5 The transport equation
While the PDE (4.11) for the Beltrami differential \(\mu\) is natural from a complex geometry point of view, it turns out to be advantageous to rephrase it as a transport equation on \(SM\). The relevant transport equation on \(SM\) can be derived using (4.11) – see Appendix A – but here we will instead take a different approach, as it leads to a more general result about thermostats having the same unparametrised geodesics, see Proposition 5.2.
Let \(g,\hat{g}\) be Riemannian metrics on \(M\). In what follows all objects defined in terms of the metric \(\hat{g}\) will be decorated with a hat symbol. There is an obvious scaling map \[\ell : SM \to \widehat{SM}, \quad (x,v) \mapsto \left(x,\frac{v}{\sqrt{\hat{g}(v,v)}}\right)\] which is a fibre-bundle isomorphism covering the identity on \(M\). As before we define \[p(x,v)=\hat{g}(v,v), \qquad r(x,v)=\hat{g}(v,Jv), \quad\text{and}\quad q(x,v)=\hat{g}(Jv,Jv).\]
The pullback of the volume form \(\hat{\Theta}\) on \(\widehat{SM}\) is \[\ell^*\hat{\Theta}=\left(\frac{pq-r^2}{p}\right)\Theta.\]
Proof. Since \[d\pi\left(X(x,v)\right)=v \quad \text{and}\quad d\pi\left(H(x,v)\right)=Jv,\] we obtain \[\pi^*\hat{g}=p\omega_1\otimes\omega_1+2r\omega_1\circ\omega_2+q\omega_2\otimes \omega_2,\] where we write \(\omega_1\circ\omega_2:=\frac{1}{2}\left(\omega_1\otimes\omega_2+\omega_2\otimes\omega_1\right)\). We first compute \[\begin{aligned} \left(X\hspace{3pt}\rule[0.5pt]{2mm}{0.5pt}\rule[0.5pt]{0.5pt}{4.5pt}\hspace{3pt}\ell^*\hat{\omega}_1\right)(x,v)&=\hat{\omega}_1\left(d\ell(X(x,v))\right)=\hat{g}\left(\ell(x,v),(d\hat{\pi}\circ d\ell)(X(x,v))\right)\\ &=\frac{1}{\sqrt{\hat{g}(v,v)}}\hat{g}(v,d\pi(X(x,v)))=\sqrt{\hat{g}(v,v)}=\sqrt{p},\end{aligned}\] where we have used that \(\hat{\pi}\circ\ell=\pi\). Likewise, we obtain \[\begin{aligned} \left(H\hspace{3pt}\rule[0.5pt]{2mm}{0.5pt}\rule[0.5pt]{0.5pt}{4.5pt}\hspace{3pt}\ell^*\hat{\omega}_1\right)(x,v)&=\hat{\omega}_1\left(d\ell(H(x,v))\right)=\hat{g}\left(\ell(x,v),(d\hat{\pi}\circ d\ell)(H(x,v))\right)\\ &=\frac{1}{\sqrt{\hat{g}(v,v)}}\hat{g}(v,d\pi(H(x,v)))=\frac{\hat{g}(v,Jv)}{\sqrt{\hat{g}(v,v)}}=\frac{r}{\sqrt{p}}.\end{aligned}\] Since \(\hat{\omega}_1\) is semibasic for the projection \(\hat{\pi}\), the pullback \(\ell^*\hat{\omega}_1\) is semibasic for the projection \(\pi\), hence \(V\hspace{3pt}\rule[0.5pt]{2mm}{0.5pt}\rule[0.5pt]{0.5pt}{4.5pt}\hspace{3pt}\ell^*\hat{\omega}_1=0\), so that we have \[\tag{5.1} \ell^*\hat{\omega}_1=\sqrt{p}\omega_1+\frac{r}{\sqrt{p}}\omega_2.\] The pullback \(\ell^*\hat{\omega}_2\) must be a multiple of \(\omega_2\). Indeed, \(\ell^*\hat{\omega}_2\) is \(\pi\)-semibasic and we obtain \[\begin{aligned} \left(X\hspace{3pt}\rule[0.5pt]{2mm}{0.5pt}\rule[0.5pt]{0.5pt}{4.5pt}\hspace{3pt}\ell^*\hat{\omega}_2\right)(x,v)&=\hat{\omega}_2\left(d\ell(X(x,v))\right)=\hat{g}\left(\hat{J}\ell(x,v),(d\hat{\pi}\circ d\ell)(X(x,v))\right)\\ &=\frac{1}{\sqrt{\hat{g}(v,v)}}\hat{g}(\hat{J}v,d\pi(X(x,v)))=\frac{\hat{g}(\hat{J}v,v)}{\sqrt{\hat{g}(v,v)}}=0.\end{aligned}\] Recall that the area form \(d\hat{\sigma}\) of \(\hat{g}\) satisfies \(\hat{\pi}^*d\hat{\sigma}=\hat{\omega}_1\wedge\hat{\omega}_2\), hence \[\ell^*(\hat{\omega}_1\wedge\hat{\omega}_2)=\pi^*d\hat{\sigma}=\sqrt{pq-r^2}\,\omega_1\wedge\omega_2.\] Thus we must have \[\tag{5.2} \ell^*\hat{\omega}_2=\frac{\sqrt{pq-r^2}}{\sqrt{p}}\omega_2.\] Since the Lie derivative of \(\pi^*\hat{g}\) with respect to \(V\) vanishes identically, we compute that \(V\sqrt{p}=r/\sqrt{p}\). Moreover, since \(\sqrt{pq-r^2}\) is the \(\pi\)-pullback of a function on \(M\), we obtain \[V\left(\frac{\sqrt{pq-r^2}}{\sqrt{p}}\right)=-\frac{r\sqrt{pq-r^2}}{p^{3/2}}.\] Pulling back the structure equation \(d\hat{\omega}_2=-\hat{\psi}\wedge\hat{\omega}_1\) whilst using (5.1) and (5.2) gives \[\begin{aligned} \ell^*(d\hat{\omega}_2)&=d(\ell^*\hat{\omega}_2)=d\left(\frac{\sqrt{pq-r^2}}{\sqrt{p}}\omega_2\right)\\ &=\left(\hat{a}\omega_1-\frac{r\sqrt{pq-r^2}}{p^{3/2}}\psi\right)\wedge\omega_2-\frac{\sqrt{pq-r^2}}{\sqrt{p}}\psi\wedge\omega_1\\ &=-\ell^*\hat{\psi}\wedge\ell^*\hat{\omega}_1=-\ell^*\hat{\psi}\wedge\left(\sqrt{p}\omega_1+\frac{r}{\sqrt{p}}\omega_2\right)\end{aligned}\] for some unique real-valued function \(\hat{a}\) on \(SM\). Comparing the coefficients in the above equations, it follows that \[\tag{5.3} \ell^*\hat{\psi}=a\omega_1+b\omega_2+\frac{\sqrt{pq-r^2}}{p}\psi\] for some unique real-valued functions \(a,b\) on \(SM\). In particular, we obtain \[\ell^*\hat{\Theta}=\ell^*\left(\hat{\omega}_1\wedge\hat{\omega}_2\wedge\hat{\psi}\right)=\left(\frac{pq-r^2}{p}\right)\omega_1\wedge\omega_2\wedge\psi,\] as claimed.
We use this lemma to derive the following observation about general thermostats:
If two thermostats determined by pairs \((g,\lambda)\) and \((\hat{g},\hat{\lambda})\) have the same unparametrised geodesics, then \[\sqrt{p}\,(\hat{V}\hat{\lambda}\circ\ell)=F\log\left(\frac{pq-r^2}{p^{3/2}}\right)+V\lambda.\]
As an immediate application we obtain the following classical fact:
Let \(g\) and \(\hat{g}\) be two Riemannian metrics on \(M\) having the same unparametrised geodesics, then \(p/(pq-r^2)^{2/3}\) is an integral for the geodesic flow of \(g\).
Proof. This special case corresponds to \(\lambda=\hat{\lambda}=0\) and hence Proposition 5.2 implies \[\begin{aligned} 0&=X\log\left(\frac{pq-r^2}{p^{3/2}}\right)=-\frac{3}{2}X\log\left(\frac{p}{(pq-r^2)^{2/3}}\right)\\ &=-\frac{3}{2}\frac{(pq-r^2)^{2/3}}{p}X\left(\frac{p}{(pq-r^2)^{2/3}}\right). \end{aligned}\]
In order to prove Proposition 5.2 we also recall a general lemma whose proof is elementary and thus omitted.
Let \(X\) be a vector field on a manifold \(M\) with volume form \(\Omega\). Let \(f\) and \(s>0\) be smooth functions. Then \[\mbox{\rm div}_{\Omega}(fX)=Xf+f\mbox{\rm div}_{\Omega}X \quad \text{and}\quad \mbox{\rm div}_{s\,\Omega}(X)=X\log s+\mbox{\rm div}_{\Omega}X.\]
Proof of Proposition 5.2. This follows from Lemma 5.1 and Lemma 5.4 and the key fact that if the thermostats have the same unparametrised geodesics then \[\tag{5.4} \ell^*\hat{F}=\frac{1}{\sqrt{p}}\,F.\] To see the last equality, note that we can rephrase the hypothesis as follows. There is a smooth function \(\tau:SM\times{\mathbb R}\to{\mathbb R}\) implementing the time change so that \[\ell\circ \phi_{\tau(x,v,t)}(x,v)=\hat{\phi}_{t}\circ\ell(x,v).\] Differentiating this with respect to \(t\) and setting \(t=0\) gives \[d\ell(fF)=\hat{F}\circ\ell,\] where \(f(x,v):=\frac{d}{dt}\tau(x,v,t)|_{t=0}\). To check that \(f\) has the desired form, apply \(d\hat{\pi}\) to the last equation to get \(f\,v=v/\sqrt{\hat{g}(v,v)}\).
Writing \(s:=(pq-r^2)/p\) and taking the divergence of (5.4) with respect to \(\ell^*\hat{\Theta}=s\Theta\) gives \[\begin{aligned} \mbox{\rm div}_{s\,\Theta}\left(\sqrt{p}\,\ell^*\hat{F}\right)&=(\ell^*\hat{F})\sqrt{p}+\sqrt{p}\,\mbox{\rm div}_{s\,\Theta}(\ell^*\hat{F})\\ &=\left(\frac{1}{\sqrt{p}}\right)F\sqrt{p}+\sqrt{p}\,\mbox{\rm div}_{\ell^*\hat{\Theta}}\left(\ell^*\hat{F}\right)\\ &=F\left(\log\sqrt{p}\right)+\sqrt{p}\left(\mbox{\rm div}_{\hat{\Theta}}\hat{F}\right)\circ \ell\\ &=\mbox{\rm div}_{s\,\Theta}F=F\log s+\mbox{\rm div}_{\Theta}F\end{aligned}\] where we have used Lemma 5.4. Since \(\mbox{\rm div}_{\Theta}F=V\lambda\) and \(\mbox{\rm div}_{\hat{\Theta}}\hat{F}=\hat{V}\hat{\lambda}\) this last equation is equivalent to \[\sqrt{p}\left(\hat{V}\hat{\lambda}\circ \ell\right)=F\log\left(\frac{s}{\sqrt{p}}\right)+V\lambda,\] which proves the claim.
Note that the crucial identity (5.4) also follows from a different argument. Since the orbits of \(F\) and \(\hat{F}\) project onto the same unparametrised curves, there must exist a smooth function \(w\) on \(SM\), so that \(\ell^*\hat{F}=wF\). From (5.1), (5.2) and (5.3), we compute that \[\ell^*\hat{X}=\frac{1}{\sqrt{p}}X-\frac{a\sqrt{p}}{\sqrt{pq-r^2}}V\quad \text{and}\quad \ell^*\hat{V}=\frac{p}{\sqrt{pq-r^2}}V\] from which one immediately obtains \(w=1/\sqrt{p}\).
A special case of Proposition 5.2 is the following:
Suppose the projective thermostat associated to the pair \((g,\lambda)=(g,a-V\theta)\) has the same unparametrised geodesics as the Weyl connection \(\mathrm{D}\) defined by \((\hat{g},\alpha)\), then \[u=\frac{3}{2}\log\left(\frac{p}{(pq-r^2)^{2/3}}\right)\] satisfies the transport equation \[\tag{5.5} Fu=Va+\beta,\] where \(\beta=\theta-\alpha\).
Proof. Applying Proposition 5.2 in the special case \(\lambda=a-V\theta\) and \(\hat{\lambda}=-\hat{V}\alpha\) gives \[-\sqrt{p}\left(\hat{V}\hat{V}\alpha\circ\ell\right)=\sqrt{p}\left(\alpha\circ\ell\right)=F\log\left(\frac{pq-r^2}{p^{3/2}}\right)+V(a-V\theta),\] the left hand side of which is simply \(\alpha\), thought of as a function on \(SM\). Hence we obtain \[\begin{aligned} -\left(Va+\theta-\alpha\right)&=F\log\left(\frac{pq-r^2}{p^{3/2}}\right)=F\left(-\frac{3}{2}\left(\log p-\frac{2}{3}\log(pq-r^2)\right)\right)\\ &=-Fu, \end{aligned}\] as claimed.