Solution (source code)

= Solution

Use the <Levi-Civita connection> to represent a tangent vector to the <unit tangent bundle> at $(x,v)$ by the initial data $(J(0),D_tJ(0))$ of the <Jacobi field> generated by a variation of the unit-speed <geodesic>. The unit-speed constraint gives $\langle D_tJ(0),v\rangle=0$. Along the <geodesic>, write
$$
J(t)=a\dot\gamma(t)+y(t)i\dot\gamma(t),\qquad D_tJ(t)=\dot y(t)i\dot\gamma(t).
$$
Here $a$ is constant, while the perpendicular <Jacobi field> equation is
$$
\ddot y+K(\gamma(t))y=0.
$$
The flow direction is represented by $(a,y,\dot y)=(1,0,0)$. Consequently $(y,z)=(y,\dot y)$ are coordinates on the transverse quotient. Define its nondegenerate <quadratic form> by $Q(y,z)=yz$. Its <signature of a quadratic form> is $(1,1)$, and
$$
\frac{d}{dt}Q(y(t),z(t))=z(t)^2-K(\gamma(t))y(t)^2.
$$
Since the <Gaussian curvature> is strictly negative and the surface is a <closed manifold>, there is $k_0>0$ with $K\le-k_0$. Hence
$$
\dot Q\ge\min(1,k_0)(y^2+z^2).
$$
The <quadratic-form criterion for an Anosov flow> now proves \b[the geodesic flow is Anosov]. This is the <geodesic hyperbolicity from negative Gaussian curvature> argument. If the surface is not oriented, the local choice of perpendicular unit vector changes both $y$ and $z$ in sign; their product is unchanged, so the <quadratic form> still exists globally.

When $K=-1$, the perpendicular solutions are $y=Ae^t+Be^{-t}$. The pure contracting data have $(y,z)=(B,-B)$ and the pure expanding data have $(y,z)=(A,A)$. In the <canonical coframe of a surface unit tangent bundle>, $H$ represents $(iv,0)$ and $V$ represents $(0,iv)$, so
$$
\boxed{E^s(x,v)=\mathbb R(H-V),\qquad E^u(x,v)=\mathbb R(H+V).}
$$
Equivalently the <Anosov stable bundle> consists of pairs $(\eta,-\eta)$ with $\eta\perp v$, and the <Anosov unstable bundle> consists of $(\eta,\eta)$. The derivative multiplies these data by $e^{-t}$ and $e^t$, respectively, using parallel transport along the <geodesic> to identify the perpendicular directions. Together with $\mathbb RX$ they give the required invariant splitting.