The Lie algebra of a Lie group

There are various ways one can understand the construction of the Lie algebra of a Lie group G. One approach uses left-invariant vector fields. A vector field X on G is said to be invariant under left translations if, for any g, h in G,

${\displaystyle (dL_{g})_{h}(X_{h})=X_{gh}}$

where ${\displaystyle L_{g}:G\to G}$ is defined by ${\displaystyle L_{g}(x)=gx}$ and ${\displaystyle (dL_{g})_{h}:T_{h}G\to T_{gh}G}$ is the differential of ${\displaystyle L_{g}}$ between tangent spaces.

Let ${\displaystyle \operatorname {Lie} (G)}$ 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., ${\displaystyle [X,Y]}$ is left-translation-invariant if X, Y are. Thus, ${\displaystyle \operatorname {Lie} (G)}$ is a Lie subalgebra of the Lie algebra of all vector fields on G and is called the Lie algebra of G. One can understand this more concretely by identifying the space of left-invariant vector fields with the tangent space at the identity, as follows: Given a left-invariant vector field, one can take its value at the identity, and given a tangent vector at the identity, one can extend it to a left-invariant vector field. This correspondence is one-to-one in both directions, so is bijective. Thus, the Lie algebra can be thought of as the tangent space at the identity and the bracket of X and Y in ${\displaystyle T_{e}G}$ can be computed by extending them to left-invariant vector fields, taking the bracket of the vector fields, and then evaluating the result at the identity.

There is also another incarnation of ${\displaystyle \operatorname {Lie} (G)}$ 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.

Matrix Lie groups

Suppose G is a closed subgroup of GL(n;C), and thus a Lie group, by the closed subgroups theorem. Then the Lie algebra of G may be computed as^[2][3]

${\displaystyle \operatorname {Lie} (G)=\left\{X\in M(n;\mathbb {C} )\mid e^{tX}\in G{\text{ for all }}t\in \mathbb {R} \right\}.}$

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

Homomorphisms

If

${\displaystyle f:G\to H}$

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

${\displaystyle df=df_{e}:\operatorname {Lie} (G)\to \operatorname {Lie} (H)}$

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

• ${\displaystyle \exp(df(X))=f(\exp(X))}$ for all X in Lie(G), where "exp" is the exponential map
• ${\displaystyle \operatorname {Lie} (\ker(f))=\ker(df)}$.^[4]
• If the image of f is closed,^[5] then ${\displaystyle \operatorname {Lie} (\operatorname {im} (f))=\operatorname {im} (df)}$^[6] and the first isomorphism theorem holds: f induces the isomorphism of Lie groups:

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

• The chain rule holds: if ${\displaystyle f:G\to H}$ and ${\displaystyle g:H\to K}$ are Lie group homomorphisms, then ${\displaystyle d(g\circ f)=(dg)\circ (df)}$.

In particular, if H is a closed subgroup^[7] of a Lie group G, then ${\displaystyle \operatorname {Lie} (H)}$ is a Lie subalgebra of ${\displaystyle \operatorname {Lie} (G)}$. 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, ${\displaystyle G/\ker(f)}$ 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)

Other properties

Let ${\displaystyle G=G_{1}\times \cdots \times G_{r}}$ be a direct product of Lie groups and ${\displaystyle p_{i}:G\to G_{i}}$ projections. Then the differentials ${\displaystyle dp_{i}:\operatorname {Lie} (G)\to \operatorname {Lie} (G_{i})}$ give the canonical identification:

${\displaystyle \operatorname {Lie} (G_{1}\times \cdots \times G_{r})=\operatorname {Lie} (G_{1})\oplus \cdots \oplus \operatorname {Lie} (G_{r}).}$

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

Let G be a connected Lie group. If H is a Lie group, then any Lie group homomorphism ${\displaystyle f:G\to H}$ is uniquely determined by its differential ${\displaystyle df}$. Precisely, there is the exponential map ${\displaystyle \exp$ :\operatorname {Lie} (G)\to G} (and one for H) such that ${\displaystyle f(\exp(X))=\exp(df(X))}$ and, since G is connected, this determines f uniquely.^[8] In general, if U is a neighborhood of the identity element in a connected topological group G, then ${\textstyle \bigcup _{n>0}U^{n}}$ coincides with G, since the former is an open (hence closed) subgroup. Now, ${\displaystyle \exp$ :\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 ${\displaystyle \operatorname {Lie} (G)}$ is the Lie algebra of real square matrices of size n and ${\textstyle \exp(X)=e^{X}=\sum _{0}^{\infty }{X^{j}/j!}}$.

Proof of Lie's third theorem

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 ${\displaystyle {\mathfrak {gl}}_{n}}$ of square matrices. The proof goes as follows: by Ado's theorem, we assume ${\displaystyle {\mathfrak {g}}\subset {\mathfrak {gl}}_{n}(\mathbb {R} )=\operatorname {Lie} (GL_{n}(\mathbb {R} ))}$ is a Lie subalgebra. Let G be the subgroup of ${\displaystyle GL_{n}(\mathbb {R} )}$ generated by ${\displaystyle e^{\mathfrak {g}}}$ and let ${\displaystyle {\widetilde {G}}}$ be a simply connected covering of G; it is not hard to show that ${\displaystyle {\widetilde {G}}}$ is a Lie group and that the covering map is a Lie group homomorphism. Since ${\displaystyle T_{e}{\widetilde {G}}=T_{e}G={\mathfrak {g}}}$, this completes the proof.

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

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

By Lie's third theorem, as ${\displaystyle \operatorname {Lie} (\mathbb {R} )=T_{0}\mathbb {R} =\mathbb {R} }$ and exp for it is the identity, this homomorphism is the differential of the Lie group homomorphism ${\displaystyle \mathbb {R} \to H}$ for some immersed subgroup H of G. This Lie group homomorphism, called the one-parameter subgroup generated by X, is precisely the exponential map ${\displaystyle t\mapsto \exp(tX)}$ and H its image. The preceding can be summarized to saying that there is a canonical bijective correspondence between ${\displaystyle {\mathfrak {g}}}$ and the set of one-parameter subgroups of G.^[17]

Proof of the homomorphisms theorem

One approach to proving the second part of the Lie group-Lie algebra correspondence (the homomorphisms theorem) is to use the Baker–Campbell–Hausdorff formula, as in Section 5.7 of Hall's book.^[18] Specifically, given the Lie algebra homomorphism ${\displaystyle \phi }$ from ${\displaystyle \operatorname {Lie} (G)}$ to ${\displaystyle \operatorname {Lie} (H)}$, we may define ${\displaystyle f:G\to H}$ locally (i.e., in a neighborhood of the identity) by the formula

${\displaystyle f(e^{X})=e^{\phi (X)},}$

where ${\displaystyle e^{X}}$ is the exponential map for G, which has an inverse defined near the identity. We now argue that f is a local homomorphism. Thus, given two elements near the identity ${\displaystyle e^{X}}$ and ${\displaystyle e^{Y}}$ (with X and Y small), we consider their product ${\displaystyle e^{X}e^{Y}}$. According to the Baker–Campbell–Hausdorff formula, we have ${\displaystyle e^{X}e^{Y}=e^{Z}}$, where

${\displaystyle Z=X+Y+{\frac {1}{2}}[X,Y]+{\frac {1}{12}}[X,[X,Y]]+\cdots ,}$

with ${\displaystyle \cdots }$ indicating other terms expressed as repeated commutators involving X and Y. Thus,

${\displaystyle f\left(e^{X}e^{Y}\right)=f\left(e^{Z}\right)=e^{\phi (Z)}=e^{\phi (X)+\phi (Y)+{\frac {1}{2}}[\phi (X),\phi (Y)]+{\frac {1}{12}}[\phi (X),[\phi (X),\phi (Y)]]+\cdots },}$

because ${\displaystyle \phi }$ is a Lie algebra homomorphism. Using the Baker–Campbell–Hausdorff formula again, this time for the group H, we see that this last expression becomes ${\displaystyle e^{\phi (X)}e^{\phi (Y)}}$, and therefore we have

${\displaystyle f\left(e^{X}e^{Y}\right)=e^{\phi (X)}e^{\phi (Y)}=f\left(e^{X}\right)f\left(e^{Y}\right).}$

Thus, f has the homomorphism property, at least when X and Y are sufficiently small. It is important to emphasize that this argument is only local, since the exponential map is only invertible in a small neighborhood of the identity in G and since the Baker–Campbell–Hausdorff formula only holds if X and Y are small. The assumption that G is simply connected has not yet been used..

The next stage in the argument is to extend f from a local homomorphism to a global one. The extension is done by defining f along a path and then using the simple connectedness of G to show that the definition is independent of the choice of path.

The adjoint representation

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

By Lie's third theorem, there exists a subgroup ${\displaystyle \operatorname {Int} ({\mathfrak {g}})}$ of ${\displaystyle GL({\mathfrak {g}})}$ whose Lie algebra is ${\displaystyle \operatorname {ad} ({\mathfrak {g}})}$. (${\displaystyle \operatorname {Int} ({\mathfrak {g}})}$ is in general not a closed subgroup; only an immersed subgroup.) It is called the adjoint group of ${\displaystyle {\mathfrak {g}}}$.^[20] If G is connected, it fits into the exact sequence:

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

where ${\displaystyle Z(G)}$ 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 ${\displaystyle \det(\operatorname {Ad} (g))=1}$ for all g in G.^[21]

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

• ${\displaystyle \operatorname {Lie} (G_{x})=\ker(d\rho (x):T_{e}G\to T_{x}X).}$
• If the orbit ${\displaystyle G\cdot x}$ is locally closed, then the orbit is a submanifold of X and ${\displaystyle T_{x}(G\cdot x)=\operatorname {im} (d\rho (x):T_{e}G\to T_{x}X)}$.^[22]

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

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

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

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

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