Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 17 / 4 / b / Solution

Use the derivative to transport the planes:The derivative is an invertible linear map, so it sends a two-dimensional subspace to a two-dimensional subspace. The flow identity ${\varphi }_t\circ {\varphi }_s={\varphi }_{t+s}$ givesHence a plane containing the generator is sent to another plane containing the generator; the displayed formula is a well-defined smooth lift to $\Lambda (SM)$. The chain rule gives ${\varphi }_{t+s}^{\ast }={\varphi }_t^{\ast }\circ {\varphi }_s^{\ast }$, and ${\varphi }_0^{\ast }$ is the identity. Thus it is a genuine lifted smooth flow, with inverse ${\varphi }_{-t}^{\ast }$, rather than merely a family of maps of individual fibres. Its generator is $F^{\ast }={{\partial }_t{\varphi }_t^{\ast }|}_{t=0}$.