Exponential Map of a Lie Group

Let $G$ be a Lie group with identity element $e$, and let $\mathfrak g=T_eG$ be its Lie algebra.

Definition (Lie group exponential). For each $X\in \mathfrak g$, there exists a unique smooth group homomorphism

such that $\gamma_X(0)=e$ and $\gamma_X'(0)=X$ (viewing $X$ as a tangent vector in $T_eG$). The exponential map of $G$ is the smooth map

Basic properties.

• $\exp_G(0)=e$ and $(d\exp_G)_0=\mathrm{id}_{\mathfrak g}$.
• The curve $t\mapsto \exp_G(tX)$ is a one-parameter subgroup: $\exp_G((s+t)X)=\exp_G(sX)\exp_G(tX)$.
• If $X,Y\in\mathfrak g$ commute (i.e. $[X,Y]=0$ for the Lie bracket), then $\exp_G(X+Y)=\exp_G(X)\exp_G(Y)$.
• If $\varphi:G\to H$ is a Lie group homomorphism, then $\varphi$ intertwines exponentials via the induced Lie algebra homomorphism:
  $$\varphi(\exp_G(X))=\exp_H(d\varphi_e(X)).$$

Examples

1. Additive group $(\mathbb R^n,+)$. Here $G=\mathbb R^n$ is a Lie group under addition, $\mathfrak g\cong \mathbb R^n$, and the one-parameter subgroup with velocity $X$ is $\gamma_X(t)=tX$. Thus $\exp_G(X)=X$ (the identity map).

1. Circle group $S^1\subset \mathbb C$. With multiplication in $\mathbb C$, $G=S^1$ has Lie algebra $\mathfrak g=T_1S^1=i\mathbb R$. The exponential map is the usual complex exponential restricted to $i\mathbb R$:
   $$\exp_{S^1}(i\theta)=e^{i\theta}.$$

1. Matrix Lie groups. If $G\subseteq GL(n,\mathbb R)$ is a matrix Lie group, then $\mathfrak g\subseteq M_n(\mathbb R)$, and $\exp_G$ is given by the matrix exponential:
   $$\exp_G(A)=\sum_{k=0}^{\infty}\frac{A^k}{k!}.$$
   For example, in $SO(2)$ this recovers rotations, and in $GL(n,\mathbb R)$ it produces invertible matrices for all $A$.