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It appears that, based on Branko Grünbaum's paper "Are Your Polyhedra the Same as My Polyhedra?" (PDF link via washington.edu), some geometric polyhedra don't have a dual that is also geometric (though it exists as an abstract polyhedra).

However, the paper doesn't seem to give any specific examples. One would think that their example of co-planer faces could be perturbed to avoid coinciding vertices (which isn't allowed for geometric polyhedra) in it's dual. But how could I construct an example that doesn't allow this?

JPI
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Kevin Kostlan
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1 Answers1

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I think you may be referencing the quoted paragraph (pg. 469 - pg. 471). In future: it is helpful to point at specific parts of the reference, since it's possible I'm barking up completely the wrong tree.

Overall, I think the confusion comes from the fact that the author is making claims about particular, (usually) highly symmetric polyhedra. While you seem concerned with whether the mentioned properties apply to all deformations of the particular polyhedra.

For a given geometric polyhedron the construction of a dual polyhedron is most often carried out by applying to its faces and vertices a polarity (that is, a reciprocation in a sphere). From properties of this operation it follows at once that the polar of a given polyhedron is a realization of the abstract polyhedron dual to the given one. However, the possibility of carrying out the polarity depends on choosing a sphere for the inversion in such a way that its center is not contained in the plane of any face. While this is easy to accomplish in any case, the resulting shape depends strongly on the position of that center. The main problem arises in connection with polyhedra with high symmetry (for example, isogonal or uniform polyhedra) if it is desired to find a dual with the same degree of symmetry: If the only position for the center is at the centroid of the polyhedron, and the polyhedron has some faces that contain the centroid – then it is not possible to find a polar polyhedron with the same symmetry.

I suspect is possible to construct a polyhedron with its centroid contained in a face such that there exists no deformation which moves the centroid out of that face (without moving it into another face, at least). Even if this is not the case, the author states that for a polyhedron which has only one valid center of inversion (at the centroid) the polarity is not well defined. Of course we could deform the polyhedron (and lose some of its symmetry) and this may fix the problem, but no claim about the (im)possibility of this is made.

I'm not qualified to say whether such deformations exist, and even my first suspicion is a bit of a stretch. Perhaps another user can add to this.

Moreover, if a polyhedron has coplanar faces [coinciding vertices] then any polar polyhedron will have coinciding vertices [coplanar faces]. All these possibilities actually occur for various interesting polyhedra. Clearly, duality-via-polarity is uninteresting for subdimensional polyhedra – it yields only trivial ones.

As well, I don't see a claim that the perturbation you are asking about does not exist, but the claim is that if a polyhedron does indeed have coplanar faces then the corresponding polar polyhedron will have coinciding verticies. This is noteworthy since many highly symmetric polyhedra of interest will have coplanar faces, and thus the polar polyhedron must have coinciding vertices unless we destroy some of the symmetry of the original.

JPI
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