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

Let $\pi :SM\to M$ be the unit tangent bundle projection. Write $X$ for the geodesic flow generator and $V$ for the vertical vector field of a surface unit tangent bundle, positively rotating $v$ toward $iv$. The acceleration prescribed in the question changes only this velocity angle, so the surface thermostat generator isIt never vanishes: $d\pi (F)=v\ne 0$. Also $F$ and $V$ are linearly independent, since $d\pi (V)=0$.

The vertical Maslov cycle of a surface thermostat isThere is exactly one such plane in each fibre, so ${\Lambda }_V$ is the image of a smooth section. A two-plane containing $F$ corresponds to a line in the rank-two quotient $T(SM)/\mathbb{R}F$; thus $\Lambda (SM)$ is a projective bundle with fibre ${\mathbb{RP}}^1$. Its total dimension is four and the cycle has dimension three, making it a smooth hypersurface. In this two-dimensional transverse space the Maslov cycle means the projective line meeting the distinguished vertical line, which here is just that distinguished line itself in each fibre.

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 $F^{\ast }={{\partial }_t{\varphi }_t^{\ast }|}_{t=0}$.

Let $H$ be the horizontal lift of $iv$ supplied by the canonical coframe of a surface unit tangent bundle. The frame $(X,H,V)$ has commutatorsSince $F=X+\lambda V$, the Lie bracket product rule givesThese formulas hold for every smooth $\lambda$, with no curvature or Anosov assumption.

Follow a tangent variation $\xi (t)=D{\varphi }_t\xi (0)$ and express it in the moving frame asPulling this identity back by $D{\varphi }_{-t}$ makes its left side constant. The derivative of the pulled-back frame vector $B$ is the pulled-back bracket $[F,B]$, so the coefficients satisfyThis derives the linearized transverse equation for a surface thermostat; its first equation in $a$ disappears on the quotient by $F$.

Near a vertical plane, $z\ne 0$, and the smooth projective coordinate $r=y/z$ describes the plane asThe vertical Maslov cycle of a surface thermostat is exactly $r=0$. Differentiating the quotient with the derived equations yieldsConsequentlySince $r$ is a local defining function for the hypersurface, the nonzero normal derivative proves that the lifted vector field is transverse to the Maslov cycle everywhere. In particular, the sign follows from the forward derivative lift and the positive vertical-rotation convention, rather than from an unspecified projective orientation.