exponential map

The exponential map

For a Lie group $G$ with Lie algebra $\mathfrak{g}$, the exponential map is

$$\begin{array}{rl}\exp :\mathfrak{g}& ⟶G\\ \xi & ⟼{\gamma }^{\xi }(1)\end{array}$$

where ${\gamma }^{\xi }(t)$ is the 1-parameter subgroup with ${{\gamma }^{\xi }}^{\prime }(0)=\xi$.

In other words, to find the exponential map, we take an element of the Lie algebra, and we convert it to the left-invariant vector field $X^{\xi }$. Then we can take the flow generated by $X^{\xi }$ and look at ${\Phi }^{(e)}(1)$ which is the integral curve starting at $\gamma (0)=e$ with ${\gamma }^{\prime }(0)=\xi$ and evaluate it at $1$.

Why 1?

Why do we evaluate on the somewhat arbitrary seeming number $1$?

It comes from the property of 1-parameter subgroup as integral curves for left-invariant vector fields. We know that

$${\gamma }^{s\xi }(t)={\gamma }^{\xi }(st)$$

Where ${\gamma }^{s\xi }$ is the integral curve starting at $e$ for a left invariant vector field identified by an element $\xi \in \mathfrak{g}$ .

So if we use this formula and evaluate where $s=1$ then we get

$${\gamma }^{\xi }(t)=\exp (t\xi )$$

Thus, the exponential function traces out integral curves of left-invariant vector fields that go through the identity element of the Lie group. We can write flows in terms of the exponential map. The flow ${\Phi }_{\xi }$ which is the flow of ${\xi }^L$, the left-invariant vector field corresponding to $\xi$ is

$$\Phi (t,g)=g\exp (-t\xi )\forall t\in \mathbb{R},g\in G$$

Properties

• If $[\xi ,\eta ]=0$ then $\exp (\xi +\eta )=\exp (\xi )\exp (\eta )$
• The exponential map respects Lie group homomorphisms. The following diagram commutes for Lie group homomorphism $\phi$

• For a matrix Lie algebra, the exponential map is the matrix exponential (proof here).