What is known about the **duals of [cyclic polytopes](https://en.wikipedia.org/wiki/Cyclic_polytope)**, in particular, their *facets* (or equivalently, the vertex-figures of cyclic polytopes)?


* In *even* dimensions, all facets of the dual are combinatorially equivalent. Are these facets themselves duals of cyclic polytopes?
* I think this cannot be true in odd dimensions $\ge 5$. For example, a 5-dimensional cyclic polytope is [2-neighborly](https://en.wikipedia.org/wiki/Neighborly_polytope), so its vertex figures are only 1-neighborly (is this true?), but 4-dimensional cyclic polytopes are 2-neighborly as well. So when does it happen that the facets are again duals of cyclic polytopes, and when are they all combinatorially equivalent?
* In general, is there some kind of classification of the combinatorial types of these facets?



What is known about the duals of cyclic polytopes?
Many properties of the vertex figures of cyclic polytopes can be obtained from Gale evenness condition. 


Let $P=C(n,d)$ be a cyclic $d$-polytope, and let $v\_1<\cdots<v\_n$ be its vertices ordered according to the moment curve. The following follows from Gale evenness condition. 


1. In even dimensions, the vertex figure of $P$ at every vertex is a cyclic $(d-1)$-polytope.
2. In odd dimensions, every facet contains $v\_1$ or $v\_n$.
3. In odd dimensions, for $d\ge 5$ and $n\ge d+2$, the vertex figures of $P$ at $v\_1$ and $v\_n$ are cyclic $(d-1)$-polytopes, but the vertex figure at some other vertex is not.


The proofs of 1) and 2) follow straight from the condition. 


For the proof of 3) use counting. First, for every $d\ge 4$ count in two different ways the vertex-facet incidences in $P$ in the case that each vertex-figure is a cyclic polytope, and obtain the following:


$$df\_{d-1}(C(n,d))=nf\_{d-2}(C(n-1,d-1))$$


Second, for odd $d\ge 5$, establish the following $$f\_{d-1}(C(n,d))=2f\_{d-2}(C(n-1,d-1))-f\_{d-3}(C(n-2,d-2))$$


Comparing the two previous expressions gives 3).


Duality gives the results you are after. 


Regards,
Guillermo


