Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 115 / 2 / ii

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- Solution ii

Solution

Using the left-invariant frame of the real Heisenberg group, horizontality gives

\overset{x}{˙} = α, \overset{y}{˙} = β, \overset{z}{˙} = x β .

For an arbitrary starting point

(x_{0}, y_{0}, z_{0})

,

x (t) = x_{0} + \int_{0}^{t} α (s) d s, y (t) = y_{0} + \int_{0}^{t} β (s) d s, z (t) = z_{0} + \int_{0}^{t} (x_{0} + \int_{0}^{s} α (r) d r) β (s) d s .

This is the control equation for the Heisenberg horizontal distribution. The final coordinate records a signed area-like integral, so changing the path in the

(x, y)

plane can change the endpoint in the central direction.

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