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Triangles in rigid Riemann surfaces
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Triangles in rigid Riemann surfaces
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Triangles in rigid Riemann surfaces
By the way, the definition with tubular neighborhoods, allows for two isometric triangles to have interiors with arbitrary topology, and so two isometric triangles homologous to zero can bound interiors with completely different area.
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Triangles in rigid Riemann surfaces
Ah, no problem. It's been an interesting discussion. Plus I'm not even sure which term is better in this situation.
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Triangles in rigid Riemann surfaces
I think when people say congruent they mean global isometry, like a ''motion" of the whole space (e.g. hyperbolic plane) that moves one triangle onto the other, so that they match precisely. It is a notion related to the isometry group of the whole space (hence Igor's answer). The term isometry, however, I think means something more general and can be used in a more flexible way: a map that preserves lengths (and angles when angles are well defined). Hence my definition of isometric triangular piece-wise geodesic curves (we call triangles) in terms of annular/tubular/collar/ neighborhoods.
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Triangles in rigid Riemann surfaces
Yeah, it seems to me that your comment confirms my assumption that what I wrote after the edit is what you are probably interested in. I basically tried to put it in a more ''formal" way. I am sorry if I have been a bit sloppy and not too precise.
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Triangles in rigid Riemann surfaces
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Triangles in rigid Riemann surfaces
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Two limit cycles which lie on the same leaf
Yes, I think $(x^2+y^2−1)(x^2+y^2−4)−1$ has a singularity at infinity, so as a compactified Riemann surface it is singular. First take $y \mapsto iy$ to rotate the points at infinity in the real projective plane and then perform the projective transformation $x \mapsto 1/x$ and $y \mapsto y/x$. Then you see two ovals touching at two points. In my example this does not occur because it is ultra-Morse: its leading highest order term is of the form $(x-a_1y)...(x-a_ny) + \text{lower order terms}$ which makes it non-singular at infinity for different $a_j$.
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Ricocheting pinball-like shot: Complexity?
I cannot give an answer to the question, but I can just mention a paper that looks at reflection problems for segments with rational angles: arXiv:1105.2972. In case it can help somehow...