The sum of angles of a triangle depends on the curvature of the surface and can deviate from $\pi$. What about the sum of angles between successive lines emanating from a given point P? Can it deviate from $2\pi$, depending on the curvature at P? (I guess not.) How is it proved?
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I think it's just a matter of definitions. No matter what curvature a line has, it must intersect a point at one angle, meaning all normal rules apply for planar lines intersecting a point. – Neil Jan 21 '13 at 13:50
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If $P$ is an ordinary point of a surface $S$ then the angles between "rays" emanating from $P$ are measured in the tangent plane at $P$ which is nothing but an ordinary euclidean plane with a distinguished point $O$. That the sum of the angles between rays emanating from $O$ is constant is a deep fact of elementary geometry, but this fact can be established without resorting to Riemannian geometry.
If $P$ is a "special" point of your surface $S$, e.g., the tip of a circular cone, where a positive amount of curvature is concentrated in one point, then the sum of the angles between rays emanating from $P$ is no longer $=2\pi$.
Christian Blatter
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The point seems to be: the angles of the triangle are not measured in the tangent plane (but on the osculating sphere). Is this the "essential" lesson to learn? – Hans-Peter Stricker Jan 21 '13 at 13:55
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1@Hans Stricker: The angles of a triangle are measured in three tangent planes at three different points. The theorem of Gauss-Bonnet's relates the sum of the three measured angles to the curvature (and area) within the triangle. – Christian Blatter Jan 21 '13 at 14:00