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LECTURE 5: LIE GROUPS and THEIR LIE ALGEBRAS 1. Lie LECTURE 5: 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) The famous Hilbert's fifth problem, solved by Gleason and Montgomery- Zippin in the 1950's, confirms that any locally Euclidean group (=topological group whose underlying space is a topological manifold) is \actually" a Lie group, and more- over, The underlying space of any Lie group is in fact an analytic manifold, and the group operations are analytic. analytic. One can define analytic maps between analytic manifolds using local charts as in the smooth case.) (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 admits a Lie group structure is T 2 = S1 × S1. There are many constraints for a manifold to be a Lie group. For example, a Lie group must be analytic manifold, and the tangent bundle of a Lie group is always trivial: TG ' G × Rn. In particular, any Lie group is orientable. 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 5: 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. Definition 1.4. A Lie group homomorphism between two Lie groups is a smooth map which is also a homomorphism of groups. For example, the map ' : R ! S1; t 7! eit is a Lie group homomorphism. Similarly one can define Lie group isomorphisms. LECTURE 5: LIE GROUPS AND THEIR LIE ALGEBRAS 3 2. Lie algebras associated to Lie groups 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 2.1. 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: Obviously g is a vector subspace of Γ1(TG). Moreover, as a vector space, g is isomor- phic to TeG. In particular, dim g = dim G. We have already seen that Γ1(TG) has a Lie bracket operation [·; ·] which makes Γ1(TG) into an infinitely dimensional Lie algebra. Proposition 2.2. 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. We first notice Y (f ◦ La)(b) = Yb(f ◦ La) = (dLa)b(Yb)f = Yabf = (Y f)(Lab) = (Y f) ◦ La(b) 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 2.3. g of is called the Lie algebra of G. As before we can define Lie algebra homomorphism. Definition 2.4. A Lie algebra homomorphism between two Lie algebras is a linear map that preserves the Lie algebra structures. 4 LECTURE 5: LIE GROUPS AND THEIR LIE ALGEBRAS Now suppose ' : 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 d', from g to h. Theorem 2.5. If ' : G ! H is a Lie group homomorphism, then the induced map d' : g ! h is a Lie algebra homomorphism. Proof. We need to show that d' preserves the Lie bracket. Since ' is a group homo- morphism, we have ' ◦ La = L'(a) ◦ '. Let X; Y be left invariant vector field on G. We denote d'(X) and d'(Y ) to be the left invariant vector fields on H that corresponding to d'e(Xe) and d'e(Ye). We need to check d'e([X; Y ]e) = [d'(X); d'(Y )]e: For any f 2 C1(H), we have d'e([X; Y ]e)f = [X; Y ]e(f ◦ ') = Xe(Y (f ◦ ')) − Ye(X(f ◦ ')) and [d'(X); d'(Y )]ef = d'e(Xe)(d'(Y )f)−d'e(Ye)(d'(X)f) = Xe(d'(Y )f◦')−Ye(d'(X)f◦'): So it is enough to check that as functions on G, Y (f ◦ ') = d'(Y )f ◦ '. In fact, for any a 2 G, we have Y (f ◦ ')(a) = Ya(f ◦ ') = dLa(Ye)(f ◦ ') = Ye(f ◦ ' ◦ La) = Ye(f ◦ L'(a) ◦ '); while d'(Y )f ◦ '(a) = d'(Y )(f)('(a)) = dL'(a) ◦ d'e(Ye)(f) = Ye(f ◦ L'(a) ◦ '): This completes the proof. 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. In other words, the Lie algebra of G = Rn is g = Rn with vanishing Lie bracket..
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