Lecture 18 - Reconstruction of a Lie Group from its Algebra (Schuller's Geometric Anatomy of Theoretical Physics)

18

Reconstruction of a Lie group from its algebra

We have seen in detail how to construct a Lie algebra from a given Lie group. We would now like to consider the inverse question, i.e. whether, given a Lie algebra, we can construct a Lie group whose associated Lie algebra is the given one and, if this is the case, whether this correspondence is bijective. 18.1

Integral curves

Definition. Let M be a smooth manifold and let Y ∈ Γ(T M ). An integral curve of Y is smooth curve γ : (−ε, ε) → M , with ε > 0, such that ∀ λ ∈ (−ε, ε) : Xγ,γ(λ) = Y |γ(λ) . It follows from the local existence and uniqueness of solutions to ordinary differential equations that, given any Y ∈ Γ(T M ) and any p ∈ M , there exist ε > 0 and a smooth curve γ : (−ε, ε) → M with γ(0) = p which is an integral curve of Y . Moreover, integral curves are locally unique. By this we mean that if γ1 and γ2 are both integral curves of Y through p, i.e. γ1 (0) = γ2 (0) = p, then γ1 = γ2 on the intersection of their domains of definition. We can get genuine uniqueness as follows. Definition. The maximal integral curve of Y ∈ Γ(T M ) through p ∈ M is the unique p integral curve γ : Imax → M of Y through p, where [ p Imax := {I ⊆ R | there exists an integral curve γ : I → M of Y through p}. p will differ from point to point. For a given vector field, in general, Imax p Definition. A vector field is complete if Imax = R for all p ∈ M .

We have the following result. Theorem 18.1. On a compact manifold, every vector field is complete. On a Lie group, even if non-compact, there are always complete vector fields. Theorem 18.2. Every left-invariant vector field on a Lie group is complete. The maximal integral curves of left-invariant vector fields are crucial in the construction of the map that allows us to go from a Lie algebra to a Lie group. 18.2

The exponential map

Let G be a Lie group. Recall that given any A ∈ Te G, we can define the uniquely determined ∼ left-invariant vector field X A := j(A) via the isomorphism j : Te G − → L(G) as X A |g := (`g )∗ (A). Then let γ A : R → G be the maximal integral curve of X A through e ∈ G.

1

Definition. Let G be a Lie group. The exponential map is defined as exp : Te G → G A 7→ exp(A) := γ A (1) Theorem 18.3. i) The map exp is smooth and a local diffeomorphism around 0 ∈ Te G, i.e. there exists an open V ⊆ Te G containing 0 such that the restriction exp |V : V → exp(V ) ⊆ G is bijective and both exp |V and (exp |V )−1 are smooth. ii) If G is compact, then exp is surjective. Note that the maximal integral curve of X 0 is the constant curve γ 0 (λ) ≡ e, and hence we have exp(0) = e. Then first part of the theorem then says that we can recover a neighbourhood of the identity of G from a neighbourhood of the identity of Te G. Since Te G is a vector space, it is non-compact (intuitively, it extends infinitely far away in every direction) and hence, if G is compact, exp cannot be injective. This is because, by the second part of the theorem, it would then be a diffeomorphism Te G → G. But as G is compact and Te G is not, they are not diffeomorphic. Proposition 18.4. Let G be a Lie group. The image of exp : Te G → G is the connected component of G containing the identity. Therefore, if G itself is connected, then exp is again surjective. Note that, in general, there is no relation between connected and compact topological spaces, i.e. a topological space can be either, both, or neither. Example 18.5. Let B : V × V be a pseudo inner product on V . Then O(V ) := {φ ∈ GL(V ) | ∀ v, w ∈ V : B(φ(v), φ(w)) = B(v, w)} is called the orthogonal group of V with respect to B. Of course, if B or the base field of V need to be emphasised, they can be included in the notation. Every φ ∈ O(V ) has determinant 1 or −1. Since det is multiplicative, we have a subgroup SO(V ) := {φ ∈ O(V ) | det φ = 1}. These are, in fact, Lie subgroups of GL(V ). The Lie group SO(V ) is connected while O(V ) = SO(V ) ∪ {φ ∈ O(V ) | det φ = −1} is disconnected. Since SO(V ) contains idV , we have so(V ) := TidV SO(V ) = TidV O(V ) =: o(V ) and exp(so(V )) = exp(o(V )) = SO(V ).

2

Example 18.6. Choosing a basis A1 , . . . , Adim G of Te G often provides a convenient coordinatisation of G near e. Consider, for example, the Lorentz group O(3, 1) ≡ O(R4 ) = {Λ ∈ GL(R4 ) | ∀ x, y ∈ R4 : B(Λ(x), Λ(y)) = B(x, y)}, where B(x, y) := ηµν xµ y ν , with 0 ≤ µ, ν ≤ 3 and  −1 0  [η µν ] = [ηµν ] :=  0 0

0 1 0 0

0 0 1 0

 0 0  . 0 1

The Lorentz group O(3, 1) is 6-dimensional, hence so is the Lorentz algebra o(3, 1). For convenience, instead of denoting a basis of o(3, 1) as {M i | i = 1, . . . , 6}, we will denote it as {M µν | 0 ≤ µ, ν ≤ 3} and require that the indices µ, ν be anti-symmetric, i.e. M µν = −M νµ . Then M µν = 0 when ρ = σ, and the set {M µν | 0 ≤ µ, ν ≤ 3}, while technically not linearly independent, contains the 6 independent elements that we want to consider as a basis. These basis elements satisfy the following bracket relation [M µν , M ρσ ] = η νσ M µρ + η µρ M νσ − η νρ M µσ − η µσ M νρ . Any element λ ∈ o(3, 1) can be expressed as linear combination of the M µν , λ = 21 ωµν M µν where the indices on the coefficients ωµν are also anti-symmetric, and the factor of 12 ensures that the sum over all µ, ν counts each anti-symmetric pair only once. Then, we have Λ = exp(λ) = exp( 12 ωµν M µν ) ∈ O(3, 1). The subgroup of O(3, 1) consisting of the the space-orientation preserving Lorentz transformations, or proper Lorentz transformations, is denoted by SO(3, 1). The subgroup consisting of the time-orientation preserving, or orthochronous, Lorentz transformations is denoted by O+ (3, 1). The Lie group O(3, 1) is disconnected: its four connected components are i) SO+ (3, 1) := SO(3, 1) ∩ O+ (3, 1), also called the restricted Lorentz group, consisting of the proper orthochronous Lorentz transformations; ii) SO(3, 1) \ O+ (3, 1), the proper non-orthochronous transformations; iii) O+ (3, 1) \ SO(3, 1), the improper orthochronous transformations; iv) O(3, 1) \ (SO(3, 1) ∪ O+ (3, 1)), the improper non-orthochronous transformations.

3

Since idR4 ∈ SO+ (3, 1), we have exp(o(3, 1)) = SO+ (3, 1). Then {M µν } provides a nice co-ordinatisation of SO+ (3, 1) since, if we choose   0 ψ1 ψ2 ψ3 −ψ 0 ϕ −ϕ   1 3 2 [ωµν ] =   −ψ2 −ϕ3 0 ϕ1  −ψ3 ϕ2 −ϕ1 0 then the Lorentz transformation exp( 21 ωµν M µν ) ∈ SO+ (3, 1) corresponds to a boost in the (ψ1 , ψ2 , ψ3 ) direction and a space rotation by (ϕ1 , ϕ2 , ϕ3 ). Indeed, in physics one often thinks of the Lie group SO+ (3, 1) as being generated by {M µν }. ∼ A representation ρ : TidR4 SO+ (3, 1) − → End(R4 ) is given by ρ(M µν )ab := η νa δbµ − η µa δbν which is probably how you have seen the M µν themselves defined in some previous course on relativity theory. Using this representation, we get a corresponding representation R : SO+ (3, 1) → GL(R4 ) via the exponential map by defining R(Λ) = exp( 21 ωµν ρ(M µν )). Then, the map exp becomes the usual exponential (series) of matrices. Definition. A one-parameter subgroup of a Lie group G is a Lie group homomorphism ξ : R → G, with R understood as a Lie group under ordinary addition. Example 18.7. Let M be a smooth manifold and let Y ∈ Γ(T M ) be a complete vector field. The flow of Y is the smooth map Θ: R × M → M (λ, p) 7→ Θλ (p) := γp (λ), where λp is the maximal integral curve of Y through p. For a fixed p, we have Θ0 = idM ,

Θλ1 ◦ Θλ2 = Θλ1 +λ2 ,

Θ−λ = Θ−1 λ .

For each λ ∈ R, the map Θλ is a diffeomorphism M → M . Denoting by Diff(M ) the group (under composition) of the diffeomorphisms M → M , we have that the map ξ : R → Diff(M ) λ 7→ Θλ is a one-parameter subgroup of Diff(M ).

4

Theorem 18.8. Let G be a Lie group. i) Let A ∈ Te G. The map ξA : R → G λ 7→ ξ A (λ) := exp(λA) is a one-parameter subgroup. ii) Every one-parameter subgroup of G has the form ξ A for some A ∈ Te G. Therefore, the Lie algebra allows us to study all the one-parameter subgroups of the Lie group. Theorem 18.9. Let G and H be Lie groups and let φ : G → H be a Lie group homomorphism. Then, for all A ∈ TeG G, we have φ(exp(A)) = exp((φ∗ )eG A). Equivalently, the following diagram commutes. TeG G

(φ∗ )eG

exp

G

TeH H exp

φ

5

H