# Continuous transformations of a triangle bound on $S_1$

Suppose we take the circle $S_1$ and three points on this circle, which defines a triangle. By moving the points continuously on $S_1$, we obtain a continuous transformation of the triangle.

I was wondering what is the structure of this group of transformations ?

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it seems like $S^1\times S^1\times S^1$ as you can move each vertex individually, unless you start identifying various configurations like $(\theta,\theta,\theta)=(0,0,0)$ – yoyo Mar 21 '11 at 21:05
@yoyo: I think that only holds if you label each vertex and allow "triangles" which have two or three overlapping points. – Alex Becker Mar 21 '11 at 22:02

If you are not allowing "degenerate" triangles and you do not want to label the vertices, then the space of triangles is the same as the space of unordered triples of distinct points on a circle. The space of ordered triples of distinct points is a subset of $S^1\times S^1\times S^1$ (namely $(S^1)^3$ minus the "diagonal"), and the set of unordered triples is the quotient of this space by the action of the symmetric group $S_3$ that permutes the coordinates (that is, it equates $(x,y,z)$ with $(x,z,y)$ and $(y,x,z)$ and so on).
Yes, more likely it is just a space with an $S^1$ action. – user641 Mar 22 '11 at 1:33