Complete negatively curved immersed ends in $\mathbb {R}^3$

Abstract

This paper extends, in a sharp way, the famous Efimov’s Theorem to immersed ends in $\mathbb {R}^3$. More precisely, let M be a non-compact connected surface with compact boundary. Then there is no complete isometric immersion of M into $\mathbb {R}^3$ satisfying that $\int _M |K|=+\infty$ and $K\le -\kappa <0$, where $\kappa$ is a positive constant and K is the Gaussian curvature of M. In particular Efimov’s Theorem holds for complete Hadamard immersed surfaces, whose Gaussian curvature K is bounded away from zero outside a compact set.

Appendix: Proof of Lemma 3.1

Consider a Riemannian smooth surface (S, g) and a connected surface with piecewise smooth boundary $D\subset S$, with internal angles at the vertices different from 0 and $2\pi$. Denote by $d_g(p,q)$ the distance induced by the Riemannian metric g, and $d_{\mathrm {int}}(p,q)$ the infimum of the g-lengths of piecewise $C^1$ curves $\gamma :[0,1]\rightarrow D$ joining p to q satisfying that $\gamma \bigl ((0,1)\bigr )\subset \mathrm {int}(D)$.

The purpose of this appendix is to prove Lemma 3.1 above, which asserts that $d_\mathrm {int}$ is a distance on D, and that the distances $d_g$ and $d_\mathrm {int}$ induce the same topology on D.

To prove Lemma 3.1 we first assume that $d_\mathrm {int}$ is a distance on D. To show that $d_g$ and $d_\mathrm {int}$ induce the same topology on D, we need to prove that, given $q\in D$ and $\epsilon >0$, there exists $\delta >0$ such that if $p\in D$ with $d_g(p,q)<\delta$ then $d_\mathrm {int}(p,q)<\epsilon$. If $q\in \mathrm {int}(D)$ the proof is trivial. Thus we will assume that $q\in \partial D$.

Fix $q\in \partial D$ and $\epsilon >0$. For some small $\lambda >0$, there exists a curve $\sigma :[-\lambda ,\lambda ]\rightarrow \partial D, \lambda >0$, parameterized by the g-arc length satisfying that $\sigma (0)=q$ and such that $\sigma |_{[-\lambda ,0]}$ and $\sigma |_{[0,\lambda ]}$ are smooth curves. Since $\partial D$ is a piecewise smooth curve and the angles at the vertices differ from 0 and $2\pi$, there exists a unit vector $v\in T_qS$ pointing to $\mathrm {int}(D)$ and transversal to both $\sigma '(0-)$ and $\sigma '(0+)$. Let $v_t$ be the parallel transport of v along the both directions on $\sigma$. If $\lambda$ is small enough, we may assume that $v_t$ is transversal to $\sigma '(t)$ and that $v_t$ points to $\mathrm {int}(D)$. Set $\sigma _s(t)=\exp _{\sigma (t)}sv_t=\gamma _t(s)$. By smoothness of the geodesic flow there exists sufficiently small $0<\eta <\min \left\rbrace \lambda ,\frac{\epsilon }{3}\right\lbrace$ such that:

• (1) for $0<s\le \eta$ and $-\eta \le t\le \eta$, the point $\sigma _s(t)\in \mathrm {int}(D)$;

• (2) $L_g(\sigma _\eta )<\frac{\epsilon }{3}$.

Given $s_0\in [0,\eta ]$ and $t_0\in [-\eta ,\eta ]$, we have that $2\eta +|t_0|-s_0> \eta +|t_0|-s_0\ge 0$. We will construct a piecewise smooth curve $\xi =\xi _{s_0t_0}:[0,2\eta +|t_0|-s_0]\rightarrow D$ from q to $\sigma _{s_0}(t_0)$ satisfying $\xi \bigl ((0,2\eta +|t_0|-s_0)\bigr )\subset \mathrm {int}(D)$ and $L_g(\xi )<\epsilon$. From q to $\sigma _{\eta }(0)=\gamma _0(\eta )$, let $\xi |_{[0,\eta ]}$ coincide with the geodesic $\gamma _0:[0,\eta ]\rightarrow D$. From $\sigma _{\eta }(0)$ to $\sigma _{\eta }(t_0)$ the curve $\xi$ follows the curve $\sigma _\eta$ in the direction that t is increasing if $t_0\ge 0$, or in the other direction if $t_0<0$. More precisely, for $0\le t\le |t_0|$, we define $\xi (\eta +t)=\sigma _\eta (t)$, if $t_0\ge 0$, and $\xi (\eta +t)=\sigma _\eta (-t)$, if $t_0<0$. Finally, from $\sigma _{\eta }(t_0)$ to $\sigma _{s_0}(t_0)$ the curve $\xi$ follows the geodesic $s\longmapsto \gamma _{t_0}(\eta -s)$. Namely, for $0\le s\le \eta -s_0$ we define $\xi (\eta +|t_0|+s)=\gamma _{t_0}(\eta -s)=\sigma _{(\eta -s)}(t_0)$. In particular we have that $\xi (2\eta +|t_0|-s_0)= \xi (\eta +|t_0|+(\eta -s_0))=\gamma _{t_0}(s_0)=\sigma _{s_0}(t_0)$. By construction we have that $L_g(\xi )\le \eta + L_g(\sigma _{\eta })+(\eta -s_0)<\epsilon$.

Given $s_1\in [0,\eta ]$ and $t_1\in [-\eta ,\eta ]$, a similar construction as above shows that $\sigma _{s_0}(t_0)$ may be connected to $\sigma _{s_1}(t_1)$ by a piecewise smooth curve $\psi :[0,1]\rightarrow D$ with $\psi ((0,1))\subset \mathrm {int}(D)$ and $L_g(\psi ) <\epsilon$.

Set $X=\{\sigma _s(t)\bigm |0\le s\le \eta ,\, -\eta \le t\le \eta \}$. Since X is a compact neighborhood of q in D, we have that $\delta =d_g(q,D-X)>0$. Now we take $p\in D$ with $d_g(p,q)<\delta$. We have that $p\in X$, hence $p=\sigma _{s_0}(t_0)$ for some $s_0\in [0,\eta ]$ and $t_0\in [-\eta ,\eta ]$. As a consequence we have that $d_{\mathrm {int}}(p,q)\le L_g(\xi _{s_0t_0})<\epsilon$.

Now we consider points $p,q,r\in D$ and we will show that $d_\mathrm {int}(p,q)+d_\mathrm {int}(q,r) \ge d_\mathrm {int}(p,r)$. We will just consider the case that $q\in \partial D$, since the other case is easier. Fix $\epsilon >0$ and consider piecewise smooth curves $\gamma :[0,1]\rightarrow D$ from p to q with $\gamma ((0,1))\subset \mathrm {int}(D)$ and $L_g(\gamma )<d_\mathrm {int}(p,q)+\epsilon$, and $\sigma :[0,1]\rightarrow D$ from q to r with $\sigma ((0,1))\subset \mathrm {int}(D)$ and $L_g(\sigma )<d_\mathrm {int}(q,r)+\epsilon$. By using a neighborhood X of q as above, it is easy to obtain a piecewise smooth curve $\varphi :[0,1]\rightarrow D$ from p to r with $\varphi ((0,1))\subset \mathrm {int}(D)$ and $L_g(\varphi )<L_g(\gamma )+L_g(\sigma )+\epsilon$. In fact, take $0<s_1<1$ such that $\gamma (s_1)\in X-\{q\}$ and $0<s_2<1$ such that $\sigma (s_2)\in X-\{q\}$. We define a curve $\varphi$ which follows $\gamma$ from $t=0$ to $t=s_1$ then follows a curve $\psi$ in $X\cap \mathrm {int}(D)$ with $L_g(\psi )<\epsilon$, and then follows $\sigma$ from $t=s_2$ to $t=1$. Thus we obtain that $d_{\mathrm {int}}(p,r)\le L_g(\varphi )< L_g(\gamma )+L_g(\sigma )+\epsilon < d_\mathrm {int}(p,q)+d_\mathrm {int}(q,r)+3\epsilon$. By making $\epsilon \rightarrow 0$ we conclude the proof of Lemma 3.1.

Remark 2

It is not difficult to see that Lemma 3.1 may be improved to assume that internal angles are just different from 0.

Remark 3

We inform that the paper [GMT] proves Theorem A, with another proof, simultaneously and independently.