Exponential map of a Lie group - Imported Knowlpedia Content

Let $G$ be a Lie group with Lie algebra $\mathfrak g=\mathrm{Lie}(G)$ (see Lie algebra of a Lie group).

For each $X\in\mathfrak g$ , there is a unique one-parameter subgroup $\gamma_X:\mathbb R\to G$ whose tangent at $0$ equals $X$ (compare exponential/one-parameter subgroup lemma). The exponential map is

\exp:\mathfrak g\to G,\qquad \exp(X):=\gamma_X(1).

Equivalently, $\gamma_X(t)=\exp(tX)$ for all $t\in\mathbb R$ .

Functoriality

If $\Phi:G\to H$ is a Lie group homomorphism, then differentiation gives a Lie algebra map $d\Phi_e$ (see differential is a Lie algebra homomorphism), and exponentials intertwine:

\Phi(\exp_G X)=\exp_H(d\Phi_e X).

Matrix groups: concrete formula

If $G\subseteq GL(n,\mathbb F)$ is a matrix Lie group, then for $X\in\mathfrak g\subseteq \mathfrak{gl}_n(\mathbb F)$ ,

\exp(X)=e^X:=\sum_{k=0}^\infty \frac{X^k}{k!}.

For example, in $U(1)$ , this recovers $t\mapsto e^{it}$ ; in $SO(3)$ , exponentials of skew-symmetric matrices are rotation matrices.

Local behavior

The exponential map is always a local diffeomorphism at $0$ (see exponential is a local diffeomorphism). The local inverse is the logarithm map.

Transporting multiplication through $\log$ gives the local group law on $\mathfrak g$ described by the Baker–Campbell–Hausdorff formula.

Global remarks

$\exp$ need not be surjective in general (not every element must lie on a 1-parameter subgroup).
For connected compact Lie groups, $\exp$ is surjective: every element lies in a maximal torus (compare maximal torus theorem) and exponentials are surjective on tori (see the torus example).

Context. The exponential map is the primary bridge between the linear object $\mathfrak g$ and the nonlinear group $G$ , converting Lie-algebraic computations into local (and sometimes global) statements about the Lie group.