5 The case of holomorphic differentials
We have seen that a triple \((g,A,\theta)\) solving (4.6) and (4.7) yields a holomorphic section of \(L_m\otimes K_M^m\) with respect to some appropriate holomorphic line bundle structure on \(L_m\). We now restrict to the case where the differential \(A\) is already holomorphic so that we obtain the coupled vortex equations \[K_g=-1+(m-1)|A|^2_g\quad \text{and} \quad \overline{\partial} A=0.\]
5.1 Anosov flows
It is possible to upgrade Corollary 4.6 in the case where \(A\) is holomorphic as follows:
Let \((g,A)\) be a pair satisfying the coupled vortex equations \(\bar{\partial}A=0\) and \(K_g=-1+(m-1)|A|^{2}_{g}\). Then the associated thermostat flow is Anosov.
Proof. We already know that there is a dominated splitting, so taking into account Remark 3.8, the strategy will be to show that \(r^{u}>0\) and \(r^{s}<0\). We will do this using the following lemma.
Let \((g,A)\) be a pair satisfying the coupled vortex equations \(\bar{\partial}A=0\) and \(K_g=-1+(m-1)|A|^{2}_{g}\). Then \(-1\leqslant K_g<0\).
Proof. The proof is quite similar to the proof of [4], the reader may also compare with [10]. The claim is obviously correct if \(A\) vanishes identically, hence we assume this not to be the case. We first prove the inequality \(K_g\leqslant 0\). As before let \(g_0\) denote the hyperbolic metric in the conformal equivalence class of \(g\) and write \(g=\mathrm{e}^{2u}g_0\) for \(u \in C^{\infty}(M)\). Using \[\tag{5.1} K_g=\mathrm{e}^{-2u}\left(-1-\Delta u\right)\quad \text{and}\quad |A|^2_{g}=\mathrm{e}^{-2mu}{|A|^2_{g_0}}\] gives \[\tag{5.2} 1+\Delta u=\mathrm{e}^{2u}-(m-1)\mathrm{e}^{-2(m-1)u}\alpha,\] where we write \(\alpha=|A|^2_{g_0}\). The inequality \(K_g\leqslant 0\) is equivalent to \[\tag{5.3} (m-1)\mathrm{e}^{-2mu}\alpha \leqslant 1\] and is clearly satisfied at the points where \(A\) vanishes. Therefore, taking the logarithm of (5.3), we see that \(K_g\leqslant 0\) follows from the non-negativity of the smooth function \[f=2mu-\log(m-1)-\log \alpha,\] which is defined on the open set \(M^{\circ}:=\left\{x \in M : A(x)\neq 0\right\}\). Note that using \(f\) the equation (5.2) becomes \[\tag{5.4} 1+\Delta u=\mathrm{e}^{2u}(1-\mathrm{e}^{-f}).\] As \(M\) is compact, the Gauss curvature \(K_g\) attains its maximum at some point \(x_0\) and moreover \(x_0 \in M^{\circ}\). Consequently, the function \(f\) attains its infimum at \(x_0\). A straightforward calculation gives \(\Delta\log \alpha=-2m\), where we use that \(A\) is holomorphic. At the minimum \(x_0\) of \(f\) we thus obtain \[\tag{5.5} 0\leqslant \Delta f(x_0)=2m\left(1+\Delta u(x_0)\right)=2m\,\mathrm{e}^{2u(x_0)}\left(1-\mathrm{e}^{-f(x_0)}\right),\] where we have used (5.4). It follows that \(f(x_0)\geqslant 0\) and hence \(f\geqslant 0\) on all of \(M^{\circ}\). This shows that \(K_g\leqslant 0\). It order to prove \(K_g<0\), we first remark that the function \(f-1+\mathrm{e}^{-f}\) is non-negative on \(M^{\circ}\). Consequently, (5.5) gives \[\Delta_g f\leqslant 2m f,\] where \(\Delta_g=\mathrm{e}^{-2u}\Delta\) denotes the Laplacian with respect to \(g\). In particular, it follows that for every point \(x \in M^{\circ}\) there exists a constant \(c>0\), an \(x\)-neighbourhood \(U_x\) and a flat metric \(g_0\) on \(U_x\) which lies in the conformal equivalence of \(g\), so that \[\left(\Delta_{g_0}-c\right)f \leqslant 0\] on \(U_x\). Therefore, by applying the strong maximum principle [16] to the operator \(\Delta_{g_0}-c\), it follows that if \(f\) vanishes at some point in \(U_x\), then it vanishes on all of \(U_x\) and consequently on \(M^{\circ}\). Since \(A\) is holomorphic, its zeros are isolated and hence \(M^{\circ}\) is dense in \(M\). Since \(K_g\) is continuous we conclude that if \(K_g\) vanishes at some point on \(M\), then it vanishes identically on \(M\), but this possibility is excluded by the Gauss–Bonnet theorem.
From (5.2) we see that \(u\) solves a PDE of the form \(\Delta u=G(x,u)\) where \[G(x,u)=-1+\mathrm{e}^{2u}-(m-1)\mathrm{e}^{-2(m-1)u}\alpha(x).\] Since \(\alpha\geqslant 0\) we have \(G(x,u)\leqslant -1+\mathrm{e}^{2u}\) and hence \(G(x,u)<0\) for \(u<0\). On the other hand, for \(u> \sup_{x \in M} \frac{1}{2}\log(1+(m-1)\alpha(x))\geqslant 0\) we get \[G(x,u)>-1+\mathrm{e}^{2u}-(m-1)\alpha(x)>0.\] Since \[\frac{\partial G}{\partial u}(x,u)=2\alpha(x)(m-1)^2\mathrm{e}^{-2(m-1)u}+2\mathrm{e}^{2u}>0\] standard quasi-linear elliptic PDE methods (see for instance [37]) imply that (5.2) has a unique smooth solution \(u\) for every smooth non-negative function \(\alpha\). Consequently, for every holomorphic differential \(A\) on \((M,[g])\) we obtain a unique solution \((g,A)\) to the coupled vortex equations \(K_g=-1+(m-1)|A|^2_g\) and \(\overline{\partial} A=0\).
We now show that \(r^u>0\) (the proof that \(r^{s}<0\) is similar). Set \(h=r^u-V(a)/m\). Then \(h\) satisfies \[F(h)+h^{2}+hB-1=0,\] where \[B:=\frac{(2-m)}{m}V(a).\] Given \((x,v)\in SM\), consider for each \(R>0\), the unique solution \(h_{R}\) to the Riccati equation along \(\phi_{t}(x,v)\): \[\dot{h}+h^2+hB-1=0\] satisfying \(h_{R}(-R)=\infty\). Using (3.5) we derive \[r^{u}(x,v)=\lim_{R\to\infty}h_{R}(0)+V(a)/m. \tag{5.6}\] Let \(c:=\max_{(x,v)}|B(x,v)|\) and \(\ell:=\frac{\sqrt{c^{2}+4}-c}{2}\). If we let \(f_{R}:=h_{R}-\ell\), then \(f_{R}\) solves \[\dot{f}+wf=q, \tag{5.7}\] where \(w:=f_{R}+B+2\ell\) and \(q:=-\ell^{2}-B\ell+1\). Observe that \(q\geqslant 0\) by our definitions of \(c\) and \(\ell\). We can solve the inhomogeneous linear equation (5.7) and use that \(q\geqslant 0\) to derive \(f_{R}(t)\geqslant 0\) and thus \(h_{R}(t)\geqslant \ell\). By taking limits, and using (5.6), we obtain \[r^{u}(x,v)\geqslant \ell +V(a)/m.\] By Lemma 5.2 we have \(c<(m-2)/\sqrt{m-1}\) and \(V(a)/m>-1/\sqrt{m-1}\). Thus \[r^{u}\geqslant \frac{\sqrt{c^{2}+4}-c}{2}-\frac{1}{\sqrt{m-1}}>0\] as desired.
As we have seen, Corollary 4.6 asserts that given a triple \((g,A,\theta)\) satisfying (4.6) and (4.7), the associated thermostat flow has a dominated splitting. When \(\theta=0\), Theorem 5.1 tells us that we can do better and in fact the thermostat flow is Anosov. At the “other end”, that is, when \(A=0\), we also know by Proposition 3.5 that the thermostat flow is also Anosov (in this case \(\mathbb{K}=K_{g}-\delta_{g}\theta=-1\)). These two “ends” are Anosov for different reasons, connected with the discussion in Remark 3.8. In the case \(\theta=0\), as we have just seen, one uses that \(r^{u}>0\), that is, the first case in (3.3). In the case \(A=0\), we use the second case in (3.3). It is conceivable that the thermostat flow is always Anosov for any triple \((g,A,\theta)\) satisfying (4.6) and (4.7), but at the time of writing it is not at all clear how to prove this. It should be noted that for the special case of the geodesic flow it is well known that a dominated splitting must be Anosov. We can see this fairly quickly using quadratic forms as follows. Suppose \(r^{u,s}:SM\to\mathbb{R}\) are two continuous functions such that \(Xr^{u,s}+[r^{s,u}]^{2}+K_{g}=0\) and \(r^{u}-r^{s}\neq 0\) everywhere. Define \[Q=2y\dot{y}-([r^{u}]^{2}+[r^{s}]^{2})y^2.\] Then a calculation shows \[\dot{Q}=(\dot{y}-r^{u}y)^{2}+(\dot{y}-r^{s}y)^{2}>0\] unless \(y=\dot{y}=0\). Hence by Proposition 3.4 the geodesic flow is Anosov.
5.2 Dissipation and volume
We will now prove the following result stated in the introduction.
Let \((g,A)\) be a pair satisfying the coupled vortex equations \(\bar{\partial}A=0\) and \(K_g=-1+(m-1)|A|^{2}_{g}\). Then the associated thermostat flow preserves an absolutely continuous measure if and only if \(A\) vanishes identically.
Proof. Since the flow is of class \(C^{\infty}\) and Anosov, an application of the smooth Livšic theorem [27] shows that \(\phi_t\) preserves an absolutely continuous measure if and only if \(\phi_t\) preserves a smooth volume form.
We write the volume form as \(\mathrm{e}^{-u}\Theta\) for some real-valued function \(u\) on \(SM\). Thus, using (2.2), we obtain \[L_{F}\left(\mathrm{e}^{-u}\Theta\right)=-\mathrm{e}^{-u}F(u)\Theta+\mathrm{e}^{-u}V(a)\Theta=(-Fu+Va)\mathrm{e}^{-u}\Theta.\] Hence the claim follows by showing that if \(u\) solves \(Fu=Va\), then \(a\) vanishes identically. In order to show this we use the following \(L^2\) identity proved in [22] which is in turn an extension of an identity in [36] 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 \[2\langle H_{c}u,VFu\rangle=\|Fu\|^{2}+\|H_{c}u\|^{2}-\langle Fc+c^{2}+K_g-H_{c}\lambda+\lambda^{2},(Vu)^{2}\rangle, \tag{5.8}\] where \(u\) is any smooth function. All norms and inner products are \(L^{2}\) with respect to the volume form \(\Theta\).
In our case \(\lambda=a\) and a calculation shows that if we pick \(c=V(a)/m\), then \[Fc+c^{2}+K_g-H_{c}\lambda+\lambda^{2}=K_g+(1-m)|A|_{g}^{2}=-1,\] hence for this choice of \(c\), (5.8) simplifies to \[2\langle H_{c}u,VFu\rangle=\|Fu\|^{2}+\|H_{c}u\|^{2}+\|Vu\|^{2}. \tag{5.9}\] If \(Fu=Va\), then \(VFu=-m^{2}a\) and we compute using that \(X\) and \(H\) preserve \(\Theta\) and that \(XVa-mHa=0\): \[\begin{aligned} 2\left\langle H_{c}u,VFu\right\rangle&=-2m^{2}\langle Hu,a\rangle-2m^{2}\langle cVu,a\rangle\\ &=2m^{2}\langle u,Ha\rangle -2m^{2}\langle cVu,a\rangle\\ &=-2m^{2}\langle Xu,V(a)/m\rangle-2m^{2}\langle cVu,a\rangle\\ &=-2m\|Va\|^{2},\end{aligned}\] where the last equation is obtained using that \(Xu=Va-aVu\) and \(c=V(a)/m\). Inserting this back into (5.9), we see that the equality obtained can only hold if \(Va\) and hence \(a\) vanishes identically.