Is there a notion of dual for a spherical polytope? I am aware of the notion of polar dual for a flat convex polytope (by a flat convex polytope, I mean the convex hull of finitely many points in $\mathbb{R}^d$). Suppose you have instead a spherical polytope. Is there a notion of duality for a spherical polytope where, preferably, the dual is also a spherical polytope? Could someone maybe point me to the definition please?
If someone wants to keep the discussion in low dimension, and talk about spherical polyhedra instead of spherical polytopes in general, then this is also fine.
 A: Let me say something about the case for polyhedra (it may generalize, but I am not sure about the details).
It is not too hard to imagine that there might be a "combinatorial duality" for spherical polyhedra, in the sense that the dual of a spherical polyhedron exists, but is only determined up to combinatorial equivalence (e.g. via dual planar graphs).
But I want to argue that there can not be a geometric duality, i.e. a duality that to every concrete spherical polyhedron gives you another one, and taking the dual again brings you back to the original one.
The reason is, that given the combinatorial type of a spherical polyedron, the realization space of that type (i.e. the space of all spherical polyhedra with this combinatorial type) has a local dimension of $2n$, where $n$ is the number of vertices.
What do I mean by that: you can describe your spherical polyhedron basically by drawing some points on the sphere, and stating between which points there should be a line. The line is then uniquely determined as the great circle arc between these points (yes, there is a choice which arc to take, but lets ignore this for now).
So if we placed our points carefully, then none of these arcs intersect, and what we have is a spherical polyhedron.
But note that we can move each point slighly, and the arcs move accordingly. And if we moved the points slightly enough, then the arcs stay disjoint, and the construct stays a spherical polyhedron.
Since each vertex moves on the surface of the 2-sphere, each vertex has two degrees of freedom, and the whole construct has $2n$ degrees of freedom.
Now, consider the spherical cube, whose dual (if our duality is meaningful in any way) is the spherical octahedron.
But the first one has $2\times 8=16$ degrees of freedom, and the latter one only $2\times 6=12$.
So not every unique realization of the spherical cube can be mapped into a unique realization of the spherical octahedron, and so the geometric duality fails.
A: A spherical polyhedron in some 3-space is equivalent to a simple graph drawn on the sphere. The dual polyhedron is just the dual graph.
Generalising graphs to appropriately-constrained CW complexes in n dimensions, a spherical polytope in some (n+1)-space is equivalent to the associated generalised graph. Its dual is again just the dual graph or complex.
Typically the dual may be obtained via polar reciprocation about the centroid of the n-sphere (i.e. of the (n+1)-ball whose surface is the n-sphere). For example the reciprocal of the spherical cube is the spherical octahedron.
