# Compact surfaces with boundary of constant negative curvature

Consider a surface (with boundary) diffeomorphic to $S^1 \times [0, 1]$ and with constant negative curvature, sitting inside $\mathbb{R}^3$. All the examples I know of such surfaces are "part of" (or "cut out of", if that makes better sense) a surface of revolution (examples are the catenoid, tractricoid, etc.; O'Neill's "Elementary Differential Geometry" has a comprehensive list on page 261, Exercise 7). I was wondering, are all constantly negatively curved $S^1 \times [0, 1]$ obtained this way (that is, cut out from a surface of revolution)?

• No, certainly not. Constant negative curvature embeddings of an annulus can be constructed extremely flexible ways. Consider taking an arbitrary knot as your embedding of $S^1 \times \{1/2\}$ and try to fatten it up to a constant negatively curved annulus. – Ryan Budney Jan 4 '15 at 7:45
• To be a bit more explicit about Ryan's comment: If you take any (smooth) immersed curve $c:S^1\to \mathbb{R}^3$ such that there exists a smooth unit normal $n:S^1\to S^2$ along $c$ (i.e., $c'(\theta)\cdot n(\theta)=0$ for all $\theta\in S^1$) satisfying the property that $c''(\theta)\cdot n(\theta)>0$ for all $\theta\in S^1$, then there exists an immersion $C:S^1\times[-1,1]\to\mathbb{R}^3$ such that $C(\theta,0)=c(\theta)$, $n(\theta)$ is the normal to the immersed surface at $c(\theta)$, and the induced metric has constant Gauss curvature $K\equiv=-1$. – Robert Bryant Jan 5 '15 at 0:38
• By the way, the catenoid has negative curvature, but it is not constant. It has zero constant mean curvature. I suspect that by curvature you meant Gaussian curvature. – Giuseppe Jan 8 '15 at 10:47
• Yes, he means Gauss curvature. – Deane Yang Jan 8 '15 at 16:15
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