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A convex polytope is projectively unique if it has a unique realization up to projective transformations. Such polytopes are somewhat mysterious but still well-studied. Examples are simplices, the square, the triangle prism, and many many more (see [1], and also this older question of mine).

Question: Are there polytopes that have precisely two realizations up to projective transformations?

I am aware of the universality theorem (UT) for polytopes, though I am not sure whether it can carry me quite that far. The UT provides polytopes with realization spaces with two connected components, but can it also provide a realization space (mod projective transformations) made of precisely two points? In the formulation that I know, it can construct a realization space up to "stable homotopy", and I am not sure whether this is strong enough for my purpose.

Lastly, I am also aware of Gale diagrams for polytopes. I am convinced that there are point-line arrangements with precisely two realizations. Do they correspond to polytopes with two realizations? I am unsure because I don't know how Gale duality interacts with projective transformations.

[1] Adiprasito, K. A., Ziegler, G. M., "Many projectively unique polytopes". Invent. math. 199, 581–652 (2015).

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  • $\begingroup$ Why is the triangular prism projectively unique? It seems like it has $14$ degrees of freedom, compared with the $12$ offered by projective transformations. (Concretely, I think the prism with vertices $\{(0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,0,1),(0,1,1)\}$ and the prism where we replace the $(0,1,1)$ vertex with $(0,1,2)$ are combinatorially but not projectively equivalent.) Am I misunderstanding the definition? $\endgroup$
    – RavenclawPrefect
    Commented Aug 26 at 16:35
  • $\begingroup$ @RavenclawPrefect The easiest way for me to see that it is projectively unique is to use that it is dual to the triangular bipyramid which has five vertices which can be arranged arbitrarily by projective transformations. The realization space of a polytope and its dual are "the same". But I am not exactly sure where the mistake lies. How do you count DOFs for the prism? $\endgroup$
    – M. Winter
    Commented Aug 26 at 17:48
  • $\begingroup$ Whoops - I miscounted and should have $15$ DOFs for both the prism and the triangular bipyramid, but this doesn't resolve my confusion. (The triangular bipyramid is easy to count DOFs for: it is combinatorially equivalent under small perturbations of any of its vertices, so we have 3 DOFs per vertex times 5 vertices.) I don't think the triangular bipyramid has a unique realization either, though? I don't think there are projective transformations which fix 4 out of 5 vertices and let the fifth move freely, at least in the generic case. $\endgroup$
    – RavenclawPrefect
    Commented Aug 26 at 21:05
  • $\begingroup$ @RavenclawPrefect Are you sure you counting correctly the DOFs of projective transformations including translations? $\endgroup$
    – M. Winter
    Commented Aug 26 at 22:37
  • $\begingroup$ I believe so? 3 DOFs for the location of the origin, then 9 more for the 3x3 matrix specifying a linear transformation, makes 12. $\endgroup$
    – RavenclawPrefect
    Commented Aug 26 at 22:59

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Yes, and there are infinitely many of them. If you have trouble visualizing Gale duality, the easiest thing may be to instead look at the way Lawrence lifts go step by step (as is explained in our paper), and think about how they interact with projective transformations.

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  • $\begingroup$ Thanks Karim. I think I have no particular trouble with "visualizing" Gale duality (I prefer it over the Lawrence lift, at least so far), but I wonder: is it sufficient to find a Gale diagram with only two realizations (up to some set of transformations)? Is it easy to state what transformations projective transformations turn into on the side of the Gale diagram? Do you know of a "simplest" polytope with precisely two realizations that has a name or easy construction, or would you just go via the "simplest" Gale diagram that comes to mind? $\endgroup$
    – M. Winter
    Commented Aug 27 at 9:20
  • $\begingroup$ Projective transformations in one correspond to projective transformations in the other one. $\endgroup$
    – Karim Adiprasito
    Commented Sep 8 at 1:46
  • $\begingroup$ I don't know how to do line break on a phone. I don't have a simplest on hand, but essentially you turn an algebraic equation with two solutions into a point configuration using the von Staudt diagrams $\endgroup$
    – Karim Adiprasito
    Commented Sep 8 at 1:48

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