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The Exponential map of a Lie group is a group homomorphism from the additive Lie algebra exactly when its identity component of a Lie group is abelian. If it is additive, its image is an abelian subgroup containing an identity neighborhood by the local exponential chart; connectedness makes that image the entire identity component. Conversely, in an abelian identity component the product $\exp (tX)\exp (tY)$ is a one-parameter subgroup with derivative $X+Y$, giving the identity by uniqueness. The disconnected Lie group $S_3\times \mathbb{R}$ shows that the full group need not be abelian.

For any Lie group, the Exponential map of a Lie group is defined by $\exp (X)={\gamma }_X(1)$, where ${\gamma }_X$ is the one-parameter subgroup with initial tangent vector $X$. For a Matrix Lie group, it is the matrix exponential $\exp X={\sum }_{n\ge 0}{X}^n/n!$. Its differential at zero is the identity, so the inverse function theorem supplies a local exponential chart. A small transformation can therefore be encoded by a vector in the Lie algebra. See MIT 18.745, Lecture 7, Definition 7.2 and Theorem 7.5.

Local invertibility makes no promise about a global inverse. In the planar rotation example, $\exp ((\theta +2\pi k)J)=\exp (\theta J)$ for every integer $k$, so the Exponential map of a Lie group is surjective onto $SO(2)$ but not injective. Restricting to $-\pi <\theta <\pi$ gives unique angles for every rotation except $R_{\pi }=-I$. The omitted half-turn exposes the cut needed for this single angle chart.

Surjectivity can also fail, even in a connected Lie group. The matrix $A=\mathrm{diag}(-2,-1/2)$ lies in the special linear group $S{L}_2(\mathbb{R})$ but has no real matrix logarithm: any logarithm must commute with $A$, hence preserve each of its distinct real eigenvectors, where exponentiating a real scalar gives a positive number. See the exponential-surjectivity obstruction from distinct negative eigenvalues. These two examples distinguish two different global obstacles: repeated parameters and unreachable elements. The Exponential map of a Lie group remains locally effective in both settings.

In a local exponential chart, the mixed second differential of the group commutator defines a bilinear map on the tangent space at the identity. Inversion after swapping the two group elements proves antisymmetry. Differentiating conjugation gives $[X,Y]=\mathrm{ad}(X)Y$; applying naturality of the Lie bracket to the Adjoint representation of a Lie group then proves the Jacobi identity. For a matrix group this recovers $XY-YX$ by expanding through the mixed term.

Choose an injective local exponential chart on a ball of radius $r$ in the Lie algebra, and let $0<\epsilon <r/2$. The neighborhood $\exp (B_{\epsilon })$ contains no nontrivial subgroup. Otherwise it contains an element $\exp X$ with $0<\Vert X\Vert <\epsilon$. Choose the first integer $m$ with $m\Vert X\Vert \ge \epsilon$. Then $\epsilon \le m\Vert X\Vert <2\epsilon <r$, so chart injectivity puts $(\exp X{)}^m=\exp (mX)$ outside $\exp (B_{\epsilon })$, a contradiction. Zero-dimensional Lie groups have the singleton identity as such an open neighborhood.

The Lie product formula extends from matrices to any finite-dimensional Lie group. In a local exponential chart, smooth multiplication and its differential give $\log (\exp (sX)\exp (sY))=s(X+Y)+O(s^2)$. Taking $s=t/m$ and using the one-parameter subgroup identity shows that the $m$th power is $\exp (t(X+Y)+O(m^{-1}))$, proving the displayed limit. This proof needs only the first-order expansion, rather than convergence of the full Baker--Campbell--Hausdorff formula.

Identify the Lie algebra $\mathfrak{g}$ with the tangent space $T_{e}G$. For $X\in \mathfrak{g}$, its left-invariant vector field is $X^L(g)=(d{L}_g{)}_{e}X$. Let ${\gamma }_X$ be its integral curve with ${\gamma }_X(0)=e$. Uniqueness of integral curves and left invariance give ${\gamma }_X(s+t)={\gamma }_X(s){\gamma }_X(t)$ wherever initially defined. Repetition of a local curve extends it to all real times. Thus ${\gamma }_X$ is the unique one-parameter subgroup with derivative $X$ at zero, and the Exponential map of a Lie group is defined bySmooth dependence of solutions of differential equations on their initial data and parameters makes this map smooth.

The derivative at zero is particularly simple:It is the identity linear map from $\mathfrak{g}$ to $T_{e}G$. The inverse function theorem therefore proves that the Exponential map of a Lie group is a local diffeomorphism at zero, producing a local exponential chart. This proves the requested local assertion.

There is a necessary qualification to the other assertion: the image of every one-parameter subgroup lies in the identity component of a Lie group $G^{\circ }$. The correct statement without a connectedness hypothesis isIndeed, if $\exp$ is a group homomorphism, its image is an abelian subgroup. It contains an identity neighborhood by the local result. An open subgroup of a connected group is the entire group: every coset is open, so its complement is also open. Thus $\exp (\mathfrak{g})=G^{\circ }$, proving that this component is abelian.

Conversely, if $G^{\circ }$ is abelian, then $\exp (tX)\exp (tY)$ is a one-parameter subgroup: its group law follows by commuting the factors. Its derivative at zero is $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$ and, at $t=1$, additivity. This proves the additivity of the Lie exponential map criterion. For connected $G$ it is exactly the stated equivalence with an abelian group. For disconnected $G$ the unrestricted claim is false: $S_3\times \mathbb{R}$ is a nonabelian Lie group, but its exponential is the homomorphism $x\mapsto (e,x)$. The original PDF does not impose connectedness in that sentence, so that qualification cannot be omitted.

Now suppose $G$ is connected and abelian, of dimension $n$. The exponential is a surjective group homomorphism, and its kernel $\Lambda$ is discrete by local injectivity at zero. The induced mapis a bijective Lie group homomorphism and a local diffeomorphism, hence a Lie group isomorphism. To identify the quotient, take linearly independent elements ${\lambda }_1,\dots ,{\lambda }_b\in \Lambda$ spanning the real linear span of $\Lambda$, and let ${\Lambda }_0={\sum }_i\mathbb{Z}{\lambda }_i$. Every coset of ${\Lambda }_0$ in $\Lambda$ has a representative in the bounded fundamental parallelepiped of these vectors. A discrete additive subgroup has finite intersection with a compact set: otherwise differences of arbitrarily close elements would approach zero, contradicting discreteness at zero. Thus $\Lambda /{\Lambda }_0$ is finite.

It follows that $\Lambda$ is a finitely generated torsion-free abelian group of rank $b$, hence has a $\mathbb{Z}$-basis of $b$ elements. These basis elements also span its $b$-dimensional real span and are linearly independent over $\mathbb{R}$. Extending them to a real basis identifies $\Lambda$ with ${\mathbb{Z}}^b\times \{0{\}}^{n-b}$. Consequently the classification of connected abelian Lie groups isThe compact circle factors record the periods of the one-parameter subgroups.

Suppose that $H$ is not a discrete subgroup. The identity is then an accumulation point of nonidentity subgroup elements: translate any failure of isolation to the identity. Choose $h_j\in H\setminus \{e\}$ with $h_j\to e$, and use a local exponential chart to write $h_j=\exp X_j$, where $X_j\ne 0$ and $X_j\to 0$ in the real Lie algebra $\mathfrak{g}$.

Choose a norm on this finite-dimensional vector space. By compactness of its unit sphere, pass to a subsequence with $X_j/\Vert X_j\Vert \to X$, where $\Vert X\Vert =1$. For an arbitrary real $t$, choose an integer $m_j$ nearest to $t/\Vert X_j\Vert$. Then $m_j\Vert X_j\Vert \to t$ andEvery left-hand element belongs to $H$, including for negative $m_j$. Closedness gives $\exp (tX)\in H$. Since $t$ was arbitrary and $X\ne 0$, the limit directions of a closed subgroup provide a nontrivial one-parameter subgroup in $H$. Its derivative at zero is $X$, so it cannot be the constant subgroup.

DefineIt contains zero, and $X\in \mathfrak{h}$ implies $aX\in \mathfrak{h}$ for every real scalar $a$, by replacing $t$ with $at$.

For $X,Y\in \mathfrak{h}$, we prove closure under addition using the Lie product formula in a Lie group. In the local exponential chart, let $Z(s)=\log (\exp (sX)\exp (sY))$. Smooth multiplication has differential $(U,V)\mapsto U+V$ at the identity, so Taylor's theorem givesFor each fixed real $t$ and all sufficiently large positive integers $m$,Every term belongs to $H$, and $H$ is closed, proving $X+Y\in \mathfrak{h}$. Thus $\mathfrak{h}$ is a real vector subspace. It is also closed: if $X_j\to X$ with $X_j\in \mathfrak{h}$, continuity of the Exponential map of a Lie group gives $\exp (tX)\in H$ for each $t$.

For the Lie bracket, use the Adjoint representation of a Lie group. Conjugating a one-parameter subgroup givesConsequently ${\mathrm{Ad}}_{\exp (sX)}Y\in \mathfrak{h}$. For $s\ne 0$, linearity puts $({\mathrm{Ad}}_{\exp (sX)}Y-Y)/s$ in $\mathfrak{h}$. Its limit as $s\to 0$ is ${\mathrm{ad}}_{X}Y=[X,Y]$. Closedness of $\mathfrak{h}$ therefore gives $[X,Y]\in \mathfrak{h}$, proving that the Lie algebra of a closed subgroup is a Lie subalgebra of $\mathfrak{g}$.

A real Lie group is a finite-dimensional real smooth manifold with a group structure whose multiplication and inversion are smooth. Its tangent space at the identity element $e$ is the vector space $\mathfrak{g}=T_{e}G$ of velocities of smooth curves through $e$. For the general linear group ${\mathrm{GL}}_n(\mathbb{R})$ it identifies with $M_n(\mathbb{R})$. The matrix exponential and local matrix logarithm areThe exponential is globally defined, and each matrix $\exp X$ is invertible with inverse $\exp (-X)$. The logarithm here is a local map near $I$; it is not a globally defined inverse on every real invertible matrix. The differential of $\exp$ at zero is the identity, and the two power series are inverse locally.

For a general $G$, the assumed Exponential map of a Lie group with $d{\exp }_0=I$ has a smooth local inverse by the inverse function theorem, giving a local exponential chart. Define the Lie bracket from local group commutators byFor small $s,t$ the group commutator lies in that chart. More explicitly, the map $F(u,v)=\log (\exp u\exp v\exp (-u)\exp (-v))$ is smooth and has $F(u,0)=F(0,v)=0$. Its mixed second differential at $(0,0)$ is a bilinear map of $X,Y$, proving bilinearity. Swapping the two group elements inverts the commutator, and $\log (g^{-1})=-\log g$ near $e$. Thus $F(v,u)=-F(u,v)$ and $[Y,X]=-[X,Y]$. In the matrix group, expansion to the mixed term $st$ gives the familiar commutator $XY-YX$.

For $g\in G$, let $c_g(h)=gh{g}^{-1}$ and define the Adjoint representation of a Lie group by ${\mathrm{Ad}}_g=d(c_g{)}_e\in \mathrm{GL}(\mathfrak{g})$. It is smooth and satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Its derived representation is $\mathrm{ad}=d(\mathrm{Ad}{)}_e:\mathfrak{g}\to \mathrm{End}(\mathfrak{g})$. Naturality of the Exponential map of a Lie group under conjugation givesAt $t=0$, the derivative of the logarithm of the commutator is ${\mathrm{Ad}}_{\exp (sX)}Y-Y$: the derivative of multiplication at $(e,e)$ adds tangent vectors, and $d{\log }_e=I$. Differentiate in $s$ to concludeTo prove the Jacobi identity without assuming it in the bracket construction, first establish that the differential of a Lie group homomorphism preserves Lie brackets. For a homomorphism $f:G\to H$, the allowed exponential identity gives ${\log }_H(f({\exp }_{G}u))=d{f}_e(u)$ locally. Apply this identity to the group commutator and take the mixed derivative to obtain $d{f}_e([X,Y])=[d{f}_{e}X,d{f}_{e}Y]$. In particular $f=\mathrm{Ad}:G\to \mathrm{GL}(\mathfrak{g})$ givesApplying both sides to $Z$ yields $[[X,Y],Z]=[X,[Y,Z]]-[Y,[X,Z]]$. Rearranging with antisymmetry provesNeither injectivity of $\mathrm{ad}$ nor the existence of a global logarithm was used.

Finally let $H$ be a normal Lie subgroup of $G$, with its corresponding Lie algebra $\mathfrak{h}\subseteq \mathfrak{g}$. For each $g\in G$, conjugation restricts to $H$, so its differential preserves $\mathfrak{h}$: ${\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}$. Differentiate along $g=\exp (tX)$ to get $[X,Y]=\mathrm{ad}(X)Y\in \mathfrak{h}$ for every $X\in \mathfrak{g}$ and $Y\in \mathfrak{h}$. Thus the Lie algebra of a normal Lie subgroup is an ideal of a Lie algebra. The assertion concerns a subgroup carrying the corresponding Lie-subgroup structure, for example any closed subgroup; no arbitrary abstract subgroup is being assigned a tangent space.