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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.