Lie algebra of a Lie group

Let $G$ be a Lie group with identity element $e$. The Lie algebra of $G$ is the real vector space

the tangent space of $G$ at the identity, equipped with a canonical Lie bracket defined as follows.

For each $X\in \mathfrak{g}$, define a vector field $\widetilde{X}$ on $G$ by left translation:

where $L_g$ is the left translation map and $(dL_g)_e$ is the pushforward (differential) at $e$. Then $\widetilde{X}$ is a left-invariant vector field , and every left-invariant vector field arises uniquely this way.

The space of left-invariant vector fields is closed under the Lie bracket of vector fields. The Lie bracket on $\mathfrak{g}$ is defined by

which makes $\mathfrak{g}$ into a Lie algebra.

The exponential map $\exp_G:\mathfrak{g}\to G$ is defined using one-parameter subgroups and relates the Lie algebra structure to the group structure near $e$. Moreover, if $\varphi:G\to H$ is a Lie group homomorphism , then its differential at the identity is a Lie algebra homomorphism, as explained in the induced map on Lie algebras .

Examples

1. For $G=GL(n,\mathbb{R})$, one has $\mathfrak{g}=\mathfrak{gl}(n,\mathbb{R})$ (all $n\times n$ real matrices), with bracket

   $$[A,B]=AB-BA.$$

2. For $G=SO(n)$, one has $\mathfrak{g}=\mathfrak{so}(n)$, the space of skew-symmetric matrices $A^T=-A$, with the same commutator bracket.

3. For the abelian Lie group $(\mathbb{R}^n,+)$, the Lie algebra is $\mathbb{R}^n$ with the zero bracket: $[X,Y]=0$ for all $X,Y\in\mathbb{R}^n$.