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 − {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 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 ∈ 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) ∈ 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) = ∞ 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 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 ϕ : 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 ∈ 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 ∈ 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 set 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 vector space, 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 algebra. 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 subalgebra of the Lie algebra 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- morphism, 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.