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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.

Malkoun
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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.

M. Winter
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  • Interesting argument. I think this does show that what I am looking for, meaning a geometric duality of the type I am thinking about does not exist, already in dimension $2$. – Malkoun Sep 23 '20 at 19:42
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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.

Guy Inchbald
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  • The correspondence between polytopes and their edge-graphs breaks down in higher dimensions. The edge-graph does no longer determine the polytope's combinatorics (I think this is still true for spherical polytopes). Also, what is the dual graph of a graph embedded in a sphere of dimension $d\ge3$? – M. Winter Sep 23 '20 at 12:09
  • Okay, clear, we are talking about some kind of complex. But it is not clear to me from the answer how to obtain a "dual complex" in any case except the usual graph case, or when we already have a convex polytope. Can you go more into the details here? – M. Winter Sep 23 '20 at 14:02
  • @M.Winter I have edited the last para to make clear that the "how to" applies to higher dimensions as well as the three in the example. Also further up to help distinguish between edge-graphs and generalised graphs. – Guy Inchbald Sep 23 '20 at 14:17
  • I think the situation becomes more complicated in higher dimensions. Also polar reciprocation works fine for "usual" polytopes, but unmodified, it does not work well for tilings of the sphere (what I was referring to as "spherical polytopes" but I guess this is perhaps not the standard name for it). – Malkoun Sep 23 '20 at 14:55
  • Thank you for the changes, but they still give not enough infomation too actually construct the dual complex. Looking up "polar reciprocation" only yields results about the 2-sphere. Can you either point to a reference explaining this in higher dimensions, or include some details in the answer? An equivalent question was asked on MO without much success, so I have to assume that the problem is non-trivial. – M. Winter Sep 23 '20 at 14:56
  • @Malkoun It works especially well for tilings of the sphere, as the surface of reciprocation is uniquely defined. – Guy Inchbald Sep 23 '20 at 16:04
  • @M.Winter Polar reciprocity is essentially a manifestation of projective duality and is best discussed in that context. Throw in the duality of graphs and CW complexes, along with higher-dimensional spherical geometry, and I know of no single unified treatment. Briefly, for polytope elements E_j of dimensionality j ( j ≤ 0 ≤ n) the incidence structure is retained but elements for j = 0 and n exchange dimensionalities, as do those for j = 1 and (n-1) and so on. Really, one has to start with 2D, move on to 3D and only then worry about higher dimensions. Would it help to add any of that? – Guy Inchbald Sep 23 '20 at 16:20
  • @M.Winter Just to add, the OP did not ask how it could be constructed, only whether it could. Maybe the how belongs to other questions. – Guy Inchbald Sep 23 '20 at 16:25