only if µ is a map of Lie algebras. Proof. We showed in Proposition 3.14 that if µ is the differential of a homomorphism then it must be a map of Lie algebras. Therefore we need to show the existence of φ, knowing that µ is a map of Lie algebras. We will deduce the theorem from Proposition 4.16, applied to G H. The Lie algebra of G H is just g h. Let Γ µ = {(X, µ(x)) X g} g h be the graph of µ. Then the fact that µ is a map of Lie algebras is equivalent to saying that Γ µ is a Lie subalgebra of g h. Let K be 27