1-parameter subgroup

For a Lie group $G$, a 1-parameter subgroup is a Lie group morphism

$$\gamma :\mathbb{R}⟶G$$

Notes

• By definition, $\gamma$ is a smooth curve on $G$.
• The set of all 1-parameter subgroups is not a subgroup itself, however, the image of $\gamma$ is a subgroup.
• $\gamma$ does not need to be injective.

Correspondence between integral curves

Note that simply from the definition, 1-parameter subgroups are not integral curves. However, we can show that there is a strong correspondence between them 1-parameter subgroups and integral curves of left-invariant vector fields.

Proposition 1

The unique integral curve ${\gamma }_X$ of $X$ with initial condition ${\gamma }_X(0)=e$ is a 1-parameter subgroup.

Proof

Let $\Phi$ be the flow generated by $X$. Then by we know that ${\Phi }_{t_1+t_2}={\Phi }_{t_1}{\Phi }_{t_2}$ so looking at ${\Phi }^{(e)}={\gamma }_X$ we see that it is a 1-parameter subgroup

Proposition 2

If $\gamma$ is a 1-parameter subgroup, then it is the integral curve of a left-invariant vector field.

Proof

We know that ${\Phi }_t(g)=g\gamma (-t)$ is a flow (we need the minus sign to ensure that it is a left action). Note that

$$h{\Phi }_t(g)=h(g\gamma (-t))=hg(\gamma (-t))={\Phi }_t(hg)$$

So the flow commutes with left multiplication so it is the flow of a left invariant vector field.
Take the integral curve of ${\Phi }_{-t}(e)=\gamma (t)$ to finish the proof.