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If $X,Y,Z$ are the distances between three points in $\mathbb{R}^{3}$ such that $X,Y,Z$ satisfy the triangle inequality. What will be the configuration space of the three points, given the translation symmetry (fix one point at origin).

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Won't the distances between any three points always satisfy the triangle inequality (hence the name)? Or am I misunderstanding the first sentence? –  Zev Chonoles Apr 29 '11 at 19:05
    
they could be collinear. –  J Verma Apr 29 '11 at 19:14
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The formulation of the triangle inequality I have usually seen is with a $\leq$, not a $<$ - perhaps you could specify in your question that you mean the strict triangle inequality. –  Zev Chonoles Apr 29 '11 at 19:20
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I don't know how configuration spaces are usually specified, but we have one point at the origin, the second point can be chosen anywhere in $\mathbb{R}^3\setminus\{0\}\cong\mathbb{S}^2\times\mathbb{R}_{>0}$, and then the third point can be chosen anywhere off the line connecting the first two points, i.e. $\mathbb{R}^3$ minus a line, which is homeomorphic to $\mathbb{S}^1\times\mathbb{R}\times\mathbb{R}_{>0}$. So the configuration space would be $$\mathbb{S}^2\times\mathbb{S}^1\times\mathbb{R}\times\mathbb{R}_{>0}^2$$ Or are you looking for a subset of $\mathbb{R}^9$? –  Zev Chonoles Apr 29 '11 at 19:29
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@Zev: it's a non-trivial bundle over $S^2\times\mathbb{R}_+$ (with the fibre $S^1\times\mathbb{R}\times\mathbb{R}_+$), not just a product –  user8268 Apr 29 '11 at 19:41
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up vote 7 down vote accepted

the question is not very clear, so I'm not sure whether this is an answer: If the distances $X,Y,Z$ are fixed and satisfy strict triangle inequality then the configuration space is (diffeomorphic to) $SO(3)\cong\mathbb{RP}^3$, as $SO(3)$ acts freely and transitively on this configuration space. If they are not fixed and you simply want to exclude configurations of collinear points then you get $SO(3)\times\{(X,Y,Z)\in\mathbb{R}_+^3;X,Y,Z$ satisfy strict triangle ineqality $\}$ which is diffeomorphic to $SO(3)\times\mathbb{R}^3$.

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can you say more about how it is diffeomorphic to $SO(3)$. –  J Verma Apr 29 '11 at 19:44
    
@J Verma: take a preferred configuration - e.g. (the 1st point is the origin) the 2nd point is on the positive half of the $x$ axis and the 3rd point is right above this half axis. Any other configuration can be obtained from this preferred one via unique rotation. That's how you identify configurations with rotations. –  user8268 Apr 29 '11 at 20:13
    
@ user8268 - thanks. –  J Verma Apr 29 '11 at 21:07
    
There is an interesting discussion (in detail) of this in a book by David Kendall et al called Shape Theory. They discuss probability distributions on the configuration space and then the idea of a random triangle. –  Tim Porter Apr 30 '11 at 11:25
    
@ Tim - Can you tell me exact chapter of the book.Thanks. –  J Verma Apr 30 '11 at 16:32
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