LECTURE 12: LIE GROUPS AND THEIR LIE ALGEBRAS 1. Lie groups Definition 1.1. A Lie group G is a smooth manifold equipped with a group structure so that the group multiplication µ : G × G ! G; (g1; g2) 7! g1 · g2 is a smooth map. Example. Here are some basic examples: • Rn, considered as a group under addition. • R∗ = R − f0g, considered as a group under multiplication. • S1, Considered as a group under multiplication. • Linear Lie groups GL(n; R), SL(n; R), O(n) etc. • If M and N are Lie groups, so is their product M × N. Remarks. (1) (Hilbert's 5th problem, [Gleason and Montgomery-Zippin, 1950's]) Any topological group whose underlying space is a topological manifold is a Lie group. (2) Not every smooth manifold admits a Lie group structure. For example, the only spheres that admit a Lie group structure are S0, S1 and S3; among all the compact 2 dimensional surfaces the only one that admits a Lie group structure is T 2 = S1 × S1. (3) Here are two simple topological constraints for a manifold to be a Lie group: • If G is a Lie group, then TG is a trivial bundle. n { Proof: We identify TeG = R . The vector bundle isomorphism is given by φ : G × TeG ! T G; φ(x; ξ) = (x; dLx(ξ)) • If G is a Lie group, then π1(G) is an abelian group. { Proof: Suppose α1, α2 2 π1(G). Define α : [0; 1] × [0; 1] ! G by α(t1; t2) = α1(t1) · α2(t2). Then along the bottom edge followed by the right edge we have the composition α1 ◦ α2, where ◦ is the product of loops in the fundamental group, while along the left edge followed by the top edge we get α2 ◦ α1. Deforming the path shows they're homotopic Now suppose G is a Lie group. For any elements a; b 2 G, there are two natural maps, the left multiplication La : G ! G; g 7! a · g and the right multiplication Rb : G ! G; g 7! g · b: −1 −1 It is obviously that La = La−1 and Rb = Rb−1 . So both La and Rb are diffeomor- phisms. Moreover, La and Rb commutes with each other: LaRb = RbLa. 1 2 LECTURE 12: LIE GROUPS AND THEIR LIE ALGEBRAS Lemma 1.2. The differential of the multiplication map µ : G × G ! G is given by dµa;b(Xa;Yb) = (dRb)a(Xa) + (dLa)b(Yb) for any (Xa;Yb) 2 TaG × TbG ' T(a;b)(G × G): Proof. Notice that as a function of a, µ(a; b) = Rb(a), and as a function of y, µ(a; b) = 1 La(b). Thus for any function f 2 C (G), (dµa;b(Xa;Yb))(f) = (Xa;Yb)(f ◦ µ(a; b)) = Xa(f ◦ Rb(a)) + Yb(f ◦ La(b)) = (dRb)a(Xa)(f) + (dLa)b(Yb)(f) As an application, we can prove Proposition 1.3. For any Lie group G, the group inversion map i : G ! G; g 7! g−1 is smooth. Proof. Consider the map f : G × G ! G × G; (a; b) 7! (a; ab): It is obviously a bijective smooth map. According to lemma above, the derivative of f is df(a;b) : TaG × TbG ! TaG × TabG; (Xa;Yb) 7! (Xa; (dRb)a(Xa) + (dLa)b(Yb)) This is a bijective linear map since dRb; dLa are. It follows by inverse function theorem that f is locally a diffeomorphism near each pair (a; b). However, since f is globally bijective, it must be a globally diffeomorphism. We conclude that its inverse, f −1 : G × G ! G × G; (a; c) 7! (a; a−1c) is a diffeomorphism. Thus the inversion map i, as the composition f −1 G,−! G × G −! G × G −!π2 G g 7−! (g; e) 7−! (g; g−1) 7−! g−1 is smooth. The following definitions are standard: Definition 1.4. Let G1;G2 be Lie groups. A map f : G1 ! G2 is called a Lie group homomorphism if it is smooth and is a group homomorphism. Definition 1.5. A subset H of a Lie group G is called a Lie subgroup if it is an immersed submanifold, is a subgroup, and the group multiplication is smooth (with respect to its own smooth structure). Note that if H is a Lie subgroup of G, then it is a Lie group, and the inclusion ι : H,! G is a Lie group homomorphism. LECTURE 12: LIE GROUPS AND THEIR LIE ALGEBRAS 3 2. Lie algebras associated to Lie groups We start with the abstract definition. Definition 2.1. A Lie algebra is a real vector space V together with a bi-linear bracket operation [·; ·]: V × V ! V; such that for any X; Y; Z 2 V , (1) (Skew-symmetry) [X; Y ] = −[Y; X], (2) (Jacobi identity) [X; [Y; Z]] + [Y; [Z; X]] + [Z; [X; Y ]] = 0. The bracket [·; ·] is called the Lie bracket. Example. Any vector space admits a trivial Lie algebra: [X; Y ] ≡ 0. Example. The set of all smooth vector fields Γ1(TM) on any smooth manifold is a Lie algebra, where the Lie bracket operation is the one we defined in lecture 9. (More generally, any associative algebra gives rise to a Lie algebra by defining [X; Y ] = XY − YX.) One can easily write down the conceptions of Lie subalgebra and Lie algebra homo- morphism. For example, a Lie algebra homomorphism is a linear map that preserves the Lie bracket. Now we will associate to any Lie group G a god-given Lie algebra. Suppose G is a Lie group. From the left translations La one can, for any vector Xe 2 TeG, define a vector field X on G by Xa = (dLa)(Xe): It is not surprising that X is invariant under any left translation: (dLa)(Xb) = dLa ◦ dLb(Xe) = dLab(Xe) = Xab: Definition 2.2. A left invariant vector field on a Lie group G is a smooth vector field X on G which satisfies (dLa)(Xb) = Xab. So any tangent vector Xe 2 TeG determines a left invariant vector field on G. Conversely, any left invariant vector field X is uniquely determined by its \value" Xe at e 2 G, since for any a 2 G, X(a) = (dLa)Xe. We will denote the set of all left invariant vector fields on Lie group G by g, i.e. g = fX 2 Γ1(TG) j X is left invariantg: So g is a vector subspace of Γ1(TG). Moreover, as a vector space, g is isomorphic to 1 TeG. In particular, dim g = dim G. As we mentioned (and as in PSet 3), Γ (TG) has a Lie bracket operation [·; ·] which makes Γ1(TG) into an 1-dimensional Lie algebra. Proposition 2.3. If X; Y 2 g, so is their Lie bracket [X; Y ]. Proof. We only need to show that [X; Y ] is left-invariant if X and Y are. First notice Y (f ◦ La)(b) = Yb(f ◦ La) = (dLa)b(Yb)f = Yabf = (Y f)(Lab) = (Y f) ◦ La(b) 4 LECTURE 12: LIE GROUPS AND THEIR LIE ALGEBRAS for any smooth function f 2 C1(G). Thus Xab(Y f) = (dLa)b(Xb)(Y f) = Xb((Y f) ◦ La) = XbY (f ◦ La): Similarly YabXf = YbX(f ◦ La). Thus dLa([X; Y ]b)f = XbY (f ◦ La) − YbX(f ◦ La) = Xab(Y f) − YabXf = [X; Y ]ab(f): It follows that the space g of all left invariant vector fields on G together with the Lie bracket operation [·; ·] is an n-dimensional Lie subalgebra of the Lie algebra of all smooth vector fields Γ1(TG). Definition 2.4. g of is called the Lie algebra of G. Now suppose φ : G ! H is a Lie group homomorphism, then its differential at e gives a linear map from TeG to TeH. Under the identification of TeG with g and TeH with h, we get an induced map, still denoted by dφ, from g to h. Theorem 2.5. If φ : G ! H is a Lie group homomorphism, then the induced map dφ : g ! h is a Lie algebra homomorphism. Proof. We need to show that dφ preserves the Lie bracket. Let X be a left invariant vec- tor field on G, then by definition dφ(X) = Xdφe(Xe). Since φ is a group homomorphism, we have φ ◦ Lg = Lh ◦ φ, where h = φ(g). It follows (dφg)(Xv)g = (dφg)(dLg)ev = d(φ◦Lg)ev = d(Lh◦φ)ev = dLh◦(dφe)v = Xdφe(v)(h) = dφ(Xv)(h): In other words, Xv and dφ(Xv) are φ-related. It follows that [Xv;Xw] and [dφ(Xv); dφ(Xw)] are φ-related. On the other hand, [Xv;Xw] is φ-related to dφ([Xv;Xw]). So [dφ(Xv); dφ(Xw)]e = dφe([Xv;Xw]e) = (dφ([Xv;Xw]))e; i.e. dφ([Xv;Xw]) = [dφ(Xv); dφ(Xw)]. We end with two examples: Example. (The Euclidean group Rn.) This is obviously a Lie group, since the group operation µ((x1; ··· ; xn); (y1; ··· ; yn)) := (x1 + y1; ··· ; xn + yn): is smooth. n Moreover, for any a 2 R , the left translation La is just the usual translation map n n n on R . So dLa is the identity map, as long as we identify TaR with R in the usual way. It follows that any left invariant vector field is in fact a constant vector field, i.e. @ @ Xv = v1 + ··· + vn @x1 @xn n @ @ for ~v = (v1; ··· ; vn) 2 T0 . Since commutes with for any pair (i; j), we R @xi @xj conclude that the Lie bracket of any two left invariant vector fields vanishes.