R^3 injection?

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13 events

when toggle format	what		by	license	comment
Jan 6, 2013 at 12:41	vote	accept	Leon Avery
S Jan 6, 2013 at 12:41	vote	accept	Leon Avery
Jan 6, 2013 at 12:41
Jan 6, 2013 at 2:32	answer	added	Gil Bor		timeline score: 2
Jan 5, 2013 at 20:48	vote	accept	Leon Avery
S Jan 6, 2013 at 12:41
Jan 5, 2013 at 19:56	comment	added	algori		Leon -- I see, so you don't allow reflections. In this case the space, call it $X$, is an open cone over the set of all triangles with the sum of the sides equal 1. The latter space, call it $Y$, is the union of two triangles with sides of one glued to the sides of the other in a bijective way, i.e., $Y=S^2$, which makes $X$ homeomorphic to $\mathbb{R}^3$.
Jan 5, 2013 at 19:54	answer	added	Eric Wofsey		timeline score: 3
Jan 5, 2013 at 19:30	comment	added	Leon Avery		SAS is an injection, but not continuous. It is discontinuous at (x1,y1) = (x2,y2) and also at (x2,y2) = (x3,y3). (This is actually what I was thinking of when I said "This is easy." in the OP.) SSS is continuous, but not an injection. As you say, "each triangle is determined, up to a composition of rotations, REFLECTIONS and translations, by the lengths of the edges". But reflections are not rigid-body motions. This is typical of all the functions I've found: the closest I can come to a continuous injection always lose one bit of information. Still, SSS is an interesting idea...
Jan 5, 2013 at 19:29	comment	added	algori		Todd -- so you did but I didn't see it when I started writing mine.
Jan 5, 2013 at 19:20	comment	added	Todd Trimble		@algori: I could have sworn I just made a similar comment...
Jan 5, 2013 at 19:08	comment	added	algori		unknown google -- if I understand you correctly, your quotient space is juet the space of all triangles in $\mathbb{R}^2$ with ordered vertices; it includes degenerate triangles, in which one of the vertices lies in the interior of the segment that joins the other two. This space is indeed a subspace of $\mathbb{R}^3$: each triangle is determined, up to a composition of rotations, reflections and translations, by the lengths of the edges (these are ordered, as the vertices are). If you do not allow reflections, then a triangle is determined by the lenghts of the sides plus orientation.
Jan 5, 2013 at 19:01	comment	added	Todd Trimble		(Or perhaps even more simply, side-side-side.)
Jan 5, 2013 at 18:55	comment	added	Todd Trimble		Maybe I'm misunderstanding, but it sounds like side-angle-side from Euclidean geometry should do the trick. The first side being the distance from point 1 to point 2, the second side being the distance from point 2 to point 3, and the angle being the oriented angle which swings the ray 1 (originating at point 2 and passing through point 1) to ray 2 (originating at point 2 and passing through point 3), swinging in the counterclockwise direction. The data (S, A, S) gives you the point in $\mathbb{R}^3$. Does this seem right?
Jan 5, 2013 at 18:38	history	asked	Leon Avery	CC BY-SA 3.0