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Use the Levi-Civita connection to represent a tangent vector to the unit tangent bundle at $(x,v)$ by the initial data $(J(0),D_{t}J(0))$ of the Jacobi field generated by a variation of the unit-speed geodesic. The unit-speed constraint gives $⟨D_{t}J(0),v⟩=0$. Along the geodesic, writeHere $a$ is constant, while the perpendicular Jacobi field equation isThe 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 $(1,1)$, andSince the Gaussian curvature is strictly negative and the surface is a closed manifold, there is $k_0>0$ with $K\le -k_0$. HenceThe quadratic-form criterion for an Anosov flow now proves 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 $y=A{e}^t+B{e}^{-t}$. 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 $(0,iv)$, soEquivalently 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 $\mathbb{R}X$ they give the required invariant splitting.