6 The tensor tomography result
In this final section we prove a vanishing theorem for the transport equation \(Fu=Va+\beta\), provided the triple \((g,A,\theta)\) defining \(F\) satisfies certain conditions. Recall that every properly convex projective structure \(\mathfrak{p}\) arises from a triple \((g,A,0)\) satisfying \[K_g=-1+2|A|^2_g \quad \text{and} \quad \overline{\partial} A=0.\] In particular, we would like to conclude that if such a \(\mathfrak{p}\) contains a Weyl connection, then \(A\) must vanish identically and hence \(\mathfrak{p}\) is hyperbolic. It turns out that one can prove a more general vanishing theorem for a class of thermostats arising from a triple \((g,A,\theta)\) where \(A\) is a differential of degree \(m\geqslant 3\) on \(M\), that is, a section of \(K^m\). Suppose \(A \in \Gamma(K^m)\). Like in the case \(m=3\) there exists a unique smooth real-valued function \(a\) on \(SM\) lying in \(\mathcal{H}_{-m}\oplus \mathcal{H}_m\), so that \(\pi^*A=(V(a)/m+ia)\omega^m\). In particular, to a triple \((g,A,\theta)\) we may associate the thermostat \(F=X+(a-V\theta)V\). We now have:
Let \(M\) be a closed oriented surface and \((g,A,\theta)\) be a triple satisfying \[\overline{\partial}A=\left(\frac{m-1}{2}\right)\left(\theta-i\star_g\theta\right)\otimes A\quad \text{and} \quad K_g-\delta_{g}\theta+(2-m)|A|^{2}_{g}\leqslant 0.\] Let \(F\) denote the vector field of the thermostat determined by \((g,A,\theta)\). Suppose there is a 1-form \(\beta\in \Omega^1(M)\) and a function \(u\in C^{\infty}(SM)\) such that \[Fu=Va+\beta.\] Then \(A=0\) and \(\beta\) is exact.
Let us first verify that this gives the desired statement.
Let \((M,\mathfrak{p})\) be a closed oriented properly convex projective surface with \(\chi(M)<0\) and with \(\mathfrak{p}\) containing a Weyl connection \(\mathrm{D}\). Then \(\mathfrak{p}\) is hyperbolic and moreover \(\mathrm{D}\) is the Levi-Civita connection of the hyperbolic metric.
Proof. By a result of Calabi [3], if \(m=3\) and \((g,A)\) satisfy \[K_g=-1+2|A|^2_g \quad \text{and} \quad \overline{\partial} A=0,\] then \(K_g\leqslant 0\). In particular, the triple \((g,A,0)\) satisfies the assumptions of Theorem 6.1 and Corollary 5.6 implies that we have a solution \(u\) to the transport equation \(Fu=Va+\beta\). Hence the theorem gives right away that \(A\) vanishes identically and hence \(\mathfrak{p}\) is hyperbolic. In particular, the Levi-Civita connection \({}^g\nabla\) of the hyperbolic metric and the connection \(\mathrm{D}\) both lie in \(\mathfrak{p}\) and hence are projectively equivalent, but this can happen if and only if \({}^{g}\nabla=\mathrm{D}\), by Corollary 4.6.
In [16] the notion of a minimal Lagrangian connection is introduced. These are torsion-free connections on \(TM\) of the form \(\nabla=\mathrm{D}+B\) where \((g,A,\theta)\) defining \(\mathrm{D}\) and \(B\) are subject to the equations \[] K_g-\delta_g\theta=-1+2|A|^2_g, \qquad \overline{\partial} A=\left(\theta-i\star_g\theta\right)\otimes A, \qquad d\theta=0.\] In particular, on a closed oriented surface of negative Euler characteristic every properly convex projective structure arises from a minimal Lagrangian connection. Another immediate consequence of Theorem 6.1 and Corollary 4.6 thus is:
Let \(M\) be a closed oriented surface of negative Euler characteristic and \(\nabla\) a minimal Lagrangian connection arising from the triple \((g,A,\theta)\). Suppose \(|A|^2_g\leqslant 1\) and that \(\nabla\) is projectively equivalent to a Weyl connection \(\mathrm{D}\). Then \(A\) vanishes identically and hence \(\nabla=\mathrm{D}\).
In order to show the theorem we use the following \(L^2\) identity proved in [10] which is in turn an extension of an identity in [22] for geodesic flows. The identity holds for arbitrary thermostats \(F=X+\lambda V\). If we let \(H_{c}:=H+cV\) where \(c:SM\to{\mathbb R}\) is any smooth function then \[2\langle H_{c}u,VFu\rangle=\|Fu\|^{2}+\|H_{c}u\|^{2}-\langle Fc+c^{2}+K_g-H_{c}\lambda+\lambda^{2},(Vu)^{2}\rangle, \tag{6.1}\] where \(u\) is any smooth function. All norms and inner products are \(L^{2}\) with respect to the volume form \(\Theta\).
We also need the following lemma whose proof is a straightforward calculation (see [18] for a proof).
We have \[\overline{\partial}A=\left(\frac{m-1}{2}\right)(\theta-i\star_g\theta)\otimes A\] if and only if \[XVa-mHa-(m-1)(\theta Va-maV\theta)=0.\]
Proof of Theorem 6.1. Without loss of generality we may assume that \(\beta\) has zero divergence. Indeed if not, a standard application of scalar elliptic PDE theory shows that we can always find a smooth function \(h\) on \(M\) such that \(\beta+dh\) has zero divergence. Now note that \(F(u+h)=Va+\beta+dh\).
A calculation shows that if we pick \(c=\theta+V(a)/m\), then \[Fc+c^{2}+K_g-H_{c}\lambda+\lambda^{2}=K_g-\delta_{g}\theta+(1-m)|A|_{g}^{2},\] where we use that \[\pi^*|A|^2_g=(Va)^2/m^2+a^2\quad \text{and} \quad \pi^*\delta_g\theta=-\left(X\theta+HV\theta\right),\] hence for this choice of \(c\), (6.1) simplifies to \[\tag{6.2} 2\langle H_{c}u,VFu\rangle-\||A|_{g}Vu\|^{2}\\ =\|Fu\|^{2}+\|H_{c}u\|^{2}-\langle K_{g}-\delta_{g}\theta+(2-m)|A|_{g}^{2},(Vu)^2\rangle.\] If \(Fu=Va+\beta\), then \(VFu=-m^{2}a+V\beta\). Using that \(X\) and \(H\) preserve \(\Theta\) and that \(XVa-mHa-(m-1)(\theta Va-maV\theta)=0\) we compute \[\begin{aligned} 2\left\langle H_{c}u,-m^{2}a\right\rangle&=-2m^{2}\langle Hu,a\rangle-2m^{2}\langle cVu,a\rangle\\ &=2m^{2}\langle u,Ha\rangle -2m^{2}\langle cVu,a\rangle\\ &=-2m^{2}\langle Xu,V(a)/m\rangle-2m(m-1)\langle u, \theta Va-maV\theta\rangle\\ &\phantom{=}\;-2m^{2}\langle cVu,a\rangle\\ &=-2m\|Va\|^{2}=-2m^3\|a\|^2,\end{aligned}\] where the last equation is obtained using that \(Xu=\beta+Va-(a-V\theta)Vu\), \(\langle \beta,Va\rangle=0\) and \(c=\theta+V(a)/m\).
Using that \(X\) and \(H\) preserve \(\Theta\) and that \(X\beta+HV\beta=0\) (\(\beta\) is assumed to have zero divergence) we compute: \[\begin{aligned} 2\left\langle H_{c}u,V\beta\right\rangle&=2\langle Hu,V\beta\rangle+2\langle cVu,V\beta\rangle\\ &=-2\langle u,HV\beta\rangle +2\langle cVu,V\beta\rangle\\ &=-2\langle Xu,\beta\rangle+2\langle cVu,V\beta\rangle\\ &=-2\|\beta\|^{2}+2\langle (a-V\theta)Vu,\beta\rangle+2\langle cVu,V\beta\rangle\\ &=-2\|\beta\|^2+2\langle aVu,\beta\rangle+2\langle (Va Vu)/m,V\beta\rangle,\end{aligned}\] where the penultimate equation is obtained using that \(Xu=\beta+Va-(a-V\theta)Vu\) and \(\langle \beta,Va\rangle=0\). The last equation uses that \(c=\theta+V(a)/m\) and \[V(\theta V\beta-V\theta\beta)=0.\]
Inserting these calculations back into (6.2), we derive \[\begin{gathered} -2m^3\|a\|^2-2\|\beta\|^2+2\langle aVu,\beta\rangle+2\langle (Va Vu)/m,V\beta\rangle-\||A|_{g}Vu\|^{2}\\ =\|Fu\|^{2}+\|H_{c}u\|^{2}-\langle K_{g}-\delta_{g}\theta+(2-m)|A|_{g}^{2},(Vu)^2\rangle.\end{gathered}\] Since \(|A|_{g}^2=a^2+(Va)^{2}/m^2\) this can be re-written as \[\begin{gathered} -2m^3\|a\|^2-\|\beta-aVu\|^2-\|V\beta-VaVu/m\|^{2}\\ =\|Fu\|^{2}+\|H_{c}u\|^{2}-\langle K_{g}-\delta_{g}\theta+(2-m)|A|_{g}^{2},(Vu)^2\rangle,\end{gathered}\] where we have used that \(\|\beta\|^2=\|V\beta\|^2\). By hypothesis the right hand side is \(\geqslant 0\) which gives right away that \(a=\beta=0\).