# How to construct the Lie bracket from a Lie group?

Suppose you have a Lie group $G$ with identity element $g$, then the Lie algebra is isomorphic to the tangent space at $g$, $T_gG$. However, to fully specify the Lie algebra, you also have to define a Lie bracket. For matrix Lie groups, the Lie bracket can be constructed from the commutator, but are there constructions that hold for general Lie groups, just from knowledge of the group $G$ and the exponential map from $T_gG$ to $G$?

To be a bit more specific, does $log(exp(x)exp(y))-log(exp(y)exp(x))$ define a Lie bracket, at least locally?

## 2 Answers

You have an adjoint action of $G$ on $\mathfrak{g}$ (by conjugation). It gives you (by differentiating) the map $ad$ from $\mathfrak{g}$ to $End(\mathfrak{g})$. Now $(ad(x))(y)=[x,y]$

Also, in terms of $exp$ you have $\log(\exp(tX)exp(tY))=t(X+Y)+t^2[X,Y]+o(t^2)$

The space of vector fields on $G$, or any smooth manifold form a Lie algebra and the bracket is given by the commutator. Sitting in side this huge Lie algebra is the space of vector fields invariant under the action of $G$ on itself by say left multiplication. Any $G$ invariant vector field on a homogeneous $G$-space is completely determined by its value at any point, in our case we will choose that point to be the identity. This gives the correspondence between the tangent space at $e$ and the space of $G$ equivariant vector fields, which have a natural Lie algebra structure.