= Solution
Identify the <Lie algebra> $\mathfrak g$ with the tangent space $T_eG$. For $X\in\mathfrak g$, its <left-invariant vector field> is $X^L(g)=(dL_g)_eX$. 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 by
$$
\boxed{\exp X=\gamma_X(1),\qquad \exp(tX)=\gamma_X(t).}
$$
Smooth dependence of solutions of differential equations on their initial data and parameters makes this map smooth.
The derivative at zero is particularly simple:
$$
(d\exp)_0(X)=\left.\frac d{dt}\right|_{t=0}\exp(tX)=X.
$$
It is the identity linear map from $\mathfrak g$ to $T_eG$. 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 is
$$
\boxed{\exp:(\mathfrak g,+)\to G\text{ is a homomorphism}
\quad\Longleftrightarrow\quad G^\circ\text{ is abelian}.}
$$
Indeed, 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 map
$$
\mathbb R^n/\Lambda\longrightarrow G
$$
is a bijective <Lie group homomorphism> and a local diffeomorphism, hence a <Lie group isomorphism>. To identify the quotient, take linearly independent elements $\lambda_1,\ldots,\lambda_b\in\Lambda$ 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> is
$$
\boxed{G\cong\mathbb R^{n-b}\times(\mathbb R/\mathbb Z)^b
=\mathbb R^a\times T^b,\qquad a+b=n.}
$$
The compact circle factors record the periods of the <one-parameter subgroups>.
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