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I do have a problem with the following question: Given a surface $M$ with $K<0$ (gaussian curvature), show that there exists no $n$-gon for $n<3$.

Up until now, I came up with the following ideas: Following from Gauss-Bonnet: A simple closed geodesic on a surface with $K \leq 0$ cannot bound a disk to either side (because such a disk has $\int K = 2\pi$).

Regarding $n=1$ and $n=2$: There cant be geodesic $1$-gon or $2$-gon (disk) on a surface with $K ≤ 0$ (because geodesics that are tangent coincide, so the exterior angles are not $π$ but strictly less).

I wanted to check if my ideas are right - and if so I need some help to phrase those into a proper proof (I think I understood the basic idea for proving this, but I cant work out the details).

Thanks in advance.

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    for n=2, a diangle? – 2017-01-08
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    i think so, yes – 2017-01-08
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    I can understand what the case $n=2$ means: if a geodesic starts from A and arrives at B, there could no geodesic from B to A. Right ? But the case $n=1$ ??? – 2017-01-08
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    From what i gathered so far, since an n-gon is a regular curve with n-smooth segments that are geodesic, in the case that n=1 the curve would have one segment which is a geodesic, in the case of n=2 it would have 2, in the case of n=0 the curve itself is a geodesic. – 2017-01-08

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Whether $K$ is $0,$greater than $0,$ or less than $0,$ the $n=2$ case is a double geodesic line segment. $n=1$ case has no meaning.

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    allow me to ask, in what way does this relate to the question? – 2017-01-08
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    In a way that what we are talking has anything to do with K<0 geodesic. And n=1 is not a closed boundary at all. – 2017-01-08