Lie groups and Lie algebras

Eckhard Meinrenken

Lecture Notes, University of Toronto, Fall 2010

1. Terminology and notation

1.1. Lie groups.

Definition 1.1. A Lie group is a group G, equipped with a manifold structure such that the group operations

Mult: G × G → G, (g_{1}, g_{2}) ↦ g_{1}g_{2}
Inv: G → G, g ↦ g^{−1}

are smooth. A morphism of Lie groups G, G′ is a morphism of groups ϕ: G → G′ that is smooth.

Remark 1.2. Using the implicit function theorem, one can show that smoothness of Inv is in fact automatic. (Exercise)

The first example of a Lie group is the general linear group

GL(n, ℝ) = {A ∈ Mat_{n}(ℝ)| det(A) ≠ 0}

of invertible n × n matrices. It is an open subset of Mat_{n}(ℝ), hence a submanifold, and the smoothness of group multiplication follows since the product map for Mat_{n}(ℝ) is obviously smooth.

Our next example is the orthogonal group

O(n) = {A ∈ Mat_{n}(ℝ)| A^{T} A = I}.

To see that it is a Lie group, it suffices to show that O(n) is an embedded submanifold of Mat_{n}(ℝ). In order to construct submanifold charts, we use the exponential map of matrices

exp: Mat_{n}(ℝ) → Mat_{n}(ℝ), B ↦ exp(B) = ∑_{n=0}^{∞} \frac{1}{n!}B^{n}

(an absolutely convergent series). One has \frac{d}{dt}|_{t=0} exp(tB) = B, hence the differential of exp at 0 is the identity id_{Mat_{n}(ℝ)}. By the inverse function theorem, this means that there is ϵ > 0 such that exp restricts to a diffeomorphism from the open neighborhood U = {B : ||B|| < ϵ} of 0 onto an open neighborhood exp(U) of I. Let

𝔬(n) = {B ∈ Mat_{n}(ℝ)| B + B^{T} = 0}.

We claim that

exp(𝔬(n) ∩ U) = O(n) ∩ exp(U),

so that exp gives a submanifold chart for O(n) over exp(U). To prove the claim, let B ∈ U. Then

exp(B) ∈ O(n) ⇔ exp(B)^{T} = exp(B)^{−1}
⇔ exp(B^{T}) = exp(−B)
⇔ B^{T} = −B
⇔ B ∈ 𝔬(n).

For a more general A ∈ O(n), we use that the map Mat_{n}(ℝ) → Mat_{n}(ℝ) given by left multiplication is a diffeomorphism. Hence, A exp(U) is an open neighborhood of A, and we have

A exp(U) ∩ O(n) = A(exp(U) ∩ O(n)) = A exp(U ∩ 𝔬(n)).

Thus, we also get a submanifold chart near A. This proves that O(n) is a submanifold. Hence its group operations are induced from those of GL(n, ℝ), they are smooth. Hence O(n) is a Lie group. Notice that O(n) is compact (the column vectors of an orthogonal matrix are an orthonormal basis of ℝ^{n}; hence O(n) is a subset of S^{n−1} × ⋯ S^{n−1} ⊂ ℝ^{n} × ⋯ ℝ^{n}).

A similar argument shows that the special linear group

SL(n, ℝ) = {A ∈ Mat_{n}(ℝ)| det(A) = 1}

is an embedded submanifold of GL(n, ℝ), and hence is a Lie group. The submanifold charts are obtained by exponentiating the subspace

𝔰𝔩(n, ℝ) = {B ∈ Mat_{n}(ℝ)| tr(B) = 0},

using the identity det(exp(B)) = exp(tr(B)).

Actually, we could have saved most of this work with O(n), SL(n, ℝ) once we have the following beautiful result of E. Cartan:

Fact: Every closed subgroup of a Lie group is an embedded submanifold, hence is again a Lie group.

We will prove this very soon, once we have developed some more basics of Lie group theory. A closed subgroup of GL(n, ℝ) (for suitable n) is called a matrix Lie group. Let us now give a few more examples of Lie groups, without detailed justifications.

Examples 1.3. (a) Any finite-dimensional vector space V over ℝ is a Lie group, with product

Mult given by addition.

(b) Let A be a finite-dimensional associative algebra over ℝ, with unit 1_{A}. Then the group A^{×} of invertible elements is a Lie group. More generally, if n ∈ ℕ we can create the algebra Mat_{n}(A) of matrices with entries in A, and the general linear group

GL(n, A) := Mat_{n}(A)^{×}

is a Lie group. If A is commutative, one has a determinant map det: Mat_{n}(A) → A, and GL(n, A) is the pre-image of A^{×}. One may then define a special linear group

SL(n, A) = {g ∈ GL(n, A)| det(g) = 1}.

(c) We mostly have in mind the cases A = ℝ, ℂ, ℍ. Here ℍ is the algebra of quaternions (due to Hamilton). Recall that ℍ = ℝ^{4} as a vector space, with elements (a, b, c, d) ∈ ℝ^{4} written as

x = a + ib + jc + kd

with imaginary units i, j, k. The algebra structure is determined by

i^{2} = j^{2} = k^{2} = −1, ij = k, jk = i, ki = j.

Note that ℍ is non-commutative (e.g. ji = −ij), hence SL(n, ℍ) is not defined. On the other hand, one can define complex conjugates

\overline{x} = a − ib − jc − kd

and

|x|^{2} := x\overline{x} = a^{2} + b^{2} + c^{2} + d^{2}.

defines a norm x ↦ |x|, with |x_{1}x_{2}| = |x_{1}||x_{2}| just as for complex or real numbers. The spaces ℝ^{n}, ℂ^{n}, ℍ^{n} inherit norms, by putting

||x||^{2} = ∑_{i=1}^{n} |x_{i}|^{2}, x = (x_{1}, …, x_{n}).

The subgroups of GL(n, ℝ), GL(n, ℂ), GL(n, ℍ) preserving this norm (in the sense that ||Ax|| = ||x|| for all x) are denoted

O(n), U(n), Sp(n)

and are called the orthogonal, unitary, and symplectic group, respectively. Since the norms of ℂ, ℍ coincide with those of ℂ ≅ ℝ^{2}, ℍ = ℂ^{2} ≅ ℝ^{4}, we have

U(n) = GL(n, ℂ) ∩ O(2n), Sp(n) = GL(n, ℍ) ∩ O(4n).

In particular, all of these groups are compact. One can also define

SO(n) = O(n) ∩ SL(n, ℝ), SU(n) = U(n) ∩ SL(n, ℂ),

these are called the special orthogonal and special unitary groups. The groups SO(n), SU(n), Sp(n) are often called the classical groups (but this term is used a bit loosely).

(d) For any Lie group G, its universal cover G̃ is again a Lie group. The universal cover { \atop SL(2, ℝ)} is an example of a Lie group that is not isomorphic to a matrix Lie group.

1.2. Lie algebras.

Definition 1.4. A Lie algebra is a vector space 𝔤, together with a bilinear map [·, ·] : 𝔤 × 𝔤 → 𝔤 satisfying anti-symmetry

[ξ, η] = −[η, ξ] for all ξ, η ∈ 𝔤,

and the Jacobi identity,

[ξ, [η, ζ]] + [η, [ζ, ξ]] + [ζ, [ξ, η]] = 0 for all ξ, η, ζ ∈ 𝔤.

The map [·, ·] is called the Lie bracket. A morphism of Lie algebras 𝔤_{1}, 𝔤_{2} is a linear map ϕ: 𝔤_{1} → 𝔤_{2} preserving brackets.

The space

𝔤𝔩(n, ℝ) = Mat_{n}(ℝ)

is a Lie algebra, with bracket the commutator of matrices. (The notation indicates that we think of Mat_{n}(ℝ) as a Lie algebra, not as an algebra.)

A Lie subalgebra of 𝔤𝔩(n, ℝ), i.e. a subspace preserved under commutators, is called a matrix Lie algebra. For instance,

𝔰𝔩(n, ℝ) = {B ∈ Mat_{n}(ℝ): tr(B) = 0}

and

𝔬(n) = {B ∈ Mat_{n}(ℝ): B^{T} = −B}

are matrix Lie algebras (as one easily verifies). It turns out that every finite-dimensional real Lie algebra is isomorphic to a matrix Lie algebra (Ado’s theorem), but the proof is not easy.

The following examples of finite-dimensional Lie algebras correspond to our examples for Lie groups. The origin of this correspondence will soon become clear.

Examples 1.5. (a) Any vector space V is a Lie algebra for the zero bracket.

(b) Any associative algebra A can be viewed as a Lie algebra under commutator. Replacing A with matrix algebras over A, it follows that 𝔤𝔩(n, A) = Mat_{n}(A), is a Lie algebra, with bracket the commutator. If A is commutative, then the subspace 𝔰𝔩(n, A) ⊂ 𝔤𝔩(n, A) of matrices of trace 0 is a Lie subalgebra.

(c) We are mainly interested in the cases A = ℝ, ℂ, ℍ. Define an inner product on ℝ^{n}, ℂ^{n}, ℍ^{n} by putting

⟨x, y⟩ = ∑_{i=1}^{n} \overline{x}_{i}y_{i},

and define 𝔬(n), 𝔲(n), 𝔰𝔭(n) as the matrices in 𝔤𝔩(n, ℝ), 𝔤𝔩(n, ℂ), 𝔤𝔩(n, ℍ) satisfying

⟨Bx, y⟩ = −⟨x, By⟩

for all x, y. These are all Lie algebras called the (infinitesimal) orthogonal, unitary, and symplectic Lie algebras. For ℝ, ℂ one can impose the additional condition tr(B) = 0, thus defining the special orthogonal and special unitary Lie algebras 𝔰𝔬(n), 𝔰𝔲(n). Actually,

𝔰𝔬(n) = 𝔬(n)

since B^{T} = −B already implies tr(B) = 0.

Exercise 1.6. Show that Sp(n) can be characterized as follows. Let J ∈ U(2n) be the unitary matrix

( \begin{matrix}0 & I_{n} \\ −I_{n} & 0 \end{matrix} )

where I_{n} is the n × n identity matrix. Then

Sp(n) = \{A ∈ U(2n)| \overline{A} = JAJ^{−1}\}.

Here \overline{A} is the componentwise complex conjugate of A.

Exercise 1.7. Let R(θ) denote the 2 × 2 rotation matrix

R(θ) = ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) .

Show that for all A ∈ SO(2m) there exists O ∈ SO(2m) such that OAO^{−1} is of the block diagonal form

( \begin{matrix}R(θ_{1}) & 0 & 0 & ⋯ & 0 \\ 0 & R(θ_{2}) & 0 & ⋯ & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & 0 & ⋯ & R(θ_{m}) \end{matrix} ) .

For A ∈ SO(2m + 1) one has a similar block diagonal presentation, with m 2 × 2 blocks R(θ_{i}) and an extra 1 in the lower right corner. Conclude that SO(n) is connected.

Exercise 1.8. Let G be a connected Lie group, and U an open neighborhood of the group unit e. Show that any g ∈ G can be written as a product g = g_{1} ⋯ g_{N} of elements g_{i} ∈ U.

Exercise 1.9. Let ϕ: G → H be a morphism of connected Lie groups, and assume that the differential d_{e}ϕ: T_{e}G → T_{e}H is bijective (resp. surjective). Show that ϕ is a covering (resp. surjective). Hint: Use Exercise ??.

2. The covering SU(2) → SO(3)

The Lie group SO(3) consists of rotations in 3-dimensional space. Let D ⊂ ℝ^{3} be the closed ball of radius π. Any element x ∈ D represents a rotation by an angle ||x|| in the direction of x. This is a 1-1 correspondence for points in the interior of D, but if x ∈ ∂D is a boundary point then x, −x represent the same rotation. Letting ∼ be the equivalence relation on D, given by antipodal identification on the boundary, we have D^{3}/ ∼= ℝP (3). Thus

SO(3) = ℝP (3)

(at least, topologically). With a little extra effort (which we’ll make below) one can make this into a diffeomorphism of manifolds.

By definition

SU(2) = {A ∈ Mat_{2}(ℂ)| A^{†} = A^{−1}, det(A) = 1}.

Using the formula for the inverse matrix, we see that SU(2) consists of matrices of the form

SU(2) = \{( \begin{matrix}z & −\overline{w} \\ w & \overline{z} \end{matrix} ) | |w|^{2} + |z|^{2} = 1\} .

That is, SU(2) = S^{3} as a manifold. Similarly,

𝔰𝔲(2) = \{( \begin{matrix}it & −\overline{u} \\ u & −it \end{matrix} ) | t ∈ ℝ, u ∈ ℂ\}

gives an identification 𝔰𝔲(2) = ℝ ⊕ ℂ = ℝ^{3}. Note that for a matrix B of this form, det(B) = t^{2} + |u|^{2}, so that det corresponds to || · ||^{2} under this identification.

The group SU(2) acts linearly on the vector space 𝔰𝔲(2), by matrix conjugation: B ↦ ABA^{−1}. Since the conjugation action preserves det, we obtain a linear action on ℝ^{3}, preserving the norm. This defines a Lie group morphism from SU(2) into O(3). Since SU(2) is connected, this must take values in the identity component:

ϕ: SU(2) → SO(3).

The kernel of this map consists of matrices A ∈ SU(2) such that ABA^{−1} = B for all B ∈ 𝔰𝔲(2). Thus, A commutes with all skew-adjoint matrices of trace 0. Since A commutes with multiples of the identity, it then commutes with all skew-adjoint matrices. But since Mat_{n}(ℂ) = 𝔲(n) ⊕ i𝔲(n) (the decomposition into skew-adjoint and self-adjoint parts), it then follows that A is a multiple of the identity matrix. Thus ker(ϕ) = {I, −I} is discrete. Since d_{e}ϕ is an isomorphism, it follows that the map ϕ is a double covering. This exhibits SU(2) = S^{3} as the double cover of SO(3).

Exercise 2.1. Give an explicit construction of a double covering of SO(4) by SU(2) × SU(2). Hint: Represent the quaternion algebra ℍ as an algebra of matrices ℍ ⊂ Mat_{2}(ℂ), by

x = a + ib + jc + kd ↦ x = ( \begin{matrix}a + ib & c + id \\ −c + id & a − ib \end{matrix} ) .

Note that |x|^{2} = det(x), and that SU(2) = {x ∈ ℍ| det(x) = 1}. Use this to define an action of SU(2) × SU(2) on ℍ preserving the norm.

3. The Lie algebra of a Lie group

3.1. Review: Tangent vectors and vector fields. We begin with a quick reminder of some manifold theory, partly just to set up our notational conventions.

Let M be a manifold, and C^{∞}(M) its algebra of smooth real-valued functions. For m ∈ M, we define the tangent space T_{m}M to be the space of directional derivatives:

T_{m}M = {v ∈ Hom(C^{∞}(M), ℝ)| v(fg) = v(f)g + v(g)f}.

Here v(f) is local, in the sense that v(f) = v(f′) if f′ − f vanishes on a neighborhood of m.

Example 3.1. If γ: J → M, J ⊂ ℝ is a smooth curve we obtain tangent vectors to the curve,

γ̇(t) ∈ T_{γ(t)}M, γ̇(t)(f) = \frac{∂}{∂t}|_{t=0}f(γ(t)).

Example 3.2. We have T_{x}ℝ^{n} = ℝ^{n}, where the isomorphism takes a ∈ ℝ^{n} to the corresponding velocity vector of the curve x + ta. That is,

v(f) = \frac{∂}{∂t}|_{t=0}f(x + ta) = ∑_{i=1}^{n} a_{i} \frac{∂f}{∂x_{i}} .

A smooth map of manifolds ϕ: M → M′ defines a tangent map:

d_{m}ϕ: T_{m}M → T_{ϕ(m)}M′, (d_{m}ϕ(v))(f) = v(f ∘ ϕ).

The locality property ensures that for an open neighborhood U ⊂ M, the inclusion identifies T_{m}U = T_{m}M. In particular, a coordinate chart ϕ: U → ϕ(U) ⊂ ℝ^{n} gives an isomorphism

d_{m}ϕ: T_{m}M = T_{m}U → T_{ϕ(m)}ϕ(U) = T_{ϕ(m)}ℝ^{n} = ℝ^{n}.

Hence T_{m}M is a vector space of dimension n = dim M. The union T M = ⋃_{m∈M} T_{m}M is a vector bundle over M, called the tangent bundle. Coordinate charts for M give vector bundle charts for T M. For a smooth map of manifolds ϕ: M → M′, the entirety of all maps d_{m}ϕ defines a smooth vector bundle map

dϕ: T M → T M′.

A vector field on M is a derivation X: C^{∞}(M) → C^{∞}(M). That is, it is a linear map satisfying

X(fg) = X(f)g + fX(g).

The space of vector fields is denoted 𝔛(M) = Der(C^{∞}(M)). Vector fields are local, in the sense that for any open subset U there is a well-defined restriction X|_{U} ∈ 𝔛(U) such that X|_{U}(f|_{U}) = (X(f))|_{U}. For any vector field, one obtains tangent vectors X_{m} ∈ T_{m}M by X_{m}(f) = X(f)|_{m}. One can think of a vector field as an assignment of tangent vectors, depending smoothly on m. More precisely, a vector field is a smooth section of the tangent bundle T M. In local coordinates, vector fields are of the form ∑_{i} a_{i} \frac{∂}{∂x_{i}} where the a_{i} are smooth functions.

It is a general fact that the commutator of derivations of an algebra is again a derivation. Thus, 𝔛(M) is a Lie algebra for the bracket

[X, Y] = X ∘ Y − Y ∘ X.

In general, smooth maps ϕ: M → M′ of manifolds do not induce maps of the Lie algebras of vector fields (unless ϕ is a diffeomorphism). One makes the following definition.

Definition 3.3. Let ϕ: M → N be a smooth map. Vector fields X, Y on M, N are called ϕ-related, written X ∼_{ϕ} Y, if

X(f ∘ ϕ) = Y (f) ∘ ϕ

for all f ∈ C^{∞}(M′).

In short, X ∘ ϕ^{∗} = ϕ^{∗} ∘ Y where ϕ^{∗}: C^{∞}(N) → C^{∞}(M), f ↦ f ∘ ϕ.

One has X ∼_{ϕ} Y if and only if Y_{ϕ(m)} = d_{m}ϕ(X_{m}). From the definitions, one checks

X_{1} ∼_{ϕ} Y_{1}, X_{2} ∼_{ϕ} Y_{2} ⇒ [X_{1}, X_{2}] ∼_{ϕ} [Y_{1}, Y_{2}].

Example 3.4. Let j: S ↪ M be an embedded submanifold. We say that a vector field X is tangent to S if X_{m} ∈ T_{m}S ⊂ T_{m}M for all m ∈ S. We claim that if two vector fields are tangent to S then so is their Lie bracket. That is, the vector fields on M that are tangent to S form a Lie subalgebra.

Indeed, the definition means that there exists a vector field X_{S} ∈ 𝔛(S) such that X_{S} ∼_{j} X. Hence, if X, Y are tangent to S, then [X_{S}, Y_{S}] ∼_{j} [X, Y], so [X_{S}, Y_{S}] is tangent.

Similarly, the vector fields vanishing on S are a Lie subalgebra.

Let X ∈ 𝔛(M). A curve γ(t), t ∈ J ⊂ ℝ is called an integral curve of X if for all t ∈ J,

γ̇(t) = X_{γ(t)}.

In local coordinates, this is an ODE \frac{dx_{i}}{dt} = a_{i}(x(t)). The existence and uniqueness theorem for ODE’s (applied in coordinate charts, and then patching the local solutions) shows that for any m ∈ M, there is a unique maximal integral curve γ(t), t ∈ J_{m} with γ(0) = m.

Definition 3.5. A vector field X is complete if for all m ∈ M, the maximal integral curve with γ(0) = m is defined for all t ∈ ℝ.

In this case, one obtains a smooth map

Φ: ℝ × M → M, (t, m) ↦ Φ_{t}(m)

such that γ(t) = Φ_{−t}(m) is the integral curve through m. The uniqueness property gives

Φ_{0} = Id, Φ_{t_{1}+t_{2}} = Φ_{t_{1}} ∘ Φ_{t_{2}}

i.e. t ↦ Φ_{t} is a group homomorphism. Conversely, given such a group homomorphism such that the map Φ is smooth, one obtains a vector field X by setting

X = \frac{∂}{∂t}|_{t=0}Φ_{−t}^{∗},

as operators on functions. That is, X(f)(m) = \frac{∂}{∂t}|_{t=0}f(Φ_{−t}(m)). ^{1}

The Lie bracket of vector fields measure the non-commutativity of their flows. In particular, if X, Y are complete vector fields, with flows Φ_{t}^{X}, Φ_{s}^{Y}, then [X, Y] = 0 if and only if

Φ_{t}^{X} ∘ Φ_{s}^{Y} = Φ_{s}^{Y} ∘ Φ_{t}^{X}.

In this case, X + Y is again a complete vector field with flow Φ_{t}^{X+Y} = Φ_{t}^{X} ∘ Φ_{t}^{Y}. (The right hand side defines a flow since the flows of X, Y commute, and the corresponding vector field is identified by taking a derivative at t = 0.)

3.2. The Lie algebra of a Lie group. Let G be a Lie group, and T G its tangent bundle. For all a ∈ G, the left,right translations

L_{a}: G → G, g ↦ ag
R_{a}: G → G, g ↦ ga

are smooth maps. Their differentials at e define isomorphisms d_{g}L_{a}: T_{g}G → T_{ag}G, and similarly for R_{a}. Let

𝔤 = T_{e}G

be the tangent space to the group unit.

A vector field X ∈ 𝔛(G) is called left-invariant if

X ∼_{L_{a}} X

for all a ∈ G, i.e. if it commutes with L_{a}^{∗}. The space 𝔛^{L}(G) of left-invariant vector fields is thus a Lie subalgebra of 𝔛(G). Similarly the space of right-invariant vector fields 𝔛^{R}(G) is a Lie subalgebra.

X.f = \frac{∂}{∂t} |_{t=0}Φ_{t}.f = \frac{∂}{∂t} |_{t=0}(Φ_{t}^{−1})^{∗}f.

1The minus sign is convention, but it is motivated as follows. Let Diff(M) be the infinite-dimensional group of diffeomorphisms of M. It acts on C^{∞}(M) by Φ.f = f ∘ Φ^{−1} = (Φ^{−1})^{∗}f. Here, the inverse is needed so that Φ_{1}.Φ_{2}.f = (Φ_{1}Φ_{2}).f. We think of vector fields as ‘infinitesimal flows’, i.e. informally as the tangent space at id to Diff(M). Hence, given a curve t ↦ Φ_{t} through Φ_{0} = id, smooth in the sense that the map ℝ × M → M, (t, m) ↦ Φ_{t}(m) is smooth, we define the corresponding vector field X = \frac{∂}{∂t} |_{t=0}Φ_{t} in terms of the action on functions: as If Φ_{t} is a flow, we have Φ_{t}^{−1} = Φ_{−t}.

Lemma 3.6. The map

𝔛^{L}(G) → 𝔤, X ↦ X_{e}

is an isomorphism of vector spaces. (Similarly for 𝔛^{R}(G).)

Proof. For a left-invariant vector field, X_{a} = (d_{e}L_{a})X_{e}, hence the map is injective. To show that it is surjective, let ξ ∈ 𝔤, and put X_{a} = (d_{e}L_{a})ξ ∈ T_{a}G. We have to show that the map G → T G, a ↦ X_{a} is smooth. It is the composition of the map G → G × 𝔤, g ↦ (g, ξ) (which is obviously smooth) with the map G × 𝔤 → T G, (g, ξ) ↦ d_{e}L_{g}(ξ). The latter map is the restriction of d Mult: T G × T G → T G to G × 𝔤 ⊂ T G × T G, and hence is smooth. □

We denote by ξ^{L} ∈ 𝔛^{L}(G), ξ^{R} ∈ 𝔛^{R}(G) the left,right invariant vector fields defined by ξ ∈ 𝔤. Thus

ξ^{L}|_{e} = ξ^{R}|_{e} = ξ

Definition 3.7. The Lie algebra of a Lie group G is the vector space 𝔤 = T_{e}G, equipped with the unique bracket such that

[ξ, η]^{L} = [ξ^{L}, η^{L}], ξ ∈ 𝔤.

Remark 3.8. If you use the right-invariant vector fields to define the bracket on 𝔤, we get a minus sign. Indeed, note that Inv: G → G takes left translations to right translations. Thus, ξ^{R} is Inv-related to some left invariant vector field. Since d_{e} Inv = − Id, we see ξ^{R} ∼_{Inv} −ξ^{L}. Consequently,

[ξ^{R}, η^{R}] ∼_{Inv} [−ξ^{L}, −η^{L}] = [ξ, η]^{L}.

But also −[ξ, η]^{R} ∼_{Inv} [ξ, η]^{L}, hence we get

[ξ^{R}, ζ^{R}] = −[ξ, ζ]^{R}.

The construction of a Lie algebra is compatible with morphisms. That is, we have a functor from Lie groups to finite-dimensional Lie algebras.

Theorem 3.9. For any morphism of Lie groups ϕ: G → G′, the tangent map d_{e}ϕ: 𝔤 → 𝔤′ is a morphism of Lie algebras. For all ξ ∈ 𝔤, ξ′ = d_{e}ϕ(ξ) one has

ξ^{L} ∼_{ϕ} (ξ′)^{L}, ξ^{R} ∼_{ϕ} (ξ′)^{R}.

Proof. Suppose ξ ∈ 𝔤, and let ξ′ = d_{e}ϕ(ξ) ∈ 𝔤′. The property ϕ(ab) = ϕ(a)ϕ(b) shows that L_{ϕ(a)} ∘ ϕ = ϕ ∘ L_{a}. Taking the differential at e, and applying to ξ we find (d_{e}L_{ϕ(a)})ξ′ = (d_{a}ϕ)(d_{e}L_{a}(ξ)) hence (ξ′)_{ϕ(a)}^{L} = (d_{a}ϕ)(ξ_{a}^{L}). That is ξ^{L} ∼_{ϕ} (ξ′)^{L}. The proof for right-invariant vector fields is similar. Since the Lie brackets of two pairs of ϕ-related vector fields are again ϕ-related, it follows that d_{e}ϕ is a Lie algebra morphism. □

Remark 3.10. Two special cases are worth pointing out.

(a) Let V be a finite-dimensional (real) vector space. A representation of a Lie group G on V is a Lie group morphism G → GL(V). A representation of a Lie algebra 𝔤 on V is a Lie algebra morphism 𝔤 → 𝔤𝔩(V). The Theorem shows that the differential of any Lie group representation is a representation of its a Lie algebra.

(b) An automorphism of a Lie group G is a Lie group morphism ϕ: G → G from G to itself, with ϕ a diffeomorphism. An automorphism of a Lie algebra is an invertible morphism from 𝔤 to itself. By the Theorem, the differential of any Lie group automorphism is an automorphism of its Lie algebra. As an example, SU(n) has a Lie group automorphism given by complex conjugation of matrices; its differential is a Lie algebra automorphism of 𝔰𝔲(n) given again by complex conjugation.

Exercise 3.11. Let ϕ: G → G be a Lie group automorphism. Show that its fixed point set is a closed subgroup of G, hence a Lie subgroup. Similarly for Lie algebra automorphisms. What is the fixed point set for the complex conjugation automorphism of SU(n)?

4. The exponential map

Theorem 4.1. The left-invariant vector fields ξ^{L} are complete, i.e. they define a flow Φ_{t}^{ξ} such that

ξ^{L} = \frac{∂}{∂t}|_{t=0}(Φ_{−t}^{ξ})^{∗}.

Letting ϕ^{ξ}(t) denote the unique integral curve with ϕ^{ξ}(0) = e. It has the property

ϕ^{ξ}(t_{1} + t_{2}) = ϕ^{ξ}(t_{1})ϕ^{ξ}(t_{2}),

and the flow of ξ^{L} is given by right translations:

Φ_{t}^{ξ}(g) = gϕ^{ξ}(−t).

Similarly, the right-invariant vector fields ξ^{R} are complete. ϕ^{ξ}(t) is an integral curve for ξ^{R} as well, and the flow of ξ^{R} is given by left translations, g ↦ ϕ^{ξ}(−t)g.

Proof. If γ(t), t ∈ J ⊂ ℝ is an integral curve of a left-invariant vector field ξ^{L}, then its left translates aγ(t) are again integral curves. In particular, for t_{0} ∈ J the curve t ↦ γ(t_{0})γ(t) is again an integral curve. Hence it coincides with γ(t_{0} + t) for all t ∈ J ∩ (J − t_{0}). In this way, an integral curve defined for small |t| can be extended to an integral curve for all t, i.e. ξ^{L} is complete.

Since ξ^{L} is left-invariant, so is its flow Φ_{t}^{ξ}. Hence

Φ_{t}^{ξ}(g) = Φ_{t}^{ξ} ∘ L_{g}(e) = L_{g} ∘ Φ_{t}^{ξ}(e) = gΦ_{t}^{ξ}(e) = gϕ^{ξ}(−t).

The property Φ_{t_{1}+t_{2}}^{ξ} = Φ_{t_{1}}^{ξ}Φ_{t_{2}}^{ξ} shows that ϕ^{ξ}(t_{1}+t_{2}) = ϕ^{ξ}(t_{1})ϕ^{ξ}(t_{2}). Finally, since ξ^{L} ∼_{Inv} −ξ^{R}, the image

Inv(ϕ^{ξ}(t)) = ϕ^{ξ}(t)^{−1} = ϕ^{ξ}(−t)

is an integral curve of −ξ^{R}. Equivalently, ϕ^{ξ}(t) is an integral curve of ξ^{R}. □

Since left and right translations commute, it follows in particular that

[ξ^{L}, η^{R}] = 0.

Definition 4.2. A 1-parameter subgroup of G is a group homomorphism ϕ: ℝ → G.

We have seen that every ξ ∈ 𝔤 defines a 1-parameter group, by taking the integral curve through e of the left-invariant vector field ξ^{L}. Every 1-parameter group arises in this way:

Proposition 4.3. If ϕ is a 1-parameter subgroup of G, then ϕ = ϕ^{ξ} where ξ = ϕ̇(0). One has

ϕ^{sξ}(t) = ϕ^{ξ}(st).

The map

ℝ × 𝔤 → G, (t, ξ) ↦ ϕ^{ξ}(t)

is smooth. Proof. Let ϕ(t) be a 1-parameter group. Then Φ_{t}(g) := gϕ(−t) defines a flow. Since this flow commutes with left translations, it is the flow of a left-invariant vector field, ξ^{L}. Here ξ is determined by taking the derivative of Φ_{−t}(e) = ϕ(t) at t = 0: ξ = ϕ̇(0). This shows ϕ = ϕ^{ξ}. As an application, since ψ(t) = ϕ^{ξ}(st) is a 1-parameter group with ψ̇^{ξ}(0) = sϕ̇^{ξ}(0) = sξ, we have ϕ^{ξ}(st) = ϕ^{sξ}(t). Smoothness of the map (t, ξ) ↦ ϕ^{ξ}(t) follows from the smooth dependence of solutions of ODE’s on parameters. □ Definition 4.4. The exponential map for the Lie group G is the smooth map defined by

exp: 𝔤 → G, ξ ↦ ϕ^{ξ}(1),

where ϕ^{ξ}(t) is the 1-parameter subgroup with ϕ̇^{ξ}(0) = ξ. Proposition 4.5. We have

ϕ^{ξ}(t) = exp(tξ).

If [ξ, η] = 0 then

exp(ξ + η) = exp(ξ) exp(η).

Proof. By the previous Proposition, ϕ^{ξ}(t) = ϕ^{tξ}(1) = exp(tξ). For the second claim, note that [ξ, η] = 0 implies that ξ^{L}, η^{L} commute. Hence their flows Φ_{t}^{ξ}, Φ_{t}^{η}, and Φ_{t}^{ξ} ∘ Φ_{t}^{η} is the flow of ξ^{L} + η^{L}. Hence it coincides with Φ_{t}^{ξ+η}. Applying to e, we get ϕ^{ξ}(t)ϕ^{η}(t) = ϕ^{ξ+η}(t). Now put t = 1. □

In terms of the exponential map, we may now write the flow of ξ^{L} as Φ_{t}^{ξ}(g) = g exp(−tξ), and similarly for the flow of ξ^{R}. That is,

ξ^{L} = \frac{∂}{∂t}|_{t=0}R_{exp(tξ)}^{∗}, ξ^{R} = \frac{∂}{∂t}|_{t=0}L_{exp(tξ)}^{∗}.

Proposition 4.6. The exponential map is functorial with respect to Lie group homomorphisms ϕ: G → H. That is, we have a commutative diagram

G ⟶^{ϕ} H
exp↑ ↑^{exp}
𝔤 ⟶_{d_{e}ϕ} 𝔥

Proof. t ↦ ϕ(exp(tξ)) is a 1-parameter subgroup of H, with differential at e given by

\frac{d}{dt}|_{t=0}ϕ(exp(tξ)) = d_{e}ϕ(ξ).

Hence ϕ(exp(tξ)) = exp(td_{e}ϕ(ξ)). Now put t = 1. □

Proposition 4.7. Let G ⊂ GL(n, ℝ) be a matrix Lie group, and 𝔤 ⊂ 𝔤𝔩(n, ℝ) its Lie algebra. Then exp: 𝔤 → G is just the exponential map for matrices,

exp(ξ) = ∑_{n=0}^{∞} \frac{1}{n!}ξ^{n}.

Furthermore, the Lie bracket on 𝔤 is just the commutator of matrices.

Proof. By the previous Proposition, applied to the inclusion of G in GL(n, ℝ), the exponential map for G is just the restriction of that for GL(n, ℝ). Hence it suffices to prove the claim for G = GL(n, ℝ). The function ∑_{n=0}^{∞} \frac{t^{n}}{n!} ξ^{n} is a 1-parameter group in GL(n, ℝ), with derivative at 0 equal to ξ ∈ 𝔤𝔩(n, ℝ). Hence it coincides with exp(tξ). Now put t = 1. □

Proposition 4.8. For a matrix Lie group G ⊂ GL(n, ℝ), the Lie bracket on 𝔤 = T_{I}G is just the commutator of matrices.

Proof. It suffices to prove for G = GL(n, ℝ). Using ξ^{L} = \frac{∂}{∂t} |_{t=0}R_{exp(tξ)}^{∗} we have

\frac{∂}{∂s}|_{s=0} \frac{∂}{∂t}|_{t=0}(R_{exp(−tξ)}^{∗}R_{exp(−sη)}^{∗}R_{exp(tξ)}^{∗}R_{exp(sη)}^{∗})

= \frac{∂}{∂s}|_{s=0}(R_{exp(−sη)}^{∗}ξ^{L}R_{exp(sη)}^{∗} − ξ^{L})

= ξ^{L}η^{L} − η^{L}ξ^{L}

= [ξ, η]^{L}.

On the other hand, write

R_{exp(−tξ)}^{∗}R_{exp(−sη)}^{∗}R_{exp(tξ)}^{∗}R_{exp(sη)}^{∗} = R_{exp(−tξ) exp(−sη) exp(tξ) exp(sη)}^{∗}.

Since the Lie group exponential map for GL(n, ℝ) coincides with the exponential map for matrices, we may use Taylor’s expansion,

exp(−tξ) exp(−sη) exp(tξ) exp(sη) = I + st(ξη − ηξ) + … = exp(st(ξη − ηξ)) + …

where … denotes terms that are cubic or higher in s, t. Hence

R_{exp(−tξ) exp(−sη) exp(tξ) exp(sη)}^{∗} = R_{exp(st(ξη−ηξ)}^{∗} + …

and consequently

\frac{∂}{∂s}|_{s=0} \frac{∂}{∂t}|_{t=0}R_{exp(−tξ) exp(−sη) exp(tξ) exp(sη)}^{∗} = \frac{∂}{∂s}|_{s=0} \frac{∂}{∂t}|_{t=0}R_{exp(st(ξη−ηξ))}^{∗} = (ξη − ηξ)^{L}.

We conclude that [ξ, η] = ξη − ηξ. □

Remark 4.9. Had we defined the Lie algebra using right-invariant vector fields, we would have obtained minus the commutator of matrices. Nonetheless, some authors use that convention.

The exponential map gives local coordinates for the group G on a neighborhood of e:

Proposition 4.10. The differential of the exponential map at the origin is d_{0} exp = id. As a consequence, there is an open neighborhood U of 0 ∈ 𝔤 such that the exponential map restricts to a diffeomorphism U → exp(U).

Proof. Let γ(t) = tξ. Then γ̇(0) = ξ since exp(γ(t)) = exp(tξ) is the 1-parameter group, we have

(d_{0} exp)(ξ) = \frac{∂}{∂t}|_{t=0} exp(tξ) = ξ.

□

Exercise 4.11. Show hat the exponential map for SU(n), SO(n) U(n) are surjective. (We will soon see that the exponential map for any compact, connected Lie group is surjective.)

Exercise 4.12. A matrix Lie group G ⊂ GL(n, ℝ) is called unipotent if for all A ∈ G, the matrix A − I is nilpotent (i.e. (A − I)^{r} = 0 for some r). The prototype of such a group are the upper triangular matrices with 1’s down the diagonal. Show that for a connected unipotent matrix Lie group, the exponential map is a diffeomorphism.

Exercise 4.13. Show that exp: 𝔤𝔩(2, ℂ) → GL(2, ℂ) is surjective. More generally, show that the exponential map for GL(n, ℂ) is surjective. (Hint: First conjugate the given matrix into Jordan normal form).

Exercise 4.14. Show that exp: 𝔰𝔩(2, ℝ) → SL(2, ℝ) is not surjective, by proving that the ma-

trices ( \begin{matrix}−1 & ±1 \\ 0 & −1 \end{matrix} ) ∈ SL(2, ℝ) are not in the image. (Hint: Assuming these matrices are of

the form exp(B), what would the eigenvalues of B have to be?) Show that these two matrices represent all conjugacy classes of elements that are not in the image of exp. (Hint: Find a classification of the conjugacy classes of SL(2, ℝ), e.g. in terms of eigenvalues.)

5. Cartan’s theorem on closed subgroups

Using the exponential map, we are now in position to prove Cartan’s theorem on closed subgroups.

Theorem 5.1. Let H be a closed subgroup of a Lie group G. Then H is an embedded submanifold, and hence is a Lie subgroup.

We first need a Lemma. Let V be a Euclidean vector space, and S(V) its unit sphere. For v ∈ V \\{0\}, let [v] = \frac{v}{||v||} ∈ S(V).

Lemma 5.2. Let v_{n}, v ∈ V \{0} with lim_{n→∞} v_{n} = 0. Then

lim_{n→∞}[v_{n}] = [v] ⇔ ∃a_{n} ∈ ℕ: lim_{n→∞} a_{n}v_{n} = v.

Proof. The implication ⇐ is obvious. For the opposite direction, suppose lim_{n→∞}[v_{n}] = [v]. Let a_{n} ∈ ℕ be defined by a_{n} − 1 < \frac{||v||}{||v_{n}||} ≤ a_{n}. Since v_{n} → 0, we have lim_{n→∞} a_{n} \frac{||v_{n}||}{||v||} = 1, and

a_{n}v_{n} = (a_{n} \frac{||v_{n}||}{||v||}) [v_{n}] ||v|| → [v] ||v|| = v.  □

Proof of E. Cartan’s theorem. It suffices to construct a submanifold chart near e ∈ H. (By left translation, one then obtains submanifold charts near arbitrary a ∈ H.) Choose an inner product on 𝔤.

We begin with a candidate for the Lie algebra of H. Let W ⊂ 𝔤 be the subset such that ξ ∈ W if and only if either ξ = 0, or ξ ≠ 0 and there exists ξ_{n} ≠ 0 with

exp(ξ_{n}) ∈ H, ξ_{n} → 0, [ξ_{n}] → [ξ].

We will now show the following:

(i) exp(W) ⊂ H, (ii) W is a subspace of 𝔤, (iii) There is an open neighborhood U of 0 and a diffeomorphism ϕ: U → ϕ(U) ⊂ G with

ϕ(0) = e such that

ϕ(U ∩ W) = ϕ(U) ∩ H.

(Thus ϕ defines a submanifold chart near e.) Step (i). Let ξ ∈ W \{0}, with sequence ξ_{n} as in the definition of W. By the Lemma, there are a_{n} ∈ ℕ with a_{n}ξ_{n} → ξ. Since exp(a_{n}ξ_{n}) = exp(ξ_{n})^{a_{n}} ∈ H, and H is closed, it follows that

exp(ξ) = lim_{n→∞} exp(a_{n}ξ_{n}) ∈ H.

Step (ii). Since the subset W is invariant under scalar multiplication, we just have to show that it is closed under addition. Suppose ξ, η ∈ W. To show that ξ + η ∈ W, we may assume that ξ, η, ξ + η are all non-zero. For t sufficiently small, we have

exp(tξ) exp(tη) = exp(u(t))

for some smooth curve t ↦ u(t) ∈ 𝔤 with u(0) = 0. Then exp(u(t)) ∈ H and

lim_{n→∞} n u(\frac{1}{n}) = lim_{h→0} \frac{u(h)}{h} = u̇(0) = ξ + η.

hence u(\frac{1}{n}) → 0, exp(u(\frac{1}{n}) ∈ H, [u(\frac{1}{n})] → [ξ + η]. This shows [ξ + η] ∈ W, proving (ii). Step (iii). Let W′ be a complement to W in 𝔤, and define

ϕ: 𝔤 ≅ W ⊕ W′ → G, ϕ(ξ + ξ′) = exp(ξ) exp(ξ′).

Since d_{0}ϕ is the identity, there is an open neighborhood U ⊂ 𝔤 of 0 such that ϕ: U → ϕ(U) is a diffeomorphism. It is automatic that ϕ(W ∩ U) ⊂ ϕ(W) ∩ ϕ(U) ⊂ H ∩ ϕ(U). We want to show that we can take U sufficiently small so that we also have the opposite inclusion

H ∩ ϕ(U) ⊂ ϕ(W ∩ U).

Suppose not. Then, any neighborhood U_{n} ⊂ 𝔤 = W ⊕ W′ of 0 contains an element (η_{n}, η′_{n}) such that

ϕ(η_{n}, η′_{n}) = exp(η_{n}) exp(η′_{n}) ∈ H

(i.e. exp(η′_{n}) ∈ H) but (η_{n}, η′_{n}) ∉ W (i.e. η′_{n} ≠ 0). Thus, taking U_{n} to be a nested sequence of neighborhoods with intersection {0}, we could construct a sequence η′_{n} ∈ W′ − {0} with η′_{n} → 0 and exp(η′_{n}) ∈ H. Passing to a subsequence we may assume that [η′_{n}] → [η] for some η ∈ W′\{0}. On the other hand, such a convergence would mean η ∈ W, by definition of W. Contradiction. □

As remarked earlier, Cartan’s theorem is very useful in practice. For a given Lie group G, the term ‘closed subgroup’ is often used as synonymous to ‘embedded Lie subgroup’.

Examples 5.3. (a) The matrix groups G = O(n), Sp(n), SL(n, ℝ), … are all closed subgroups of some GL(N, ℝ), and hence are Lie groups.

(b) Suppose that ϕ: G → H is a morphism of Lie groups. Then ker(ϕ) = ϕ^{−1}(e) ⊂ G is a closed subgroup. Hence it is an embedded Lie subgroup of G.

(c) The center Z(G) of a Lie group G is the set of all a ∈ G such that ag = ga for all a ∈ G. It is a closed subgroup, and hence an embedded Lie subgroup.

(d) Suppose H ⊂ G is a closed subgroup. Its normalizer N_{G}(H) ⊂ G is the set of all a ∈ G such that aH = Ha. (I.e. h ∈ H implies aha^{−1} ∈ H.) This is a closed subgroup, hence a Lie subgroup. The centralizer Z_{G}(H) is the set of all a ∈ G such that ah = ha for all h ∈ H, it too is a closed subgroup, hence a Lie subgroup.

The E. Cartan theorem is just one of many ‘automatic smoothness’ results in Lie theory. Here is another.

Theorem 5.4. Let G, H be Lie groups, and ϕ: G → H be a continuous group morphism. Then ϕ is smooth.

As a corollary, a given topological group carries at most one smooth structure for which it is a Lie group. For profs of these (and stronger) statements, see the book by Duistermaat-Kolk.

6. The adjoint representations

6.1. Automorphisms. The group Aut(𝔤) of automorphisms of a Lie algebra 𝔤 is closed in the group End(𝔤)^{×} of vector space automorphisms, hence it is a Lie group. To identify its Lie algebra, let D ∈ End(𝔤) be such that exp(tD) ∈ Aut(𝔤) for t ∈ ℝ. Taking the derivative of the defining condition exp(tD)[ξ, η] = [exp(tD)ξ, exp(tD)η], we obtain the property

D[ξ, η] = [Dξ, η] + [ξ, Dη]

saying that D is a derivation of the Lie algebra. Conversely, if D is a derivation then

D^{n}[ξ, η] = ∑_{k=0}^{n} \binom{n}{k}[D^{k}ξ, D^{n−k}η]

by induction, which then shows that exp(D) = ∑_{n} \frac{D^{n}}{n!} is an automorphism. Hence the Lie algebra of Aut(𝔤) is the Lie algebra Der(𝔤) of derivations of 𝔤.

Exercise 6.1. Using similar arguments, verify that the Lie algebra of SO(n), SU(n), Sp(n), … are 𝔰𝔬(n), 𝔰𝔲(n), 𝔰𝔭(n), . . ..

6.2. The adjoint representation of G. Recall that an automorphism of a Lie group G is an invertible morphism from G to itself. The automorphisms form a group Aut(G). Any a ∈ G defines an ‘inner’ automorphism Ad_{a} ∈ Aut(G) by conjugation:

Ad_{a}(g) = aga^{−1}

Indeed, Ad_{a} is an automorphism since Ad_{a}^{−1} = Ad_{a^{−1}} and

Ad_{a}(g_{1}g_{2}) = ag_{1}g_{2}a^{−1} = ag_{1}a^{−1}ag_{2}a^{−1} = Ad_{a}(g_{1}) Ad_{a}(g_{2}).

Note also that Ad_{a_{1}a_{2}} = Ad_{a_{1}} Ad_{a_{2}}, thus we have a group morphism

Ad: G → Aut(G)

into the group of automorphisms. The kernel of this morphism is the center Z(G), the image is (by definition) the subgroup Int(G) of inner automorphisms. Note that for any ϕ ∈ Aut(G), and any a ∈ G,

ϕ ∘ Ad_{a} ∘ϕ^{−1} = Ad_{ϕ(a)} .

That is, Int(G) is a normal subgroup of Aut(G). (I.e. the conjugate of an inner automorphism by any automorphism is inner.) It follows that Out(G) = Aut(G)/ Int(G) inherits a group structure; it is called the outer automorphism group.

Example 6.2. If G = SU(2) the complex conjugation of matrices is an inner automorphism, but for G = SU(n) with n ≥ 3 it cannot be inner (since an inner automorphism has to preserve the spectrum of a matrix). Indeed, one know that Out(SU(n)) = ℤ_{2} for n ≥ 3.

The differential of the automorphism Ad_{a}: G → G is a Lie algebra automorphism, denoted by the same letter: Ad_{a} = d_{e} Ad_{a}: 𝔤 → 𝔤. The resulting map

Ad: G → Aut(𝔤)

is called the adjoint representation of G. Since the Ad_{a} are Lie algebra/group morphisms, they are compatible with the exponential map,

exp(Ad_{a} ξ) = Ad_{a} exp(ξ).

Remark 6.3. If G ⊂ GL(n, ℝ) is a matrix Lie group, then Ad_{a} ∈ Aut(𝔤) is the conjugation of matrices

Ad_{a}(ξ) = aξa^{−1}.

This follows by taking the derivative of Ad_{a}(exp(tξ)) = a exp(tξ)a^{−1}, using that exp is just the exponential series for matrices.

6.3. The adjoint representation of 𝔤. Let Der(𝔤) be the Lie algebra of derivations of the Lie algebra 𝔤. There is a Lie algebra morphism,

ad: 𝔤 → Der(𝔤), ξ ↦ [ξ, ·].

The fact that ad_{ξ} is a derivation follows from the Jacobi identity; the fact that ξ ↦ ad_{ξ} it is a Lie algebra morphism is again the Jacobi identity. The kernel of ad is the center of the Lie algebra 𝔤, i.e. elements having zero bracket with all elements of 𝔤, while the image is the Lie subalgebra Int(𝔤) ⊂ Der(𝔤) of inner derivations. It is a normal Lie subalgebra, i.e [Der(𝔤), Int(𝔤)] ⊂ Int(𝔤), and the quotient Lie algebra Out(𝔤) are the outer automorphims.

Suppose now that G is a Lie group, with Lie algebra 𝔤. We have remarked above that the Lie algebra of Aut(𝔤) is Der(𝔤). Recall that the differential of any G-representation is a 𝔤-representation. In particular, we can consider the differential of G → Aut(𝔤).

Theorem 6.4. If 𝔤 is the Lie algebra of G, then the adjoint representation ad: 𝔤 → Der(𝔤) is the differential of the adjoint representation Ad: G → Aut(𝔤). One has the equality of operators

exp(ad_{ξ}) = Ad(exp ξ)

for all ξ ∈ 𝔤.

Proof. For the first part we have to show \frac{∂}{∂t} |_{t=0} Ad_{exp(tξ)} η = ad_{ξ} η. This is easy if G is a matrix Lie group:

\frac{∂}{∂t}|_{t=0} Ad_{exp(tξ)} η = \frac{∂}{∂t}|_{t=0} exp(tξ)η exp(−tξ) = ξη − ηξ = [ξ, η].

For general Lie groups we compute, using

exp(s Ad_{exp(tξ)} η) = Ad_{exp(tξ)} exp(sη) = exp(tξ) exp(sη) exp(−tξ),
\frac{∂}{∂t}|_{t=0}(Ad_{exp(tξ)} η)^{L} = \frac{∂}{∂t}|_{t=0} \frac{∂}{∂s}|_{s=0}R_{exp(s Ad_{exp(tξ)} η)}^{∗}

= \frac{∂}{∂t}|_{t=0} \frac{∂}{∂s}|_{s=0}R_{exp(tξ) exp(sη) exp(−tξ)}^{∗}

= \frac{∂}{∂t}|_{t=0} \frac{∂}{∂s}|_{s=0}R_{exp(tξ)}^{∗}R_{exp(sη)}^{∗}R_{exp(−tξ)}^{∗}

= \frac{∂}{∂t}|_{t=0}R_{exp(tξ)}^{∗} η^{L} R_{exp(−tξ)}^{∗}

= [ξ^{L}, η^{L}]

= [ξ, η]^{L} = (ad_{ξ} η)^{L}.

This proves the first part. The second part is the commutativity of the diagram

G ⟶^{Ad} Aut(𝔤)
exp↑  ↑^{exp}
𝔤 ⟶_{ad} Der(𝔤)

which is just a special case of the functoriality property of exp with respect to Lie group morphisms. □

Remark 6.5. As a special case, this formula holds for matrices. That is, for B, C ∈ Mat_{n}(ℝ),

e^{B} C e^{−B} = ∑_{n=0}^{∞} \frac{1}{n!}[B, [B, ⋯ [B, C] ⋯]].

The formula also holds in some other contexts, e.g. if B, C are elements of an algebra with B nilpotent (i.e. B^{N} = 0 for some N). In this case, both the exponential series for e^{B} and the series on the right hand side are finite. (Indeed, [B, [B, ⋯ [B, C] ⋯]] with n B’s is a sum of terms B^{j}CB^{n−j}, and hence must vanish if n ≥ 2N.)

7. The differential of the exponential map

We had seen that d_{0} exp = id. More generally, one can derive a formula for the differential of the exponential map at arbitrary points ξ ∈ 𝔤,

d_{ξ} exp: 𝔤 = T_{ξ}𝔤 → T_{exp ξ}G.

Using left translation, we can move T_{exp ξ}G back to 𝔤, and obtain an endomorphism of 𝔤.

Theorem 7.1. The differential of the exponential map exp: 𝔤 → G at ξ ∈ 𝔤 is the linear operator d_{ξ} exp: 𝔤 → T_{exp(ξ)}𝔤 given by the formula,

d_{ξ} exp = (d_{e}L_{exp ξ}) ∘ \frac{1 − exp(− ad_{ξ})}{ad_{ξ}}.

Here the operator on the right hand side is defined to be the result of substituting ad_{ξ} in the entire holomorphic function \frac{1−e^{−z}}{z}. Equivalently, it may be written as an integral

\frac{1 − exp(− ad_{ξ})}{ad_{ξ}} = ∫_{0}^{1} ds exp(−s ad_{ξ}).

Proof. We have to show that for all ξ, η ∈ 𝔤,

(d_{ξ} exp)(η) ∘ L_{exp(−ξ)}^{∗} = ∫_{0}^{1} ds (exp(−s ad_{ξ})η)

as operators on functions f ∈ C^{∞}(G). To compute the left had side, write (d_{ξ} exp)(η) ∘ L_{exp(−ξ)}^{∗}(f) = \frac{∂}{∂t}|_{t=0}(L_{exp(−ξ)}^{∗}(f))(exp(ξ + tη)) = \frac{∂}{∂t}|_{t=0}f(exp(−ξ) exp(ξ + tη)). We think of this as the value of \frac{∂}{∂t} |_{t=0}R_{exp(−ξ)}^{∗}R_{exp(ξ+tη)}^{∗}f at e, and compute as follows: ^{2}

\frac{∂}{∂t}|_{t=0}R_{exp(−ξ)}^{∗}R_{exp(ξ+tη)}^{∗} = ∫_{0}^{1} ds \frac{∂}{∂t}|_{t=0} \frac{∂}{∂s}R_{exp(−sξ)}^{∗}R_{exp(s(ξ+tη)}^{∗}

= ∫_{0}^{1} ds \frac{∂}{∂t}|_{t=0}R_{exp(−sξ)}^{∗}(tη)^{L}R_{exp(s(ξ+tη)}^{∗}

= ∫_{0}^{1} ds R_{exp(−sξ)}^{∗} η^{L} R_{exp(s(ξ))}^{∗} = ∫_{0}^{1} ds (Ad_{exp(−sξ)} η)^{L} = ∫_{0}^{1} ds (exp(−s ad_{ξ})η)^{L}.

Applying this result to f at e, we obtain ∫_{0}^{1} ds (exp(−s ad_{ξ})η)(f) as desired. □ Corollary 7.2. The exponential map is a local diffeomorphism near ξ ∈ 𝔤 if and only if ad_{ξ} has no eigenvalue in the set 2πiℤ\{0}. Proof. d_{ξ} exp is an isomorphism if and only if \frac{1−exp(− ad_{ξ})}{ad_{ξ}} is invertible, i.e. has non-zero determinant. The determinant is given in terms of the eigenvalues of ad_{ξ} as a product, ∏_{λ} \frac{1−e^{−λ}}{λ}. This vanishes if and only if there is a non-zero eigenvalue λ with e^{λ} = 1. □

2We will use the identities \frac{∂}{∂s} R_{exp(sζ)}^{∗} = R_{exp(sζ)}^{∗} ζ^{L} = ζ^{L} R_{exp(sζ)}^{∗} for all ζ ∈ 𝔤. Proof: \frac{∂}{∂s} R_{exp(sζ)}^{∗} = \frac{∂}{∂u} |_{u=0}R_{exp((s+u)ζ)}^{∗} = \frac{∂}{∂u} |_{u=0}R_{exp(uζ)}^{∗}R_{exp(sζ)}^{∗} = ζ^{L}R_{exp(sζ)}^{∗}.

As an application, one obtains a version of the Baker-Campbell-Hausdorff formula. Let g ↦ log(g) be the inverse function to exp, defined for g close to e. For ξ, η ∈ 𝔤 close to 0, the function

log(exp(ξ) exp(η))

The BCH formula gives the Taylor series expansion of this function. The series starts out with

log(exp(ξ) exp(η)) = ξ + η + \frac{1}{2}[ξ, η] + ⋯

but gets rather complicated. To derive the formula, introduce a t-dependence, and let f(t, ξ, η) be defined by exp(ξ) exp(tη) = exp(f(t, ξ, η)) (for ξ, η sufficiently small). Thus

exp(f) = exp(ξ) exp(tη)

We have, on the one hand,

(d_{e}L_{exp(f)})^{−1} \frac{∂}{∂t} exp(f) = d_{e}L_{exp(tη)}^{−1} \frac{∂}{∂t} exp(tη) = η.

On the other hand, by the formula for the differential of exp,

(d_{e}L_{exp(f)})^{−1} \frac{∂}{∂t} exp(f) = (d_{e}L_{exp(f)})^{−1}(d_{f} exp)(\frac{∂f}{∂t}) = \frac{1 − e^{− ad_{f}}}{ad_{f}} (\frac{∂f}{∂t}).

Hence

\frac{df}{dt} = \frac{ad_{f}}{1 − e^{− ad_{f}}} η.

Letting χ be the function, holomorphic near w = 1,

χ(w) = \frac{log(w)}{1 − w^{−1}} = 1 + ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n(n + 1)}(w − 1)^{n},

we may write the right hand side as χ(e^{ad^{f}})η. By Applying Ad to the defining equation for f we obtain e^{ad^{f}} = e^{ad^{ξ}}e^{t ad_{η}}. Hence

\frac{df}{dt} = χ(e^{ad^{ξ}}e^{t ad_{η}})η.

Finally, integrating from 0 to 1 and using f(0) = ξ, f(1) = log(exp(ξ) exp(η)), we find:

log(exp(ξ) exp(η)) = ξ + ( ∫_{0}^{1} χ(e^{ad^{ξ}}e^{t ad_{η}})dt)η.

To work out the terms of the series, one puts

w − 1 = e^{ad^{ξ}}e^{t ad_{η}} − 1 = ∑_{i+j≥1} \frac{t^{j}}{i!j!} ad_{ξ}^{i} ad_{η}^{j}

in the power series expansion of χ, and integrates the resulting series in t. We arrive at:

Theorem 7.3 (Baker-Campbell-Hausdorff series). Let G be a Lie group, with exponential map exp: 𝔤 → G. For ξ, η ∈ 𝔤 sufficiently small we have the following formula

log(exp(ξ) exp(η)) = ξ + η + ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n(n + 1)}( ∫_{0}^{1} dt ( ∑_{i+j≥1} \frac{t^{j}}{i!j!} ad_{ξ}^{i} ad_{η}^{j} )^{n})η.

An important point is that the resulting Taylor series in ξ, η is a Lie series: all terms of the series are of the form of a constant times ad_{ξ}^{n_{1}} ad_{η}^{m_{2}} ⋯ ad_{ξ}^{n_{r}} η. The first few terms read,

log(exp(ξ) exp(η)) = ξ + η + \frac{1}{2}[ξ, η] + \frac{1}{12}[ξ, [ξ, η]] − \frac{1}{12}[η, [ξ, η]] + \frac{1}{24}[η, [ξ, [η, ξ]]] + ….

Exercise 7.4. Work out these terms from the formula.

There is a somewhat better version of the BCH formula, due to Dynkin. A good discussion can be found in the book by Onishchik-Vinberg, Chapter I.3.2.

8. Actions of Lie groups and Lie algebras

8.1. Lie group actions.

Definition 8.1. An action of a Lie group G on a manifold M is a group homomorphism

A: G → Diff(M), g ↦ A_{g}

into the group of diffeomorphisms on M, such that the action map

G × M → M, (g, m) ↦ A_{g}(m)

is smooth.

We will often write g.m rather than A_{g}(m). With this notation, g_{1}.(g_{2}.m) = (g_{1}g_{2}).m and e.m = m. A map Φ: M_{1} → M_{2} between G-manifolds is called G-equivariant if g.Φ(m) = Φ(g.m) for all m ∈ M, i.e. the following diagram commutes:

G × M_{1} ⟶ M_{1}
↓^{id ×Φ}  ↓^{Φ}
G × M_{2} ⟶ M_{2}

where the horizontal maps are the action maps.

Examples 8.2. (a) An ℝ-action on M is the same thing as a global flow.

(b) The group G acts M = G by right multiplication, A_{g} = R_{g^{−1}}, left multiplication, A_{g} = L_{g}, and by conjugation, A_{g} = Ad_{g} = L_{g} ∘ R_{g^{−1}}. The left and right action commute, hence they define an action of G×G. The conjugation action can be regarded as the action of the diagonal subgroup G ⊂ G × G.

(c) Any G-representation G → End(V) can be regarded as a G-action, by viewing M as a manifold.

(d) For any closed subgroup H ⊂ G, the space of right cosets G/H = {gH| g ∈ G} has a unique manifold structure such that the quotient map G → G/H is a smooth submersion, and the action of G by left multiplication on G descends to a smooth G-action on G/H. (Some ideas of teh proof will be explained below.)

(e) The defining represenation of the orthogonal group O(n) on ℝ^{n} restricts to an action on the unit sphere S^{n−1}, which in turn descends to an action on the projective space ℝP (n − 1). One also has actions on the Grassmann manifold Gr_{ℝ}(k, n) of k-planes in ℝ^{n}, or on the flag manifold Fl(n) (consisting of sequences {0} = V_{0} ⊂ V_{1} ⊂ ⋯ V_{n−1} ⊂ V_{n} = ℝ^{n} with dim V_{i} = i). These examples are all of the form O(n)/H for various choices of H. (E.g, for Gr(k, n) one takes H to be the subgroup preserving ℝ^{k} ⊂ ℝ^{n}.)

8.2. Lie algebra actions.

Definition 8.3. An action of a finite-dimensional Lie algebra 𝔤 on M is a Lie algebra homomorphism 𝔤 → 𝔛(M), ξ ↦ A_{ξ} such that the action map

𝔤 × M → T M, (ξ, m) ↦ A_{ξ}|_{m}

is smooth.

We will often write ξ_{M} =: A_{ξ} for the vector field corresponding to ξ. Thus, [ξ_{M}, η_{M}] = [ξ, η]_{M} for all ξ, η ∈ 𝔤. A smooth map Φ: M_{1} → M_{2} between to 𝔤-manifolds is called equivariant if ξ_{M_{1}} ∼_{Φ} ξ_{M_{2}} for all ξ ∈ 𝔤.

Examples 8.4. (a) Any vector field X defines an action of the Abelian Lie algebra ℝ, by

λ ↦ λX.

(b) Any Lie algebra representation ϕ: 𝔤 → gl(V) may be viewed as a Lie algebra action. Indeed, if f ∈ C^{∞}(V) we have d_{v}f ∈ V^{∗}, and

(A_{ξ}f)(v) = \frac{d}{dt}|_{t=0}f(v − tξ.v)

defines a 𝔤-action. (Exercise)

(c) For any Lie group G, we have actions of its Lie algebra 𝔤 by A_{ξ} = ξ^{L}, A_{ξ} = −ξ^{R} and A_{ξ} = ξ^{L} − ξ^{R}.

(d) Given a closed subgroup H ⊂ G, the vector fields −ξ^{R} ∈ 𝔛(G), ξ ∈ 𝔤 are invariant under the right multiplication, hence they are related under the quotient map to vector fields on G/H. That is, there is a unique 𝔤-action on G/H such that the quotient map G → G/H is equivariant.

Definition 8.5. Let G be a Lie group with Lie algebra 𝔤. Given a G-action g ↦ A_{g} on M, one defines its generating vector fields by

A_{ξ} = \frac{d}{dt}|_{t=0}A_{exp(−tξ)}^{∗}.

Example 8.6. The generating vector field for the action by right multiplication

A_{a} = R_{a^{−1}} ∈ Diff(G)

are the left-invariant vector fields,

A(ξ) = \frac{∂}{∂t}|_{t=0}R_{exp(tξ)}^{∗} = ξ^{L}.

Similarly, the generating vector fields for the action by left multiplication are −ξ^{R}, and those for the conjugation action are ξ^{L} − ξ^{R}.

Observe that if Φ: M_{1} → M_{2} is an equivariant map of G-manifolds, then the generating vector fields for the action are Φ-related.

Theorem 8.7. The generating vector fields of any G-action g → A_{g} define a 𝔤-action ξ → A_{ξ}.

Proof. Write ξ_{M} := A_{ξ} for the generating vector fields of a G-action on M. We have to show that ξ ↦ ξ_{M} is a Lie algebra morphism. Note that the map

Φ: G × M → M, (a, m) ↦ a^{−1}.m

is G-equivariant if we take the G-action on G×M to be g.(a, m) = (ag^{−1}, m). Hence ξ_{G×M} ∼_{Φ} ξ_{M}. But ξ_{G×M} = ξ^{L} (viewed as vector fields on the product G × M), hence ξ ↦ ξ_{G×M} is a Lie algebra morphism. It follows that

0 = [(ξ_{1})_{G×M}, (ξ_{1})_{G×M}] − ([ξ_{1}, ξ_{2}])_{G×M} ∼ [(ξ_{1})_{M}, (ξ_{2})_{M}] − [ξ_{1}, ξ_{2}]_{M}.

Since Φ is a surjective submersion (i.e. the differential dΦ: T (G × M) → T M is surjective), this shows that [(ξ_{1})_{M}, (ξ_{2})_{M}] − [ξ_{1}, ξ_{2}]_{M} = 0. □

8.3. Integrating Lie algebra actions. Let us now consider the inverse problem: For a Lie group G with Lie algebra 𝔤, integrating a given 𝔤-action to a G-action. The construction will use some facts about foliations.

Let M be a manifold. A k-dimensional distribution on M is a linear subspace ℜ ⊂ 𝔛(M) of the space of vector fields such that at any point m ∈ M, the subspace E_{m} ⊂ T_{m}M spanned by all X_{m}, X ∈ ℜ is of dimension k. The subspaces E_{m} define a rank k vector bundle E ⊂ T M with ℜ = Γ(E), hence a distribution is equivalently given by this subbundle E. An integral submanifold of the distribution ℜ is a k-dimensional submanifold S such that all X ∈ ℜ are tangent to S. In terms of E, this means that T_{m}S = E_{m} for all m ∈ S. The distribution is called integrable if for all m ∈ M there exists an integral submanifold containing m. In this case, there exists a maximal such submanifold, L_{m}. The decomposition of M into maximal integral submanifolds is called a k-dimensional foliation of M, the maximal integral submanifolds themselves are called the leaves of the foliation.

Not every distribution is integrable. Recall that if two vector fields are tangent to a submanifold, then so is their Lie bracket. Hence, a necessary condition for integrability of a distribution is that ℜ is a Lie subalgebra. Frobenius’ theorem gives the converse:

Theorem 8.8 (Frobenius theorem). A rank k distibution ℜ ⊂ 𝔛(M) is integrable if and only if ℜ is a Lie subalgebra.

The idea of proof is to show that if ℜ is a Lie subalgebra, then ℜ is spanned, near any m ∈ M, by k commuting vector fields. one then uses the flow of these vector fields to construct integral submanifold.

Exercise 8.9. Prove Frobenius’ theorem for distributions ℜ of rank k = 2. (Hint: If X ∈ ℜ with X_{m} ≠ 0, one can choose local coordinates such that X = \frac{∂}{∂x_{1}} . Given a second vector field Y ∈ ℜ, such that [X, Y] ∈ ℜ and X_{m}, Y_{m} are linearly independent, show that one can replace Y by some Z = aX + bY ∈ ℜ such that b_{m} ≠ 0 and [X, Z] = 0 on a neighborhood of m.)

Given a Lie algebra of dimension k and an effective 𝔤-action on M (i.e. ξ_{M} = 0 implies ξ = 0), one obtains an integrable rank k distribution ℜ as the span (over C^{∞}(M)) of the ξ_{M}’s. We use this to prove:

Theorem 8.10. Let G be a connected, simply connected Lie group with Lie algebra 𝔤. A Lie algebra action 𝔤 → 𝔛(M), ξ ↦ ξ_{M} integrates to an action of G if and only if the vector fields ξ_{M} are all complete.

Proof of the theorem. The idea of proof is as follows. Let M̂ = G × M, and pr_{1}, pr_{2} the projections to the two factors. A G-action on M defines a foliation of M̂ = G × M, with leafs the orbits of the diagonal action (where G acts on itself by left multiplication). Equivalently, the leaves are the fibers of the map M̂ → M, (g, m) ↦ g^{−1}.m. Hence they are indexed by the elements of m, as follows

L_{m} = {(g, g.m)| g ∈ G}.

pr_{1} restricts to diffeomorphisms π_{m}: L_{m} → G, and we recover the action as

g.m = pr_{2}(π_{m}^{−1}(g)).

Given a 𝔤-action, our plan is to construct the foliation from an integrable distribution. Consider the Lie algebra action on M̂ = G × M, given by

ξM̂ = (−ξ^{R}, ξ_{M}) ∈ 𝔛(G × M).

Note that the vector fields ξM̂ are complete, since ξ_{M} are by assumption complete: If Φ_{t}^{ξ} is the flow of ξ_{M}, the flow of ξM̂ = (−ξ^{R}, ξ_{M}) is given by

Φ̂_{t}^{ξ} = (L_{exp(tξ)}, Φ_{t}^{ξ}) ∈ Diff(G × M).

The action ξ ↦ ξM̂ is effective, hence it defines an integable dim G-dimensional distribution ℜ ⊂ 𝔛(M̂). Let L_{m} ↪ G × M be the unique leaf containing the point (e, m). Projection to the first factor induces a smooth map π_{m} : L_{m} → G.

The map π_{m} is surjective: Given g ∈ G write g = g_{r} … g_{1} where g_{i} = exp(ξ_{i}). The path Φ̂_{t}^{ξ_{1}}(e, m), t ∈ [0, 1] lies in L_{m}, and has end point (g_{1}, m_{1}) where m_{1} = Φ_{1}^{ξ_{1}}(m). Concatenation with the path Φ̂_{t}^{ξ_{2}}(g_{1}, m_{1}), t ∈ [0, 1] gives a (piecewise smooth) path from (m, e) to (g_{2}g_{1}, m_{2}) where m_{2} = Φ_{1}^{ξ_{2}}Φ_{1}^{ξ_{1}}(m). Proceeding in this manner, we obtain a piecewise smooth path in L_{m} from (e, m) to (g_{r} ⋯ g_{1}, m_{r}) = (g, m_{r}). This shows π_{m}^{−1}(g) ≠ ∅.

For any (g, x) ∈ L_{m} the tangent map d_{(g,x)}π_{m} is an isomorphism. Hence π_{m}: L_{m} → G is a (surjective) covering map. Since G is simply connected by assumption, we conclude that π_{m}: L_{m} → G is a diffeomorphism. We now define A_{g}(m) = pr_{2}(π_{m}^{−1}(g)). Concretely, the construction above shows that if g = exp(ξ_{r}) ⋯ exp(ξ_{1}) then

A_{g}(m) = (Φ_{1}^{ξ_{r}} ∘ ⋯ ∘ Φ_{1}^{ξ_{1}})(m).

From this description it is clear that A_{gh} = A_{g} ∘ A_{h}. □

Let us remark that, in general, one cannot drop the assumption that G is simply connected. Consider for example G = SU(2), with 𝔰𝔲(2)-action ξ ↦ −ξ^{R}. This exponentiates to an action of SU(2) by left multiplication. But 𝔰𝔲(2) ≅ 𝔰𝔬(3) as Lie algebras, and the action does not exponentiate to an action of the group SO(3).

As an important special case, we obtain:

Theorem 8.11. Let H, G be Lie groups, with Lie algebras 𝔥 → 𝔤. If H is connected and simply connected, then any Lie algebra morphism ϕ: 𝔥 → 𝔤 integrates uniquely to a Lie group morphism ψ: H → G.

Proof. Define an 𝔥-action on G by ξ ↦ −ϕ(ξ)^{R}. Since the right-invariant vector fields are complete, this action integrates to a Lie group action A: H → Diff(G). This action commutes with the action of G by right multiplication. Hence, A_{h}(g) = ψ(h)g where ψ(h) = A_{h}(e). The action property now shows ψ(h_{1})ψ(h_{2}) = ψ(h_{1}h_{2}), so that ψ: H → G is a Lie group morphism integrating ϕ. □

Corollary 8.12. Let G be a connected, simply connected Lie group, with Lie algebra 𝔤. Then any 𝔤-representation on a finite-dimensional vector space V integrates to a G-representation on V.

Proof. A 𝔤-representation on V is a Lie algebra morphism 𝔤 → End(V), hence it integrates to a Lie group morphism G → End(V)^{×}. □

By a Lie subgroup of a Lie group H, we mean a Lie group G together with an injective Lie group morphism G ↪ H. That is, the inclusion map need not be an embedding.

Lemma 8.13. Let 𝔤 ⊂ 𝔥 be a Lie subalgebra of a finite-dimensional Lie algebra, and H a Lie group integrating 𝔥. Then there exists a unique connected Lie subgroup G ⊂ H integrating 𝔤.

Proof. Consider the distribution on H spanned by the vector fields −ξ^{R}, ξ ∈ 𝔤. It is integrable, hence it defines a foliation of H. The leaves of any foliation carry a unique manifold structure such that the inclusion map is an immersion. Take G ⊂ H to be the leaf through e ∈ H, with this manifold structure. Explicitly, G consists of products exp(ξ_{r}) ⋯ exp(ξ_{1}) where ξ_{i} ∈ 𝔤. From this description it follows that G is a Lie group. □

By Ado’s theorem, any finite-dimensional Lie algebra 𝔤 is isomorphic to a matrix Lie algebra. We will skip the proof of this important (but relatively deep) result, since it involves a considerable amount of structure theory of Lie algebras. Given such a presentation 𝔤 ⊂ 𝔤𝔩(n, ℝ), the Lemma gives a Lie subgroup G ⊂ GL(n, ℝ) integrating 𝔤. Replacing G with its universal covering, this proves:

Theorem 8.14 (Lie’s third theorem). For any finite-dimensional real Lie algebra 𝔤, there exists a connected, simply connected Lie group G, unique up to isomorphism, having 𝔤 as its Lie algebra.

The book by Duistermaat-Kolk contains a different, more conceptual proof of Cartan’s theorem. This new proof has found important generalizations to the integration of Lie algebroids. In conjunction with the previous Theorem, Lie’s third theorem gives an equivalence between the categories of finite-dimensional Lie algebras 𝔤 and connected, simply-connected Lie groups G.

8.4. Proper actions. Let us quickly list some terminology for Lie group actions A: G → Diff(M). For any m ∈ M, the set G.m := {(g, m) g ∈ G} is called the orbit of m. The space M/G = {G.m| m ∈ M} is called the orbit space for the given action. It inherits a topology as a quotient space of M, but can be a very singular space. The action A is called transitive if there is only one orbit, i.e. M/G = pt. In this case, M is called a homogeneous space.

The subgroup G_{m} = {g ∈ G| g.m = m} is called the stabilizer of m. From the definition, it is clear that stabilizer subgroups are closed subgroups of G, hence are embedded Lie subgroups. In particular, the orbit G/G_{m} inherits a manifold structure. The inclusion of the orbit is smooth relative to this manifold structure. For any g ∈ G, the stabilizers of a point m and of its translate g.m are related by the adjoint action:

G_{g.m} = Ad_{g}(G_{m}).

The action is free if all stabilizers G_{m} are trivial. For instance, the actions of G by left or right multiplication on G are both free, but the conjugation action is not. The action A is effective if ker(A) = {e}, i.e. A_{g} = id_{M} implies g = e. For instance, the conjugation action of G on itself is effective if and only if the center of G is trivial.

The action A is called proper if the action map G × M → M is proper (i.e. pre-images of compact sets are compact). For example, the left or right actions of G on itself are proper. Note that for a proper G-action, the action of any closed subgroup H ⊂ G is still proper. Also, for G compact any G-action is proper.

For a proper action, the stabilizer groups G_{m} are compact since G_{m} may be viewed as the intersection of the closed subspace G×{m} ⊂ G×M with the preimage of {m} ∈ M under the action map. One can use this fact to construct slices for the action, i.e. G_{m}-invariant embedded submanifolds S ⊂ M with m ∈ S such that G.S is an open neighborhood of the orbit G.m, and such that gS ∩ S ≠ ∅ ⇔ g ∈ G_{m}. Slices give models a neighborhood of G.m in the orbit space, since (G.S)/G = S/G_{m}. In particular, if G_{m} is trivial, we see that a neighborhood of G.m ⊂ M/G is a manifold (modeled by S).

Theorem 8.15. For a free, proper action on a manifold M, the orbit space M/G inherits a manifold structure such that the quotient map M → M/G is a smooth submersion. Given an H-action on M that commutes with the G-action, the orbit space M/G inherits an H-action.

Example 8.16. Let H be a closed subgroup of G, acting on G by right multiplication. This action is proper, hence G/H is a manifold. The action of G by left multiplication commutes with the actuion of H, hence it descends to a smooth action on G/H.

Exercise 8.17. Show that the conjugation action of G is proper if and only if G is compact.

9. Universal cover

10. The universal enveloping algebra

11. Tori