Properly embedded minimal annuli in $\mathbb {H}^2 \times \mathbb {R}$

Abstract

In this paper we study the moduli space of properly Alexandrov-embedded, minimal annuli in $\mathbb {H}^2 \times \mathbb {R}$ with horizontal ends. We say that the ends are horizontal when they are graphs of ${\mathcal {C}}^{2, \alpha }$ functions over $\partial _\infty \mathbb {H}^2$. Contrary to expectation, we show that one can not fully prescribe the two boundary curves at infinity, but rather, one can prescribe one of the boundary curves, but the other one only up to a translation and a tilt, along with the position of the neck and the vertical flux of the annulus. We also prove general existence theorems for minimal annuli with discrete groups of symmetries.

Appendix A: Two tall rectangles are not area-minimizing

The purpose of this appendix is to prove that a couple of tall rectangles in $\mathbb {H}^2 \times \mathbb {R}$ of the same height with the vertical segments in common are not an area-minimizing surface. Recall that for a complete surface area-minimizing means that any compact piece minimizes the area among all the surfaces with the same boundary. To prove this, we are going to see that, in general, a couple of horizontal disks connected by vertical totally geodesic planes near infinity (see Fig. 11) are better competitors for the area functional.

Fig. 11

The tall rectangles, the horizontal disks and the vertical totally geodesic planes

The simplest analytic representation of these surfaces is obtained using the upper half-space model for $\mathbb {H}^2$. Thus we use standard coordinates $(x,y) \in \mathbb {H}^2$ with $y > 0$, and hence coordinates (x, y, t) on $\mathbb {H}^2 \times {\mathbb {R}}$. We shall place the vertical segments of the tall rectangles over the points (0, 0) and infinity. For a given $r >0$ we denote $\Gamma _r$ the geodesic in $\mathbb {H}^2$ with end points (r, 0) and $(-r,0)$. Let $P_r:=\Gamma _r \times \mathbb {R}$ be the corresponding totally geodesic plane. For simplicity we will write $\Gamma =\Gamma _1$ and $P=P_1$. Finally, we denote $\Lambda$ as the geodesic given by $\{x=0, y>0\}$. The point in the intersection $\Gamma \cap \Lambda$ is denoted by $o=(0,1)$.

On the geodesic $\Gamma$ we consider $\rho$ the signed distance to the point o. Then the induced metric in P is given by $d \rho ^2 +dt^2.$ We write a curve in P in coordinates $(\rho ,t)$ of the form

$$\begin{aligned} \rho \mapsto (\rho ,\lambda (\rho )). \end{aligned}$$

If we impose that the surface generated by horizontal translations along the geodesic $\Lambda$ of the above curve is minimal, then we get (see [16, page 316]) that the function $\lambda (\rho )$ satisfies the equation

$$\begin{aligned} \lambda '(\rho )= \frac{d}{\sqrt{\cosh ^2 \rho -d^2}}, \quad d\ge 0. \end{aligned}$$

(A.1)

If $d>1$ we have

$$\begin{aligned} \lambda (\rho )= \int _{\cosh ^{-1}(d)}^\rho \frac{d}{\sqrt{\cosh ^2 u -d^2}} du. \end{aligned}$$

(A.2)

We parametrize $\Gamma$ as $\theta \mapsto \mathrm{e}^{\mathrm{i}\theta },$$\theta \in (0,\pi ).$ Then it is straightforward to check that

$$\begin{aligned} \rho = \log \left( \tan \frac{\theta }{2}\right) . \end{aligned}$$

Combining the above expression and (A.1) then we obtain:

$$\begin{aligned} \lambda '(\theta )= \frac{d \csc (\theta )}{\sqrt{\csc ^2(\theta ) - d^2}}, \end{aligned}$$

(A.3)

and

$$\begin{aligned} \lambda (\theta )= \int _{\theta }^{\theta _0} \frac{d \csc ( t)}{\sqrt{\csc ^2(t) -d^2}} dt, \quad \theta _0=\csc ^{-1}(d). \end{aligned}$$

So the upper half part of the tall rectangle is given by the parametrization

$$\begin{aligned} \phi (r,\theta )=(r \cos (\theta ), r \sin (\theta ), \lambda (\theta )), \quad r \in (0,\infty ), \quad \theta \in (0,\theta _0). \end{aligned}$$

(A.4)

The lower half part is parametrized by $(r,\theta ) \mapsto (r \cos (\theta ), r \sin (\theta ),- \lambda (\theta )).$ Let $R_1$ be the complete tall rectangle described above.

The height of the tall rectangle $R_1$ is given by $2 \lambda (0) >\pi$. Moreover, notice that the intersection of the tall rectangle with the slice $t=\lambda (\theta )$ consists of the straight line $\{ r \mathrm{e}^{\mathrm{i} \theta } \; : \; r>0\}.$

Let $R_2$ be the tall rectangle obtained from $R_1$ applying the symmetry $(x,y,t) \mapsto (-x,y,t).$

Fix $r \in (0,1)$ and $\theta \in (0,\theta _0)$. We consider the two horizontal slices, $S_{\lambda (\theta )}$ and $S_{-\lambda (\theta )}$, at height $\lambda (\theta )$ and $- \lambda (\theta )$, respectively, and the two vertical planes $P_r$ and $P_{1/r}.$ We cut the tall rectangles $R_1 \cup R_2$ with the union

$$\begin{aligned} S_{\lambda (\theta )} \cup S_{-\lambda (\theta )} \cup P_r \cup P_{1/r}. \end{aligned}$$

The resulting curve is a compact curve in $R_1 \cup R_2$ with two connected components: $\beta _i \subset R_i$, $i=1,2.$ These two curves are symmetric with respect to the plane $x=0$. Let $\Sigma _i$ the disk in $R_i$ spanned by $\beta _i$, $i=1,2$. Observe that $\Sigma _i$ is a symmetric bi-graph over the annular sector $\Delta _i$ (see Fig. 12). Moreover we will denote:

• $D_1=\{ (u \cdot \mathrm{e}^{\mathrm{i} v}, \lambda (\theta )) \; : \; (u,v) \in (r,1/r) \times (\theta ,\pi -\theta )\}.$

• $D_2=\{ (u \cdot \mathrm{e}^{\mathrm{i} v} , - \lambda (\theta )) \; : \; (u,v) \in (r,1/r) \times (\theta ,\pi -\theta )\}.$

• $B_1$ the region of $P_r$ bounded by $\beta _1 \cup \beta _2$ and the curves $D_1 \cap P_r$ and $D_2 \cap P_r$.

• $B_2$ the region of $P_{1/r}$ bounded by $\beta _1 \cup \beta _2$ and the curves $D_1 \cap P_{1/r}$ and $D_2 \cap P_{1/r}$.

Fig. 12

The annular sectors $\Delta _1$ and $\Delta _2$

We want to prove that

Lemma A.1

The area of the region $D_1 \cup D_2 \cup B_1 \cup B_2$ is smaller than the area of $\Sigma _1 \cup \Sigma _2$. In particular, $R_1 \cup R_2$ is not area-minimizing.

Proof

By the symmety of the surface it is enough to prove that for some $\theta$ and r

$$\begin{aligned} \frac{Area(\Sigma _1)}{Area( D_1)+Area( B_1 )}>1. \end{aligned}$$

From (A.4) it is not difficult to see that

$$\begin{aligned}&A_1(r,\theta ):=Area(\Sigma _1)\\&\quad = -2 \log (r)\left( \cot (\theta ) \sqrt{4-2 d^2+2 d^2\cos (2 \theta )} + 2 E(\theta | d^2)-2F(\theta | d^2)\right. \\&\qquad - \left. \cot (\theta _0) \sqrt{4-2 d^2+2 d^2\cos (2 \theta _0)}-2 E(\theta _0|d^2)+2F(\theta _0 |d^2)\right) , \end{aligned}$$

where E and F are the classical elliptic functions given by

$$\begin{aligned} E(\Phi |m)=\int _0^\Phi \sqrt{1-m \sin ^2(s)} ds, \quad F(\Phi |m)=\int _0^\Phi \frac{1}{\sqrt{1-m \sin ^2(s)}} ds. \end{aligned}$$

Similarly, we have

$$\begin{aligned} A_2(r,\theta ):=Area(D_1)=- 4 \log (r) \cot (\theta ). \end{aligned}$$

Finally, observe that $Area(B_1)$ is less than the area of the region in $P_r$ given by

$$\begin{aligned} \{ (r \cdot \mathrm{e}^{\mathrm{i} v} , t) \; : \; v \in (\theta ,\pi -\theta ), t \in (-\lambda (0),\lambda (0))\}. \end{aligned}$$

So,

$$\begin{aligned} Area(B_1) < A_3(\theta ):=2 \lambda (0) \log \left( \cot ^2 \left( \frac{\theta }{2}\right) \right) . \end{aligned}$$

Now we take $r(\theta )=\tan ^n \left( \frac{\theta }{2}\right)$, where $n \in \mathbb {N}$ will be taken big enough in terms of d. Notice that for this choice of r we have

$$\begin{aligned} \frac{Area(\Sigma _1)}{Area( D_1)+Area( B_1 )}> f(\theta ):=\frac{A_1(r(\theta ),\theta )}{A_2(r(\theta ),\theta )+A_3(\theta )}. \end{aligned}$$

It is straitforward that

$$\begin{aligned} \lim _{\theta \rightarrow 0} f(\theta )=1 ,\quad \lim _{\theta \rightarrow 0} f'(\theta )=-\frac{\lambda (0)}{n}+F\left( \theta _0\left| \csc ^2(\theta _0)\right. \right) -E\left( \theta _0\left| \csc ^2(\theta _0)\right. \right) . \end{aligned}$$

As $F\left( \theta _0\left| \csc ^2(\theta _0)\right. \right) -E\left( \theta _0\left| \csc ^2(\theta _0)\right. \right) >0$, then

$$\begin{aligned} \lim _{\theta \rightarrow 0} f'(\theta )>0 \end{aligned}$$

for n big enough. So, we have that for $\theta$ small enough $f(\theta )>1$, which proves our lemma. $\square$