Lie group–Lie algebra correspondence

In mathematics, Lie group–Lie algebra correspondence allows one to study Lie groups, which are geometric objects, in terms of Lie algebras, which are linear objects. In this article, a Lie group refers to a real Lie group. For the complex and p-adic cases, see complex Lie group and p-adic Lie group.

In this article, manifolds (in particular Lie groups) are assumed to be second countable; in particular, they have at most countably many connected components.

Basics

Let G be a Lie group. A vector field X on G is said to be invariant under left translations if, for any g, h in G,

$(dl_g)_h(X_h) = X_{gh}$

where $l_g: G \to G, x \mapsto gx$ and $(dl_g)_h: T_h G \to T_{gh} G$ is the differential of $l_g$ between tangent spaces. (In other words, it is $l_g$-related to itself for any g in G.)

Let $\operatorname{Lie}(G)$ be the set of all left-translation-invariant vector fields on G. It is a real vector space. Moreover it is closed under Lie bracket; i.e., $[X, Y]$ is left-translation-invariant if X, Y are. Thus, $\operatorname{Lie}(G)$ is a Lie subalgebra of the Lie algebra of all vector fields on G and is called the Lie algebra of G.

The left-invariance amounts to the fact that the vector bundle map over G

$\phi: G \times \operatorname{Lie}(G) \to TG$

given by $\phi(g, X) = X_g$ is an isomorphism. It follows that the canonical map

$\operatorname{Lie}(G) \to T_e G, \, X \mapsto X_e$

is an isomorphism of vector spaces and one usually identifies $\operatorname{Lie}(G)$ with $T_e G$. In particular, the dimension of G as a real manifold is the dimension of the vector space $\operatorname{Lie}(G)$, and $\operatorname{Lie}(G) = \operatorname{Lie}(G^0)$ where $G^0$ is the connected component of the identity element. (Note φ makes the tangent bundle $TG$ a Lie group isomorphic to $G \times \operatorname{Lie}(G)$.)

(There is also a third incarnation of $\operatorname{Lie}(G)$ as the Lie algebra of primitive elements of the Hopf algebra of distributions on G with support at the identity element; for this, see #Related constructions below.)

If

$f: G \to H$

is a Lie group homomorphism, then its differential at the identity element

$df = df_e: \operatorname{Lie}(G) \to \operatorname{Lie}(H)$

is a Lie algebra homomorphism (brackets go to brackets), which has the following properties:

• $\operatorname{Lie}(\operatorname{ker}(f)) = \operatorname{ker}(df)$.
• If the image of f is closed, then $\operatorname{Lie}(\operatorname{im}(f)) = \operatorname{im}(df)$ and the first isomorphism theorem holds: f induces the isomorphism of Lie groups:

$G/\operatorname{ker}(f) \to \operatorname{im}(f)$.

• The product rule holds: if $f, g: G \to H$ are Lie group homomorphisms, then $d(fg)(X) = df(X) + dg(X).$
• The chain rule holds: if $f: G \to H$ and $g: H \to K$ are Lie group homomorphisms, then $d(g \circ f) = (dg) \circ (df).$
• The differential of the (multiplicative) inverse is the additive inverse: $d(g \mapsto g^{-1})(X) = -X.$

In particular, if H is a closed subgroup of a Lie group G, then $\operatorname{Lie}(H)$ is a Lie subalgebra of $\operatorname{Lie}(G)$. Also, if f is injective, then f is an immersion and so G is said to be an immersed (Lie) subgroup of H. For example, $G/\operatorname{ker}(f)$ is an immersed subgroup of H. If f is surjective, then f is a submersion and if, in addition, G is compact, then f is a principal bundle with the structure group its kernel. (Ehresmann's lemma)

Let $G = G_1 \times \cdots \times G_r$ be a direct product of Lie groups and $p_i: G \to G_i$ projections. Then the differentials $dp_i: \operatorname{Lie}(G) \to \operatorname{Lie}(G_i)$ give the canonical identification:

$\operatorname{Lie}(G_1 \times \cdots \times G_r) = \operatorname{Lie}(G_1) \oplus \cdots \oplus \operatorname{Lie}(G_r)$.

If $H, H'$ are Lie subgroups of a Lie group, then $\operatorname{Lie}(H \cap H') = \operatorname{Lie}(H) \cap \operatorname{Lie}(H').$

Let G be a connected Lie group. If H is a Lie group, then any Lie group homomorphism $f: G \to H$ is uniquely determined by its differential $df$. Precisely, there is the exponential map $\operatorname{exp}: \operatorname{Lie}(G) \to G$ (and one for H) such that $f(\operatorname{exp}(X)) = \operatorname{exp}(df(X))$ and, since G is connected, this determines f uniquely. In general, if U is a neighborhood of the identity element in a connected topological group G, then $\bigcup_{n > 0} U^n$ coincides with G, since the former is an open (hence closed) subgroup. Now, $\operatorname{exp}: \operatorname{Lie}(G) \to G$ defines a local homeomorphism from a neighborhood of the zero vector to the neighborhood of the identity element. For example, if G is the Lie group of invertible real square matrices of size n (general linear group), then $\operatorname{Lie}(G)$ is the Lie algebra of real square matrices of size n and $\displaystyle \exp(X) = e^X = \sum_0^\infty {X^j / j!}$.

The next criterion is frequently used to compute the Lie algebra of a given Lie group. Let G be a Lie group and H an immersed subgroup. Then

$\operatorname{Lie}(H) = \{ X \in \operatorname{Lie}(G) | \operatorname{exp}(tX) \in H \text{ for all } t \text{ in } \mathbb{\mathbb{R}} \}.$

For example, one can use the criterion to establish the correspondence for classical compact groups (cf. the table in "compact Lie groups" below.)

The correspondence

$\operatorname{Lie}$ defines a functor from the category of Lie groups to the category of finite-dimensional real Lie algebras. Lie's third theorem states that $\operatorname{Lie}$ defines an equivalence from the subcategory of simply connected Lie groups to the category of finite-dimensional real Lie algebras. Explicitly, the theorem contains the following two statements:

• Every finite-dimensional real Lie algebra is the Lie algebra of some simply connected Lie group.
• If $\phi: \operatorname{Lie}(G) \to \operatorname{Lie}(H)$ is a Lie algebra homomorphism and if G is simply connected, then there exists a (unique) Lie group homomorphism $f: G \to H$ such that $\phi = df$.

The assumption that G is simply connected cannot be omitted. For example, the Lie algebras of SO(3) and SU(2) are isomorphic, but there is no corresponding homomorphism of SO(3) into SU(2). Rather, the homomorphism goes from the simply connected group SU(2) to the non-simply connected group SO(3). If G and H are both simply connected and have isomorphic Lie algebras, the above result allows one to show that G and H are isomorphic. One method to construct f is to use the Baker–Campbell–Hausdorff formula.

Perhaps, the most elegant proof of the first result above uses Ado's theorem, which says any finite-dimensional Lie algebra (over a field of any characteristic) is a Lie subalgebra of the Lie algebra $\mathfrak{gl}_n$ of square matrices. The proof goes as follows: by Ado's theorem, we assume $\mathfrak{g} \subset \mathfrak{gl}_n(\mathbb{R}) = \operatorname{Lie}(GL_n(\mathbb{R}))$ is a Lie subalgebra. Let G be the subgroup of $GL_n(\mathbb{R})$ generated by $e^{\mathfrak{g}}$ and let $\widetilde{G}$ be a simply connected covering of G; it is not hard to show that $\widetilde{G}$ is a Lie group and that the covering map is a Lie group homomorphism. Since $T_e \widetilde{G} = T_e G = \mathfrak{g}$, this completes the proof.

Example: Each element X in the Lie algebra $\mathfrak{g} = \operatorname{Lie}(G)$ gives rise to the Lie algebra homomorphism

$\mathbb{R} \to \mathfrak{g}, \, t \mapsto tX.$

By Lie's third theorem, as $\operatorname{Lie}(\mathbb{R}) = T_0 \mathbb{R} = \mathbb{R}$ and exp for it is the identity, this homomorphism is the differential of the Lie group homomorphism $\mathbb{R} \to H$ for some immersed subgroup H of G. This Lie group homomorphism, called the one-parameter subgroup generated by X, is precisely the exponential map $t \mapsto \operatorname{exp}(tX)$ and H its image. The preceding can be summarized to saying that there is a canonical bijective correspondence between $\mathfrak{g}$ and the set of one-parameter subgroups of G.

Another aspect of the correspondence between Lie groups and Lie algebras is the following "subgroups/subalgebras theorem":

• If G is a Lie group and $\mathfrak{h}$ is a Lie subalgebra of $\operatorname{Lie}(G)$, then there is a unique connected Lie subgroup (not necessarily closed) H of G with Lie algebra $\mathfrak{h}$.

Lie group representations

A special case of Lie correspondence is a correspondence between finite-dimensional representations of a Lie group and representations of the associated Lie algebra.

The general linear group $GL_n(\mathbb{C})$ is a (real) Lie group and any Lie group homomorphism

$\pi: G \to GL_n(\mathbb{C})$

is called a representation of the Lie group G. The differential

$d\pi: \mathfrak{g} \to \mathfrak{gl}_n(\mathbb{C})$,

is then a Lie algebra homomorphism called a Lie algebra representation. (The differential $d \pi$ is often simply denoted by $\pi$.)

Lie's third theorem mentioned early then says that if $G$ is the simply connected Lie group whose Lie algebra is $\mathfrak{g}$ then we have a natural bijection between Lie group representations of $G$ and Lie algebra representations of $\mathfrak{g}$ for each dimension.

An example of a Lie group representation is the adjoint representation of a Lie group G; each element g in a Lie group G defines an automorphism of G by conjugation: $c_g(h) = ghg^{-1}$; the differential $d c_g$ is then an automorphism of the Lie algebra $\mathfrak{g}$. This way, we get a representation $\operatorname{Ad}: G \to GL(\mathfrak{g}), \, g \mapsto dc_g$, called the adjoint representation. The corresponding Lie algebra homomorphism $\mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ is called the adjoint representation of $\mathfrak{g}$ and is denoted by $\operatorname{ad}$. One can show $\operatorname{ad}(X)(Y) = [X, Y]$, which in particular implies that the Lie bracket of $\mathfrak{g}$ is determined by the group law on G.

By Lie's third theorem, there exists a subgroup $\operatorname{Int}(\mathfrak{g})$ of $GL(\mathfrak{g})$ whose Lie algebra is $\operatorname{ad}(\mathfrak{g})$. ($\operatorname{Int}(\mathfrak{g})$ is in general not a closed subgroup; only an immersed subgroup.) It is called the adjoint group of $\mathfrak{g}$. If G is connected, it fits into the exact sequence:

$0 \to Z(G) \to G \overset{\operatorname{Ad}}\to \operatorname{Int}(\mathfrak{g}) \to 0$

where $Z(G)$ is the center of G. If the center of G is discrete, then Ad here is a covering map.

Let G be a connected Lie group. Then G is unimodular if and only if $\operatorname{det}(\operatorname{Ad}(g)) = 1$ for all g in G.

Let G be a Lie group acting on a manifold X and G_x the stabilizer of a point x in X. Let $\rho(x): G \to X, \, g \mapsto g \cdot x$. Then

• $\operatorname{Lie}(G_x) = \operatorname{ker}(d \rho(x): T_eG \to T_x X)$.
• If the orbit $G \cdot x$ is locally closed, then the orbit is a submanifold of X and $T_x (G \cdot x) = \operatorname{im}(d \rho(x): T_eG \to T_x X)$.

For a subset A of $\mathfrak{g}$ or G, let

$\mathfrak{z}_{\mathfrak{g}}(A) = \{ X \in \mathfrak{g} | \operatorname{ad}(a)X = 0 \text{ or } \operatorname{Ad}(a)X = 0 \text{ for all } a \text{ in } A\}$

$Z_G(A) = \{ g \in G | \operatorname{Ad}(g)a = 0 \text{ or } ga = ag \text{ for all } a \text{ in } A \}$

be the Lie algebra centralizer and the Lie group centralizer of A. Then $\operatorname{Lie}(Z_G(A)) = \mathfrak{z}_{\mathfrak{g}}(A)$.

If H is a closed connected subgroup of G, then H is normal if and only if $\operatorname{Lie}(H)$ is an ideal and in such a case $\operatorname{Lie}(G/H) = \operatorname{Lie}(G)/\operatorname{Lie}(H)$.

Abelian Lie groups

Let G be a connected Lie group. Since the Lie algebra of the center of G is the center of the Lie algebra of G (cf. the previous §), G is abelian if and only its Lie algebra is abelian.

If G is abelian, then the exponential map $\operatorname{exp}: \mathfrak{g} \to G$ is a surjective group homomorphism. The kernel of it is a discrete group (since the dimension is zero) called the integer lattice of G and is denoted by $\Gamma$. By the first isomorphism theorem, $\operatorname{exp}$ induces the isomorphism $\mathfrak{g}/\Gamma \to G$.

By the rigidity argument, the fundamental group $\pi_1(G)$ of a connected Lie group G is a central subgroup of a simply connected covering $\widetilde{G}$ of G; in other words, G fits into the central extension

$1 \to \pi_1(G) \to \widetilde{G} \overset{p}\to G \to 1.$

Equivalently, given a Lie algebra $\mathfrak{g}$ and a simply connected Lie group $\widetilde{G}$ whose Lie algebra is $\mathfrak{g}$, there is a one-to-one correspondence between quotients of $\widetilde{G}$ by discrete central subgroups and connected Lie groups having Lie algebra $\mathfrak{g}$.

For the complex case, complex tori are important; see complex Lie group for this topic.

Compact Lie groups

Main article: Compact Lie group

Let G be a connected Lie group with finite center. Then the following are equivalent.

• G is compact.
• (Weyl) The simply connected covering $\widetilde{G}$ of G is compact.
• The adjoint group $\operatorname{Int}\mathfrak{g}$ is compact.
• There exists an embedding $G \hookrightarrow O(n, \mathbb{R})$ as a closed subgroup.
• The Killing form on $\mathfrak{g}$ is negative definite.
• For each X in $\mathfrak{g}$, $\operatorname{ad}(X)$ is diagonalizable and has zero or purely imaginary eigenvalues.
• There exists an invariant inner product on $\mathfrak{g}$.

Compact Lie group	Complexification of associated Lie algebra	Root system
SU(n+1) $= \{ A \in M_{n+1}(\mathbb{C}) | {\overline{A}}^T A = I, \operatorname{det}(A) = 1 \}$	$\mathfrak{sl}(n+1, \mathbb{C})$ $= \{ X \in M_{n+1}(\mathbb{C}) | \operatorname{tr} X = 0 \}$	An
SO(2n+1) $= \{ A \in M_{2n+1}(\mathbb{R}) | A^T A = I, \operatorname{det}(A) = 1 \}$	$\mathfrak{so}(2n+1, \mathbb{C})$ $= \{ X \in M_{2n+1}(\mathbb{C}) | X^T + X = 0 \}$	Bn
Sp(n) $= \{ A \in U(2n) | A^T J A = J \}, \, J = \begin{bmatrix} 0 & I_n \\ - I_n & 0\end{bmatrix}$	$\mathfrak{sp}(n, \mathbb{C})$ $= \{ X \in M_{2n}(\mathbb{C}) | X^T J + JX = 0 \}$	Cn
SO(2n) $= \{ A \in M_{2n}(\mathbb{R}) | A^T A = I, \operatorname{det}(A) = 1 \}$	$\mathfrak{so}(2n, \mathbb{C})$ $= \{ X \in M_{2n}(\mathbb{C}) | X^T + X = 0 \}$	Dn

If G is a compact Lie group, then

$H^k(\mathfrak{g}; \mathbb{R}) = H_{\text{dR}}(G)$

where the left-hand side is the Lie algebra cohomology of $\mathfrak{g}$ and the right-hand side is the de Rham cohomology of G. (Roughly, this is a consequence of the fact that any differential form on G can be made left invariant by the averaging argument.)

Related constructions

Let G be a Lie group. The associated Lie algebra $\operatorname{Lie}(G)$ of G may be alternatively defined as follows. Let $A(G)$ be the algebra of distributions on G with support at the identity element with the multiplication given by convolution. $A(G)$ is in fact a Hopf algebra. The Lie algebra of G is then $\mathfrak{g} = \operatorname{Lie}(G) = P(A(G))$, the Lie algebra of primitive elements in $A(G)$. By the Milnor–Moore theorem, there is the canonical isomorphism $U(\mathfrak{g}) = A(G)$ between the universal enveloping algebra of $\mathfrak{g}$ and $A(G)$.

See also

• compact Lie algebra
• Milnor–Moore theorem
• Formal Lie group
• Malcev Lie algebra

References

• .

External links

• Notes for Math 261A Lie groups and Lie algebras
• Formal Lie theory in characteristic zero, a blog post by Akhil Mathew