Lie Groups, Invariant Fields, and the Exponential Map

A finite-dimensional real Lie group is a group $G$, in the sense of Group and abelian group, together with the structure of a finite-dimensional real smooth manifold such that the maps

$m:G\times G⟶G,m(g,h)=gh,$

and

$\mathrm{inv}:G⟶G,\mathrm{inv}(g)=g^{-1},$

are smooth in the sense of Smooth maps between manifolds with boundary. Its identity element is denoted by $e$ (or occasionally $1$).

Unless an item explicitly says otherwise, every Lie group on this page is real, finite-dimensional, and has no manifold boundary. Dimension zero is allowed; for example, any countable discrete group with its discrete zero-dimensional smooth structure is a Lie group. A Lie group cannot be empty because a group contains its identity. The definition applies unchanged in dimension one, uses no metric or nondegeneracy hypothesis, and makes no choice from a family of sets.

Let $G$ and $H$ be Lie groups in the sense of Lie group. A Lie-group homomorphism $F:G\to H$ is a group homomorphism in the sense of Monoid homomorphism and group homomorphism that is also smooth. Thus

$F(gh)=F(g)F(h)$

for all $g,h\in G$, while preservation of identity and inverses follows from the group-homomorphism laws.

A Lie-group isomorphism is a bijective Lie-group homomorphism whose inverse is smooth. Its set-theoretic inverse is automatically a group homomorphism by The inverse of a bijective group homomorphism is a group homomorphism, so the smoothness clause makes that inverse a Lie-group homomorphism as well. A Lie-group automorphism is a Lie-group isomorphism from $G$ to itself.

Smoothness is part of the homomorphism definition on this page. The separate automatic-regularity theorem for merely continuous group homomorphisms is not being assumed. These definitions apply without alteration in dimensions zero and one. Their Lie-group inputs are nonempty and boundaryless by the page convention, and the definitions use neither nondegeneracy nor any choice principle.

Let $G$ be a Lie group and fix $g\in G$. The left translation by $g$ and right translation by $g$ are respectively

$L_g:G⟶G,L_g(h)=gh,$

and

$R_g:G⟶G,R_g(h)=hg\mathrm{.}$

Both maps are smooth: each is obtained from the smooth multiplication $m:G\times G\to G$ in Lie group by holding one argument fixed. The notation here is the ordinary right-translation convention. Kirillov writes a right action as $h\mapsto h{g}^{-1}$; therefore this page's $R_g$ is that source's right action by $g^{-1}$.

The definition applies in dimensions zero and one and uses no metric or nondegeneracy condition. A Lie group is nonempty, its manifold is boundaryless by the page convention, and the one supplied element $g$ involves no choice from a family.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. For every $g$ in a Lie group $G$, the maps $L_g$ and $R_g$ are diffeomorphisms with respective inverses $L_{g^{-1}}$ and $R_{g^{-1}}$. Moreover,

${\mathrm{\Phi }}_L:G\times T_{e}G⟶TG,{\mathrm{\Phi }}_L(g,X)=d(L_g{)}_{e}X,$

and

${\mathrm{\Phi }}_R:G\times T_{e}G⟶TG,{\mathrm{\Phi }}_R(g,X)=d(R_g{)}_{e}X,$

are smooth vector-bundle isomorphisms over ${\mathrm{id}}_G$.

The countable-choice assumption is used exactly through the supplied theorem that equips tangent bundles and global differentials with their smooth structures.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$, and let $X$ be a smooth vector field on a Lie group $G$. The field $X$ is left invariant if

$d(L_g{)}_h(X_h)=X_{gh}$

for every $g,h\in G$. It is right invariant if

$d(R_g{)}_h(X_h)=X_{hg}$

for every $g,h\in G$, where $R_g(h)=hg$ is the ordinary right translation.

In the left-invariant case, setting $h=e$ gives $X_g=d(L_g{)}_e(X_e)$. Conversely, if that identity-value formula holds for every $g$, then the chain rule and $L_g\circ L_h=L_{gh}$ give

$d(L_g{)}_h(X_h)=d(L_g{)}_{h}d(L_h{)}_e(X_e)=d(L_{gh}{)}_e(X_e)=X_{gh}\mathrm{.}$

Thus left invariance is equivalent to the displayed identity-value formula. Likewise, $R_g\circ R_h=R_{hg}$ shows that right invariance is equivalent to $X_g=d(R_g{)}_e(X_e)$. This agrees with Kirillov's Definition 2.26 after translating the source's inverse-parametrized right action into the ordinary right-translation convention used here.

The assumption ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited exactly through the supplied definition of a smooth vector field on the canonical smooth tangent bundle and through the supplied translation trivializations. The chain-rule calculation is pointwise and makes no further choice. A Lie group is nonempty; in dimension zero the tangent values are all zero, and the same definition applies in dimension one. No metric or nondegeneracy hypothesis occurs, and Lie groups are boundaryless by the page convention.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let ${\mathfrak{X}}_L(G)$ and ${\mathfrak{X}}_R(G)$ be the real vector spaces of left- and right-invariant smooth vector fields on a Lie group $G$. Evaluation at the identity gives linear isomorphisms

${\mathrm{ev}}_e^L:{\mathfrak{X}}_L(G)⟶T_{e}G$

and

${\mathrm{ev}}_e^R:{\mathfrak{X}}_R(G)⟶T_{e}G\mathrm{.}$

Their respective inverses send $v\in T_{e}G$ to

$v_g^L=d(L_g{)}_e(v)\text{and}{v}_g^R=d(R_g{)}_e(v)\mathrm{.}$

The countable-choice assumption is used exactly through the supplied invariant-field and smooth translation-trivialization results.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. If $X$ and $Y$ are left-invariant smooth vector fields on a Lie group $G$, then their Lie bracket $[X,Y]$ is left invariant.

The countable-choice assumption is used exactly through the supplied invariant-field and smooth translation-trivialization results.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$, let $G$ be a Lie group with identity $e$, and write $\mathfrak{g}=T_{e}G$. For $u,v\in \mathfrak{g}$, let $u^L$ and $v^L$ be their unique left-invariant smooth extensions. Define the Lie-group tangent bracket by

$[u,v{]}_G=[u^L,v^L{]}_e\mathrm{.}$

The evaluation isomorphism Left-invariant vector fields evaluate isomorphically at the identity makes both extensions unique, so the definition is unambiguous. Moreover, The Lie bracket of left-invariant fields is left invariant makes $[u^L,v^L]$ left invariant, and hence its identity value determines it:

$[u^L,v^L]=[u,v{]}_G^L\mathrm{.}$

The bracket on fields is the library's fixed commutator $[X,Y]f=X(Yf)-Y(Xf)$ from The Lie bracket of smooth vector fields. This explicitly fixes the sign for every later occurrence of $\mathrm{ad}$, the Maurer--Cartan equation, and right-invariant fields. In particular, no opposite vector-field commutator convention is being imported from a source.

The assumption ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited exactly through the supplied invariant-extension and bracket-closure results; evaluating the supplied vector-field bracket at $e$ adds no choice. A Lie group is nonempty. If $\dim G=0$, then $\mathfrak{g}=0$ and this is the unique zero bracket; the definition applies unchanged in dimension one. No metric or nondegeneracy hypothesis occurs, and the group is boundaryless by convention.

Let $\mathbb{F}$ be either $\mathbb{R}$ or $\mathbb{C}$. A finite-dimensional Lie algebra over $\mathbb{F}$ is a finite-dimensional $\mathbb{F}$-vector space $\mathfrak{g}$ together with an $\mathbb{F}$-bilinear map

$[,]:\mathfrak{g}\times \mathfrak{g}⟶\mathfrak{g}$

such that, for all $X,Y,Z\in \mathfrak{g}$,

$[X,X]=0$

and

$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0.$

The first identity is alternation and the second is the Jacobi identity. Over $\mathbb{R}$ or $\mathbb{C}$, alternation is equivalent to skew-symmetry. Indeed, bilinearity and alternation give $0=[X+Y,X+Y]=[X,Y]+[Y,X]$, while skew-symmetry gives $2[X,X]=0$ and hence $[X,X]=0$ because both fields have characteristic zero. For a complex Lie algebra the bracket is required to be complex-bilinear, not merely real-bilinear.

The definition permits the zero bracket. The zero vector space, with its unique bracket, is a zero-dimensional Lie algebra. On a one-dimensional space, every alternating bilinear bracket is zero: any two vectors are scalar multiples of one vector and bilinearity reduces their bracket to $[X,X]=0$. A vector space is nonempty because it contains zero. No basis is selected by asserting finite-dimensionality, so the definition is choice-free. It is purely algebraic and has no metric, nondegeneracy, manifold-boundary, or endpoint condition.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. If $G$ is a finite-dimensional real Lie group with identity $e$, then $\mathfrak{g}=T_{e}G$, equipped with the bracket transported from left-invariant smooth vector fields, is a finite-dimensional real Lie algebra. It is denoted

$\mathrm{Lie}(G)=\mathfrak{g}\mathrm{.}$

The countable-choice assumption is used exactly through the supplied invariant-field and smooth-vector-field results.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. For $X,Y\in \mathfrak{g}=T_{e}G$, let $X^R,Y^R$ be their ordinary right-invariant smooth extensions, and let $[X,Y{]}_G$ be the tangent bracket defined through left-invariant fields. Then

$[X^R,Y^R]=-[X,Y{]}_G^R\mathrm{.}$

Thus evaluation identifies ordinary right-invariant fields with the opposite Lie algebra ${\mathfrak{g}}^{\mathrm{o}\mathrm{p}}$, not with the left-invariant bracket. The countable-choice assumption is used exactly through the supplied invariant-extension and tangent-bracket results.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ as in The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$), let $G$ be a Lie group Lie group with identity $e$, and write $\mathfrak{g}=T_{e}G$. A $\mathfrak{g}$-valued one-form on $G$ means a smooth map $\alpha :TG\to \mathfrak{g}$ whose restriction ${\alpha }_g:T_{g}G\to \mathfrak{g}$ is linear for every $g$. Equivalently, $V\mapsto ({\pi }_G(V),\alpha (V))$ is a smooth vector-bundle map $TG\to G\times \mathfrak{g}$ over ${\mathrm{id}}_G$ in the sense of Vector bundle maps over a smooth base map.

The left Maurer--Cartan form is the $\mathfrak{g}$-valued one-form $\theta :TG\to \mathfrak{g}$ whose fibre map at $g$ is

${\theta }_g=d(L_{g^{-1}}{)}_g:T_{g}G⟶T_{e}G=\mathfrak{g}\mathrm{.}$

This formula is well-defined because left translation by $g^{-1}$ sends $g$ to $e$. It is smooth and fibrewise linear because Translations are diffeomorphisms and their differentials trivialize the tangent bundle identifies

$TG⟶G\times \mathfrak{g},V_g⟼(g,{\theta }_g{V}_g)$

as the inverse vector-bundle isomorphism to left trivialization. In particular, ${\theta }_e=d(L_e{)}_e={\mathrm{id}}_{\mathfrak{g}}$.

The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is used exactly through that supplied smooth left-trivialization theorem; the pointwise formula and passage to its second component add no choice. A Lie group is nonempty. When $\dim G=0$, $\theta$ is the unique map between zero tangent fibres, and the definition is unchanged in dimension one. Lie groups are boundaryless by convention, and no metric, nondegeneracy, endpoint, or biconditional occurs.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let $\theta$ be the left Maurer--Cartan form of a Lie group $G$. Every fibre map ${\theta }_g:T_{g}G\to \mathfrak{g}$ is a linear isomorphism, with inverse $d(L_g{)}_e$. For $h,g\in G$ and $V\in T_{g}G$, define the pullback here by

$(L_h^{\ast }\theta {)}_g(V)={\theta }_{hg}(d(L_h{)}_{g}V)\mathrm{.}$

Then $L_h^{\ast }\theta =\theta$ for every $h\in G$. Moreover, for every $X\in \mathfrak{g}$ and its left-invariant extension $X^L$,

${\theta }_g(X_g^L)=X$

at every $g\in G$. The countable-choice assumption is used exactly through the supplied Maurer--Cartan definition and invariant-extension theorem.

Let $M$ be a smooth manifold, let $V$ be a finite-dimensional real vector space, and let $k\ge 0$. A smooth $V$-valued differential $k$-form on $M$ is a family of alternating $k$-linear maps

${\alpha }_p:(T_{p}M{)}^k⟶V$

such that $\lambda \circ \alpha$ is a scalar smooth $k$-form in the sense of A smooth differential $k$-form for every $\lambda \in V^{\ast }$. Equivalently, for one (and hence every) basis $(v_1,\dots ,v_r)$ of $V$, the unique components in $\alpha ={\sum }_i{v}_i{\alpha }^i$ all belong to ${\mathrm{\Omega }}^k(M)$.

For such an $\alpha$, its componentwise exterior derivative is the unique $V$-valued $(k+1)$-form $d_V\alpha$ characterized by

$\lambda \circ d_V\alpha =d(\lambda \circ \alpha )$

for every $\lambda \in V^{\ast }$. Indeed, in a basis with dual basis $({\lambda }^1,\dots ,{\lambda }^r)$, set

$d_V\alpha ={\sum }_i{v}_{i}d{\alpha }^i,{\alpha }^i={\lambda }^i\circ \alpha \mathrm{.}$

The scalar formula defining $d$ is real-linear term by term The exterior derivative by the invariant vector-field formula, and its output is a smooth form The invariant exterior-derivative formula is $C^{\mathrm{\infty }}$-multilinear. Thus the displayed construction has the stated characterization. It is independent of the chosen basis because linear functionals separate points of $V$, and the characterization also proves uniqueness.

The two smoothness descriptions are equivalent in both directions: testing all $\lambda$ includes the dual basis components, while every $\lambda \circ \alpha$ is a fixed real linear combination of the components in one basis. If $M$ is empty, these assignments and identities are vacuous. If $V=0$, there is only the zero-valued form and its derivative is zero; if $\dim V=1$, the definition is exactly the scalar definition after choosing one nonzero basis vector. No metric, nondegeneracy, manifold boundary, or endpoint is involved. A single finite basis exists by finite-dimensionality; the definition is basis-independent and chooses no basis or family, so no choice axiom is used.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let $\theta$ be the left Maurer--Cartan form of a Lie group $G$, with values in $\mathfrak{g}=T_{e}G$. Write $d\theta$ for its componentwise exterior derivative from Finite-dimensional vector-valued forms and their exterior derivative. For vector fields $A,B$, define the bracket-valued wedge by

$[\theta \wedge \theta ](A,B)=[\theta (A),\theta (B){]}_G-[\theta (B),\theta (A){]}_G=2[\theta (A),\theta (B){]}_G\mathrm{.}$

Then the normalized left Maurer--Cartan structure equation is

$d\theta +\frac{1}{2}[\theta \wedge \theta ]=0.$

The countable-choice assumption is used exactly through the supplied Maurer--Cartan, invariant-field, and tangent-bracket results.

The standard coordinate on $\mathbb{R}$ makes addition $(s,t)\mapsto s+t$ and inversion $t\mapsto -t$ smooth, so $(\mathbb{R},+)$ is a one-dimensional Lie group in the sense of Lie group. A one-parameter subgroup of a Lie group $G$ is a Lie-group homomorphism

$\gamma :(\mathbb{R},+)⟶G$

in the sense of Lie-group homomorphism, isomorphism, and automorphism. Thus $\gamma$ is smooth, is defined for every real parameter, and satisfies

$\gamma (s+t)=\gamma (s)\gamma (t),\gamma (0)=e,\gamma (-t)=\gamma (t{)}^{-1}\mathrm{.}$

The last two identities are consequences of the group-homomorphism law, not extra data. A smooth curve defined only on an interval around $0$ is therefore not yet a one-parameter subgroup, even if it satisfies the product law whenever all displayed parameters remain in that interval. The term also does not assert that the image is embedded or closed.

Both the domain and every Lie-group codomain are nonempty and boundaryless. For a zero-dimensional codomain the definition still permits, for example, the constant homomorphism; for a one-dimensional codomain it is unchanged. No metric, nondegeneracy, or finite endpoint occurs, and all maps and group operations are supplied explicitly, so no choice axiom is used. This is a definition, not a biconditional characterization.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Every left-invariant smooth vector field on a finite-dimensional real Lie group is complete. Equivalently, its maximal flow is defined on all of $\mathbb{R}\times G$. The countable-choice assumption is used exactly through the supplied invariant-field and smooth-tangent-bundle framework.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let $G$ be a finite-dimensional real Lie group with Lie algebra $\mathfrak{g}=T_{e}G$.

If $\gamma :\mathbb{R}\to G$ is a one-parameter subgroup and $X={\gamma }^{\mathrm{\prime }}(0)$, then

${\gamma }^{\mathrm{\prime }}(t)=d(L_{\gamma (t)}{)}_{e}X=X_{\gamma (t)}^L$

for every $t\in \mathbb{R}$; thus $\gamma$ is the integral curve through $e$ of the left-invariant field $X^L$. Conversely, the global integral curve through $e$ of $X^L$ is a one-parameter subgroup. Consequently every $X\in \mathfrak{g}$ determines a unique one-parameter subgroup with initial velocity $X$.

The countable-choice assumption is used exactly through the supplied smooth invariant-field construction and completeness theorem.