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A perpendicular Jacobi field on a Riemannian surface has scalar equation $\ddot{y}+Ky=0$. For strictly negative curvature on a closed manifold, the displayed derivative is uniformly positive definite, so the quadratic-form criterion for an Anosov flow makes the geodesic flow Anosov. When $K=-1$, the stable and unstable lines in the canonical frame are respectively $\mathbb{R}(H-V)$ and $\mathbb{R}(H+V)$, with factors $e^{-t}$ and $e^t$.

Here is the quadratic-form criterion for an Anosov flow. Assume that the generator $F$ is nowhere zero and form the transverse vector bundleSince $D{\varphi }_{t}F(x)=F({\varphi }_{t}x)$, the derivative induces a linear cocycle $P_t:{\mathcal{E}}_x\to {\mathcal{E}}_{{\varphi }_{t}x}$. Suppose there is a continuous nondegenerate indefinite quadratic form $Q_x$ on each fibre, with constant signature of a quadratic form, such that $t\mapsto Q_{{\varphi }_{t}x}(P_t\xi )$ is differentiable andis continuous and positive definite on every transverse fibre. Then ${\varphi }_t$ is an Anosov flow. An equivalent way to specify $Q$ is a quadratic form on $TN$ whose radical is exactly $\mathbb{R}F$ and whose induced form and derivative satisfy these conditions on the quotient.

On the closed manifold, compactness makes the positivity uniform: for some $c>0$, $\mathcal{D}{Q}_x(\xi )\ge c\Vert \xi {\Vert }^2$. Thus the positive and negative cones are strictly separated by the linearized flow, yielding uniform unstable and stable transverse subbundles and their invariant lifts. The positive index is the unstable dimension and the negative index is the stable dimension. Strict positivity must be imposed transversely; imposing it on the flow direction would be impossible because that direction is carried into itself.

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.