Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 32 / 1 / iii / Solution

Euclidean Lipschitz continuity does not imply Carnot-Carathéodory Lipschitz continuity. In the exponential coordinates above take the vertical path $\gamma (t)=(0,0,t)$. It is Euclidean Lipschitz continuous, butIndeed left invariance reduces to the central increment, graded dilation gives the square-root factor, and reversal changes the sign without changing the distance.

One can also see the scale directly. A horizontal curve starting at the identity satisfies $\dot{z}=(x\dot{y}-y\dot{x})/2$. If it ends on the vertical axis, its planar projection is a closed loop andFor a projection of length $L$ starting at zero, $\sup |(x,y)|\le L$, so $|z|\le L^2/2$. Thus reaching a central displacement $z$ requires length at least $\sqrt{2|z|}$, while a square loop with the appropriate signed area gives length at most $4\sqrt{|z|}$. In particular $d_{CC}(\gamma (s),\gamma (t))/|t-s|$ is unbounded as $t\to s$. Vertical displacement records area, which scales quadratically with horizontal length.