1. Lie groups Definition 1.1. A Lie G is a smooth equipped with a group structure so that the group

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 . • R∗ = R − {0}, 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 (= whose underlying is a ) is “actually” a , 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 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 ∈ 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.


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) ∈ TaG × TbG ' T(a,b)(G × G).

Proof. Notice that as a of a, µ(a, b) = Rb(a), and as a function of y, µ(a, b) = ∞ La(b). Thus for any function f ∈ 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 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 between two Lie groups is a smooth map which is also a homomorphism of groups.

For example, the map ϕ : R → S1, t 7→ eit is a Lie . Similarly one can define Lie group . 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 ∈ TeG, define a vector field X on G by

Xa = (dLa)(Xe). It is not surprising that X is under any left :

(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 ∈ 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 ∈ G, since for any a ∈ G, X(a) = (dLa)Xe. We will denote the of all left invariant vector fields on Lie group G by g, i.e. g = {X ∈ Γ∞(TG) | X is left invariant}. Obviously g is a vector subspace of Γ∞(TG). Moreover, as a , g is isomor- phic to TeG. In particular, dim g = dim G. We have already seen that Γ∞(TG) has a Lie bracket operation [·, ·] which makes Γ∞(TG) into an infinitely dimensional Lie . Proposition 2.2. If X,Y ∈ 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 ∈ C∞(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 of the of all smooth vector fields Γ∞(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 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. Since ϕ is a group homo- , we have ϕ ◦ La = Lϕ(a) ◦ ϕ. Let X,Y be left invariant vector field on G. We denote dϕ(X) and dϕ(Y ) to be the left invariant vector fields on H that corresponding to dϕe(Xe) and dϕe(Ye). We need to check dϕe([X,Y ]e) = [dϕ(X), dϕ(Y )]e. For any f ∈ C∞(H), we have

dϕe([X,Y ]e)f = [X,Y ]e(f ◦ ϕ) = Xe(Y (f ◦ ϕ)) − Ye(X(f ◦ ϕ)) and

[dϕ(X), dϕ(Y )]ef = dϕe(Xe)(dϕ(Y )f)−dϕe(Ye)(dϕ(X)f) = Xe(dϕ(Y )f◦ϕ)−Ye(dϕ(X)f◦ϕ). So it is enough to check that as functions on G, Y (f ◦ ϕ) = dϕ(Y )f ◦ ϕ. In fact, for any a ∈ G, we have

Y (f ◦ ϕ)(a) = Ya(f ◦ ϕ) = dLa(Ye)(f ◦ ϕ) = Ye(f ◦ ϕ ◦ La) = Ye(f ◦ Lϕ(a) ◦ ϕ), while

dϕ(Y )f ◦ ϕ(a) = dϕ(Y )(f)(ϕ(a)) = dLϕ(a) ◦ dϕe(Ye)(f) = Ye(f ◦ Lϕ(a) ◦ ϕ). 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 ∈ 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) ∈ 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.