On essential-selfadjointness of differential operators on closed manifolds

Let us recall some classical results which can be found in [R-S-75], sec. X. Let $({\cal H},<.|.>)$ be an Hilbert space. A linear operator $P:D(P)\rightarrow{\cal H}$ with $D(P)$ a dense subspace of ${\cal H}$ is said to be symmetric if, for all $x,y\in D(P)$, $<Px|y>=<x|Py>$. The closure of $P$ is the operator $\bar{P}$ whose graph is the closure of the graph of $P$ in ${\cal H}\times{\cal H}$. The adjoint $P^{\star}$ of $P$ is defined as follows: the domain $D(P^{\star})$ is the set of $x\in{\cal H}$ so that $y\rightarrow<Py|x>$ as defined on $D(P)$ extends continously to ${\cal H}$. We have then $<Py|x>\equiv<y|z>$ and we define $P^{\star}x=z$. The operator $P$ is ESA if $P^{\star}$ is the closure of $P$. In other words, $P$ has an unique self-adjoint extension. A useful property is the following one

Recall that a differential operator with smooth coefficients acts on Schwartz distributions and in particular on $L^{2}$ functions. In particular, we see that any symmetric elliptic operator $P$ on a closed manifold is ESA: if $(P\pm i)u=0$, $u$ is smooth and the result follows from the symmetry of $P$. This is why we are only interested here to non elliptic operators.

Let $X:=[\alpha,\beta]$ be a compact interval. We consider a differential operator $P$ of degree $2$ whose coefficients are smooth up to the boundary. Assuming that $P$ is symmetric on $C_{c}^{\infty}(]\alpha,\beta[,|dx|)$ (it is usually called formally symmetric), then $P$ is given by the equation (1).

We want to describe the Dirichlet boundary conditions. For that, we assume that $P$ is elliptic near the boundary, i.e. that $a$ does not vanish at the points of $\partial X$. We will take for domain of $P$ the space $D(P):=C^{\infty}(\bar{X},{\mathbb{C}})\cap\{f|f|_{\partial X}=0\}$.

Proof.–We get first that $Pg\in L^{2}$ by looking at $\int_{X}Pf\bar{g}|dx|$ with $f\in C_{c}^{\infty}(]\alpha,\beta[)$. It follows that $g$ is continuous near the boundary. Then we have, if $f\in D(P)$,

$$\int_{X}(Pf\bar{g}-f\overline{Pg})|dx|=[af^{\prime}g]_{\alpha}^{\beta}$$

We have to control the righthandside in terms of the $L^{2}$ norm of $f$ which is clearly not possible if $g$ does not vanish on the boundary because $a$ does not vanish at on $\partial X$. $\square$

Let us prove the following localization

Proof.–

Let us first prove that, if $P_{\Omega}$ is ESA, $P$ is ESA: let us take a $\rho\in C_{c}^{\infty}(\Omega)$ with $\rho\equiv 1$ near $Z$. Then, if $Pu=v$ with $u,v\in L^{2}(X)$, $(1-\rho)u$ belongs to the Sobolev space $H^{2}(X)$ by ellipticity of $P$ on the support of $1-\rho$. In particular $P((1-\rho)u)\in L^{2}$. There exists $(u^{\prime}_{n},v_{n}^{\prime}=Pu_{n}^{\prime})$ a sequence of smooth functions converging to $((1-\rho)u,P((1-\rho)u)$ in $L^{2}$ by density of $C^{\infty}(X)$ in $H^{2}(X)$. We have now $P(\rho u)=w$ with $\rho u,w\in L^{2}$ and ${\rm support}(\rho u)\subset\Omega$. ESA of $P_{\Omega}$ allows to approximate $(\rho u,w)$ by smooth functions $(u_{n}^{\prime\prime},v_{n}^{\prime\prime}=Pu_{n}^{\prime\prime})$ and we can assume that $u_{n}^{\prime\prime}$ vanishes near the boundary because $\rho u$ does. Then $(u^{\prime}_{n}+u^{\prime\prime}_{n},v^{\prime}_{n}+v^{\prime\prime}_{n})$ are smooth, converge to $(u,v)$ in $L^{2}\times L^{2}$ and $P(u^{\prime}_{n}+u^{\prime\prime}_{n})=v^{\prime}_{n}+v^{\prime\prime}_{n}$. This allows to conclude that $P$ is ESA.

Let us now prove that if $P$ is ESA, $P_{\Omega}$ is ESA: let us start with $P_{\Omega}u=v$ with $(u,v)\in L^{2}(\Omega)$, $u\in H^{2}$ near the boundary and $u(\partial\Omega)=0$. We choose $\rho\in C_{c}^{\infty}(\Omega)$ with $\rho=1$ near $Z$. Similarly to the previous argument, we decompose $u=\rho u+(1-\rho)u$. And by ellipticity of $P$ near $\partial\Omega$ we get that $(1-\rho)u$ belongs to the Sobolev space $H_{0}^{2}(\Omega)$ of distributions which are in $H^{2}(\Omega)$ and vanish at the boundary. By density of smooth functions vanishing at the boundary in the Sobolev space $H_{0}^{2}$, we get an approximating sequence to $((1-\rho)u,P((1-\rho)u)$. Now we are left with $\rho u$ with $P(\rho u)\in L^{2}$ and we use the fact that $P$ is ESA to get an approximating sequence $(u^{\prime}_{n},Pu^{\prime}_{n})$ on $X$. Choosing $\rho_{1}\in C_{c}^{\infty}(\Omega)$ with $\rho_{1}=1$ on the support of $\rho$, we take $u^{\prime\prime}_{n}=\rho_{1}u^{\prime}_{n}$ and we get an approximating sequence for $(\rho u,P(\rho u)$. This allows to conclude. $\square$

Note that the previous localization result extends probably to higher dimensional manifolds, but this extension is much less simple.

Any symmetric operator on the circle $S^{1}={\mathbb{R}}/{\mathbb{Z}}$ equipped with the Lebesgue measure $|dx|$ can be written as

$$P=d_{x}a(x)d_{x}-ib(x)d_{x}-i\frac{1}{2}b^{\prime}(x)+c(x)$$

(1)

where $d_{x}:=d/dx$, $a,~{}b,~{}c$ are smooth real valued periodic functions of period $1$. We assume always in what follows that the zeroes of $a$ are of finite multiplicities. The symbol $p$ of $P$ is

$$p=-a(x)\xi^{2}+b(x)\xi$$

The term $c(x)$ plays no role in the essential self-adjointeness, so we will forget it in what follows.

Our main result is

Our proof consists in describing the properties of $a$ and $b$ leading to classical completeness and to study the quantum completeness in the corresponding cases.

Note that the result in the case where the zeroes of $a$ are non degenerate is also proved using some microlocal analysis in [Tai-20].

We have the

Proof.– Recall that the Hamiltonian differential equation writes

$$dx/dt=-2a(x)\xi+b(x),~{}d\xi/dt=a^{\prime}(x)\xi^{2}-b^{\prime}(x)\xi.$$

The function $p$ is constant along the integral curves. We will denote by $(x_{0},\xi_{0})$ the data at time $0$ of the integral curves that we will consider.

The proof splits into three cases:

1. 1.

   Assume that $a(0)=0$ and $b(0)>0$. We will show that the flow is not null complete. Let us look at the set $p=0$ near $x=0$. This set is the union of the disjoint curves $\{\xi=0\}$ and $C:=\{a(x)\xi-b(x)=0\}$. The curve $C\cap\{x>0\}$ is oriented by the flow so that $x$ is decaying because then $dx/dt=-b(x)$. Let us start on $C$, with $x_{0}>0$ small enough and $\xi_{0}$ so that $-a(x_{0})\xi_{0}+b(x_{0})=0$. Then $-a(x(t))\xi(t)+b(x(t))=0$ for all $t$. Along $C$, we have $dx/dt=-b(x)$. Hence, there exists $t_{0}>0$ so that $x(t_{0})=0$ and $\xi(t_{0})=+\infty$. The flow is not null complete: the maximal integral curve is only defined up to $t_{0}^{-}$.

2. 2.

   Assume that $0$ is a non degenerate zero of $a$ and $b(0)=0$. Let us start with $x_{0}=0$ and $\xi_{0}\neq 0$. We have $x(t)=0$ for all $t$ and $d\xi/dt=a^{\prime}(0)\xi^{2}-b^{\prime}(0)\xi$. The solution of this differential equation is not defined for all $t$’s because $a^{\prime}(0)\neq 0$. The flow is not null complete.

3. 3.

   Assume now that zeroes of $a$ are degenerate and $b$ vanishes on $a^{-1}(0)$. We want to prove that the flow is complete. We have, by conservation of $p$, for any integral curve, there exists $E$ so that $-a(x)\xi^{2}+b(x)\xi\equiv E$. It follows that $\xi$ stays bounded on any compact interval in $x$ disjoint from $a^{-1}(0)$. We need only to consider what happens when $x(t)$ comes close to a zero of $a$, says $x=0$.

   Let us first assume that $x_{0}=0$, then $x(t)\equiv 0$ for all $t$ and $d\xi/dt=-b^{\prime}(0)\xi$. The trajectory is complete.

   If $x_{0}\neq 0$ is close to $0$, we get $dx/dt=\pm\sqrt{-4a(x)E-b(x)^{2}}=O(|x|)$. It follows that $x(t)$ does not reach $0$ in finite time and hence the integral curve ie defined for all times.

$\square$

We will show in this section that if $a(0)=0$ is a simple zero of $a$ then $P$ is not ESA.

Let $I:=[-\alpha,\alpha]$ with no other zeroes of $a$ inside $I$. The point $0$ is a regular singular point (see Appendix A) of the differential equation $(P-i)u=0$.

The indicial equation writes $Ar^{2}-iBr=0$ with $A:=a^{\prime}(0),~{}B=b(0)$. Hence the solutions of this equation near $0$ writes, for $x>0$, $y(x)=f(x)+x_{+}^{iB/A}g(x)$ if $B\neq 0$ and $y(x)=f(x)+g(x)\log x$ if $B=0$ with $f,g$ smooth up to $x=0$ (see Appendix A) and similarly for $x<0$ with $x_{-}$ and $\log(-x)$.

Let $y_{+}$ be the unique solution of $(P-i)y_{+}=0,~{}y^{\prime}_{+}(\alpha)=0,~{}y^{\prime}_{+}(\alpha)=1$ on $]0,\alpha]$, so that $y_{+}$ satisfies the Dirichlet boundary condition at $\alpha$. And define $y_{-}$ similarly with $y_{-}(-\alpha)=0$. If we extend $y_{+}$ by zero for $x<0$, we get a Schwartz distribution $Y_{+}$ and $(P-i)Y_{+}$ is supported by the origine. We have $PY_{+}=d_{x}d_{x}aY_{+}+d_{x}[a,d_{x}]Y_{+}-ibd_{x}Y_{+}-ib^{\prime}Y_{+}/2$. We check that $d_{x}aY_{+}$ is in $L^{2}_{\rm loc}$. So that $(P-i)Y_{+}$ is near $0$ in the Sobolev space $H^{-1}$, because $d_{x}aY_{+}\in L^{2}$. The derivatives $\delta^{\prime}(0),\cdots$ of the Dirac distribution are not in $H^{-1}$. We have the same result for $Y_{-}$. It follows that $(P-i)Y_{\pm}=\mu_{\pm}\delta(0)$.

Hence there is a non zero linear combination $Y$ of $Y_{+}$ and $Y_{-}$ which satisfies $(P-i)Y=0$ and the Dirichlet boundary conditions at $\pm\alpha$. This proves that $P_{I}$ is not ESA and hence $P$ is not ESA by Lemma 2.2.

Let us assume that all zeroes of $a$ are degenerate. Then if $I=]c,d[$ is an interval between two zeroes of $a$ and assume $a>0$ on $I$, we will show that there is an explicit unitary map from $L^{2}(I,|dx|)$ onto $L^{2}({\mathbb{R}},|dy|)$ sending $C_{c}^{\infty}(I)$ (the set of operator with compact support on $I$) into $C_{c}^{\infty}({\mathbb{R}})$ (set of operator with compact support on $\mathbb{R}$) and sending $P$ to $Q=d_{y}^{2}+V$ with an explicit $V$.

First step: a gauge transform. Let us consider $P_{S}:=e^{-iS}Pe^{iS}$ where $S$ is smooth and real valued. We get, by an easy calculation,

$$P_{S}=d_{x}ad_{x}-i(b-2aS^{\prime})d_{x}-i(b^{\prime}/2-a^{\prime}S^{\prime}-aS")-aS^{\prime 2}+bS^{\prime}$$

Choosing $S$ so that $S^{\prime}=b/2a$, we get

$$P_{S}=d_{x}ad_{x}+b^{2}/4a$$

Second step: a change of variable.

Let us choose $x_{0}\in I$. Let us define $y=\phi(x)=\int_{x_{0}}^{x}a^{-\frac{1}{2}}(t)dt$. The map $\phi$ is smooth diffeomorphism in $I$ onto ${\mathbb{R}}$. Let us introduce the unitary transform $\Omega:L^{2}(I,|dx|)\rightarrow L^{2}({\mathbb{R}},|dy|)$ defined by $\Omega f(\phi(x))=a(x)^{1/4}f(x)$. We compute $P_{\Omega}:=\Omega P_{S}\Omega^{\star}$. We get $P_{\Omega}=d_{y}^{2}+V(y)$ with

$$V(y)=\left(\frac{b^{2}}{2a}+\frac{a^{\prime 2}}{16a}-\frac{a^{\prime\prime}}{4}\right)(\phi^{-1}(y))$$

$(a^{\prime 2}/16a+a^{\prime\prime}/4)$ is bounded. The derivative of $(b/a^{1/2})\circ\phi^{-1}$ is

$$\left(b^{\prime}-\frac{a^{\prime}b}{2a}\right)\circ\phi^{-1}$$

which is also bounded. So $(b/a^{1/2})\circ\phi^{-1}$ is bounded by $C(1+|y|)$ and we get that $V(y)\leq C(y^{2}+1)$.

It follows then from the Farine-Lavis Theorem (Theorem X.38 of [R-S-75]) that $P_{\Omega}$ is ESA and hence $P$ is ESA on $C_{c}^{\infty}(I)$. It follows that $P$ is ESA on $C_{c}^{\infty}(S^{1}\setminus a^{-1}(0))$ and a fortiori on $C^{\infty}(S^{1})$.

Finally, we study the case where all zeroes of $a$ are degenerate and $b$ does not vanish at least at one of these, say $x=0$. We will need the following

It follows that the functions $a(x)d_{x}u_{j}$ are in $L^{2}$ and that $P$ is not ESA by the same argument than in Section 4.3.

Proof.– (of Lemma) We check first the existence of $u_{1}$ in an elementary way by showing the existence of a full Taylor expansion directly: we start with the Ansatz $u_{1}(x)=1+a_{1}x+a_{2}x^{2}+\cdots$. We get $b(0)a_{1}+(b^{\prime}(0)/2-1)=0$. Hence $a_{1}$. Then, inductively, we get an expression for $a_{k}$ as a function of the $a_{l}$ for $l<k$. Applying Malgrange’s Theorem 7.1 in [Ma-74], we get a smooth solution $u_{1}$ with $u_{1}(0)=1$.

Then we make the Ansatz $u_{2}=u_{1}v$ and we get the following differential equation for $v$:

$$\left(d_{x}+\frac{a^{\prime}}{a}+2\frac{u^{\prime}_{1}}{u_{1}}-i\frac{b}{a}\right)d_{x}v=0$$

It follows that, we can choose

$$d_{x}v=\frac{1}{au_{1}^{2}}e^{iE}$$

If $k$ is the order of the zero of $a$ at $x=0$, we can choose local coordinates near $0$ so that $E=1/y^{k-1}$, we get $d_{y}v=A(y)y^{-k}{\rm exp}(i/y^{k-1})$. We can integrate by part and we get

$$v(y)={A_{0}(y)}e^{i/y^{k-1}}-\int_{y}^{1}A_{1}(y)e^{i/y^{k-1}}$$

with $A_{0}$ and $A_{1}$ smooth. We can iterate the integration by part and get a formal solution $v(x)\equiv(v_{0}+v_{1}x+\cdots)e^{iE(x)}$. Again we can apply Malgrange’s Theorem. $\square$

We will consider for $X$ a 2-torus with a smooth Lorentzian metric $g$. Recall that a Lorentzian metric on a surface is a smooth non degenerate symmetric 2-form of signature $(1,1)$. There is, as in the Riemannian case, an associated geodesic flow (the Hamiltonian flow of the dual metric), a canonical volume form and a Laplace operator, which is an hyperbolic operator.

The null curves: a smooth curve $\gamma$ is said to be null if, at every point $x$ of $\gamma$, and any tangent vector $V$ to $\gamma$ at the point $x$, we have $g_{x}(V,V)=0$, i.e. the tangent spaces to $\gamma$ are isotropic. Locally the null curves are the leaves of two transverse foliations. This is not always true globally.

A closed null leaf $\gamma$ is a simple closed curve which is null. There is then a neighborhood of $\gamma$ with two null foliations: ${\cal F}_{+}$ is close to the tangent space to $\gamma$ and ${\cal F}_{-}$ is transversal to $\gamma$. We can then define a Poincaré map $P_{\gamma}$ as follows. Take a point $x_{0}$ on $\gamma$ and a germ of leaf of ${\cal F}_{-}$, $C$ at $x_{0}$. Then $P_{\gamma}$ is a germ of diffeomorphism $(C,x_{0})$ into itself obtained by following the null leaves of ${\cal F}_{+}$. The map $P_{\gamma}$ is uniquely defined modulo smooth conjugation by germs of diffeomorphisms. Note that $P_{\gamma}$ is orientation preserving because $X$ is orientable.

It is known that Lorentzian metrics on the 2-torus are not always geodesically complete. It is the case for example for the Clifton-Pohl torus:

Let $T$ be the quotient of ${\mathbb{R}}^{2}\setminus 0$ by the group generated by the homothety of ratio $2$. On $T$, the Clifton-Pohl Lorentzian metric is $g:=dxdy/(x^{2}+y^{2})$. The associated Laplacian $\Box_{g}=(x^{2}+y^{2})\partial^{2}/\partial x\partial y$ is formally self-adjoint on $L^{2}(T,|dxdy|/(x^{2}+y^{2}))$.

There is also a much simpler example, namely the quotient on $({\mathbb{R}}_{x}^{+}\times{\mathbb{R}}_{y},dxdy)$ by the group generated by $(x,y)\rightarrow(2x,y/2)$. The manifold is not closed, but non completeness sits already in a compact region.

It is known these metric are not geodesically complete. What about ESA of $\Box_{g}$?

We will prove a rather general result:

For the proof of Theorem 5.1, we will need two lemmas:

Proof.– If $g=e^{\phi}g_{0}$, the dual metric satisfy $g^{\star}=e^{-\phi}g^{\star}_{0}$ and hence the geodesic flow restricted to $g^{\star}=0$ are conformal with a bounded ratio. $\square$

Proof.–If $\Box_{g}u=v$, we have also $\Box_{g_{0}}u=e^{\phi}v$ and $e^{\phi}v$ is in $L^{2}$ as soon as $v$ is. Hence, if $\Box_{g_{0}}$ is ESA, there exists a sequence $(u_{n},w_{n}=\Box_{g_{0}}u_{n})_{n\in{\mathbb{N}}}$ converging in $L^{2}$ to $(u,e^{\phi}v)$ and $e^{-\phi}v_{n}$ converges to $v$. $\square$

This proves part 2 of Theorem 5.1.

It is well known and due to Sternberg [St-57] that a smooth germ of map $({\mathbb{R}},0)\rightarrow({\mathbb{R}},0)$ whose differential at the origin is in $]0,1[\cup]1,+\infty[$ is smoothly conjugated to $y\rightarrow\lambda y$ and hence is the time 1 flow of the vector field $\mu y\partial_{y}$ with $\lambda=e^{\mu}$.

A similar result hold for more degenerate diffeomorphisms: we assume that $g$ admits a closed null-leaf so that the Poincaré map is of the form $P(y)=y+y^{k}R(y)$ where $R(0)\neq 0$ and $k\geq 2$. It is proved in [Ta-73] (see Theorem 4)

Let $\gamma$ be closed null-leaf of $g$ and $U$ a neighbourhood  of $\gamma$ so that we have the two null foliations ${\cal F}_{+}$ and ${\cal F}_{-}$. We have the:

Proof.– Let us parametrize the closed leaf $\gamma$ by $x\in{\mathbb{R}}/{\mathbb{Z}}$ and extend the coordinate $x$ in some neighbourhood  $U$ of $\gamma$ so that the null foliation ${\cal F}_{-}$ is given by $dx=0$. Choose then for $y$ any coordinate in $U$ so that $y=0$ on $\gamma$. We introduce the differential equation $dy/dx=A(x,y)$ associated to the foliation ${\cal F}_{+}$ close to $\gamma$. Note that $A(x,0)=0$. Let $\phi_{x}(y)$ be the flow of this differential equation. The map $y\rightarrow\phi_{1}(y)$ is the Poincaré map of $\gamma$. By Theorem 5.2, we can choose a vector field $a(y)\partial y$ so that the time one flow is the same Poincaré map; and denote by $(\phi_{0})_{x}(y)$ this flow. Let us consider the germ of diffeomorphism near $\gamma$ defined by

$$F:(x,y)\rightarrow\big{(}x,y^{\prime}=(\phi_{0})_{x}\circ\phi_{x}^{-1}(y)\big{)}.$$

The map $F$ sends the integral curves of $dy-bdx$ onto the integral curves of $dy-adx$ and is periodic of period $1$ because the time 1 flows are the same. Hence the two null foliation are given respectively by $dx=0$ and $dy^{\prime}-a(y^{\prime})dx=0$. The Theorem follows.

$\square$

The idea is to use the normal form which, being invariant by translation in $x$, allows a separation of variables and hence application of Theorem 4.1.

Let us first prove the null incompleteness. Using the normal form and the conformal invariance of null completeness, we have to study near $y=0$ the Hamiltonian $h=-\eta(a(y)\eta+\xi)$. The function $\xi$ is a constant of the motion. Let us take initial conditions with $y_{0}>0$, $\xi_{0}>0$, and $a(y_{0})\eta_{0}+\xi_{0}=0$. We have, using that $a(y)\eta+\xi_{0}$ stays at $0$, $dy/dt=2a(y)\eta=-2\xi_{0}$. Hence $y(t)$ vanishes for a finite time $t_{0}$ and, we have then $\eta(t_{0})=\infty$. Null incompleteness follows.

The Lorentzian Laplacian associated to $g=dx(dy-a(y)dy)$ is given by

$$\Box=\partial_{y}a(y)\partial_{y}+\partial^{2}_{xy}$$

Let us look at solutions of $\Box u=v$ of the form $u(x,y)=e^{2\pi ix}v(y)$ with $v$ compactly supported near $0$. We have

$$\Box u(x,y)=e^{2\pi ix}\left(\partial_{y}a(y)\partial_{y}+2i\pi\partial_{y}\right)v(y)$$

The operator $P:=\partial_{y}a(y)\partial_{y}+2i\pi\partial_{y}$ is a Sturm-Liouville operator already studied in section 3. $P$ is not ESA. It follows then that there exists $v$ compactly supported near $0$ and $L^{2}$ so that $Pv=w\in L^{2}$ and there is no sequences $(v_{n},w_{n}=Pv_{n})$ converging in $L^{2}\times L^{2}$ to $(v,w)$. The result follows.

The goal of this section is to show that, for a generic Lorentzian metric on the 2-torus, there exists at least one null closed curve $\gamma$ whose Poincaré map is hyperbolic, i.e. the differential of $P_{\gamma}$ at the point $x_{0}$ of $\gamma$ is not tangent to the identity. It follows that, for a generic metric on the 2-torus, the geodesic flow is not complete and the Lorentzian Laplacian is not ESA.

A $C^{\infty}-$generic property is a property which holds for an open dense subset of the metrics in the $C^{\infty}$ topology.

We have the

The following argument is due to Etienne Ghys.

Proof.– The openness of the set of metric with a closed null hyperbolic geodesic is evident.

We say that the metric $g$ splits if the null leaves belongs to two distincts foliations ${\cal F}_{+}$ and ${\cal F}_{-}$. We say that the metric $g$ is orientable if there is a smooth non vanishing vector field $V$ on $X$ so that $g(V,V)$ is strictly positive everywhere. Any orientable metric splits: the two foliations are the boundaries of the connected component $C_{+}$ of the cone $g>0$ containing $V$, indeed using an orientation of $X$, we choose ${\cal F}_{+}$ so that the frame generated by ${\cal F}_{+}$ and $V$ is positively oriented. We now study the two different cases. 1. Case where $g$ splits: the genericity then follows from the fact that having an hyperbolic closed leaf is a generic property for a foliation of a torus (see [PdM-82]), here for ${\cal F}_{+}$.

2. Case where $g$ do not split: we introduce in this case a two-fold cover $Y$ of $X$ for which the lift $G$ of the metric $g$ is orientable. This cover $Y$ is equipped with an involution $J$ exchanging the two null foliations. Let us take a null closed curve $\gamma$ of one of these foliations. Then $J(\gamma)$ is a null closed curve of the other. They cannot cross: they have the same rotation number, because $J$ is homotopic to the identity. Moreover all intersections have the same sign because both foliations as well as $Y$ are orientable. It follows that the projection of $\gamma$ onto $X$ is simple. Still having a closed hyperbolic leaf for the foliation ${\cal F}_{+}$ of $G$ is generic and ${\cal F}_{-}=J({\cal F}_{+})$. $\square$