Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 15 / 1

An integral curve of a vector field $X$ through $p$ is a smooth curve $c:I\to M$, on an interval containing zero, satisfying $c(0)=p$ and $\dot{c}(t)=X_{c(t)}$. The local flow is the smoothly varying family of these integral curves of a vector field:Here $D$ is an open neighborhood of $\{0\}\times M$; for the maximal local flow, its fibre over $p$ is the maximal existence interval of $c_p$. Local existence, uniqueness and smooth dependence for a smooth ordinary differential equation give this family. Uniqueness also gives ${\varphi }_{t+s}(p)={\varphi }_t({\varphi }_s(p))$ whenever the relevant curves and compositions are defined, and ${\varphi }_{-t}={\varphi }_t^{-1}$ on their common domains.

Reparametrizing the integral curves of a vector field givesThus the flow of $2X$ is $\boxed{{\tilde{\varphi }}_t={\varphi }_{2t}}$, with domain $\{(t,p):(2t,p)\in D\}$.

Commutation of the two local flows means ${\varphi }_t{\psi }_s={\psi }_s{\varphi }_t$ for all sufficiently small independent parameters $s,t$, on their common local domains. Differentiate this identity in $s$ at zero. The chain rule givesIn a coordinate chart, differentiating again in $t$ at zero gives $D{X}_p{Y}_p=D{Y}_p{X}_p$. But the Lie bracket of vector fields has coordinate expressionso commuting local flows imply $[X,Y]=0$.

For the converse, suppose the Lie bracket of vector fields vanishes. Write $u(t)={\varphi }_t(p)$ in a coordinate chart, and put $J(t)=d{\varphi }_t{|}_p$. Differentiating the local flow equation with respect to the initial point gives the variational ordinary differential equationConsequently $W(t)=Y_{u(t)}-J(t)Y_p$ satisfieswhere the zero Lie bracket of vector fields was used in the second equality. Uniqueness for this linear ordinary differential equation gives $W=0$. This is a local argument and can be continued through successive coordinate charts, proving $d{\varphi }_t(Y)=Y\circ {\varphi }_t$ wherever the local flow is defined.

For fixed $t$, the curve $s\mapsto {\varphi }_t({\psi }_s(p))$ therefore has derivative $Y$ at its current point and starts at ${\varphi }_t(p)$. Uniqueness of the integral curve of a vector field $Y$ identifies it with $s\mapsto {\psi }_s({\varphi }_t(p))$. Hence $\boxed{{\varphi }_t{\psi }_s={\psi }_s{\varphi }_t⟺[X,Y]=0}$ locally. No complete vector field hypothesis is needed.

Using the coordinate formula for the Lie bracket of vector fields, the two components areThus $\boxed{[X,Y]=-(x^2/2){\partial }_x+xy{\partial }_y}$, which is not the zero vector field.

The integral curves of a vector field $X$ solve $\dot{x}=y$, $\dot{y}=0$, whereas those of $Y$ solve $\dot{x}=0$, $\dot{y}=x^2/2$. Their local flows areBoth formulas exist for all real parameters, so both are complete vector fields. Direct composition givesAt $(1,0)$ their first coordinates differ by $ts/2$. Hence these local flows do not commute when $s,t$ are nonzero.

For the Lie bracket of vector fields, start at $(2,0)$. An integral curve of a vector field $[X,Y]$ isIt solves $\dot{x}=-x^2/2$, $\dot{y}=xy$, but $x(t)$ diverges as $t\downarrow -1$. It cannot extend through that finite time as a curve in ${\mathbb{R}}^2$. Complete vector fields are therefore not closed under the Lie bracket.

For the sum $X+Y$, its integral curves of a vector field obey $\dot{x}=y$, $\dot{y}=x^2/2$. Substitution of $x=(1+\alpha t{)}^{-2}$ and $y=\beta (1+\alpha t{)}^{-3}$ requiresChoosing $\alpha =-1/(2\sqrt{3})$ gives the integral curve of a vector fieldIt starts at $(1,1/\sqrt{3})$ and escapes to infinity at the finite time $2\sqrt{3}$. Complete vector fields are not closed under addition either. Both failures use the same pair of complete vector fields.