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It is well-known that a pair of pants can be given a constant negative curvature metric. Indeed, one way to see that the genus $g$ surface can be given a constant negative curvature metric for $g \geq 2$ is to use the fact that each such surface can be partitioned into pants. Here's a picture of a pants-decomposition of the genus-3 surface. Image credit: this blog post of Elizabeth Denne.

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Now, when I draw a pair of pants embedded in Euclidean 3-space, it always looks to me as though there is some point, around the area where the three legs are joined, where the curvature becomes non-negative.

Question: Can we embed a pair of pants in $\mathbb{R}^3$ in such a way that the inherited metric has everywhere-negative curvature?

Added: The formulation of this question still needs some work. At the moment it admits a rather "cheap" solution. One can take, for instance a negatively curved cylinder (easy to construct as a surface of revolution) and then delete a disk from that. I think I want to add some hypothesis along the lines:

Each of the boundary circles of the pants is a geodesic.

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    $\begingroup$ I have never heard of three-legged pants. In general, people can inherit pants, but I do not think pants are legally able to inherit anything: estate.findlaw.com/wills/inheritance-law-and-your-rights.html $\endgroup$ – Michael Apr 22 '17 at 23:14
  • $\begingroup$ I also notice the very first picture of the Denne blog you link to gives a picture of pair-of-pants as topologically similar to a disc with two holes. So that disc (minus the holes) can be embedded to the outside or inside of any surface you like, such as a sphere, to have any curvature you like. $\endgroup$ – Michael Apr 22 '17 at 23:20
  • $\begingroup$ I think it is unknown. $\endgroup$ – Moishe Kohan Apr 23 '17 at 3:01

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