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Home , Diffeomorphism, Euclidean group, Group homomorphism, Lie algebra, Lie group

LECTURE 12: LIE GROUPS AND THEIR LIE

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, ) 7→ g1 · g2 is a smooth . 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) (Hilbert’s 5th problem, [Gleason and Montgomery-Zippin, 1950’s]) Any whose underlying space is a is a . (2) Not every smooth manifold admits a Lie group structure. For example, the only 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 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 . – Proof: Suppose α1, α2 ∈ 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 , 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 ∈ 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) ∈ 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 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 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 if it is smooth and is a . Definition 1.5. A H of a Lie group G is called a Lie if it is an immersed , 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 is a real V together with a bi-linear operation [·, ·]: V × V → V, such that for any X,Y,Z ∈ V , (1) (Skew-) [X,Y ] = −[Y,X], (2) () [X, [Y,Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0. The bracket [·, ·] is called the Lie bracket. Example. Any vector space admits a trivial : [X,Y ] ≡ 0. Example. The of all smooth vector fields Γ∞(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 gives rise to a Lie algebra by defining [X,Y ] = XY − YX.)

One can easily write down the conceptions of Lie and Lie algebra homo- . 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 ∈ 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 2.2. 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 ∈ 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}. So g is a vector subspace of Γ∞(TG). Moreover, as a vector space, g is isomorphic to ∞ TeG. In particular, dim g = dim G. As we mentioned (and as in PSet 3), Γ (TG) has a Lie bracket operation [·, ·] which makes Γ∞(TG) into an ∞-dimensional Lie algebra. Proposition 2.3. 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. 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 ∈ 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 2.4. 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 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. LECTURE 12: LIE GROUPS AND THEIR LIE ALGEBRAS 5

Example. (The general ) GL(n, R) = {X ∈ M(n, R) | det X 6= 0}. Obviously GL(n, R) is n2-dimensional noncompact Lie group. Moreover, it is not con- nected, but consists of exactly two connected components,

GL+(n, R) = {X ∈ M(n, R) | det X > 0} and GL−(n, R) = {X ∈ M(n, R) | det X < 0}. 2 The fact that GL(n, R) is an open subset of M(n, R) ' Rn implies that the Lie algebra of GL(n, R), as the at e = In, is the set M(n, R) itself, i.e. gl(n, R) = {A | A is an n × n }.

To figure out the Lie bracket operation, we take a matrix A = (Aij)n×n ∈ g, and take the global coordinate system by (Xij). Then the corresponding tangent vector P ∂ at TIn GL(n, R) is Aij ∂Xij , and the corresponding left-invariant vector on G at the ij P ik ∂ matrix X = (X ) is X Akj ∂Xij . It follows that the Lie bracket [A, B] between matrices A, B ∈ g is the matrix corresponding to X ∂ X ∂  X ∂ X ∂ XikA , XpqB = XikA B − XpqB A kj ∂Xij qr ∂Xpr kj jr ∂Xir qr rj ∂Xpj X ∂ = Xik (A B − B A ) . kr rj kr rj ∂Xij In other words, the Lie bracket operation on g is the matrix [A, B] = AB − BA.