# All the tetrahedra that can be inscribed in the unit sphere up to isometry

I'm interested in restricting the set, $A$, of all tetrahedra whose vertices lie on the unit sphere to a set, $B$, such that every element of $A$ can be rotated/reflected into a unique element of $B$. Everything I've got is based on spatial intuition, and I want to know how to formalize it.

The two dimensional case, with the unit circle and a triangle, seems easy; just fix one of the vertices of the triangle at (1, 0) (this should take care of rotations) and demand that the vertex closer to (1, 0) has a positive y-coordinate (this should take care of reflections).

The three dimensional case seems somewhat trickier. I start by fixing one of the vertices at (1, 0, 0). There is still a bunch of tetrahedra that could be obtained by rotating the other vertices about the x-axis, so I demand that the second vertex has a y-coordinate of zero. To nullify a possible reflection I also demand that this vertex has a z-coordinate greater than or equal to 0. There are still different tetrahedra that could be obtained by reflecting across the xz-plane so I demand that the third point has a positive y-coordinate. Also, since a tetrahedron is really characterized by a set of vertices, not a tuple, I should say that the nth vertex is the is the nth closest to the fixed vertex (1, 0, 0), if that makes sense.

As you can see, my arguments are completely heuristic. I want to know if they are accurate and how to formalize them.

UPDATE:

Perhaps we're in n dimensions with the unit sphere, $S^{n-1}$, and some n-simplex inscribed in it. We could rotate our simplex in such a way that minimizes the sum of the distances between the vertices and each of $\frac{n(n-1)}{2}$ points on $S^{n-1}$.

The dimension of the space of all the simplices is $(n+1)(n-1)=n^2-1$ because of the $n+1$ vertices each one having $n-1$ degrees of freedom. Requiring the distance to a single point be minimized is a single constraint and so reduces the number of degrees of freedom to $n^2-2$. The dimension of the rotation group is $\frac{n(n-1)}{2}$, so I think the quotient space we're after will have the dimension $\frac{n^2}{2}+\frac{n}{2} - 1$. To get this dimension we need to set $\frac{n(n-1)}{2}$ constraints, which is what I was aiming for by minimizing the distance to $\frac{n(n-1)}{2}$ points. I don't think it accounts for reflections, but its a start.

• For what it's worth, the set of all subsets of a given cardinality in a given topological space is called a configuration space. What you are looking at in particular is the quotient of a conf. space by the action of $O(3)$. An introduction can be found here mimuw.edu.pl/~sjack/prosem/… – Arnaud Mortier Apr 8 '18 at 20:13
• Also: "demand that the vertex closer to (1, 0) has a positive y-coordinate (this should take care of reflections)." has the problem that the two remaining vertices might both be equally close. (In 2D this works out OK. In 3D, there may be chirality problems.) – John Hughes Apr 8 '18 at 20:27