Recall that on a differentiable manifold, it is not possible to use a tangent vector \({v}\) to “transport a point in the direction \({v}\)” in a coordinate-independent way, since there is no special curve on \({M}\) among the many that have \({v}\) as a tangent. On a Lie group this is possible, since the left-invariant vector fields provide a unique flow in the direction of \({v}\).

A **one-parameter subgroup** of \({G}\) is a homomorphism \({\phi\colon\mathbb{R}\to G}\). Given a left-invariant vector field \({A}\), there is a unique one-parameter subgroup \({\phi_{A}}\) such that \({\phi_{A}\left(0\right)=e}\) and \({\dot{\phi}_{A}\left(t\right)=A}\) for all \({t}\) (i.e. \({\phi_{A}\left(t\right)}\) is the local flow from the Lie derivative, but being defined for all \({t}\) it is called simply the **flow** of \({A}\)). We can then define the **exponential map** \({\textrm{exp}\colon\mathfrak{g}\to G}\) by

\(\displaystyle \mathrm{exp}(A)\equiv e^{A}\equiv \phi_{A}\left(1\right). \)

Since scaling the parametrization scales the tangent vectors, we have \({\phi_{A}\left(t\right)=\phi_{tA}\left(1\right)=\mathrm{exp}(tA)}\).

In particular, the elements of \({G}\) infinitesimally close to the identity can be written \({e+\varepsilon A}\). The exponential map is a generalization of familiar exponential functions: if \({G=\mathbb{R}^{+}}\), the positive reals under multiplication, \({\mathfrak{g}=\mathbb{R}}\) and \({\textrm{exp}}\) is the normal exponential function for real numbers; if \({G}\) is the non-zero complex numbers under multiplication, \({\mathfrak{g}=\mathbb{C}}\) and \({\textrm{exp}}\) is the normal complex exponential function; and if \({G=GL(n,\mathbb{R})}\), the real invertible \({n\times n}\) matrices under matrix multiplication, \({\mathfrak{g}=gl(n,\mathbb{R})}\), the real \({n\times n}\) matrices, and \({\textrm{exp}}\) is matrix exponentiation, defined by

\(\displaystyle e^{A}\equiv\overset{\infty}{\underset{k=0}{\sum}}\frac{1}{k!}A^{k}. \)

The multiplication of matrix exponentials does not follow the scalar rule, instead being given by the **Baker-Campbell-Hausdorff formula**:

\(\displaystyle e^{A}e^{B}\equiv e^{A+B+\frac{1}{2}\left[A,B\right]+\dotsb} \)

The terms that continue the series are all expressed in terms of Lie commutators. The terms shown above comprise the entire series if both matrices commute with the commutator, i.e. if \({\left[A,\left[A,B\right]\right]=\left[B,\left[A,B\right]\right]=0}\). This formula is valid for any associative algebra.