consider a s-dimensional simplex in d-dimensional space. of course, s <= d. the simplex does not need to be regular, but it cannot be degenerate. an s-dimensional simplex has s+1 vertices. fix the position of s vertices, and let the one remaining vertex move around while not altering the shape of the simplex. what path or hypersurface does the remaining vertex cover? parameterize that surface with a function that takes a vector as input so that the zero vector as input puts the free vertex in its original position.

in 2D, a 1D simplex is a line segment. fixing the location of one of the two endpoints, the other can spin around in a circle. the circle can be parameterized by a function of angle, measuring from the original location of the free vertex.

a 2D simplex is a triangle. after fixing 2 vertices in 2D, it could be argued that the triangle cannot move at all. however, it is possible to mirror-image it across the fixed edge, "not altering the shape" by a definition that permits mirror image. the free vertex can therefore occupy two discrete positions. it's a little annoying but possible to compute the mirror image coordinates of a given point across the mirror line through the two other points.

a 1D simplex in 3D is a line segment pivoting around a ball-and-socket joint. the free vertex traces out a sphere. parameterize location of the free point on a sphere with spherical coordinates. even after fixing (0,0) there are a bunch of ways to set up the coordinates.

we call a sphere in 3D space a 2-sphere because its surface is a 2D manifold. the free vertex of a 1D simplex in 3D traces out a 2-sphere. in general, the free vertex of a s-simplex in d-space I think traces out a (d-s)-sphere. note that a 0-sphere is 2 isolated points, corresponding to mirror imaging. parameterizing the (d-s)-sphere with a vector of (d-s) components seems messy.

a triangle in 3D with an edge fixed in place is free to rotate around that edge. the free vertex traces a circle (1-sphere). finding a formula for that circle seems annoying but doable.

in 4D, the free vertex of a tetrahedron with its base nailed in place traces out a circle. what is a formula for that circle? 4D is impossible to visualize, but tetrahedra and circles can be visualized.

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