${\overline{Sas}_{n+1}}(X,{\mathrm{J}}Y)={\overline{Sas}_{n+1}}(w_{i}^{\mathcal{H}},w_{j}^{\mathcal{H}})=0$
since $w_{i}$ and $w_{j}$ are orthogonal. Let us now apply Example 9.1.10, for
instance by taking $Q$ the geodesic going through $x$ with speed $w_{i}$. The
normal bundle $\mathrm{N}^{1}Q$ is a submanifold orthogonal to the fibers of
$\mathrm{p}$ and tangent to $w_{i}^{\mathcal{H}}$ and $w_{j}^{\mathcal{V}}$.
So $d\omega(X,Y)=0$ by the usual argument.

  * •

Finally, we have to deal with the case $X=w_{i}^{\mathcal{H}}$ and
$Y=w_{i}^{\mathcal{V}}$. Here
${\overline{Sas}_{n+1}}(X,{\mathrm{J}}Y)={\overline{Sas}_{n+1}}(w_{i}^{\mathcal{H}},w_{i}^{\mathcal{H}})=1$.
For this computation, we may assume $n=1$, up to restricting to the totally
geodesic 2-plane spanned by $v$ and $w$, which is a copy of $\mathbb{H}^{2}$.
Hence we will simply denote $w_{i}=w$, and moreover we can assume (up to
changing the sign) that $w=x\boxtimes v$, where $\boxtimes$ denotes the
Lorentzian cross product in ${\mathbb{R}}^{2,1}$. In other words, $(x,v,w)$
forms an oriented orthonormal triple in ${\mathbb{R}}^{2,1}$.

Now, let us extend $X$ and $Y$ to globally defined vector fields on
$T^{1}\mathbb{H}^{2}$, by means of the assignment $(x,v)\mapsto(x\boxtimes
v)^{\mathcal{H}}$ and $(x,v)\mapsto(x\boxtimes v)^{\mathcal{V}}$. By this
definition, $X$ and $Y$ are orthogonal to $\chi$; we claim that $[X,Y]=-\chi$.
This will conclude the proof, since

$d\omega(X,Y)=\partial_{X}(\omega(Y))-\partial_{Y}(\omega(X))-\omega[X,Y]=-\omega[X,Y]={\overline{Sas}_{2}}(\chi,\chi)=1~{}.$

For the claim about the Lie bracket, let us use the hyperboloid model. Then
$X=(x\boxtimes v,0)$ and $Y=(0,x\boxtimes v)$. We can consider $X$ and $Y$ as
globally defined (by the same expressions) in the ambient space
${\mathbb{R}}^{2,1}\times{\mathbb{R}}^{2,1}$, so as to compute the Lie bracket
in ${\mathbb{R}}^{2,1}\times{\mathbb{R}}^{2,1}$, which will remain tangential
to $T^{1}\mathbb{H}^{2}$ since $T^{1}\mathbb{H}^{2}$ is a submanifold. Using
the formula $[X,Y]=Jac_{X}Y-Jac_{Y}X$, where $Jac_{X}$ denotes the Jacobian of
the vector field thought as a map from ${\mathbb{R}}^{3}$ to
${\mathbb{R}}^{3}$, we obtain

$[X,Y]=(x\boxtimes(x\boxtimes v),0)-(0,(x\boxtimes v)\boxtimes
v)=-(v,x)=-\chi_{(x,v)}$

by the standard properties of the cross-product.

In summary, we have shown that $d\omega$ and $\mathrm{p}^{*}\Omega$ coincide
on the basis
$\\{\chi,w_{1}^{\mathcal{H}},\ldots,w_{n}^{\mathcal{H}},w_{1}^{\mathcal{V}},\ldots,w_{n}^{\mathcal{V}}\\}$
of $T_{(x,v)}T^{1}\mathbb{H}^{n+1}$, and therefore the desired identity holds.
∎

We get as an immediate consequence the closedness of the fundamental form
$\Omega$, a fact whose proof has been deferred from Section 5.2.

######  Corollary 10.2.3.

The fundamental form $\Omega={\mathbbm{g}}(\cdot,\mathbbm{J}\cdot)$ is closed.

###### Proof.

Using Proposition 10.2.2 we have

$\mathrm{p}^{*}(d\Omega)=d(\mathrm{p}^{*}\Omega)=d(d\omega)=0~{}.$

Since $\mathrm{p}$ is surjective, it follows that $d\Omega=0$. ∎

####  10.3 Lagrangian immersions and local integrability

We have now all the ingredients to relate the Gauss maps of immersed
hypersurfaces in $\mathbb{H}^{n+1}$ with the Lagrangian condition for the
symplectic geometry of $\mathbb{G}_{n+1}$.

######  Corollary 10.3.1.

Given an oriented manifold $M^{n}$ and an immersion
$\sigma:M\to\mathbb{H}^{n+1}$, $\nobreak{G_{\sigma}:M\to\mathbb{G}_{n+1}}$ is
a Lagrangian immersion.

###### Proof.

By Proposition 10.1.2, the pull-back by $G_{\sigma}$ of the principal
${\mathbb{R}}$-bundle $\mathrm{p}$ is flat because there exists
$\widehat{G}_{\sigma}=\zeta_{\sigma}:M\to T^{1}\mathbb{H}^{n+1}$ orthogonal to
$\chi$ such that $G_{\sigma}=\mathrm{p}\circ\widehat{G}_{\sigma}$. Hence
$(\widehat{G}_{\sigma})^{*}d\omega$ vanishes identically and
$\widehat{G}_{\sigma}$ induces a flat section of $G_{\sigma}^{*}\mathrm{p}$.
But by Proposition 10.2.2,
$(\widehat{G}_{\sigma})^{*}d\omega=(\widehat{G}_{\sigma})^{*}(\mathrm{p}^{*}\Omega)=(G_{\sigma})^{*}\Omega$,
therefore $G_{\sigma}$ is Lagrangian. ∎

Observe that in Corollary 10.3.1 we only use the flatness property of
Proposition 10.1.2, and not the triviality of the pull-back principal bundle.
When $M$ is simply connected, we can partially reverse Corollary 10.3.1,
showing that the Lagrangian condition is essentially the only local
obstruction.

######  Corollary 10.3.2.

Given an orientable simply connected manifold $M^{n}$ and a Lagrangian
immersion $G:M\to\mathbb{G}_{n+1}$, there exists an immersion $\zeta:M\to
T^{1}\mathbb{H}^{n+1}$ orthogonal to the fibers of $\mathrm{p}$ such that
$G=\mathrm{p}\circ\zeta$. Moreover, if $\pi\circ\zeta:M\to\mathbb{H}^{n+1}$ is
an immersion, then $G$ coincides with its Gauss map.

###### Proof.

Since $G$ is Lagrangian, by Proposition 10.1.2 the $G$-pull-back bundle of
$\mathrm{p}$ is flat, and it is therefore a trivial flat bundle since $M$ is
simply connected. Hence it admits a global flat section, which provides the
map $\zeta:M\to T^{1}\mathbb{H}^{n+1}$ orthogonal to $\chi$. Assuming moreover
that $\sigma:=\pi\circ\zeta$ is an immersion, by Proposition 9.1.3,
$G=\mathrm{p}\circ\zeta$ is the Gauss map of $\sigma$. ∎

Clearly the map $\zeta$ is not uniquely determined, and the different choices
differ by the action of $\varphi_{t}$. Lemma 10.3.3, and the following
corollary, show that (by post-composing with $\varphi_{t}$ if necessary) one
can always find $\zeta$ such that $\pi\circ\zeta$ is _locally_ an immersion.

######  Lemma 10.3.3.

Let $M$ be a $n$-manifold and $\zeta:M\to T^{1}\mathbb{H}^{n+1}$ be an
immersion orthogonal to $\chi$. Suppose that the differential of
$\pi\circ\zeta$ is singular at $p\in M$. Then there exists $\epsilon>0$ such
that the differential of $\pi\circ\varphi_{t}\circ\zeta$ at $p$ is non-
singular for all $t\in(-\epsilon,\epsilon)\setminus\\{0\\}$.

###### Proof.

Let us denote $\zeta_{t}:=\varphi_{t}\circ\zeta$, $\sigma:=\pi\circ\zeta$, and
$\sigma_{t}:=\pi\circ\zeta_{t}$. Assume also $\zeta(p)=(x,v)$. Let
$\\{V_{1},\ldots,V_{k}\\}$ be a basis of the kernel of $d_{p}\sigma$ and let
us complete it to a basis $\\{V_{1},\ldots,V_{n}\\}$ of $T_{p}M$. Hence if we
denote $w_{j}:=d_{p}\sigma(V_{j})$ for $j>k$, $\\{w_{k+1},\ldots,w_{n}\\}$ is
a basis of the image of $d_{p}\sigma$. Exactly as in the proof of Proposition
9.1.3, we have $w_{k+1},\ldots,w_{n}\in v^{\perp}\subset
T_{x}\mathbb{H}^{n+1}$. Hence we have:

$d\zeta(V_{1})=u_{1}^{\mathcal{V}},\ldots,d\zeta(V_{k})=u_{k}^{\mathcal{V}},d\zeta(V_{k+1})=w_{k+1}^{\mathcal{H}}+u_{k+1}^{\mathcal{V}},\ldots,d\zeta(V_{n})=w_{n}^{\mathcal{H}}+u_{n}^{\mathcal{V}}$

for some $u_{1},\ldots,u_{n}\in v^{\perp}$.

On the one hand, since $\zeta$ is an immersion, $u_{1},\ldots,u_{k}$ are
linearly independent. On the other hand, $\zeta$ is orthogonal to $\chi$,
hence by Remark 10.2.1 we have $\zeta^{*}d\omega=0$. Using Equation (10.1), it
follows that:

${\overline{Sas}_{n+1}}(d\zeta(V_{i}),{\mathrm{J}}\circ d\zeta(V_{j}))=0$

for all $i,j=1,\dots n$. Applying this to any choice of $i\leq k$ and $j>k$,
we find $\langle u_{i},w_{j}\rangle=0$. Hence
$\\{u_{1},\ldots,u_{k},w_{k+1},\ldots,w_{n}\\}$ is a basis of $v^{\perp}$.

We are now ready to prove the statement. By Equations (5.11) and (5.12), we
have

$d\sigma_{t}(V_{i})=d\pi\circ d\varphi_{t}(u_{i}^{\mathcal{V}})=\sinh(t)u_{i}$

for $1\leq i\leq k$, while

$d\sigma_{t}(V_{j})=d\pi\circ
d\varphi_{t}(w_{j}^{\mathcal{H}}+u_{j}^{\mathcal{V}})=\cosh(t)w_{j}+\sinh(t)u_{j}$

for $k+1\leq j\leq n$. The proof will be over if we show that
$\\{d\sigma(V_{1}),\ldots,d\sigma(V_{n})\\}$ are linearly independent for
$t\in(-\epsilon,\epsilon)$, $t\neq 0$. In light of the above expressions,
dividing by $\sinh(t)$ (which is not zero if $t\neq 0$) or $\cosh(t)$, this is
equivalent to showing that

$\\{u_{1},\ldots,u_{k},w_{k+1}+\tanh(t)u_{k+1},\ldots,w_{n}+\tanh(t)u_{n}\\}$

are linearly independent for small $t$. This is true because we have proved
above that $\\{u_{1},\ldots,u_{k},w_{k+1},\ldots,w_{n}\\}$ is a basis, and
linear independence is an open condition. ∎

######  Theorem 10.3.4.

Let $G\colon M^{n}\to\mathbb{G}_{n+1}$ be an immersion. Then $G$ is Lagrangian
if and only if for all $p\in M$ there exists a neighbourhood $U$ of $p$ and an
immersion $\sigma\colon U\to\mathbb{H}^{n+1}$ such that $G_{\sigma}=G|_{U}$.

###### Proof.

The “if” part follows from Corollary 10.3.1. Conversely, let $U$ be a simply
connected neighbourhood of $p$. By Corollary 10.3.2, there exists an immersion
$\zeta\colon U\to T^{1}\mathbb{H}^{n+1}$ orthogonal to the fibers of
$\mathrm{p}$ such that $G=\mathrm{p}\circ\zeta$. If the differential of
$\pi\circ\zeta$ is non-singular at $p$, then, up to restricting $U$, we can
assume $\sigma:=\pi\circ\zeta$ is an immersion of $U$ into $\mathbb{H}^{n+1}$.
By the second part of Corollary 10.3.2, $G|_{U}$ is the Gauss map of $\sigma$.
If the differential of $\pi\circ\zeta$ is instead singular at $p$, by Lemma
10.3.3 it suffices to replace $\zeta$ by $\zeta_{t}$ for small $t$ and we
obtain the same conclusion. ∎

#####  10.3.1 Lagrangian condition in ${\mathbb{G}}$ in terms of complex
metrics

In Proposition 9.2.1 we showed that the pseudo-Riemannian metric of
${\mathbbm{g}}$ on $\mathbb{G}_{3}$ is the real part of the hRm
$\langle\cdot,\cdot\rangle_{\mathbb{G}}$. Moreover, by Theorem 6.5.2, we know
that a complex metric $g$ with constant curvature $-1$ on a surface $S$
corresponds to an equivariant immersion $\widetilde{S}\to{\mathbb{G}}$, unique
up to post-composition with $\mathrm{Isom}(\mathbb{H}^{3})$.

We show that there is a characterization of Lagrangian (i.e. locally
integrable) immersions $G\colon S\to\mathbb{G}_{3}$ in terms of the induced
complex metric.

Before that, let us show that the symplectic form on $\mathbb{G}_{3}$ can be
expressed in terms of the complex structure of ${\mathbb{G}}$.

######  Proposition 10.3.5.

Let $(U,z)$ be an affine chart on $\mathbb{C}\mathbb{P}^{1}$. Then, on the
local chart $(U\times U\setminus\Delta,z\times z)$ for $\mathbb{G}_{3}$, the
symplectic form $\Omega$ satisfies

$\Omega=-\frac{1}{2}Re\bigg{(}\frac{dz_{1}\wedge
dz_{2}}{(z_{1}-z_{2})^{2}}\bigg{)}.$ (10.2)

###### Proof.

Let us first note that, using exactly the same techniques as in the proof of
Theorem 5.3.1, one can prove that the RHS of Equation 10.2 does not depend on
the local affine chart and that it is
${\mathrm{PSL}(2,\mathds{C})}$-invariant. Hence it is sufficient to check that
Equation 10.2 holds in a point of ${\mathbb{G}}$

Let us recall that in Equation (6.12) we computed the map
$d\mathrm{p}\colon\chi_{(x,v)}^{\perp}\to T_{\ell}{\mathbb{G}}$, where, in the
hyperplane model ${\mathbb{C}}\times{\mathbb{R}}^{+}$ for $\mathbb{H}^{3}$, we
denote $x=(0,1)$, $v=(1,0)$,
$\ell=(1,-1)\in{\mathbb{C}}\times{\mathbb{C}}\setminus\Delta$.

With notations as in the proof of Proposition 10.2.2, define
$w_{1}^{\mathcal{H}},w_{2}^{\mathcal{H}},w_{1}^{\mathcal{V}},w_{2}^{\mathcal{V}}\in
v^{\perp}\oplus v^{\perp}=\chi_{(x,v)}^{\bot}$ as

$\displaystyle w_{1}^{\mathcal{H}}$ $\displaystyle=((i,0),(0,0))\ ,$
$\displaystyle w_{2}^{\mathcal{H}}$ $\displaystyle=((0,1),(0,0))\ ,$
$\displaystyle w_{1}^{\mathcal{V}}$ $\displaystyle=((0,0),(i,0))\ ,$
$\displaystyle w_{2}^{\mathcal{V}}$ $\displaystyle=((0,0),(0,1))\ .$

The images via $d\mathrm{p}$ of these vectors define a orthonormal basis for
$T_{\ell}{\mathbb{G}}$: explicitly via Equation (6.12) we have

$\displaystyle d\mathrm{p}{(w_{1}^{\mathcal{H}})}$ $\displaystyle=(i,i)\ ,$
$\displaystyle d\mathrm{p}{(w_{2}^{\mathcal{H}})}$ $\displaystyle=(-1,1)\ ,$
$\displaystyle d\mathrm{p}{(w_{1}^{\mathcal{V}})}$ $\displaystyle=(-i,i)\ ,$
$\displaystyle d\mathrm{p}{(w_{2}^{\mathcal{V}})}$ $\displaystyle=(1,1)\ .$

We computed in the proof of Proposition 10.2.2 that
$\Omega(d\mathrm{p}(w_{1}^{\mathcal{H}}),d\mathrm{p}(w_{2}^{\mathcal{H}}))=\Omega(d\mathrm{p}(w_{1}^{\mathcal{V}}),d\mathrm{p}(w_{2}^{\mathcal{V}}))=0$,
and that
$\Omega(d\mathrm{p}(w_{j}^{\mathcal{H}}),d\mathrm{p}(w_{k}^{\mathcal{V}}))=\delta_{j,k}$.
Using the explicit descriptions above, one gets directly that the RHS of
Equation 10.2 coincides on this basis of $T_{\ell}{\mathbb{G}}$. ∎

######  Corollary 10.3.6.

An immersion $G\colon S\to\langle\cdot,\cdot\rangle{\mathbb{G}}$ is Lagrangian
if and only if the (local) area form $dA_{g}$ of
$g=G^{*}\langle\cdot,\cdot\rangle_{\mathbb{G}}$ is real, i.e. it restricts to
a real-valued $2$-form on $TS$.

###### Proof.

First of all, observe that clearly the statement does not depend on the sign
of the area form.

Let $G=(f_{1},f_{2})$, hence $g=\frac{-4}{(f_{1}-f_{2})^{2}}df_{1}\cdot
df_{2}$, with $f_{1},f_{2}\colon S\to\mathbb{C}\mathbb{P}^{1}$.

Let us show that

$dA_{g}=8i\frac{1}{(f_{1}-f_{2})^{2}}df_{1}\wedge df_{2}\ .$ (10.3)

In order to prove it, take a basis $(V_{1},V_{2})$ for ${\mathbb{C}}T_{p}S$
such that each $V_{k}$ is a non-zero isotropic vector for $g$ with $V_{k}\in
Ker(d_{p}f_{k}\colon{\mathbb{C}}T_{p}S\to{\mathbb{C}})$. Then, the vectors

$W_{1}=\frac{V_{1}+iV_{2}}{\sqrt{2ig(V_{1},V_{2})}}\ ,\quad
W_{2}=\frac{V_{1}-iV_{2}}{i\sqrt{2ig(V_{1},V_{2})}}$

define a $g$-orthonormal basis, and $dA_{g}(W_{1},W_{2})=1$ leads to

$dA_{g}(V_{1},V_{2})=2ig(V_{1},V_{2}):$

as a result, one gets that Equation $\eqref{eq forma d area di g}$ holds since
it holds in the basis $(V_{1},V_{2})$.

Finally, by Proposition 10.3.5 and Equation (10.3), one gets

$\displaystyle G^{*}\Omega$ $\displaystyle=-\frac{1}{2}Re\
\Big{(}G^{*}\Big{(}\frac{dz_{1}\wedge
dz_{2}}{(z_{1}-z_{2})^{2}}\Big{)}\Big{)}=$
$\displaystyle=-\frac{1}{2}Re\Big{(}\frac{df_{1}\wedge
df_{2}}{(f_{1}-f_{2})^{2}}\Big{)}=$
$\displaystyle=-\frac{1}{2}Im\Big{(}i\frac{df_{1}\wedge
df_{2}}{(f_{1}-f_{2})^{2}}\Big{)}=$
$\displaystyle=\frac{1}{16}Im\Big{(}dA_{g}\Big{)}\ ,$

and the thesis follows. ∎

####  10.4 Global integrability

Let us now approach the problem of global integrability. We provide an example
to show that in general $\pi\circ\varphi_{t}\circ\zeta$ might fail to be
_globally_ an immersion for _all_ $t\in{\mathbb{R}}$, as we already mentioned
after Proposition 9.1.3.

######  Example 10.4.1.

Let us construct a curve $G:(-T,T)\to\mathbb{G}_{2}$ with the property of
being locally integrable111As a matter of fact, any curve in $\mathbb{G}_{2}$
is locally integrable by Theorem 10.3.4, since the domain is simply connected
and it is trivially Lagrangian. However, in this example, we will see by
construction that $G$ is locally integrable. but not globally integrable.

Fix $r>0$ and a maximal geodesic $\ell$ in $\mathbb{H}^{2}$. Let us consider a
smooth curve $\sigma_{+}\colon(-\varepsilon,T)\to\mathbb{H}^{2}$, for some
small enough $\varepsilon$ and big enough $T$, so that:

  * •

$\sigma_{+}$ is an immersion and is parameterized by arclength;

  * •

$(\sigma_{+})|_{(-\varepsilon,\varepsilon)}$ lies on the $r$-cap equidistant
from $\ell$, oriented in such a way that the induced unit normal vector field
$(\nu_{+})|_{(-\varepsilon,\varepsilon)}$ is directed towards $\ell$;

  * •

$(\sigma_{+})|_{(T_{0},T)}$ lies on the metric circle
$\\{x\in\mathbb{H}^{2}\,|\,d_{\mathbb{H}^{2}}(x,x_{0})=r\\}$ for some
$x_{0}\in\mathbb{H}^{2}$ and some $\varepsilon<T_{0}<T$, oriented in such a
way that the induced unit normal vector field $(\nu_{+})|_{(T_{0},T)}$ is
directed towards $x_{0}$.

Figure 10.1: On the left, the construction of the curve $\sigma_{+}$, which is
an arclenght parameterization of a portion of $r$-cap equidistant from $\ell$
for $s\in(-\varepsilon,\varepsilon)$, and of a circle of radius $r$ for
$s\in(T_{0},T)$. On the right, the curve $\sigma_{-}$ whose image is obtained
by an order-two elliptic isometry from the image of $\sigma_{+}$.

See Figure 10.1. By a simple computation, for instance using Equation (9.11),
the curvature of $\sigma_{+}$ equals $\tanh(r)$ on
$(-\varepsilon,\varepsilon)$ and $\frac{1}{\tanh(r)}$ on $(T_{0},T)$. Hence by
the intermediate value theorem, the image of the curvature function
$k\colon(-\varepsilon,T)\to{\mathbb{R}}$ contains the interval
$[\tanh(r),\frac{1}{\tanh(r)}]$. An important consequence of this observation
is that $(\sigma_{+})_{t}$ fails to be an immersion when $t\geq r$. More
precisely, if we denote $\zeta_{\sigma_{+}}$ the lift to $T^{1}\mathbb{H}^{2}$
as usual, using Equation (5.5) analogously as for Equation (9.6), we obtain
that:

$d(\pi\circ\varphi_{t}\circ\zeta_{\sigma_{+}})(V)=(\cosh(t)-\sinh(t)k)d\sigma(V)~{}.$

This shows that the differential of
$\pi\circ\varphi_{t}\circ\zeta_{\sigma_{+}}$ at a point $s\in(-\varepsilon,T)$
vanishes if and only if

$\tanh(t)=\frac{1}{k(s)}~{}.$ (10.4)

Since the image of the function $k$ contains the interval
$[\tanh(r),\frac{1}{\tanh(r)}]$, if $t\geq r$ then there exists $s$ such that
(10.4) is satisfied and therefore $\pi\circ\varphi_{t}\circ\zeta_{\sigma_{+}}$
is not an immersion at $s$.

Now, let $y_{0}\in\ell$ be the point at distance $r$ from $\sigma_{+}(0)$ and
let ${R}_{0}\colon\mathbb{H}^{2}\to\mathbb{H}^{2}$ be the symmetry at $y_{0}$,
i.e. ${R}_{0}$ is the isometry of $\mathbb{H}^{2}$ such that
${R}_{0}(y_{0})=y_{0}$ and $d_{y_{0}}{R}_{0}=-\mathrm{id}$. Define
$\sigma_{-}\colon(-T,\varepsilon)\to\mathbb{H}^{2}$ as

$\sigma_{-}(s):={R}_{0}(\sigma_{+}(-s)).$

As a result, the normal vector field $\nu_{-}$ of $\sigma_{-}$ is such that
$dR_{0}({\nu_{+}}(s))={-}{\nu_{-}}(-s)$ and the curvature of $\sigma_{-}$
takes any value in the interval $[-\frac{1}{\tanh(r)},-{\tanh(r)}]$. Hence
$(\sigma_{-})_{t}$ fails to be an immersion for $t\leq-r$.

Finally, consider the two lifts $\zeta_{\sigma_{+}}$ and $\zeta_{\sigma_{-}}$
in $T^{1}\mathbb{H}^{2}$. By construction one has that for all
$s\in(-\varepsilon,\varepsilon)$

$\varphi_{r}\circ\zeta_{\sigma_{+}}(s)=\varphi_{-r}\circ\zeta_{\sigma_{-}}(s).$

As a result, we can define our counterexample $\zeta\colon(-T,T)\to
T^{1}\mathbb{H}^{2}$ as

$\zeta(s)=\begin{cases}\varphi_{r}\circ\zeta_{\sigma_{+}}(s)\quad&\text{ if
$s>-\varepsilon$}\\\ \varphi_{-r}\circ\zeta_{\sigma_{-}}(s)\quad&\text{ if
$s<\varepsilon$}\end{cases}.$

By construction, we have that
$\mathrm{p}\circ\zeta_{\sigma_{+}}=\mathrm{p}\circ\zeta|_{(-\varepsilon,T)}$
and
$\mathrm{p}\circ\zeta_{\sigma_{-}}=\mathrm{p}\circ\zeta|_{(-T,\varepsilon)}$,
therefore $\mathrm{p}\circ\zeta$ is an immersion into $\mathbb{G}_{2}$ and
clearly it is locally integrable. However, by the above discussion,
$\pi\circ\varphi_{t}\circ\zeta$ fails to be an immersion for every
$t\in{\mathbb{R}}$: for $t\geq 0$ because
$\pi\circ\varphi_{t}\circ\zeta_{\sigma_{+}}$ has vanishing differential at
some $s\geq-\varepsilon$, and for $t\leq 0$ because the differential of
$\pi\circ\varphi_{t}\circ\zeta_{\sigma_{-}}$ vanishes at some
$s\leq\varepsilon$. See Figure 10.2.

Figure 10.2: The curve $\zeta:(-T,T)\to T^{1}\mathbb{H}^{2}$ projects to a map
$\pi\circ\zeta$ into $\mathbb{H}^{2}$ which is not an immersion, because it is
constantly equal to $x_{0}$ for $s\in(T_{0},T)$ and to $R_{0}(x_{0})$ for
$s\in(-T,-T_{0})$. Moreover the curvature of the regular part takes all the
values in $(-\infty,+\infty)$. For this reason, each immersion
$\varphi_{t}\circ\zeta$ into $T^{1}\mathbb{H}^{2}$ does not project to an
immersion in $\mathbb{H}^{2}$.

Corollary 10.3.2 and Theorem 10.3.4 can be improved under the additional
assumption that the immersion $G$ is Riemannian. More precisely, we provide an
improved characterization of immersions into $\mathbb{G}_{n+1}$ that are Gauss
maps of immersions with small principal curvatures in terms of the Lagrangian
and Riemannian condition, again for when $M$ is simply connected.

######  Theorem 10.4.2.

Given a simply connected manifold $M^{n}$ and an immersion
$G:M^{n}\to\mathbb{G}_{n+1}$, $G$ is Riemannian and Lagrangian if and only if
there exists an immersion $\sigma\colon M\to\mathbb{H}^{n+1}$ with small
principal curvatures such that $G_{\sigma}=G$, i.e. if and only if $G$ is
globally integrable.

If in addition $\sigma$ is complete, then it is a proper embedding,
$G_{\sigma}$ is an embedding and $M$ is diffeomorphic to ${\mathbb{R}}^{n}$.

###### Proof.

We know from Corollary 10.3.1 and Proposition 9.3.2 that the Riemannian and
Lagrangian conditions on $G$ are necessary. To see that they are also
sufficient, by Corollary 10.3.2 there exists $\zeta\colon
M\to\mathbb{G}_{n+1}$ orthogonal to the fibers of $\mathrm{p}$ such that
$\mathrm{p}\circ\zeta=G$. We claim that $\pi\circ\zeta$ is an immersion, which
implies that $G=G_{\sigma}$ for $\sigma=\pi\circ\zeta$ by the second part of
Corollary 10.3.2. Indeed, if $X\in T_{p}M$ is such that
$d\zeta(X)\in\mathcal{V}_{\zeta(x)}=\ker(d\pi_{\zeta(x)})$, then
$d\zeta(X)=w^{\mathcal{V}}$ for some $w\in T_{\sigma(p)}\mathbb{H}^{n+1}$.
Hence by Definition 5.1.2 and the construction of the metric ${\mathbbm{g}}$,
${\mathbbm{g}}(X,X)=-\langle w,w\rangle\leq 0$: since ${\mathbbm{g}}$ is
Riemannian, necessarily $w=0$ and therefore $X=0$.

By Proposition 9.3.2 $\sigma$ has small principal curvatures. The “in
addition” part follows by Proposition 9.3.15 and Proposition 9.3.16. ∎

As another consequence of Proposition 9.3.15 and Proposition 9.3.16, we obtain
the following result.

######  Theorem 10.4.3.

Let $M^{n}$ be a manifold. If $G\colon M\to\mathbb{G}_{n+1}$ is a complete
Riemannian and Lagrangian immersion, then $M$ is diffeomorphic to
${\mathbb{R}}^{n}$ and there exists a proper embedding
$\sigma:M\to\mathbb{H}^{n+1}$ with small principal curvatures such that
$G=G_{\sigma}$.

###### Proof.

Let us lift $G$ to the universal cover $\widetilde{M}$, obtaining a Riemannian
and Lagrangian immersion $\widetilde{G}:\widetilde{M}\to\mathbb{G}_{n+1}$
which is still complete. By Theorem 10.4.2, $\widetilde{G}$ is the Gauss map
of an immersion $\sigma$ with small principal curvatures. We claim that
$\sigma$ is complete. Indeed by Equation (9.4) the first fundamental form of
$\widetilde{G}$, which is complete by hypothesis, equals
$\mathrm{I}-\mathrm{I}\mathrm{I}\mathrm{I}$, hence it is complete since
$\mathrm{I}\mathrm{I}\mathrm{I}$ is positive semi-definite.

It follows from Proposition 9.3.16 that $\widetilde{G}$ is injective. Hence
$\widetilde{M}=M$ and $\widetilde{G}=G$, and therefore $G$ is the Gauss map of
$\sigma$, which is complete. By the “in addition” part of Theorem 10.4.2,
$\sigma$ is properly embedded and $M$ is diffeomorphic to ${\mathbb{R}}^{n}$.
∎

In summary, the Lagrangian condition is essentially the only _local_
obstruction to realizing an immersion $G:M\to\mathbb{G}_{n+1}$ as the Gauss
map of an immersion into $\mathbb{H}^{n+1}$, up to the subtlety that this
might be an immersion only when lifted to $T^{1}\mathbb{H}^{n+1}$. This
subtlety however never occurs in the small principal curvatures case. In the
remainder of the paper, we will discuss the problem of characterizing
immersions into $\mathbb{G}_{n+1}$ which are Gauss maps of _equivariant_
immersions into $\mathbb{H}^{n+1}$ with small principal curvatures.

####  10.5 Equivariant integrability of Riemannian immersions

In this section, we provide the first characterization of _equivariant_
immersions in $\mathbb{G}_{n+1}$ which arise as the Gauss maps of
_equivariant_ immersions in $\mathbb{H}^{n+1}$, in the Riemannian case. This
is the content of Theorem 10.5.5. We first try to motivate the problem,
introduce the obstruction, namely the Maslov class, and study some of its
properties. See for instance [Ars00] for a discussion on the Maslov class in
more general settings.

#####  10.5.1 Motivating examples

Given an $n$-manifold $M$, a representation
$\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$, and a $\rho$-equivariant
immersion $\widetilde{\sigma}:\widetilde{M}\to\mathbb{H}^{n+1}$, it is
immediate to see that the Gauss map
$G_{\widetilde{\sigma}}:\widetilde{M}\to\mathbb{G}_{n+1}$ is
$\rho$-equivariant (recall also Remark 5.2.3). Moreover, if
${\widetilde{\sigma}}$ has small principal curvatures, it follows from the
discussion of the previous sections that $G_{\widetilde{\sigma}}$ is a
Lagrangian and Riemannian immersion.

However, a $\rho$-equivariant Lagrangian immersion (even with the additional
assumptions of being Riemannian and being an embedding) does not always arise
as the Gauss map associated to a _$\rho$ -equivariant_ immersion in
$\mathbb{H}^{n+1}$, as the following simple example shows for $n=1$.

######  Example 10.5.1.

Let us construct a coordinate system for a portion of $\mathbb{G}_{2}$. Let
$\gamma:{\mathbb{R}}\to\mathbb{H}^{2}$ be a geodesic parameterized by
arclength, and let us define a map
$\eta:{\mathbb{R}}\times(0,\pi)\to\mathbb{G}_{2}$ by defining $\eta(t,\theta)$
as the oriented geodesic that intersects $\gamma$ at $\gamma(t)$ with an angle
$\theta$ (measured counterclockwise with respect to the standard orientation
of $\mathbb{H}^{2}$). We can lift $\eta$ to a map
$\widehat{\eta}:{\mathbb{R}}\times(0,\pi)\to T^{1}\mathbb{H}^{2}$, which will
however _not_ be orthogonal to the fibers of the projection
$T^{1}\mathbb{H}^{2}\to\mathbb{G}_{2}$. The lift is simply defined as

$\widehat{\eta}(t,\theta)=(\gamma(t),\cos(\theta)\gamma^{\prime}(t)+\sin(\theta)w)~{},$

where $w$ is the vector forming an angle $\frac{\pi}{2}$ with
$\gamma^{\prime}(t)$. We omitted the dependence of $w$ on $t$ since, in the
hyperboloid model, $w$ is actually a constant vector in ${\mathbb{R}}^{2,1}$.

Let us compute the pull-back of the metric ${\mathbbm{g}}$ on $\mathbb{G}_{2}$
by the map $\eta$. We have already observed in Example 9.1.9 that the
restriction of ${\mathbbm{g}}$ on the image of
$\theta\mapsto\eta(t_{0},\theta)$ is minus the standard metric of
$\mathbb{S}^{1}$. Indeed in this simple case,
$d\widehat{\eta}_{(t,\theta)}(\partial_{\theta})=(0,w)$ is in the vertical
subspace $\mathcal{V}^{0}$ and by Definition 5.1.2 its squared norm is $-1$.
On the other hand, since the vector field
$\cos(\theta)\gamma^{\prime}(t)+\sin(\theta)w$ is parallel along $\gamma$,
when we differentiate in $t$ we obtain, by applying the definition of
horizontal lift:

$d\widehat{\eta}_{(t,\theta)}(\partial_{t})=\cos(\theta)(\gamma^{\prime}(t))^{\mathcal{H}}+\sin(\theta)w^{\mathcal{H}}~{}.$
(10.5)

Moreover, Equation (10.5) gives the decomposition of
$d\widehat{\eta}_{(t,\theta)}(\partial_{t})$ in
$T_{\widehat{\eta}_{(t,\theta)}}T^{1}\mathbb{H}^{2}={\mathrm{Span}(\chi)\oplus\chi^{\perp}}$
as in Equation (5.8), since the first term is a multiple of the generator of
the geodesic flow, and the second term is in $\mathcal{H}^{0}$. This shows, by
definition of the metric ${\mathbbm{g}}$, that
$d\eta_{(t,\theta)}(\partial_{t})$ has squared norm $\sin^{2}(\theta)$ and
that $d\eta_{(t,\theta)}(\partial_{t})$ and
$d\eta_{(t,\theta)}(\partial_{\theta})$ are orthogonal. In conclusion, we have
showed:

$\eta^{*}{\mathbbm{g}}=-d\theta^{2}+\sin^{2}(\theta)dt^{2}~{}.$

We are now ready to produce our example of $\rho$-equivariant embedding
$G:\widetilde{M}\to\mathbb{G}_{2}$ which is not $\rho$-integrable. Consider
$M=\mathbb{S}^{1}$, $\widetilde{M}={\mathbb{R}}$, and the representation
$\rho:{\mathds{Z}}\to\mathrm{Isom}(\mathbb{H}^{2})$ which is a hyperbolic
translation along $\gamma$. The induced action on $\mathbb{G}_{2}$ preserves
the image of $\eta$ and its generator acts on the $(t,\theta)$-coordinates
simply by $(t,\theta)\mapsto(t+c,\theta)$. Hence the map

$G:{\mathbb{R}}\to\mathbb{G}_{2}\qquad G(s)=\eta(s,\theta_{0})$

for some constant $\theta_{0}\in(0,\pi)$ is a $\rho$-equivariant Riemannian
embedding by the above expression of $\eta^{*}{\mathbbm{g}}$. Of course the
Lagrangian condition is trivially satisfied since $n=1$. By Theorem 10.4.2 $G$
coincides with the Gauss map $G_{\sigma}$ associated to some embedding
$\sigma:{\mathbb{R}}\to\mathbb{H}^{2}$ with small curvature. It is easy to see
that any such embedding $\sigma$ is not $\rho$-equivariant unless
$\theta_{0}=\frac{\pi}{2}$, see Figure 10.3.

Figure 10.3: On the left, the family of geodesics forming a fixed angle
$\theta$ with the horizontal geodesic (in the Poincaré disc model of
$\mathbb{H}^{2}$). On the right, the metric
$-d\theta^{2}+\sin^{2}(\theta)dt^{2}$ represents a portion of Anti-de Sitter
space of dimension 2. Here are pictured some lines defined by $\theta=c$, in
the projective model of Anti-de Sitter space.

We also briefly provide an example of a locally integrable, but not globally
integrable immersion in $\mathbb{G}_{3}$ for $M$ not simply connected (lifting
to the universal cover $\widetilde{M}$ this corresponds to $\rho$ being the
trivial representation). This example in particular motivates Corollary
10.5.6, which is a direct consequence of Theorem 10.5.5.

######  Example 10.5.2.

First, let us consider a totally geodesic plane $\mathcal{P}$ in
$\mathbb{H}^{3}$ and an annulus $\mathcal{A}$ contained in $\mathcal{P}$. Let
$c:\widetilde{\mathcal{A}}\to\mathcal{A}$ be the universal covering. Then $c$
is an immersion in $\mathbb{H}^{3}$ with small principal curvatures (in fact,
totally geodesic), and is clearly not injective. See Figure 10.4 on the left.
Of course, in light of Proposition 9.3.15, this is possible because the
immersion $c$ is not complete.

Now, let us deform $\mathcal{A}$ in the following way. We cut $\mathcal{A}$
along a geodesic segment $s\subset\mathcal{P}$ to obtain a rectangle
$\mathcal{R}$ having two (opposite) geodesic sides, say $r_{1}$ and $r_{2}$.
Then we deform such rectangle to get an immersion
$c^{\prime}:\mathcal{R}\to\mathbb{H}^{3}$, so that one geodesic side remains
unchanged (say $c^{\prime}(r_{1})=s$), while the other side $r_{2}$ is mapped
to an $r$-cap equidistant from $\mathcal{P}$, for small $r$, in such a way
that it projects to $s$ under the normal evolution. We can also arrange $c$ so
that a neighbourhood of $r_{1}$ is mapped to $\mathcal{P}$, while a
neighbourhood of $r_{2}$ is mapped to the $r$-cap equidistant from
$\mathcal{P}$. See Figure 10.4 on the right.

By virtue of this construction, the Gauss map of $c^{\prime}$ coincides on the
edges $r_{1}$ and $r_{2}$ of $\mathcal{R}$, and therefore induces an immersion
$G^{\prime}:\mathcal{A}\to\mathbb{G}_{3}$. Clearly $G^{\prime}$ is locally
integrable, but not globally integrable. In other words, the lift to the
universal cover of $G^{\prime}$ is a $\rho$-equivariant immersion of
$\widetilde{A}$ to $\mathbb{G}_{3}$ which is not $\rho$-integrable, for $\rho$
the trivial representation.

Figure 10.4: On the left, a totally geodesic annulus $\mathcal{A}$ in a plane
$\mathcal{P}$. On the right, an embedded rectangle with the property that a
neighbourhood of one side lies in $\mathcal{P}$, while a neighbourhood of the
opposite side lies on an $r$-cap equidistant from $\mathcal{P}$. Such
rectangle induces an embedding of $\mathcal{A}$ in $\mathbb{G}_{3}$ which is
locally, but not globally, integrable.

Motivated by the previous examples, we introduce the relevant definition for
our problem.

######  Definition 10.5.3.

Given an $n$-manifold $M$ and a representation
$\rho\colon\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$, a $\rho$-equivariant
immersion $G\colon\widetilde{M}\to\mathbb{G}_{n+1}$ is _$\rho$ -integrable_ if
there exists a $\rho$-equivariant immersion
${\widetilde{\sigma}}\colon\widetilde{M}\to\mathbb{H}^{n+1}$ whose Gauss map
is $G$.

#####  10.5.2 The Maslov class and the characterization

Let us now introduce the obstruction which will permit us to classify
$\rho$-integrable Lagrangian immersions under the Riemannian assumption,
namely the Maslov class. For this purpose, let
$G:\widetilde{M}\to\mathbb{G}_{n+1}$ be a Riemannian immersion. The _second
fundamental form_ of $G$ is a symmetric bilinear form on $M$ with values in
the normal bundle of $G$, defined as

$\overline{\mathrm{I}}\mathrm{I}(V,W)=({\mathbbm{D}}_{dG(V)}(dG(W)))^{\perp}$

for vector fields $V,W$, where $\mathbbm{D}$ denotes the ambient Levi-Civita
connection of ${\mathbbm{g}}$ and $\perp$ the projection to the normal
subspace of $G$. One can prove that $\overline{\mathrm{I}}\mathrm{I}(V,W)$ is
a tensor, i.e. that it depends on the value of $V$ and $W$ pointwise. The
_mean curvature_ is then

$\overline{\mathrm{H}}=\frac{1}{n}\mathrm{tr}_{\overline{\mathrm{I}}}\overline{\mathrm{I}}\mathrm{I}~{},$

that is, it is the trace of $\overline{\mathrm{I}}\mathrm{I}$ with respect to
the first fundamental form $\overline{\mathrm{I}}$ of $G$, and is therefore a
section of the normal bundle of $G$.

Consider now the 1-form on $\widetilde{M}$ given by
$G^{*}(\Omega(\overline{\mathrm{H}},\cdot))$. It will follow from Proposition
10.5.7 (see Corollary 10.5.9) that this is a closed 1-form. Since
$\mathrm{Isom}(\mathbb{H}^{n+1})$ acts by automorphisms of the para-Kähler
manifold $(\mathbb{G}_{n+1},{\mathbbm{g}},\mathbb{J},\Omega)$, if $G$ is
$\rho$-equivariant, then the form $G^{*}(\Omega(\overline{\mathrm{H}},\cdot))$
is $\pi_{1}(M)$-invariant: as a result, it defines a well-posed closed 1-form
on $M$. Its cohomology class is the so-called Maslov class:

######  Definition 10.5.4.

Given an $n$-manifold $M$, a representation
$\rho\colon\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$ and a
$\rho$-equivariant Lagrangian and Riemannian immersion
$G\colon\widetilde{M}\to\mathbb{G}_{n+1}$, the _Maslov class_ of $G$ is the
cohomology class

$\mu_{G}:=[G^{*}(\Omega(\overline{\mathrm{H}},\cdot))]\in H_{dR}^{1}(M)~{}.$

The main result of this section is the following, and it will be deduced as a
consequence of Theorem 10.5.14.

######  Theorem 10.5.5.

Given an orientable $n$-manifold $M$ and a representation
$\rho\colon\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$, a $\rho$-equivariant
Riemannian and Lagrangian immersion $G\colon\widetilde{M}\to\mathbb{G}_{n+1}$
is $\rho$-integrable if and only if $\mu_{G}=0$ in $H_{dR}^{1}(M)$.

We immediately obtain the following characterization of global integrability
for $\pi_{1}(M)\neq\\{1\\}$.

######  Corollary 10.5.6.

Given an orientable $n$-manifold $M$ and an immersion $G\colon
M\to\mathbb{G}_{n+1}$, $G$ is the Gauss map of an immersion
$\sigma:M\to\mathbb{H}^{n+1}$ of small principal curvatures if and only if $G$
is Riemannian and Lagrangian and $\mu_{G}=0$ in $H_{dR}^{1}(M)$.

###### Proof.

Denote $\rho$ the trivial representation. Given $G:M\to\mathbb{G}_{n+1}$,
precomposing with the universal covering map we obtain an immersion
$\widetilde{G}:\widetilde{M}\to\mathbb{G}_{n+1}$ which is $\rho$-equivariant.
Observe that $\widetilde{G}$ is the Gauss map of some immersion
$\widetilde{\sigma}:\widetilde{M}\to\mathbb{H}^{n+1}$ by Theorem 10.4.2. Then
$G$ is the Gauss map of some immersion in $\mathbb{H}^{n+1}$ if and only if
$\widetilde{\sigma}$ descends to the quotient $M$, i.e. it is
$\rho$-integrable. Hence this is equivalent to the vanishing of the Maslov
class by Theorem 10.5.5. ∎

#####  10.5.3 Mean curvature of Gauss maps

Recall that, given an embedding $\sigma:M\to\mathbb{H}^{n+1}$ with small
principal curvatures, we introduced in (9.9) the function
$\kappa_{\sigma}:M\to{\mathbb{R}}$ which is the mean of the hyperbolic
arctangents of the principal curvatures of $\sigma$. This function is strictly
related to the mean curvature of the Gauss map of $\sigma$, as in the
following proposition.

######  Proposition 10.5.7.

Let $M^{n}$ be an oriented manifold, $\sigma:M\to\mathbb{H}^{n+1}$ an
embedding with small principal curvatures, and
$G_{\sigma}:M\to\mathbb{G}_{n+1}$ its Gauss map. Then

$G_{\sigma}^{*}(\Omega(\overline{\mathrm{H}},\cdot))=d(\kappa_{\sigma})=d\left({\frac{1}{n}}\sum_{i=1}^{n}\mathrm{arctanh}\lambda_{i}\right)~{},$

where $\lambda_{1},\ldots,\lambda_{n}$ denote the principal curvatures of
$\sigma$.

The essential step in the proof of Proposition 10.5.7 is the following
computation for the mean curvature vector of the Gauss map $G_{\sigma}$:

$\overline{\mathrm{H}}=-\mathbbm{J}(dG_{\sigma}(\overline{\nabla}\kappa_{\sigma}))~{},$
(10.6)

where $\overline{\nabla}$ denotes the gradient with respect to the first
fundamental form $\overline{\mathrm{I}}$ of $G_{\sigma}$. Indeed, once
Equation (10.6) is established, Proposition 10.5.7 follows immediately since

$\Omega(\overline{\mathrm{H}},dG_{\sigma}(V))=-{\mathbbm{g}}(\mathbbm{J}(\overline{\mathrm{H}}),dG_{\sigma}(V))={\mathbbm{g}}(dG_{\sigma}(\overline{\nabla}\kappa_{\sigma}),dG_{\sigma}(V))=d\kappa_{\sigma}(V)~{}.$

The computations leading to Equation (10.6) will be done in
$T^{1}\mathbb{H}^{n+1}$ equipped with the metric ${\overline{Sas}_{n+1}^{\
*}}$ defined in Remark 5.2.5, which is the restriction of the flat pseudo-
Riemannian metric (5.19) of ${\mathbb{R}}^{n+1,1}\times{\mathbb{R}}^{n+1,1}$
to $T^{1}\mathbb{H}^{n+1}$, seen as a submanifold as in (5.2). This approach
is actually very useful: the Levi-Civita connection of
${\overline{Sas}_{n+1}^{\ *}}$ on $T^{1}\mathbb{H}^{n+1}$, that we denote by
$\widehat{D}$, will be just the normal projection of the flat connection
$\mathrm{d}$ of ${\mathbb{R}}^{n+1,1}\times{\mathbb{R}}^{n+1,1}$ to
$T^{1}\mathbb{H}^{n+1}$.

Indeed, the following lemma will be useful to compute the Levi-Civita
connection $\mathbbm{D}$ of $\mathbb{G}_{n+1}$. Given a vector $X\in
T_{\ell}\mathbb{G}_{n+1}$ and $(x,v)\in\mathrm{p}^{-1}(\ell)$, we define the
_horizontal lift_ of $X$ at $(x,v)$ as the unique vector $\widetilde{X}\in
T_{(x,v)}T^{1}\mathbb{H}^{n+1}$ such that

$\widetilde{X}\in\chi_{(x,v)}^{\perp}\qquad\text{and}\qquad
d\mathrm{p}(\widetilde{X})=X~{}.$ (10.7)

For a vector field $X$ on an open set $U$ of $\mathbb{G}_{n+1}$, we will also
refer to the vector field $\widetilde{X}$ on $\mathrm{p}^{-1}(U)$, defined by
the conditions (10.7), as the _horizontal lift_ of the vector field $X$.

######  Lemma 10.5.8.

Given two vector fields $X,Y$ on ${\mathbbm{g}}$,

$\mathbbm{D}_{X}Y=d\mathrm{p}(\widehat{D}_{\widetilde{X}}\widetilde{Y})$

###### Proof.

By the well-known characterization of the Levi-Civita connection, it is
sufficient to prove that the expression
$\mathbbm{A}_{X}Y:=d\mathrm{p}(\widehat{D}_{\widetilde{X}}\widetilde{Y})$ is a
well-posed linear connection which is torsion-free and compatible with the
metric of ${\mathbbm{g}}$. We remark that this is not obvious because,
although the metric of ${\mathbbm{g}}$ is the restriction of the metric
${\overline{Sas}_{n+1}^{\ *}}$ to $\chi^{\perp}$, there is no flat section of
the bundle projection $p:T^{1}\mathbb{H}^{n+1}\to\mathbb{G}_{n+1}$, hence
$\mathbb{G}_{n+1}$ cannot be seen as an isometrically embedded submanifold of
$T^{1}\mathbb{H}^{n+1}$.

First, observe that the expression
$(\mathbbm{A}_{X}Y)_{|\ell}=(d_{(x,v)}\mathrm{p})(\widehat{D}_{\widetilde{X}}\widetilde{Y})$
does not depend on the choice of the point $(x,v)\in\mathrm{p}^{-1}(\ell)$.
Indeed, given two points $(x_{1},v_{1})$ and $(x_{2},v_{2})$ in
$\mathrm{p}^{-1}(\ell)$, there exists $t$ such that
$(x_{2},v_{2})=\varphi_{t}(x_{1},v_{1})$. By a small adaptation of Lemma
5.1.3, the geodesic flow $\varphi_{t}$ acts by isometries of the metric
${\overline{Sas}_{n+1}^{\ *}}$ (see also Remarks 5.2.4 and 5.2.5), hence it
also preserves the horizontal lifts $\widetilde{X}$ and $\widetilde{Y}$ and
the Levi-Civita connection $\widehat{D}$. Hence
$d\mathrm{p}(\widehat{D}_{\widetilde{X}}\widetilde{Y})=\mathbbm{A}_{X}Y$ is a
well-defined vector field on $\mathbb{G}_{n+1}$ whose horizontal lift is the
projection of $\widehat{D}_{\widetilde{X}}\widetilde{Y}$ to $\chi^{\bot}$.

We check that $\mathbbm{A}$ is a linear connection. It is immediate to check
the additivity in $X$ and $Y$. Moreover we have the $C^{\infty}$-linearity in
$X$ since:

$\mathbbm{A}_{fX}Y=d\mathrm{p}\left(\widehat{D}_{(f\circ\mathrm{p})\widetilde{X}}\widetilde{Y}\right)=d\mathrm{p}\left((f\circ\mathrm{p})\hat{D}_{\widetilde{X}}\widetilde{Y}\right)=f\cdot(\mathbbm{A}_{X}Y)~{},$

and the Leibnitz rule in $Y$, for:

$\mathbbm{A}_{X}(fY)=d\mathrm{p}\left(\partial_{\widetilde{X}}(f\circ\mathrm{p})\
\widetilde{Y}+(f\circ\mathrm{p})\widehat{D}_{\widetilde{X}}\widetilde{Y}\right)=\partial_{X}f\
Y+f\cdot(\mathbbm{A}_{X}Y)~{}.$

The connection $\mathbbm{A}$ is torsion-free:

$\mathbbm{A}_{X}Y-\mathbbm{A}_{Y}X=d\mathrm{p}(\widehat{D}_{\widetilde{X}}\widetilde{Y})-d\mathrm{p}(\widehat{D}_{\widetilde{Y}}\widetilde{X})=d\mathrm{p}([\widetilde{X},\widetilde{Y}])=[X,Y]~{}.$

Finally, we show that $\mathbbm{A}$ is compatible with the metric
${\mathbbm{g}}$:

$\displaystyle{\mathbbm{g}}(\mathbbm{A}_{X}Y,Z)+{\mathbbm{g}}(Y,\mathbbm{A}_{X}Z)$
$\displaystyle={\overline{Sas}_{n+1}^{\
*}}(\widetilde{\mathbbm{A}_{X}Y},\widetilde{Z})+{\overline{Sas}_{n+1}^{\
*}}(\widetilde{Y},\widetilde{\mathbbm{A}_{X}Z})=$
$\displaystyle={\overline{Sas}_{n+1}^{\
*}}(\widehat{D}_{\widetilde{X}}\widetilde{Y},\widetilde{Z})+{\overline{Sas}_{n+1}^{\
*}}(\widetilde{Y},\widehat{D}_{\widetilde{X}}\widetilde{Z})=$
$\displaystyle=\partial_{\widetilde{X}}({\overline{Sas}_{n+1}^{\
*}}(\widetilde{Y},\widetilde{Z}))=\partial_{X}({\mathbbm{g}}(Y,Z))~{},$

where in the first line we used the definition of ${\mathbbm{g}}$, and in the
second line the fact that the horizontal lift of $\mathbbm{A}_{X}Y$ is the
orthogonal projection (with kernel spanned by $\chi$) of
$\widehat{D}_{\widetilde{X}}\widetilde{Y}$. ∎

We are now ready to provide the proof of Proposition 10.5.7.

###### Proof of Proposition 10.5.7.

As already observed after Equation (10.6), it suffices to prove that
$\overline{\mathrm{H}}=-\mathbbm{J}(dG_{\sigma}(\overline{\nabla}\kappa_{\sigma}))$.
So we shall compute the mean curvature vector of $G_{\sigma}$ in
$\mathbb{G}_{n+1}$. For this purpose, let $\\{e_{1},\ldots,e_{n}\\}$ be a
local frame on $M$ which is orthonormal with respect to the first fundamental
form $\overline{\mathrm{I}}=G_{\sigma}^{*}{\mathbbm{g}}$. To simplify the
notation, let us denote $\epsilon_{i}:=dG_{\sigma}(e_{i})$. Then
$\\{\mathbbm{J}\epsilon_{1},\ldots,\mathbbm{J}\epsilon_{n}\\}$ is an
orthonormal basis for the normal bundle of $G_{\sigma}$, on which the metric
${\mathbbm{g}}$ is negative definite since $G_{\sigma}$ is Riemannian. The
mean curvature vector can be computed as:

$\overline{\mathrm{H}}=\frac{1}{n}\sum_{i=1}^{n}\overline{\mathrm{I}}\mathrm{I}(\epsilon_{i},\epsilon_{i})=-\frac{1}{n}\sum_{i=1}^{n}\sum_{k=1}^{n}{\mathbbm{g}}(\overline{\mathrm{I}}\mathrm{I}(\epsilon_{i},\epsilon_{i}),\mathbbm{J}\epsilon_{k})\mathbbm{J}\epsilon_{k}=-\frac{1}{n}\sum_{i=1}^{n}\sum_{k=1}^{n}{\mathbbm{g}}(\mathbbm{D}_{\epsilon_{i}}\epsilon_{i},\mathbbm{J}\epsilon_{k})\mathbbm{J}\epsilon_{k}~{},$

where in the last equality we used that $\overline{\mathrm{I}}\mathrm{I}(V,W)$
is the normal projection of $\mathbbm{D}_{V}W$.

Let us now apply this expression to a particular
$\overline{\mathrm{I}}$-orthonormal frame $\\{e_{1},\ldots,e_{n}\\}$ obtained
in the following way. Pick a local $\mathrm{I}$-orthogonal frame on $M$ of
eigenvectors for the shape operator $B$ of $\sigma$, and normalize each of
them so as to have unit norm for $\overline{\mathrm{I}}$. Hence each $e_{i}$
is an eigenvector of $B$, whose corresponding eigenvalue $\lambda_{i}$ are the
principal curvatures of $\sigma$. We claim that, with this choice,
${\mathbbm{g}}(\mathbbm{D}_{\epsilon_{i}}\epsilon_{i},\mathbbm{J}\epsilon_{k})=d(\mathrm{arctanh}\lambda_{i})(e_{k})$.
This will conclude the proof, for

$\overline{\mathrm{H}}=\frac{1}{n}\mathbbm{J}\left(\sum_{i=1}^{n}\sum_{k=1}^{n}d(\mathrm{arctanh}\lambda_{i})(e_{k})\epsilon_{k}\right)=\mathbbm{J}\left(\sum_{k=1}^{n}\partial_{e_{k}}\kappa_{\sigma}\
dG_{\sigma}(e_{k})\right)=\mathbbm{J}(dG_{\sigma}(\overline{\nabla}\kappa_{\sigma})))~{}.$

To show the claim, we will first use Lemma 10.5.8 to get

${\mathbbm{g}}(\mathbbm{D}_{\epsilon_{i}}\epsilon_{i},\mathbbm{J}\epsilon_{k})={\overline{Sas}_{n+1}^{\
*}}(\widehat{D}_{\widetilde{\epsilon}_{i}}\widetilde{\epsilon}_{i},J\widetilde{\epsilon}_{k})~{},$

where $\widehat{D}$ is the Levi-Civita connection of ${\overline{Sas}_{n+1}^{\
*}}$ and $\widetilde{\epsilon}_{i}$ is the horizontal lift of $\epsilon_{i}$.
As in Equation (9.2), we can write

$d\zeta_{\sigma}(e_{i})=(d\sigma(e_{i}),-\lambda_{i}d\sigma(e_{i}))$

and the Levi-Civita connection $\widehat{D}$ is the normal projection with
respect to the metric (5.19) of the flat connection $\mathrm{d}$ of
${\mathbb{R}}^{n+1,1}\times{\mathbb{R}}^{n+1,1}$. Hence we can compute:

$\displaystyle{\mathbbm{g}}(\mathbbm{D}_{\epsilon_{i}}\epsilon_{i},\mathbbm{J}\epsilon_{k})$
$\displaystyle=-\lambda_{k}\langle\mathrm{d}_{d\sigma(e_{i})}d\sigma(e_{i}),d\sigma(e_{k})\rangle+\lambda_{i}\langle\mathrm{d}_{d\sigma(e_{i})}d\sigma(e_{i}),d\sigma(e_{k})\rangle+\partial_{e_{i}}\lambda_{i}\langle
d\sigma(e_{i}),d\sigma(e_{k})\rangle$
$\displaystyle=(\lambda_{i}-\lambda_{k})\
\mathrm{I}(\nabla_{e_{i}}e_{i},e_{k})+(\partial_{e_{i}}\lambda_{i})\
\mathrm{I}(e_{i},e_{k})~{}.$

We recall that $g$ denotes the first fundamental form of $\sigma$, and
$\nabla$ its Levi-Civita connection, and in the last equality we used that the
Levi-Civita connection of $\mathbb{H}^{n+1}$ is the projection to the
hyperboloid in Minkowski space ${\mathbb{R}}^{n+1,1}$ of the ambient flat
connection.

Now, when $i=k$ we obtained the desired result:

${\mathbbm{g}}(\mathbbm{D}_{\epsilon_{i}}\epsilon_{i},\mathbbm{J}\epsilon_{i})=\frac{\partial_{e_{i}}\lambda_{i}}{1-\lambda_{i}^{2}}=d(\mathrm{arctanh}\lambda_{i})(e_{i})$

since $e_{i}$ is a unit vector for the metric $\overline{\mathrm{I}}$, hence
using the expression
$\overline{\mathrm{I}}=\mathrm{I}-\mathrm{I}\mathrm{I}\mathrm{I}$ from
Equation (9.4) its squared norm for the metric $\mathrm{I}$ is
$(1-\lambda_{i}^{2})^{-1}$. When $i\neq k$, the latter term disappears since
$\\{e_{1},\ldots,e_{n}\\}$ is an orthogonal frame for $g$, and we are thus
left with showing that

$(\lambda_{i}-\lambda_{k})\
\mathrm{I}(\nabla_{e_{i}}e_{i},e_{k})=d(\mathrm{arctanh}\lambda_{i})(e_{k})~{}.$

For this purpose, using the compatibility of $\nabla$ with the metric, namely
$\partial_{e_{i}}(\mathrm{I}(e_{i},e_{k}))=\mathrm{I}(\nabla_{e_{i}}e_{i},e_{k})+\mathrm{I}(e_{i},\nabla_{e_{i}}e_{k})$,
that $\mathrm{I}(e_{i},e_{k})=0$, and that $\nabla$ is torsion-free, we get:

$(\lambda_{i}-\lambda_{k})\
\mathrm{I}(\nabla_{e_{i}}e_{i},e_{k})=-(\lambda_{i}-\lambda_{k})\
\mathrm{I}(e_{i},\nabla_{e_{i}}e_{k})=\lambda_{k}\
\mathrm{I}(e_{i},\nabla_{e_{i}}e_{k})-\lambda_{i}\
\mathrm{I}(e_{i},\nabla_{e_{k}}e_{i})-\lambda_{i}\
\mathrm{I}(e_{i},[e_{i},e_{k}])~{}.$

Now, recall that the Codazzi equation for $\sigma$ is $d^{\nabla}B=0$.
Applying it to the vector fields $e_{i}$ and $e_{k}$, we obtain

$d^{\nabla}B(e_{i},e_{k})=\nabla_{e_{i}}(\lambda_{k}e_{k})-\nabla_{e_{k}}(\lambda_{i}e_{i})-B([e_{i},e_{k}])=0,$

from which we derive

$\lambda_{k}\nabla_{e_{i}}e_{k}-\lambda_{i}\nabla_{e_{k}}e_{i}=(\partial_{e_{k}}\lambda_{i})e_{i}-(\partial_{e_{i}}\lambda_{k})e_{k}+B([e_{i},e_{k}])~{}.$
(10.8)

Using Equation (10.8) in the previous expression, we finally obtain:

$\displaystyle(\lambda_{i}-\lambda_{k})\
\mathrm{I}(\nabla_{e_{i}}e_{i},e_{k})=$
$\displaystyle(\partial_{e_{k}}\lambda_{i})\
\mathrm{I}(e_{i},e_{i})-(\partial_{e_{i}}\lambda_{k})\
\mathrm{I}(e_{i},e_{k})+\mathrm{I}(e_{i},B[e_{i},e_{k}])-\mathrm{I}(B(e_{i}),[e_{i},e_{k}])$
$\displaystyle=$
$\displaystyle\frac{\partial_{e_{k}}\lambda_{i}}{1-\lambda_{i}^{2}}=d(\mathrm{arctanh}\lambda_{i})(e_{k})$

where the cancellations from the first to the second line are due to the fact
that $B$ is $\mathrm{I}$-self adjoint and that $\mathrm{I}(e_{i},e_{k})=0$.
This concludes the proof. ∎

######  Corollary 10.5.9.

Given an $n$-manifold $M$, a representation
$\rho\colon\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$ and a
$\rho$-equivariant Lagrangian and Riemannian immersion
$G\colon\widetilde{M}\to\mathbb{G}_{n+1}$, the Maslov class $\mu_{G}$ is a
well-defined cohomology class in $H^{1}_{dR}(M,{\mathbb{R}})$.

###### Proof.

By Theorem 10.4.2, $G$ is the Gauss map of a (in general non equivariant)
immersion $\sigma:\widetilde{M}\to\mathbb{H}^{n+1}$. By Proposition 10.5.7,
the 1-form $G^{*}(\Omega(\overline{\mathrm{H}},\cdot))$ on $\widetilde{M}$ is
exact, and $\rho$-equivariant, hence it induces a closed 1-form on $M$ whose
cohomology class is $\mu_{G}$ as in Definition 10.5.4. ∎

#####  10.5.4 Holonomy of the pull-back bundle and proof of Theorem 10.5.5

An immediate consequence of Proposition 10.5.7 is that the vanishing of the
Maslov class is a necessary condition for a $\rho$-equivariant Lagrangian and
Riemannian embedding $G:\widetilde{M}\to\mathbb{G}_{n+1}$ to be
$\rho$-integrable (Definition 10.5.3). Indeed, if
$\widetilde{\sigma}:\widetilde{M}\to\mathbb{H}^{n+1}$ is a $\rho$-equivariant
embedding with $G_{\widetilde{\sigma}}=G$ (hence necessarily with small
principal curvatures), then the function $f_{\widetilde{\sigma}}$ descends to
a well-defined function on $M$, hence by Proposition 10.5.7
$G^{*}(\Omega(\overline{\mathrm{H}},\cdot))$ is an exact 1-form, i.e. the
Maslov class $\mu_{G_{\widetilde{\sigma}}}$ vanishes. We will now see that
this condition is also sufficient, which will be a consequence of a more
general result, Theorem 10.5.14.

Let $G:\widetilde{M}\to\mathbb{G}_{n+1}$ be a $\rho$-equivariant Lagrangian
and Riemannian embedding. We have already used that the $G$-pull-back bundle
$\widetilde{\mathrm{p}}_{G}\colon\widetilde{P}\to\widetilde{M}$ of
$\mathrm{p}:T^{1}\mathbb{H}^{n+1}\to\mathbb{G}_{n+1}$ is a flat trivial
${\mathbb{R}}$-principal bundle over $\widetilde{M}$, namely, it is
isomorphic, as a flat principal bundle, to the trivial bundle
$\widetilde{M}\times{\mathbb{R}}\to\widetilde{M}$ with flat sections
$\widetilde{M}\times\\{\ast\\}$. Moreover, $G$ being $\rho$-equivariant, the
fundamental group $\pi_{1}(M)$ acts freely and properly discontinously on
$\widetilde{P}$, thus inducing a flat ${\mathbb{R}}$-principal bundle
structure $\mathrm{p}_{G}\colon P\to M$, where $P$ is the quotient of
$\widetilde{P}$ by the action of $\pi_{1}(M)$. However the bundle
$\mathrm{p}_{G}\colon P\to M$ is not trivial in general. The obstruction to
triviality is represented by the _holonomy_ of the bundle, which can be
defined, in our setting, as follows.

######  Definition 10.5.10.

Let $P\to M$ be a flat principal ${\mathbb{R}}$-bundle that is isomorphic to
the quotient of the trivial bundle
$\widetilde{M}\times{\mathbb{R}}\to\widetilde{M}$ by an equivariant (left)
action of $\pi_{1}(M)$. The _holonomy representation_ is the representation
$\mathrm{hol}:\pi_{1}(M)\to{\mathbb{R}}$ such that the action of $\pi_{1}(M)$
is expressed by:

$\alpha\cdot(m,s)=(\alpha\cdot m,\mathrm{hol}(\alpha)+s)~{}.$

######  Remark 10.5.11.

Fix $p\in M$ and $\alpha$ a closed $C^{1}$ loop based at $p$. Then pick a
horizontal lift $\widehat{\alpha}$ to the total space of $\mathrm{p}_{G}$,
namely with $\frac{d\widehat{\alpha}}{dt}$ orthogonal to the fibers, so that
$\mathrm{p}\circ\widehat{\alpha}=\alpha$. (The lift is uniquely determined by
its initial point in $\mathrm{p}_{G}^{-1}(p)$.) It follows from Definition
10.5.10 that

$\mathrm{hol}_{G}(\alpha)\cdot\widehat{\alpha}(1)=\widehat{\alpha}(0).$

In the identification $\pi_{1}(M)=\pi_{1}(M,[p])$, this allows to give an
alternative definition of $\mathrm{hol}_{G}$ through homotopy classes of
closed paths in $M$.

######  Remark 10.5.12.

We remark that in general, for flat principal $G$-bundles, the holonomy
representation is only defined up to conjugacy, but in out case
$G={\mathbb{R}}$ is abelian and therefore $\mathrm{hol}$ is uniquely
determined by the isomorphism class of the flat principal bundle.

Also observe that, since ${\mathbb{R}}$ is abelian, $\mathrm{hol}_{G}$ induces
a map from $H_{1}(M,{\mathds{Z}})$ to ${\mathbb{R}}$, where
$H_{1}(M,{\mathds{Z}})$ is the first homology group of $M$ and we are using
that there is a canonical isomorphism between $H_{1}(M,{\mathds{Z}})$ and the
abelianization of the fundamental group of $M$ in a point. Equivalently,
$\mathrm{hol}_{G}$ is identified to an element of $H^{1}(M,{\mathbb{R}})$.

We can interpret the holonomy of the principal bundle $\mathrm{p}_{G}$ in
terms of the geometry of $\mathbb{H}^{n+1}$. Global flat sections of the
trivial bundle $\widetilde{\mathrm{p}}_{G}\colon\widetilde{P}\to\widetilde{M}$
correspond to Riemannian embeddings $\zeta:\widetilde{M}\to
T^{1}\mathbb{H}^{n+1}$ as in Corollaries 10.3.1 and 10.3.2. By Theorem 10.4.2,
such a $\zeta$ is the lift to $T^{1}\mathbb{H}^{n+1}$ of an embedding
$\sigma:\widetilde{M}\to\mathbb{H}^{n+1}$ with small principal curvatures.

Now, let $\alpha\in\pi_{1}(M)$. By equivariance of $G$, namely
$G\circ\alpha=\rho(\alpha)\circ G$, it follows that
$\mathrm{p}\circ\zeta\circ\alpha=\mathrm{p}\circ\rho(\alpha)\circ\zeta$, hence
$\rho(\alpha)\circ\zeta\circ\alpha^{-1}:\widetilde{M}\to
T^{1}\mathbb{H}^{n+1}$ provides another flat section of the pull-back bundle
$\widetilde{\mathrm{p}}_{G}$. Therefore there exists
$t_{\alpha}\in{\mathbb{R}}$ such that

$\varphi_{t_{\alpha}}\circ\zeta=\rho(\alpha)\circ\zeta\circ\alpha^{-1}~{}.$
(10.9)

Then the value $t_{\alpha}$ is precisely the holonomy of the quotient bundle
$\mathrm{p}_{G}$, namely the group representation

$\mathrm{hol}_{G}\colon\pi_{1}(M)\to{\mathbb{R}}\qquad\mathrm{hol}_{G}(\alpha)=t_{\alpha}$

A direct consequence of this discussion is the following:

######  Lemma 10.5.13.

Given an $n$-manifold $M$ and a representation
$\rho\colon\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$, a $\rho$-equivariant
Lagrangian and Riemannian embedding $G\colon\widetilde{M}\to\mathbb{G}_{n+1}$
is $\rho$-integrable if and only if the ${\mathbb{R}}$-principal flat bundle
$\mathrm{p}_{G}$ is trivial.

###### Proof.

The bundle $\mathrm{p}_{G}$ is trivial if and only if its holonomy
$\mathrm{hol}_{G}$ vanishes identically, that is, if and only if
$t_{\alpha}=0$ for every $\alpha\in\pi_{1}(M)$. By the above construction,
this is equivalent to the condition that
$\zeta\circ\alpha=\rho(\alpha)\circ\zeta$ for all $\alpha$, which is
equivalent to $\sigma\circ\alpha=\rho(\alpha)\circ\sigma$, namely that
$\sigma$ is $\rho$-equivariant. ∎

We are ready to prove the following.

######  Theorem 10.5.14.

Given an $n$-manifold $M$, a representation
$\rho\colon\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$ and a
$\rho$-equivariant Lagrangian and Riemannian embedding
$G\colon\widetilde{M}\to\mathbb{G}_{n+1}$, the holonomy of $\mathrm{p}_{G}$ is
given by

$\mathrm{hol}_{G}(\alpha)=\int_{\alpha}\mu_{G}.$

for all $\alpha\in\pi_{1}(M)$.

Observe that Theorem 10.5.5 follows immediately from Theorem 10.5.14 since, by
the standard de Rham Theorem, there exists an isomorphism

$\begin{split}H_{dR}^{1}(M,{\mathbb{R}})&\xrightarrow{\sim}H^{1}(M,{\mathbb{R}})\\\
\eta&\mapsto\bigg{(}\xi\mapsto\int_{\xi}\eta\bigg{)},\end{split}$

hence $\mathrm{hol}_{G}\equiv 0$ if and only if $\mu_{G}=0$.

###### Proof of Theorem 10.5.14.

Let $\zeta:\widetilde{M}\to T^{1}\mathbb{H}^{n+1}$ be a map such that
$\mathrm{p}\circ\zeta=G$, so as to induce a global section of the pull-back
bundle $\widetilde{\mathrm{p}}_{G}$. Then by Equation (10.9) the holonomy
$t_{\alpha}=\mathrm{hol}_{G}(\alpha)$ satisfies
$\varphi_{t_{\alpha}}\circ\zeta\circ\alpha=\rho(\alpha)\circ\zeta$. By
Proposition 9.1.5, this gives the following equivariance relation for
$\sigma=\pi\circ\zeta$:

$(\sigma\circ\alpha)_{t_{\alpha}}=\rho(\alpha)\circ\sigma~{}.$

Let now $\kappa_{\sigma}$ denote the mean of the hyperbolic arctangents of the
principal curvatures, as in Equation (9.9). Lemma 9.3.5 and the fact that
$\rho(\alpha)$ acts isometrically imply:

$f_{\sigma\circ\alpha}=\kappa_{\sigma}+t_{\alpha}~{}.$

Now, by Proposition 10.5.7 and the definition of the Maslov class, we have:

$\int_{\alpha}\mu_{G}=\int_{\alpha}d\kappa_{\sigma}=\kappa_{\sigma}(\alpha(p))-\kappa_{\sigma}(p)=t_{\alpha}$

for any point $p\in M$. This concludes the proof. ∎

####  10.6 Minimal Lagrangian immersions

We prove here two direct corollaries of Theorem 10.5.5. Let us first recall
the definition of minimal Lagrangian (Riemannian) immersions.

######  Definition 10.6.1.

A Riemannian immersion of an $n$-manifold into $\mathbb{G}_{n+1}$ is _minimal
Lagrangian_ if:

  * •

its mean curvature vector vanishes identically;

  * •

it is Lagrangian with respect to the symplectic form $\Omega$.

Our first corollary is essentially a consequence of Theorem 10.5.5.

######  Corollary 10.6.2.

Let $M^{n}$ be a closed orientable manifold and
$\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$ a representation. If
$G:\widetilde{M}\to\mathbb{G}_{n+1}$ is a $\rho$-equivariant Riemannian
minimal Lagrangian immersion, then $G$ is the Gauss map of a
$\rho$-equivariant embedding
$\widetilde{\sigma}:\widetilde{M}\to\mathbb{H}^{n+1}$ with small principal
curvatures such that

$\kappa_{\widetilde{\sigma}}={\frac{1}{n}}\sum_{i=1}^{n}\mathrm{arctanh}\lambda_{i}=0~{}.$

In particular, $\rho$ is a nearly-Fuchsian representation and $G$ is an
embedding.

We remark that if $n=2$, then the condition $\kappa_{\widetilde{\sigma}}=0$ is
equivalent to $\lambda_{1}+\lambda_{2}=0$ since $\mathrm{arctanh}$ is an odd
and injective function. That is, in this case $\widetilde{\sigma}$ is a
_minimal_ embedding in $\mathbb{H}^{3}$.

###### Proof.

Suppose $G$ is a $\rho$-equivariant minimal Lagrangian immersion. Since its
mean curvature vector vanishes identically, we have $\mu_{G}=0$ and therefore
$G$ is $\rho$-integrable by Theorem 10.5.5. That is, there exists a
$\rho$-equivariant immersion
$\widetilde{\sigma}:\widetilde{M}\to\mathbb{H}^{n+1}$ such that
$G=G_{\widetilde{\sigma}}$. By Proposition 9.3.2, $\widetilde{\sigma}$ has
small principal curvatures, hence $\rho$ is nearly-Fuchsian. By Proposition
10.5.7, we have that $\kappa_{\widetilde{\sigma}}$ is constant. By Lemma
9.3.5, up to taking the normal evolution, we can find $\widetilde{\sigma}$
such that $\kappa_{\widetilde{\sigma}}$ vanishes identically.

Finally, $\widetilde{\sigma}$ is complete by cocompactness, and therefore both
$\widetilde{\sigma}$ and $G$ are embeddings by Proposition 9.3.15 and
Proposition 9.3.16. ∎

The following is a uniqueness result for $\rho$-equivariant minimal Lagrangian
immersions.

######  Corollary 10.6.3.

Given a closed orientable manifold $M^{n}$ and a representation
$\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$, there exists at most one
$\rho$-equivariant Riemannian minimal Lagrangian immersion
$G:\widetilde{M}\to\mathbb{G}_{n+1}$ up to reparametrization. If such a $G$
exists, then $\rho$ is nearly-Fuchsian and $G$ induces a minimal Lagrangian
embedding of $M$ in ${\mathbb{G}}_{\rho}$.

###### Proof.

Suppose that $G$ and $G^{\prime}$ are $\rho$-equivariant minimal Lagrangian
immersions. By Corollary 10.6.2, there exist $\rho$-equivariant embeddings
$\widetilde{\sigma},\widetilde{\sigma}^{\prime}:\widetilde{M}\to\mathbb{H}^{n+1}$
with small principal curvatures such that $G=G_{\widetilde{\sigma}}$ and
$G^{\prime}=G_{\widetilde{\sigma}^{\prime}}$, with
$\kappa{\widetilde{\sigma}}=\kappa{\widetilde{\sigma}^{\prime}}=0$. Moreover,
$G$ and $G^{\prime}$ induce embeddings in ${\mathbb{G}}_{\rho}$ by Corollary
9.4.8.

By Remark 9.4.6, both $\sigma$ and $\sigma^{\prime}$ induce embeddings of $M$
in the nearly-Fuchsian manifold $\faktor{\mathbb{H}^{n+1}}{\rho(\pi_{1}(M))}$;
let us denote with $\Sigma$ and $\Sigma^{\prime}$ the corresponding images. We
claim that $\Sigma=\Sigma^{\prime}$, which implies the uniqueness in the
statement.

To see this, consider the signed distance from $\Sigma$, which is a proper
function

$r:\faktor{\mathbb{H}^{n+1}}{\rho(\pi_{1}(M))}\to{\mathbb{R}}~{}.$

Since $\Sigma^{\prime}$ is closed, $r|_{\Sigma^{\prime}}$ admits a maximum
value $r_{\max}$ achieved at some point $x_{\max}\in\Sigma^{\prime}$. This
means that at the point $x_{\max}$, $\Sigma^{\prime}$ is tangent to a
hypersurface $\Sigma_{r_{\max}}$ at signed distance $r_{\max}$ from $\Sigma$,
and $\Sigma^{\prime}$ is contained in the side of $\Sigma_{r_{\max}}$ where
$r$ is decreasing. This implies that, if $B^{\prime}$ denotes the shape
operator of $\Sigma^{\prime}$ and $B_{r_{\max}}$ that of $\Sigma_{r_{\max}}$,
both computed with respect to the unit normal vector pointing to the side of
increasing $r$, then $B_{r_{\max}}-B^{\prime}$ is positive semi-definite at
$x_{\max}$.

Let us now denote by $\lambda_{1},\ldots,\lambda_{n}$ the eigenvalues of
$B_{r_{\max}}$ and $\lambda^{\prime}_{1},\ldots,\lambda_{n}^{\prime}$ those of
$B^{\prime}$. Let us moreover assume that
$\lambda_{1}\leq\ldots\leq\lambda_{n}$ and similarly for the
$\lambda_{i}^{\prime}$. By Weyl’s monotonicity theorem,
$\lambda_{i}\geq\lambda_{i}^{\prime}$ at $x_{\max}$ for $i=1,\ldots,n$. Since
$\mathrm{arctanh}$ is a monotone increasing function, this implies that

$\sum_{i=1}^{n}\mathrm{arctanh}\lambda_{i}(x_{\max})\geq\sum_{i=1}^{n}\mathrm{arctanh}\lambda_{i}^{\prime}(x_{\max})~{}.$

Since $\kappa{\widetilde{\sigma}^{\prime}}=0$, the right-hand side vanishes.
On the other hand, since $\kappa{\widetilde{\sigma}}=0$, from Lemma 9.3.5 the
left-hand side is identically equal to $-r_{\max}$. Hence $r_{\max}\leq 0$.
Repeating the same argument replacing the maximum point of $r$ on
$\Sigma^{\prime}$ by the minimum point, one shows $r_{\min}\geq 0$. Hence
$r|_{\Sigma^{\prime}}$ vanishes identically, which proves that
$\Sigma=\Sigma^{\prime}$ and thus concludes the proof. ∎

###  Chapter 11 Hamiltonian symplectomorphisms of $\mathbb{G}_{\rho}$ and
deformations of immersions

In this chapter we will provide the second characterization of
$\rho$-integrability, in the case of a nearly-Fuchsian representation
$\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$. We first introduce the
terminology and state the result (Theorem 11.1.4); then we introduce the so-
called _Lagrangian Flux map_ which will play a central role in the proof of
Theorem 11.1.4.

####  11.1 Hamiltonian group and nearly-Fuchsian manifolds

We will restrict hereafter to the case of nearly-Fuchsian representations
$\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$. Let
$G:\widetilde{M}\to\mathbb{G}_{n+1}$ be a $\rho$-integrable immersion as in
Definition 10.5.3. Since $\rho$ is nearly-Fuchsian, we showed in Corollary
9.4.8 that $G$ induces an embedded submanifold in the para-Kähler manifold
${\mathbb{G}}_{\rho}$, defined in Definition 9.4.7. This motivates the
following definition in the spirit of Definition 10.5.3.

######  Definition 11.1.1.

Given a closed orientable $n$-manifold $M$ and a nearly-Fuchsian
representation $\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$, an
embedding $M\to{\mathbb{G}}_{\rho}$ is $\rho$-_integrable_ if it is induced in
the quotient from a $\rho$-integrable embedding
${G}\colon\widetilde{M}\to\mathbb{G}_{n+1}$. Similarly, an embedded
submanifold $\mathcal{L}\subset{\mathbb{G}}_{\rho}$ is $\rho$-_integrable_ if
it is the image of a $\rho$-integrable embedding.

Theorem 11.1.4 below gives a description of the set of $\rho$-integrable
submanifolds $\mathcal{L}\subset{\mathbb{G}}_{\rho}$ which are induced by
immersions $G$ with small principal curvatures. Clearly, as we have previously
shown, a necessary condition on $\mathcal{L}$ is that of being Lagrangian and
Riemannian. To state the theorem, we need to recall the notion of Hamiltonian
symplectomorphism.

######  Definition 11.1.2.

Given a symplectic manifold $(\mathcal{X},\Omega)$, a compactly supported
symplectomorphism $\Phi$ is _Hamiltonian_ if there exists a compactly
supported smooth function $F_{\bullet}:\mathcal{X}\times[0,1]\to{\mathbb{R}}$
such that $\Phi=\Phi_{1}$, where $\Phi_{s_{0}}$ is the flow at time $s_{0}$ of
the (time-dependent) vector field $X_{s}$ defined by:

$dF_{s}=\Omega(X_{s},\cdot)~{}.$ (11.1)

The isotopy $\Phi_{\bullet}:\mathcal{X}\times[0,1]\to\mathcal{X}$ is called
_Hamiltonian isotopy_.

######  Remark 11.1.3.

If $\Phi_{\bullet}$ is a Hamiltonian isotopy as in Definition 11.1.2, then
$\Phi_{s}$ is a symplectomorphism for every $s\in[0,1]$. Indeed

$\mathcal{L}_{X_{s}}\Omega=\iota_{X_{s}}d\Omega+d(\iota_{X_{s}}\Omega)=0$

as a consequence of Cartan’s formula and Equation (11.1), and $\Phi_{s}$ is
clearly Hamiltonian.

Compactly supported Hamiltonian symplectomorphisms form a group which we will
denote by $\mathrm{Ham}_{c}(\mathcal{X},\Omega)$.

The aim of this chapter is to prove the following result.

######  Theorem 11.1.4.

Let $M$ be a closed orientable $n$-manifold,
$\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$ be a nearly-Fuchsian
representation and $\mathcal{L}\subset{\mathbb{G}}_{\rho}$ a Riemannian
$\rho$-integrable submanifold. Then a Riemannian submanifold
$\mathcal{L}^{\prime}$ is $\rho$-integrable if and only if there exists
$\Phi\in\mathrm{Ham}_{c}({\mathbb{G}}_{\rho},\Omega)$ such that
$\Phi(\mathcal{L})=\mathcal{L}^{\prime}$.

Of course, although not stated in Theorem 11.1.4, both $\mathcal{L}$ and
$\mathcal{L}^{\prime}$ are necessarily Lagrangian as a consequence of
Corollary 10.3.1.

####  11.2 The Lagrangian Flux

We shall now define the Flux map for Lagrangian submanifolds, which was
introduced in [Sol13], and relate it to the holonomy of
${\mathbb{R}}$-principal bundles.

######  Definition 11.2.1.

Let $(\mathcal{X},\Omega)$ be a symplectic manifold and let
$\Upsilon_{\bullet}:M\times[0,1]\to\mathcal{X}$ be a smooth map such that each
$\Upsilon_{t}$ is a Lagrangian embedding of $M$. Then we define:

$\mathrm{Flux}(\Upsilon_{\bullet})=\int_{0}^{1}\Upsilon_{s}^{*}(\Omega(X_{s},\cdot))ds\in
H^{1}_{dR}(M,{\mathbb{R}})~{},$

where

$X_{s_{0}}(\Upsilon_{s_{0}}(p))=\left.\frac{d}{ds}\right|_{s=s_{0}}\Upsilon_{s}(p)\in
T_{\Upsilon_{s_{0}}(p)}\mathcal{X}~{}.$

Observe that by Cartan’s formula the integrand
$\Upsilon_{s}^{*}(\Omega(X_{s},\cdot))$ is a closed 1-form for every $s$,
hence $\mathrm{Flux}(\Upsilon_{\bullet})$ is well-defined as a cohomology
class in $H^{1}_{dR}(M,{\mathbb{R}})$.

Now, let $\mathcal{L}$ be a Lagrangian embedded submanifold in
${\mathbb{G}}_{\rho}$, which is induced by a $\rho$-equivariant immersion
$G:\widetilde{M}\to\mathbb{G}_{n+1}$. Recall that in Subsection 10.5.4 we
defined the principal ${\mathbb{R}}$-bundle $\mathrm{p}_{G}$ as the quotient
of $\widetilde{\mathrm{p}}_{G}=G^{*}\mathrm{p}$ by the action of $\pi_{1}(M)$.
Moreover in Theorem 10.5.14 we computed the holonomy

$\mathrm{hol}_{G}:\pi_{1}(M)\to{\mathbb{R}}$

of $\mathrm{p}_{G}$. The key relation between Lagrangian flux and
$\mathrm{hol}_{G}$ is stated in the following proposition.

######  Proposition 11.2.2.

Let $M$ be a closed orientable $n$-manifold and
$\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$ be a nearly-Fuchsian
representation. If $\Upsilon_{s}$ is as in Definition 11.2.1 and
$\Upsilon_{0}(M)=\mathcal{L}$, $\Upsilon_{1}(M)=\mathcal{L}^{\prime}$, then

$\mathrm{hol}_{\Upsilon_{1}}(\alpha)-\mathrm{hol}_{\Upsilon_{0}}(\alpha)=\int_{\alpha}\mathrm{Flux}(\Upsilon_{\bullet})~{}.$

In particular, $\mathrm{Flux}(\Upsilon_{\bullet})(\alpha)$ depends uniquely on
the endpoints of $\Upsilon_{\bullet}$.

To prove Proposition 11.2.2, we will make use of the following expression for
the holonomy representation.

######  Proposition 11.2.3.

Let $G\colon\widetilde{M}\to\mathbb{G}_{n+1}$ be a $\rho$-equivariant
Lagrangian embedding and $\mathrm{p}_{G}$ be the associated
${\mathbb{R}}$-principal bundle over $M$. If $\alpha:[0,1]\to M$ is a smooth
loop and $\overline{\alpha}$ a smooth loop in the total space of
$\mathrm{p}_{G}$ such that $\alpha=\mathrm{p}_{G}(\overline{\alpha})$, then

$\mathrm{hol}_{G}(\alpha)=\int_{\overline{\alpha}}\omega$

where $\omega$ is the principal connection of $\mathrm{p}_{G}$.

###### Proof.

Say $\alpha(0)=\alpha(1)=x_{0}$. Recalling Remark 10.5.11, let
$\widehat{\alpha}$ be the horizontal lift of $\alpha$ starting at
$\overline{\alpha}(0)$. We apply Stokes’ theorem. Define a smooth map $f$ from
$[0,1]\times[0,1]$ to the total space of $\mathrm{p}_{G}$ so that

  * •

$f(x,0)=\overline{\alpha}(x)$,

  * •

$f(x,1)=\widehat{\alpha}(x)$,

  * •

$f(0,y)\equiv\overline{\alpha}(0)=\overline{\alpha}(1)$,

  * •

$y\mapsto f(1,y)$ parametrizes the interval from $\overline{\alpha}(1)$ to
$\widehat{\alpha}(1)$ in $\mathrm{p}_{G}^{-1}(x_{0})\approx{\mathbb{R}}$.

By Stokes’ Theorem and the flatness of $\mathrm{p}_{G}$, one gets that

$\displaystyle 0=\int_{[0,1]\times[0,1]}f^{*}d\omega$
$\displaystyle=\int_{\overline{\alpha}}\omega+\int_{f(1,\cdot)}\omega-\int_{\widehat{\alpha}}\omega-\int_{f(0,\cdot)}\omega$
$\displaystyle=\int_{\overline{\alpha}}\omega+\int_{f(1,\cdot)}\omega.$

By Remark 10.5.11,
$\widehat{\alpha}(1)=(-\mathrm{hol}_{G}(\alpha))\cdot\overline{\alpha}(1)$.
Since $\omega={\overline{Sas}_{n+1}}(\chi,\cdot)$ and $f(1,\cdot)$ is
contained in $\mathrm{p}_{G}^{-1}(x_{0})$, one gets that

$\int_{f(1,\cdot)}\omega=-\mathrm{hol}_{G}(\alpha)$

and the proof follows. ∎

###### Proof of Proposition 11.2.2.

Define $\Theta:[0,1]\times S^{1}\to M$ by
$\Theta(s,t)=\Upsilon_{s}(\alpha(t))$. Since the bundle $\mathrm{p}_{G}$ has
contractible fibre, there always exists a smooth global section. In
particular, there exists $\overline{\Theta}$ such that
$\Theta=\mathrm{p}_{G}\circ\overline{\Theta}$. By Proposition 11.2.3,
recalling that $d\omega=\mathrm{p}^{*}\Omega$, and applying Stokes’ Theorem,
we obtain:

$\mathrm{hol}_{\Upsilon_{1}}(\alpha)-\mathrm{hol}_{\Upsilon_{0}}(\alpha)=\int_{\overline{\Theta}(1,\cdot)}\omega-\int_{\overline{\Theta}(0,\cdot)}\omega=\int_{[0,1]\times
S^{1}}\overline{\Theta}^{*}d\omega=\int_{[0,1]\times S^{1}}\Theta^{*}\Omega$

and the last term equals $\int_{\alpha}\mathrm{Flux}(\Upsilon_{\bullet})$. ∎

We conclude this section by proving one (easy) implication of Theorem 11.1.4.
As mentioned in the introduction, this implication does not need the
hypothesis that $\mathcal{L}$ and $\mathcal{L}^{\prime}$ are Riemannian.

###### Proof of the “if” part of Theorem 11.1.4.

Suppose there exists a Hamiltonian symplectomorphism $\Phi=\Phi_{1}$, endpoint
of a Hamiltonian isotopy $\Phi_{\bullet}$, such that
$\Phi(\mathcal{L})=\mathcal{L}^{\prime}$. Then define the map
$\Upsilon_{\bullet}:M\times[0,1]\to{\mathbb{G}}_{\rho}$ in such a way that
$\Upsilon_{0}:M\to\mathcal{G}$ is an embedding with image $\mathcal{L}$ and

$\Upsilon_{s}=\Phi_{s}\circ\Upsilon_{0}~{}.$

By Remark 11.1.3, $\Phi_{s}$ is a (Hamiltonian) symplectomorphism for every
$s\in[0,1]$, hence $\Upsilon_{s}$ is a Lagrangian embedding for all $s$. We
claim that $\mathrm{Flux}(\Upsilon_{\bullet})$ vanishes in
$H^{1}_{dR}(M,{\mathbb{R}})$. Indeed, for every $s$ we have

$\Upsilon_{s}^{*}(\Omega(X_{s},\cdot))=\Upsilon_{s}^{*}dF_{s}=df_{s}$

by Equation (11.1), where $X_{s}$ is the vector field generating the
Hamiltonian isotopy (and hence $\Upsilon_{\bullet}$) and
$f_{s}=F_{s}\circ\Upsilon_{s}$. Therefore
$\int_{0}^{1}\Upsilon_{s}^{*}(\Omega(X_{s},\cdot))ds$ is exact, namely
$\mathrm{Flux}(\Upsilon_{\bullet})=0$.

Using Proposition 11.2.2, we have
$\mathrm{hol}_{\Upsilon_{0}}=\mathrm{hol}_{\Upsilon_{1}}$. By Lemma 10.5.13,
this shows that $\mathcal{L}$ is $\rho$-integrable if and only if
$\mathcal{L}^{\prime}$ is $\rho$-integrable, and this concludes the proof of
the first implication in Theorem 11.1.4. ∎

####  11.3 Conclusion of Theorem 11.1.4

We are left with the other implication in Theorem 11.1.4. Given two Riemannian
$\rho$-integrable submanifolds
$\mathcal{L},\mathcal{L}^{\prime}\subset{\mathbb{G}}_{\rho}$, we shall produce
$\Phi\in\mathrm{Ham}_{c}({\mathbb{G}}_{\rho},\Omega)$ mapping $\mathcal{L}$ to
$\mathcal{L}^{\prime}$. We remark here that the results and methods of [Sol13]
use stronger topological hypothesis, hence do not apply under our assumptions.

Roughly speaking, the idea is to reduce the problem to finding a deformation
in the nearly-Fuchsian manifold $\faktor{\mathbb{H}^{n+1}}{\rho(\pi_{1}(M))}$
which interpolates between two hypersurfaces of small principal curvatures
corresponding to $\mathcal{L}$ to $\mathcal{L}^{\prime}$. For technical
reasons, it will be easier to deal with convex hypersurfaces that we defined
in Definition 9.3.8

######  Lemma 11.3.1.

Let $M^{n}$ be a closed oriented manifold,
$\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$ be a nearly-Fuchsian
representation and $\widetilde{\sigma}:\widetilde{M}\to\mathbb{H}^{n+1}$ be a
$\rho$-equivariant embedding. If $\widetilde{\sigma}$ is convex, then the
Gauss map $G_{\widetilde{\sigma}}^{+}$ is an equivariant diffeomorphism
between $\widetilde{M}$ and the connected component $\Omega_{+}$ of
$\partial\mathbb{H}^{n+1}\setminus\Lambda_{\rho}$.

###### Proof.

By the same argument as in Section 9.4 (see the discussion between Proposition
9.4.2 and Proposition 9.4.4), $\widetilde{\sigma}$ extends to a continuous
injective map of the visual boundary of $\widetilde{M}$ with image
$\Lambda_{\rho}$. We can now repeat wordly the argument of Proposition 9.3.16
to show that, if $B$ is negative semi-definite, then
$G_{\widetilde{\sigma}}^{+}$ is a diffeomorphism onto its image. To show that
$G_{\widetilde{\sigma}}^{+}(\widetilde{M})=\Omega_{+}$, we repeat instead the
proof of Proposition 9.4.4. More precisely, one first shows (using tangent
horospheres) that every $x\in\Omega_{+}$ is in the image of
$G_{\widetilde{\sigma}}^{+}$. Then, by continuity, it suffices to show that
every $x\in\Lambda_{\rho}$ is not on the image of
$G_{\widetilde{\sigma}}^{+}$. To see this, the last paragraph of the proof of
Proposition 9.4.4 applies unchanged, and when considering tangent $r$-caps we
can even take $r=0$, that is, replace $r$-caps by totally geodesic
hyperplanes. See Figure 9.8. ∎

######  Lemma 11.3.2.

Let $M^{n}$ be a closed oriented manifold and
$\rho:\pi_{1}(M)\to\mathrm{Isom}(\mathbb{H}^{n+1})$ be a nearly-Fuchsian
representation. Given two closed hypersurfaces $\Sigma_{0}$ and $\Sigma_{1}$
of small principal curvatures in the nearly-Fuchsian manifold
$\faktor{\mathbb{H}^{n+1}}{\rho(\pi_{1}(M))}$, there exists an isotopy

$\upsilon_{\bullet}\colon
M\times[0,1]\to\faktor{\mathbb{H}^{n+1}}{\rho(\pi_{1}(M))}$

such that:

  * •

$\upsilon_{s}$ is a convex embedding for all $s\in[0,1]$;

  * •

$\upsilon_{0}(M)$ is a hypersurface equidistant from $\Sigma_{0}$;

  * •

$\upsilon_{1}(M)$ is a hypersurface equidistant from $\Sigma_{1}$.

###### Proof.

First of all, let us observe that we can find hypersurfaces equidistant from
$\Sigma_{0}$ and $\Sigma_{1}$ which are convex. Indeed, by Corollary 9.3.4 and
Remark 9.4.6, $t$-equidistant hypersurfaces are embedded for all
$t\in{\mathbb{R}}$. Moreover, by compactness, the principal curvatures of
$\Sigma_{0}$ and $\Sigma_{1}$ are in $(-\epsilon,\epsilon)$ for some
$0<\epsilon<1$, and applying Equation (9.11) we may find $t_{0}$ such that the
principal curvatures of the $t$-equidistant hypersurfaces are negative for
$t\geq t_{0}$ – namely, the equidistant hypersurfaces are convex.

Abusing notation, up to taking equidistant hypersurfaces as explained above,
we will now assume that $\Sigma_{0}$ and $\Sigma_{1}$ are convex, and our goal
is to produce $\upsilon_{\bullet}$ such that $\upsilon_{s}$ is a convex
embedding for all $s\in[0,1]$, $\upsilon_{0}(M)=\Sigma_{0}$ and
$\upsilon_{1}(M)=\Sigma_{1}$. Up to replacing $\Sigma_{0}$ and $\Sigma_{1}$
again with equidistant hypersurfaces, we can also assume that
$\Sigma_{0}\cap\Sigma_{1}=\emptyset$, that $\Sigma_{1}$ is in the concave side
of $\Sigma_{0}$, and that the equidistant surfaces from $\Sigma_{1}$ which
intersect $\Sigma_{0}$ are all convex. We call $\mathcal{A}$ the region of
$\faktor{\mathbb{H}^{n+1}}{\rho(\pi_{1}(M))}$ bounded by $\Sigma_{0}$ and
containing $\Sigma_{1}$.

Let us now consider the (signed) distance functions $r_{0}$ and $r_{1}$ from
$\Sigma_{0}$ and $\Sigma_{1}$ respectively, chosen in such a way that both
$r_{0}$ and $r_{1}$ are positive functions on the concave side of $\Sigma_{0}$
and $\Sigma_{1}$ respectively. Again by Corollary 9.3.4 and Remark 9.4.6,
these functions are smooth and have nonsingular differential everywhere. Let
us denote by $\nu_{i}$ the gradient of $r_{i}$. The vector field $\nu_{i}$ has
unit norm and is tangent to the orthogonal foliations of $\Sigma_{i}$ which
have been described in the proof of Proposition 9.4.4. (Proposition 9.4.4
describes the foliation in the universal cover, but it clearly descends to the
quotient $\faktor{\mathbb{H}^{n+1}}{\rho(\pi_{1}(M))}$.)

We claim that both $r_{i}$’s are convex functions in the region $\mathcal{A}$,
i.e. that their Hessians are positive semi-definite, as a consequence of the
fact that the level sets of $r_{i}$ in $\mathcal{A}$ are all convex. Recall
that the Riemannian Hessian of a smooth function
$f:\mathcal{A}\to{\mathbb{R}}$ is the symmetric 2-tensor defined as

$\nabla^{2}f(X,Y)=\partial_{X}(\partial_{Y}f)-\partial_{D_{X}Y}f~{},$ (11.2)

where $X,Y$ are local vector fields and $D$ is the ambient Levi-Civita
connection as usual. Clearly $\nabla^{2}r_{i}(\nu_{i},\nu_{i})=0$ since
$r_{i}$ is linear along the integral curves of $\nu_{i}$ and such integral
curves, which are the leaves of the orthogonal foliation described above, are
geodesics. Moroever, if $X$ is a vector field tangent to the level sets of
$r_{i}$, then $\nabla^{2}r_{i}(X,\nu_{i})=0$: indeed the first term in the RHS
of Equation (11.2) vanishes because $r_{i}$ is linear along the integral
curves, and the second term as well, because $D_{X}\nu_{i}=-B_{i}(X)$ is
tangential to the level sets of $r_{i}$ and thus
$\partial_{D_{X}\nu_{i}}r_{i}=0$.

To conclude that $\nabla^{2}r_{i}$ is positive semi-definite, it remains to
show that $\nabla^{2}r_{i}(X,X)\geq 0$ for all $X$ tangent to the level sets.
It is more instructive to perform this computation in the general setting of a
smooth function $f:\mathcal{A}\to{\mathbb{R}}$. Since the unit normal vector
field to the level set of $f$ is $\nu=\frac{Df}{\|Df\|}$,with $Df$ being the
gradient of $f$, for all $X,Y$ vector fields tangent to the fibers, we get
that:

$\nabla^{2}f(X,Y)=-\partial_{D_{X}Y}f=-\langle
D_{X}Y,\nu\rangle\partial_{\nu}f=-\|Df\|\mathrm{I}\mathrm{I}(X,Y)~{},$ (11.3)

where in the last step we used that $\partial_{\nu}f=\langle
Df,\nu\rangle=\|Df\|$, and $\mathrm{I}\mathrm{I}$ denotes the second
fundamental form of the level sets of $f$. When $f=r_{i}$, in the region
$\mathcal{A}$ the level sets of $r_{i}$ are convex, hence
$\mathrm{I}\mathrm{I}$ is negative semi-definite and $\nabla^{2}r_{i}(X,X)\geq
0$.

We remark that Equation (11.3) also shows that, if $f$ is a convex function,
then its level sets are convex hypersurfaces as long as $Df\neq 0$. We shall
now apply this remark to the zero set of the function
$f_{s}=(1-s)r_{0}+sr_{1}$ for $s\in[0,1]$. The differential of $f_{s}$ never
vanishes, for $\|Dr_{0}\|=\|Dr_{1}\|=1$, hence $Df_{s}=0$ is only possible for
$s=\frac{1}{2}$ if $Dr_{0}=-Dr_{1}$: nevertheless, this cannot happen since
the geodesics with initial vector $Dr_{0}=\nu_{0}$ and $Dr_{1}=\nu_{1}$ both
have final endpoint in $\Omega_{+}$ and initial endpoint in $\Omega_{-}$ by
(the proof of) Proposition 9.4.4. Hence $\\{f_{s}=0\\}$ is an embedded
hypersurface for all $s$. Observe moreover that

$\\{f_{s}=0\\}=\bigg{\\{}\frac{r_{0}}{r_{0}-r_{1}}=s\bigg{\\}}~{}.$

Since $\Sigma_{0}\cap\Sigma_{1}=\emptyset$, $r_{0}-r_{1}$ never vanishes, and
this shows that the hypersurfaces $\\{f_{s}=0\\}$ provide a foliation of the
region between $\Sigma_{0}$ and $\Sigma_{1}$, which is contained in
$\mathcal{A}$. Since both $r_{0}$ and $r_{1}$ are convex functions in
$\mathcal{A}$,

$\nabla^{2}f_{s}(X,X)=(1-s)\nabla^{2}r_{0}(X,X)+s\nabla^{2}r_{1}(X,X)\geq 0$

for every $X$, hence $f_{s}$ is convex. As remarked just after Equation
(11.3), since $\|Df_{s}\|\neq 0$, $\\{f_{s}=0\\}$ is a convex hypersurface.

It is not hard now to produce
$\upsilon_{\bullet}:M\times[0,1]\to\faktor{\mathbb{H}^{n+1}}{\rho(\pi_{1}(M))}$
such that $\upsilon_{s}(M)=\\{f_{s}=0\\}$. For instance one can flow along the
vector field $\frac{DF}{\|DF\|^{2}}$ where $F=\frac{r_{0}}{r_{0}-r_{1}}$.
Alternatively one can apply Lemma 11.3.1 to infer that the Gauss maps
$G^{+}_{\widetilde{\sigma}}$ in the universal cover induce diffeomorphisms of
each hypersurface $\\{f_{s}=0\\}$ with
$\faktor{\Omega_{+}}{\rho(\pi_{1}(M))}\cong M$, and define $\upsilon_{s}$ as
the inverse map. ∎

###### Proof of the “only if” part of Theorem 11.1.4.

Suppose $\mathcal{L}$ and $\mathcal{L}^{\prime}$ are $\rho$-integrable
Riemannian submanifolds in ${\mathbb{G}}_{\rho}$. Then there exists
hypersurfaces $\Sigma$ and $\Sigma^{\prime}$ in
$\faktor{\mathbb{H}^{n+1}}{\rho(\pi_{1}(M))}$ whose Gauss map image induce
$\mathcal{L}$ and $\mathcal{L}^{\prime}$ respectively. We now apply Lemma
11.3.2 and find $\upsilon_{\bullet}$ such that $\upsilon_{s}$ is convex for
every $s$, and the images of $\upsilon_{0}$ and $\upsilon_{1}$ are equidistant
hypersurfaces from $\Sigma$ and $\Sigma^{\prime}$ respectively. Define
$\Upsilon_{\bullet}:M\times[0,1]\to{\mathbb{G}}_{\rho}$ so that $\Upsilon_{s}$
is the map into ${\mathbb{G}}_{\rho}$ induced by the Gauss map of the lifts on
the universal cover
$\widetilde{\upsilon}_{s}:\widetilde{M}\to\mathbb{H}^{n+1}$. As a consequence
of Lemma 11.3.1 the Gauss map of each $\widetilde{\upsilon}_{s}$ is an
embedding with image in
$\Omega_{+}\times\partial\mathbb{H}^{n+1}\setminus\Delta$ in
$\mathbb{G}_{n+1}$. Repeating the same argument of Remark 9.4.6,
$\Upsilon_{s}:M\to{\mathbb{G}}_{\rho}$ is an embedding for every $s$. As a
particular case, by Lemma 9.1.6, $\Upsilon_{0}(M)=\mathcal{L}$ and
$\Upsilon_{1}(M)=\mathcal{L}^{\prime}$.

By construction for every $s\in[0,1]$ the image of $\Upsilon_{s}$ is a
$\rho$-integrable embedded submanifold in ${\mathbb{G}}_{\rho}$. Let us denote
by $\mathcal{L}_{s}$ such submanifold. It follows (Lemma 10.5.13) that
$\mathrm{hol}_{\Upsilon_{s}}$ is trivial for all $s$. By Proposition 11.2.2,
together with the definition of $\mathrm{Flux}$, we have that

$\int_{0}^{s}\Upsilon_{r}^{*}(\Omega(X_{r},\cdot))dr=0\in
H^{1}_{dR}(M,{\mathbb{R}})$

for all $s$, where $X_{s}$ is the vector field generating $\Upsilon_{s}$.
Hence necessarily the cohomology class of
$\Upsilon_{s}^{*}(\Omega(X_{s},\cdot))$ in $H^{1}_{dR}(M,{\mathbb{R}})$ is
trivial for all $s$. We can therefore find a smooth function
$f_{\bullet}:M\times[0,1]\to{\mathbb{R}}$ such that
$df_{s}=\Upsilon_{s}^{*}(\Omega(X_{s},\cdot))$ for all $s$. Pushing forward
$f_{s}$ by means of $\Upsilon_{s}$, we have defined smooth functions on
$\mathcal{L}_{s}$ whose differential equals $\Omega(X_{s},\cdot)$. Let us
extend them to $F_{\bullet}:{\mathbb{G}}_{\rho}\to{\mathbb{R}}$ so that
$F_{\bullet}$ is compactly supported and $F_{s}\circ\Upsilon_{s}=f_{s}$.

Let $\widehat{X}_{s}$ be the symplectic gradient of $F_{s}$, namely

$dF_{s}=\Omega(\widehat{X}_{s},\cdot)~{},$

and let $\Phi_{s}$ be the flow generated by $\widehat{X}_{s}$. From
$d(F_{s}\circ\Upsilon_{s})=df_{s}$ we see that

$\Omega(\widehat{X}_{s},d\Upsilon_{s}(V))=\Omega(X_{s},d\Upsilon_{s}(V))$

for all $V\in T_{p}M$. This implies that $\Omega(\widehat{X}_{s}-X_{s},\cdot)$
vanishes identically along the Lagrangian submanifold $\mathcal{L}_{s}$. By
non-degeneracy of the symplectic form $\Omega$, $\widehat{X}_{s}-X_{s}$ is
tangential to $\mathcal{L}_{s}$. Therefore $\Phi_{s}\circ\Upsilon_{0}$ and
$\Upsilon_{s}$ differ by pre-composition with a diffeomorphism $\phi_{s}$ of
$M$ (which is indeed obtained as the flow on $M$ of the vector field
$\Upsilon_{s}^{*}(\widehat{X}_{s}-X_{s})$). This shows that
$\Phi_{s}(\mathcal{L})=\mathcal{L}_{s}$. In particular, $\Phi=\Phi_{1}$ is the
desired compactly supported Hamiltonian symplectomorphism of
${\mathbb{G}}_{\rho}$ such that $\Phi(\mathcal{L})=\mathcal{L}^{\prime}$. ∎

####  11.4 Evolution by geometric flows

The aim of this last section is to provide a relationship between certain
geometric flows for hypersurfaces in $\mathbb{H}^{n+1}$ and their induced
flows in $T^{1}\mathbb{H}^{n+1}$ and in $\mathbb{G}_{n+1}$.

Let $M=M^{n}$ be an oriented manifold. Let $\sigma\colon
M\times(-\varepsilon,\varepsilon)\to\mathbb{H}^{n+1}$ be a smooth map such
that $\sigma_{t}=\sigma(\cdot,t)$ is an immersion with small principal
curvatures for all $t$, and let $\nu=\nu(x,t)$ be the normal vector field.

######  Proposition 11.4.1.

Let $f\colon M\times(-\varepsilon,\varepsilon)\to{\mathbb{R}}$ be a smooth map
such that

$\frac{d}{dt}\sigma_{t}=f_{t}\nu_{t},$

and let $\zeta_{t}:=\zeta_{\sigma_{t}}:M\to T^{1}\mathbb{H}^{n+1}$ be the lift
to $T^{1}\mathbb{H}^{n+1}$, $G_{t}:=G_{\sigma_{t}}:M\to\mathbb{G}_{n+1}$ be
the Gauss map. Then,

$\displaystyle\frac{d}{dt}\zeta_{t}$
$\displaystyle=-d\zeta_{t}(B_{t}(\overline{\nabla}^{t}f_{t}))-{\mathrm{J}}(d\zeta_{t}(\overline{\nabla}^{t}f_{t}))+f_{t}\chi$
(11.4) $\displaystyle\frac{d}{dt}G_{t}$
$\displaystyle=-dG_{t}(B_{t}(\overline{\nabla}^{t}f_{t}))-\mathbb{J}(dG_{t}(\overline{\nabla}^{t}f_{t}))$
(11.5)

where $\overline{\nabla}^{t}f_{t}$ is the gradient of $f_{t}$ with respect to
the first fundamental form $\overline{\mathrm{I}}_{t}=G_{t}^{*}{\mathbbm{g}}$
and $B_{t}$ is the shape operator of $\sigma_{t}$.

As a preliminary step to prove Proposition 11.4.1, we compute the variation in
time of the normal vector field. Recalling that $D$ denotes the Levi-Civita
connection on $\mathbb{H}^{n+1}$, we show:

$D_{\frac{d\sigma_{t}}{dt}}{\nu_{t}}=-d\sigma(\nabla^{t}f_{t})~{},$ (11.6)

where now $\nabla^{t}$ denotes the gradient with respect to the first
fundamental form $\mathrm{I}_{t}$ of $\sigma_{t}$. On the one hand, by metric
compatibility,

$\langle
D_{\frac{d\sigma_{t}}{dt}}\nu_{t},\nu_{t}\rangle=\frac{1}{2}\partial_{\frac{d\sigma_{t}}{dt}}\langle\nu_{t},\nu_{t}\rangle=0$

hence $D_{\frac{d\sigma_{t}}{dt}}\nu_{t}$ is tangent to the hypersurface.

On the other hand, let $X$ be any vector field over $M$. Since $X$ and
$\frac{\partial}{\partial t}$ commute on $M\times(-\varepsilon,\varepsilon)$,

$\displaystyle\langle D_{\frac{d\sigma_{t}}{dt}}\nu_{t},d\sigma_{t}(X)\rangle$
$\displaystyle=\partial_{\frac{d\sigma_{t}}{dt}}\langle\nu_{t},d\sigma_{t}(X)\rangle-\langle\nu_{t},D_{\frac{d\sigma_{t}}{dt}}(d\sigma_{t}(X))\rangle$
$\displaystyle=0-\langle\nu_{t},D_{\frac{d\sigma_{t}}{dt}}(d\sigma_{t}(X))\rangle$
$\displaystyle=-\langle\nu_{t},D_{d\sigma_{t}(X)}(f_{t}\nu_{t})\rangle$
$\displaystyle=-X(f_{t})-f_{t}\langle\nu_{t},D_{d\sigma_{t}(X)}\nu_{t}\rangle$
$\displaystyle=-X(f_{t})=-\mathrm{I}_{t}(\nabla^{t}f_{t},X)~{}.$

This shows Equation (11.6). As a result, in the hyperboloid model (5.2) we
have:

$\frac{d}{dt}\zeta_{t}=\bigg{(}\frac{d}{dt}\sigma_{t},D_{\frac{d\sigma_{t}}{dt}}\nu\bigg{)}=(f_{t}\nu_{t},-d\sigma_{t}(\nabla^{t}f_{t}))$
(11.7)

###### Proof of Proposition 11.4.1.

Let $e_{1;t},\dots,e_{n;t}$ be a local $\overline{\mathrm{I}}_{t}$-orthonormal
frame diagonalizing $B_{t}$, so $B_{t}(e_{k;t})=\lambda_{k;t}e_{k;t}$. By
definition of ${\overline{Sas}_{n+1}}$ and of ${\mathrm{J}}$,

$\big{(}d\zeta_{t}(e_{1;t}),\dots,d\zeta_{t}(e_{n;t}),\chi,{\mathrm{J}}d\zeta_{t}(e_{1;t}),\dots,{\mathrm{J}}d\zeta_{t}(e_{n;t})\big{)}$
(11.8)

defines at each point of the image an orthonormal basis for the tangent space
of $T^{1}\mathbb{H}^{n+1}$, with the former $n+1$ vectors having norm $1$ and
the latter $n$ vectors having norm $-1$.

We prove Equation (11.4), then Equation (11.5) follows after observing that

$\frac{d}{dt}G_{t}=(dG_{t})\bigg{(}\frac{\partial}{\partial
t}\bigg{)}=(d\mathrm{p}\circ d\zeta_{t})\bigg{(}\frac{\partial}{\partial
t}\bigg{)}=d\mathrm{p}\bigg{(}\frac{d}{dt}\zeta_{t}\bigg{)}.$

We show that LHS and RHS of (11.4) have the same coordinates with respect to
the basis (11.8). By Equations (9.2) and (11.7),

$\displaystyle{\overline{Sas}_{n+1}}\bigg{(}\frac{d}{dt}\zeta_{t},{\mathrm{J}}d\zeta_{t}(e_{k;t})\bigg{)}$
$\displaystyle=f_{t}\langle\nu_{t},-d\sigma_{t}(B_{t}(e_{k;t}))\rangle-\langle-d\sigma_{t}(\nabla^{t}f_{t}),d\sigma_{t}(e_{k;t})\rangle$
$\displaystyle=\langle
d\sigma_{t}(\nabla^{t}f_{t}),d\sigma_{t}(e_{k;t})\rangle=\partial_{e_{k;t}}f_{t}$
$\displaystyle=\overline{\mathrm{I}}_{t}(\overline{\nabla}^{t}f_{t},e_{k;t})={\overline{Sas}_{n+1}}(-{\mathrm{J}}d\zeta_{t}(\overline{\nabla}^{t}f_{t}),{\mathrm{J}}d\zeta_{t}(e_{k;t}))~{}.$

Similarly, recalling that $B_{t}$ is self-adjoint with respect to both
$\mathrm{I}_{t}$ and $\overline{\mathrm{I}}_{t}$, one has

$\displaystyle{\overline{Sas}_{n+1}}\bigg{(}\frac{d}{dt}\zeta_{t},d\zeta_{t}(e_{k;t})\bigg{)}$
$\displaystyle=\langle
f_{t}\nu_{t},d\sigma_{t}(e_{k;t})\rangle-\langle-d\sigma_{t}(\nabla^{t}f_{t}),-d\sigma_{t}(B_{t}(e_{k;t}))\rangle$
$\displaystyle=-\langle
d\sigma_{t}(\nabla^{t}f_{t}),d\sigma_{t}(B_{t}(e_{k;t}))\rangle$
$\displaystyle=-\mathrm{I}_{t}(\nabla^{t}f_{t},B_{t}(e_{k;t}))=-\overline{\mathrm{I}}_{t}(\overline{\nabla}^{t}f_{t},B_{t}(e_{k;t}))$
$\displaystyle=-\overline{\mathrm{I}}_{t}(B_{t}(\overline{\nabla}^{t}f_{t}),e_{k;t})={\overline{Sas}_{n+1}}(-d\zeta_{t}(B_{t}(\overline{\nabla}^{t}f_{t})),d\zeta_{t}(e_{k;t}))~{}.$

Finally,

$\displaystyle{\overline{Sas}_{n+1}}\bigg{(}\frac{d}{dt}\zeta_{t},\chi\bigg{)}=f_{t}\langle\nu_{t},\nu_{t}\rangle=f_{t}={\overline{Sas}_{n+1}}(f_{t}\chi,\chi)$

and the proof follows. ∎

An interesting corollary of Proposition 11.4.1 involves mean curvature flow.
Directly by Proposition 10.5.7, one has the following.

######  Corollary 11.4.2.

The flow in $\mathbb{H}^{n+1}$ defined by

$\frac{d}{dt}\sigma_{t}=\frac{1}{n}\sum_{k=1}^{n}\mathrm{arctanh}(\lambda_{k;t}),$

on hypersurfaces of small principal curvatures, induces in $\mathbb{G}_{n+1}$
the mean curvature flow up to a tangent factor, namely

$\frac{d}{dt}G_{t}=\mathrm{\overline{H}_{t}}+B_{t}(\mathbb{J}(\mathrm{\overline{H}_{t}})).$

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