The half space property for cmc 1/2 graphs in $\mathbb {E}(-1,\tau )$

Abstract

In this paper, we prove a half-space theorem with respect to constant mean curvature 1/2 entire graphs in $\mathbb {E}(-1,\tau )$. If $\Sigma$ is such an entire graph and $\Sigma '$ is a properly immersed constant mean curvature 1/2 surface included in the mean convex side of $\Sigma$ then $\Sigma '$ is a vertical translate of $\Sigma$. We also have an equivalent statement for the non mean convex side of $\Sigma$.

Appendices

Appendix 1: Some computations in Killing Riemannian submersions

In this appendix we recall some definitions about Killing Riemannian submersions and make some computations concerning graphs in such an ambient space (see [5, 10]).

Let $(M^{n+1},\bar{g})$ and $(B^n,(\cdot ,\cdot ))$ be two complete Riemannian manifolds. Let $\pi :M\rightarrow B$ be a submersion. The tangent space $T_pM$ at $p$ then splits in $\ker d\pi \oplus (\ker d\pi )^\perp$ where $\ker d\pi$ is the 1-dimensional space of vertical vectors and $(\ker d\pi )^\perp$ is the space of horizontal vectors. The submersion $\pi$ is called Riemannian if $d\pi$ is an isometry from $(\ker d\pi )^\perp$ to $T_{\pi (p)}B$.

Definition 1

A Riemannian submersion $\pi :M\rightarrow B$ is a Killing submersion if it admits a complete vertical unit Killing vector field.

If $\pi :M\rightarrow B$ is a Killing Riemannian submersion, we denote by $\xi$ this unit Killing vector field. Besides, if $X$ is a vector field in $B$, we denote by $\widetilde{X}$ its horizontal lift by $\pi$.

Using this notation there exists a 2-form $\omega$ on $B$ such that for any vector fields $X,Y$ on $B$ we have:

$$\begin{aligned}{}[\widetilde{X},\widetilde{Y}]=\widetilde{[X,Y]}+\omega (X,Y)\xi \end{aligned}$$

We notice that $[\widetilde{X},\xi ]=0$. When $X$ is a tangent vector to $B$, we denote $X^\omega$ the vector such that $(X^\omega ,Y)=\omega (X,Y)$ for any $Y$. With this notation the Levi-Civita connection of $M$ and $B$ are relied by

$$\begin{aligned} \overline{\nabla }_{\widetilde{X}}\widetilde{Y}&= \widetilde{\nabla _XY}+\frac{1}{2}\omega (X,Y)\xi \end{aligned}$$

(11)

$$\begin{aligned} \overline{\nabla }_{\widetilde{X}}\xi&= -\frac{1}{2}\widetilde{X^\omega } \end{aligned}$$

(12)

$$\begin{aligned} \overline{\nabla }_\xi \xi&= 0 \end{aligned}$$

(13)

In a Killing Riemannian submersion, we are interested by surfaces that are the image of sections $\sigma$. These surfaces are called vertical graphs in $M$ and the image of $\sigma$ is also called the graph of $\sigma$. If $\sigma$ is a section defined over $\Omega \subset B$, we define on $\Omega$ a vector field $G\sigma$ by the following property:

$$\begin{aligned} (G\sigma ,X)=\bar{g}(\hbox {d}\sigma (X),\xi ) \end{aligned}$$

for any $X$ tangent to $B$. This vector field $G\sigma$ plays the role of the gradient of a function.

First, the upward pointing unit normal to the graph of $\sigma$ is given by the following expression:

$$\begin{aligned} N=\frac{-\widetilde{G\sigma }+\xi }{\sqrt{1+\Vert G\sigma \Vert ^2}} \end{aligned}$$

In the following, we denote $\sqrt{1+\Vert G\sigma \Vert ^2}$ by $W$. In fact the expression of $N$ is defined in the whole $\pi ^{-1}(\Omega )$ and the mean curvature of the graph of $\sigma$ is given by

$$\begin{aligned} nH=-{{\mathrm{div}}}_{\bar{g}}\left( \frac{-\widetilde{G\sigma }+\xi }{W}\right) \end{aligned}$$

So if $(E_i)$ is an orthonormal frame of $T_pB$ we have

$$\begin{aligned} nH&=-\sum _i\bar{g}\left( \overline{\nabla }_{\widetilde{E}_i} \frac{-\widetilde{G\sigma }+\xi }{W}, \widetilde{E}_i\right) -\bar{g}\left( \overline{\nabla }_\xi \frac{-\widetilde{G\sigma }+\xi }{W}, \xi \right) \\&=\sum _i\bar{g}\left( \widetilde{\nabla _{E_i}\frac{G\sigma }{W}},\widetilde{E_i}\right) + \frac{1}{2W}\bar{g}\left( \widetilde{E_i^\omega },\widetilde{E_i}\right) \\&={{\mathrm{div}}}\left( \frac{G\sigma }{W}\right) +\sum _i\frac{1}{2W}\omega (E_i,E_i)\\&={{\mathrm{div}}}\left( \frac{G\sigma }{W}\right) \end{aligned}$$

where ${{\mathrm{div}}}$ denote the divergence operator on $B$. So the mean curvature is given by

$$\begin{aligned} nH={{\mathrm{div}}}\left( \frac{G\sigma }{\sqrt{1+\Vert G\sigma \Vert ^2}}\right) \end{aligned}$$

(14)

Let us denote by $\Sigma$ the graph of $\sigma$. The map $\sigma : \Omega \rightarrow \Sigma$ is a chart, so let us make some computation using this system of coordinates. We have:

$$\begin{aligned} \hbox {d}\sigma (X)=\widetilde{X}+(G\sigma ,X)\xi \end{aligned}$$

so the induced metric is $g(X,Y)=(X,Y)+(G\sigma ,X)(G\sigma ,Y)$. If $u$ is a function on $\Omega$, we get

$$\begin{aligned} (\nabla u,X)=\hbox {d}u(X)=g(\nabla _g u,X)=(\nabla _g u,X)+(G\sigma ,\nabla _g u) (G\sigma ,X) \end{aligned}$$

Thus $\nabla u=\nabla _g u+(G\sigma ,\nabla _g u)G\sigma$; this implies that

$$\begin{aligned} \nabla _g u=\nabla u-(\chi _\sigma ,\nabla u)\chi _\sigma \end{aligned}$$

where $\chi _\sigma =\frac{G\sigma }{\sqrt{1+\Vert G\sigma \Vert ^2}}$. As a consequence, we have:

$$\begin{aligned} g(\nabla _g v,\nabla _g w)&=(\nabla v-(\chi _\sigma ,\nabla v)\chi _\sigma , \nabla w-(\chi _\sigma ,\nabla w)\chi _\sigma )\\&\quad \quad \quad +(G\sigma , \nabla v-(\chi _\sigma , \nabla v)\chi _\sigma )(G\sigma , \nabla w-(\chi _\sigma , \nabla w)\chi _\sigma )\\&=(\nabla v,\nabla w)-(2-\Vert \chi _\sigma \Vert ^2)(\chi _\sigma , \nabla v)(\chi _\sigma , \nabla w)\\&\quad \quad \quad +W^2(1-\Vert \chi _\sigma \Vert ^2)^2(\chi _\sigma , \nabla v)(\chi _\sigma , \nabla w)\\&=(\nabla v,\nabla w)-(\chi _\sigma , \nabla v)(\chi _\sigma , \nabla w) \end{aligned}$$

Besides the divergence operator for the metric $g$ have the following expression:

$$\begin{aligned} {{\mathrm{div}}}_g X=\frac{1}{\sqrt{1+\Vert G\sigma \Vert ^2}}{{\mathrm{div}}}\left( \sqrt{1+\Vert G\sigma \Vert ^2} X\right) \end{aligned}$$

If $\Sigma$ has constant mean curvature, the function $\nu =(N,\xi )$ is a Jacobi function so $0=\Delta _\Sigma \nu +(Ric_{\bar{g}}(N,N)+\Vert S\Vert ^2)\nu$ with $S$ the shape operator of $\Sigma$. Let $v$ be a function on $\Sigma$, we then have

$$\begin{aligned} \Delta _\Sigma (\nu v) +(Ric_{\bar{g}}(N,N)+\Vert A\Vert ^2) (\nu v)&= (\Delta _\Sigma \nu )v+2(\nabla _\Sigma \nu ,\nabla _\Sigma v)+\nu (\Delta _\Sigma v)\\&\quad \quad \quad + (Ric_{\bar{g}}(N,N)+\Vert S\Vert ^2) (\nu v)\\&=\nu \left( \Delta _\Sigma v-2\nu \left( \nabla _\Sigma \frac{1}{\nu },\nabla _\Sigma v\right) \right) \end{aligned}$$

Thus $\nu v$ is a Jacobi function if and only if $\Delta _\Sigma v-2\nu (\nabla _\Sigma \frac{1}{\nu },\nabla _\Sigma v)=0$. Looking at $v$ as a function defined on $\Omega$, since $\nu =W^{-1}$, it gives:

$$\begin{aligned} 0&=\Delta _g v-2W^{-1}g(\nabla _g W,\nabla _g v)\\&=W^{-1}\big ({{\mathrm{div}}}(W(\nabla v-(\chi _\sigma ,\nabla v)\chi _\sigma ))-2((\nabla W,\nabla v)-(\chi _\sigma , \nabla W) (\chi _\sigma , \nabla v))\big )\\&=W^{-1}\left( {{\mathrm{div}}}\left( W^2\frac{\nabla v-(\chi _\sigma ,\nabla v)\chi _\sigma }{W}\right) -\left( \nabla W^2, \frac{\nabla v-(\chi _\sigma ,\nabla v)\chi _\sigma }{W}\right) \right) \\&=W\left( {{\mathrm{div}}}\left( \frac{\nabla v-(\chi _\sigma ,\nabla v)\chi _\sigma }{W}\right) \right) \end{aligned}$$

So $v\nu$ is a Jacobi function if and only if

$$\begin{aligned} 0={{\mathrm{div}}}\left( \frac{\nabla v-(\chi _\sigma ,\nabla v)\chi _\sigma }{W}\right) \end{aligned}$$

(15)

Appendix 2: Some barriers

In this appendix, we construct some barriers from above and below on the exterior boundary component of an annulus for (3).

We use the model for $\mathbb {E}(-1,\tau )$ but we consider hyperbolic polar coordinates on $D_{-1}$, so $x=\tanh (\rho /2)\cos \theta$ and $y=\tanh (\rho /2)\sin \theta$.

Let $0<\rho _1<\rho _0$ be two radii and $f$ be a smooth function on $\{\rho =\rho _0\}$. For $M$ a constant, we construct a smooth function $h$ on $\{\rho _1\le \rho \le \rho _0\}$ such that

• $h=f$ on $\{\rho =\rho _0\}$, $h\ge M$ in $\{\rho =\rho _1\}$ and

• $\displaystyle {{\mathrm{div}}}_{\mathbb {H}^2}\left( \frac{Gh}{\sqrt{1+\Vert Gh\Vert ^2}}\right) \le 1$

We see $f$ as a function of $\theta$ and we define on $\{\rho _1\le \rho \le \rho _0\}$ the function $h(\rho ,\theta )=f(\theta )+\alpha (\rho _0-\rho )$. Let us prove that for $\alpha$ sufficiently large $h$ satisfies the expected properties. We have $h(\rho _0,\theta )=f(\theta )$ and $h(\rho _1,\theta )=f(\theta )+\alpha (\rho _0-\rho _1)\ge M$ if $\alpha \ge (M-\min f)/(\rho _0-\rho _1)$. In the polar coordinates, we have

$$\begin{aligned} Gh=-\alpha \partial _\rho + \frac{1}{\sinh \rho }\left( \frac{f'(\theta )}{\sinh \rho }- 2\tau \tanh \frac{\rho }{2}\right) \partial _\theta \end{aligned}$$

Thus

$$\begin{aligned} {{\mathrm{div}}}_{\mathbb {H}^2}\left( \frac{Gh}{\sqrt{1+\Vert Gh\Vert ^2}}\right)&= -\frac{\cosh \rho }{\sinh \rho }\frac{\alpha }{W}+\alpha \frac{\partial _\rho \left( \frac{f'(\theta )}{\sinh \rho }- 2\tau \tanh \frac{\rho }{2}\right) ^2}{2W^3}+\frac{1}{\sinh \rho } \frac{\partial _\theta \left( \frac{f'(\theta )}{\sinh \rho }\right) }{W}\\&\quad \quad -\frac{1}{\sinh \rho } \frac{\left( \frac{f'(\theta )}{\sinh \rho }- 2\tau \tanh \frac{\rho }{2}\right) \partial _\theta \left( \frac{f'(\theta )}{\sinh \rho }- 2\tau \tanh \frac{\rho }{2}\right) ^2}{2W^3} \end{aligned}$$

where

$$\begin{aligned} W=\sqrt{1+\alpha ^2+\left( \frac{f'(\theta )}{\sinh \rho }- 2\tau \tanh \frac{\rho }{2}\right) ^2} \end{aligned}$$

with $\alpha >0$, there exists a positive constant $m$ that only depends on $f, \rho _0$ and $\rho _1$ such that:

$$\begin{aligned} {{\mathrm{div}}}_{\mathbb {H}^2}\left( \frac{Gh}{\sqrt{1+\Vert Gh\Vert ^2}}\right) \le \frac{m}{\alpha ^2}+\frac{m}{\alpha }+\frac{m}{\alpha ^3} \end{aligned}$$

So when $\alpha$ is sufficently large, the mean curvature of the graph of $h$ satisfies the expected estimate.

Now let us define $k(\rho ,\theta )=f(\theta )+\alpha (\rho _0-\rho )$ with $\alpha <0$ By choosing $\alpha$ small, we can ensure that $k(\rho _1,\theta )\le M$. Moreover, because of the above computation, there is a $m>0$ such that

$$\begin{aligned} {{\mathrm{div}}}_{\mathbb {H}^2}\left( \frac{Gk}{\sqrt{1+\Vert Gk\Vert ^2}}\right) \ge \frac{\cosh \rho _0}{\sinh \rho _0}\frac{-\alpha }{\sqrt{\alpha ^2+m}}- \frac{m}{\alpha ^2}-\frac{m}{|\alpha |}-\frac{m}{|\alpha |^3} \end{aligned}$$

So choosing $\alpha$ sufficiently close from $-\infty , k$ satisfies to

• $k=f$ on $\{\rho =\rho _0\}$, $k\le M$ in $\{\rho =\rho _1\}$ and

• $\displaystyle {{\mathrm{div}}}_{\mathbb {H}^2}\left( \frac{Gk}{\sqrt{1+\Vert Gk\Vert ^2}}\right) \ge 1$