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Jan 8, 2015 at 16:37 comment added user9072 This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.
Jan 8, 2015 at 16:15 comment added Deane Yang Yes, he means Gauss curvature.
Jan 8, 2015 at 10:47 comment added Giuseppe 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.
Jan 5, 2015 at 0:38 comment added Robert Bryant 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 $K\equiv=-1$.
Jan 4, 2015 at 19:21 answer added Igor Rivin timeline score: 2
Jan 4, 2015 at 7:45 comment added Ryan Budney 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.
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Jan 4, 2015 at 5:37 history asked user64605 CC BY-SA 3.0