For any three points in a spacetime of constant positive (constant negative) curvature plausibly regarded as vertices of a triangle, is it always the case that the interior angles of the resulting triangle sum to greater than (less than) 180 degrees? I thought the answer to this question was straightforwardly 'yes'. But, given the following reason, I'm no longer sure about this. It was recently pointed out to me that, with respect to flat (i.e., Minkowski) spacetime, only triangles whose vertices are spacelike separated have interior angles summing to 180 degrees, and when they aren't so separated, the angles are meaningless.

So I'm wondering if the same situation arises in spacetimes of constant positive (constant negative) curvature. That is, in such spacetimes is it also the case that the interior angles of triangles sum to greater than (less than) 180 degrees only when the points of the vertices are spacelike separated, and when they aren't so separated, the angles are meaningless?


1 Answer 1


Well even on the surface of a sphere an equilateral triangle occupying 1/8 of the surface has three angles of 90 degrees, total 270 degrees. So I think you should hold the same view, that thge answer is yes.

Can two corners of a triangle be separated in time but not in space - I suppose so. A geometric line with one end existing instantaineously and the other for a finite period is such a triangle.

What then do the angles mean; The angles describe the ration of the geometric length to temporal duration of the long point. I wouldn't say that is meaningless.

I don't see a real reason to think the assertion doesn't hold absolutely even when time is considered a 4th geometric dimension.


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