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I'm having a bit of trouble formally proving the following question from differential geometry:

Show that every smooth embedded torus in $\mathbb R^3$ has a point with negative curvature.

I know intuitively that this is true. If we look at the "inner" portion of the torus it is basically something similar to the saddle surface, which is a surface whose curves from the tangent plane go into two different directions. Hence it would make sense why there must exist a single point say $p$ in our torus $T$ that has negative curvature.

I am just a unsure on how to go about the proof formally. Any advice is appreciated.

richie
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    See https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem#Interpretation_and_significance , if you can use the Gauss-Bonnet theorem and Euler characteristic. – aschepler May 18 '22 at 18:18
  • That wikipedia link is sort of vague, but as far as I can tell it certainly does not address every smooth embedded torus in $\mathbb R^3$. – Lee Mosher May 18 '22 at 18:47
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    Yeah, that link isn't quite sufficient. But $\chi=0$ and Gauss-Bonnet tell you that if there are points of positive curvature, there must be points of negative curvature. Now show that every compact surface in $\Bbb R^3$ must have a point of positive curvature. – Ted Shifrin May 18 '22 at 20:52

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