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A contact form $\alpha$ defines the hyperplane distribution $\xi =\ker \alpha$. Multiplying $\alpha$ by a nowhere-zero function leaves this distribution unchanged. The contact distribution symplectic form distinguishes it from an integrable hyperplane distribution.

For a contact form on a $(2r+1)$-manifold, $d\alpha {|}_{\ker \alpha }$ is nondegenerate: the contact condition says its top exterior power is nowhere zero on the hyperplanes. This fiberwise symplectic form identifies contact-plane vectors with covectors on that plane and determines the horizontal generator of contact stability.

A smooth interval family of contact forms on a compact manifold without boundary is carried back to its initial contact distribution by a smooth isotopy. The pullback of each form is a positive smooth multiple of the initial form. Solve for the horizontal generator of contact stability and integrate the resulting scalar equation for the contact conformal factor along an isotopy.

For a family of contact forms, require $Y_t\in \ker {\alpha }_t$ and solve $({\iota }_{Y_t}d{\alpha }_t){|}_{\ker {\alpha }_t}=-{\dot{\alpha }}_t{|}_{\ker {\alpha }_t}$. The contact distribution symplectic form gives a unique smooth solution. Then ${\dot{\alpha }}_t+{\mathcal{L}}_{Y_t}{\alpha }_t=h_t{\alpha }_t$, with $h_t={\dot{\alpha }}_t(R_t)$ and $R_t$ the Reeb vector field.

We use the following precise degree-zero/one kernel of the magnetic X-ray transform in degrees zero and one theorem, as permitted here: for a smooth Anosov magnetic flow on a closed oriented surface, ifwith $q$ a smooth base function and $\xi$ a smooth one-form, then $q=0$ and $\xi$ is exact. Exact one-forms give the converse, because $F(w\circ \pi )=dw(v)$. No assumption of pointwise negative Gaussian curvature is needed for this theorem. We now derive the contact obstruction from it rather than assume the desired rigidity conclusion.

Use the usual connected-surface convention. Since ${\int }_{M}f{\Omega }_a=0$, the top-degree de Rham cohomology class of $f{\Omega }_a$ vanishes: integration identifies $H_{\mathrm{dR}}^2(M)$ with $\mathbb{R}$. Choose a smooth one-form $\eta$ on $M$ satisfying $d\eta =f{\Omega }_a$, and setThe structure equations and the previous volume calculation giveThis primitive exists because of the zero-flux hypothesis, and need not itself be contact.

A useful elementary fact is that every smooth invariant function of an Anosov flow is constant on a connected manifold. If $Fh=0$, then $d{h}_x(v_s)=d{h}_{{\varphi }_{t}x}(D{\varphi }_t{v}_s)$, and boundedness of $dh$ with exponential contraction makes this zero as $t\to \infty$. Negative time does the same for an unstable vector. Also $dh(F)=0$. The Anosov splitting spans the tangent space, so $dh=0$. This avoids any extra ergodicity assumption.

Suppose a smooth contact form $\tau$ is preserved. Invariance and contraction imply $\tau (v_s)=\tau (D{\varphi }_t{v}_s)=0$ in the limit; negative time gives the same on unstable vectors. Thus $\tau$ annihilates both transverse bundles. Its value $\tau (F)$ cannot vanish anywhere, since otherwise $\tau$ would be zero at that point, contradicting the contact condition. It is a smooth invariant function and therefore a nonzero constant. Divide $\tau$ by that constant to arrange $\tau (F)=1$.

The Cartan formula for the Lie derivative now gives ${\iota }_{F}d\tau =0$. In dimension three, all two-forms annihilating $F$ are scalar multiples of ${\iota }_F\mu$, soThe scalar $k$ is smooth and nowhere zero by the contact condition. Both sides without $k$ are flow-invariant; hence $Fk=0$ and the preceding Anosov argument makes $k$ constant. It follows thatis a closed differential one-form.

We require the Liouville mean of such a closed differential form, and derive it explicitly. Let $R_s$ be fibre rotation and average $\zeta$ over $0\le s\le 2\pi$. Since $d\zeta =0$, the Cartan formula for the Lie derivative givesConsequently its rotational average differs from $\zeta$ by an exact form. For a closed rotationally invariant one-form $\overline{\zeta }$, the identity $d(\overline{\zeta }(V))={\mathcal{L}}_V\overline{\zeta }-{\iota }_{V}d\overline{\zeta }=0$ makes its vertical value a constant $a$. Subtracting $a\omega$ leaves a rotationally invariant horizontal form, which descends to a base form $\xi$. We have therefore proved the averaging closed one-forms on a surface unit tangent bundle decompositionThe last relation follows from closedness and $d\omega =-K{\pi }^{\ast }{\Omega }_a$.

Evaluation at $F$ gives $\zeta (F)={\xi }_x(v)+af+Fh$. The average of the linear expression ${\xi }_x(v)$ around each unit circle is zero. The second term has total integral $2\pi a{\int }_{M}f{\Omega }_a=0$. Finally ${\int }_{SM}Fhd\mu =0$ by volume preservation and Stokes theorem. Thusfor every closed differential one-form $\zeta$ in this zero-flux system. Apply this to $\tau +k\sigma$: the directional average of ${\eta }_x(v)$ also vanishes, so its integral is $(1+k){\mathrm{Vol}}_{\mu }(SM)$. Positivity of the volume forces $k=-1$. Hence $d\tau =d\sigma$ and $\zeta =\tau -\sigma$ is closed.

Using its proved decomposition and $\tau (F)=1$, we obtainThe stated magnetic transform kernel theorem gives $af=0$ and exactness of $\eta -\xi$, hence $d\eta =d\xi$. If $a=0$, then $d\xi =0$ and $f{\Omega }_a=d\eta =0$, so $f=0$. If $a\ne 0$, the equation $af=0$ already gives $f=0$. Necessity is proved.

Conversely, if $f=0$, then $F=X$. The canonical form $\alpha$ is contact because $\alpha \wedge d\alpha =-\mu$ never vanishes, andThus the flow preserves $\alpha$, provingIf disconnected surfaces are admitted, the flux condition must hold on each component for the exactness step; a single cancelling total flux does not imply componentwise exactness. The result above uses the standard connected interpretation of a surface.

Put ${\xi }_t=\ker {\alpha }_t$, the contact distribution. If $\dim M=2r+1$, the contact form condition is ${\alpha }_t\wedge (d{\alpha }_t{)}^r\ne 0$. It implies that $d{\alpha }_t{|}_{{\xi }_t}$ is a symplectic form on each contact hyperplane: inserting a vector transverse to ${\xi }_t$ in that volume form leaves the nonzero top exterior power $(d{\alpha }_t{|}_{{\xi }_t}{)}^r$. Hence there is a unique smooth time-dependent vector field $Y_t\in {\xi }_t$ satisfyingIts smoothness follows by inverting the smoothly varying nondegenerate matrix of this contact distribution symplectic form. This is the horizontal generator of contact stability.

Let $R_t$ be the Reeb vector field of ${\alpha }_t$. The one-form ${\dot{\alpha }}_t+{\iota }_{Y_t}d{\alpha }_t$ vanishes on ${\xi }_t$, so it is $h_t{\alpha }_t$. Evaluating on $R_t$, using ${\alpha }_t(R_t)=1$ and ${\iota }_{R_t}d{\alpha }_t=0$, identifies the coefficient:By Cartan's magic formula, the Lie derivative of a differential form isbecause $Y_t$ lies in the contact hyperplane.

Let ${\rho }_t$ solve ${\partial }_t{\rho }_t=Y_t\circ {\rho }_t$, with ${\rho }_0={\mathrm{id}}_M$. Since $M$ is compact without boundary and $Y_t$ is smooth on the closed time interval, its solutions cannot escape and exist for the entire interval. Reversing the time-dependent equation supplies a smooth inverse for each ${\rho }_t$. Thus these flow maps give a smooth isotopy. Differentiating the pullback of a differential form along this isotopy yieldsFor each base point this is a scalar linear equation for a covector, initially ${\alpha }_0$. Its solution isBy finite-time flow completeness on a compact manifold, this construction gives the whole time interval. The contact conformal factor along an isotopy is smooth in $(t,x)$ and nowhere zero, proving Gray stability theorem. The precise domains here are ${\rho }_t:M\to M$ and $\rho :M\times [0,1]\to M$; the extra time factor attached to the already indexed ${\rho }_t$ in the source is a notational slip. The proof produces the required family on its whole specified interval. If a parameter domain of all $\mathbb{R}$ is intended, extend the smooth field slightly beyond $[0,1]$, multiply it by a time cutoff, and use compactness to obtain the extended isotopy $\rho :M\times \mathbb{R}\to M$. Its restriction to $[0,1]$ is unchanged.

Use polar coordinates $z_j=r_j{e}^{i{\theta }_j}$ on each complex coordinate plane. On $S^3$, where $r_1^2+r_2^2=1$,Direct substitution shows that ${\alpha }_s\wedge d{\alpha }_s$ is nowhere zero for $s>0$, so ${\alpha }_s$ is a contact form.

The vector fieldsatisfies ${\alpha }_s(R_s)=r_1^2+r_2^2=1$ and ${\iota }_{R_s}d{\alpha }_s=0$ on tangent vectors to the sphere. It is therefore the Reeb vector field. Its flow isIf both coordinates are nonzero, an orbit closes only if the two angular frequencies have rational ratio, equivalently if $s\in \mathbb{Q}$. For irrational $s$, the only closed orbits are $\{z_2=0\}$ and $\{z_1=0\}$, the two coordinate circles. This is the irrational contact ellipsoid flow on the three-sphere.