Let $P$ be a (convex, bounded) polytope with the following property: for every vertex $v$, there are exactly two facets which do not contain $v$. Does it follow that $P$ is (combinatorially) a Cartesian product of two simplices?


Some remarks


* by "facet" I mean "face of codimension $1$"
* a Cartesian product of two simplices has the desired property
* this is trivially true for $\dim P = 2$
* this is true for $\dim P = 3$: by playing with the Euler formula and the list of polyhedra with small $f$-vectors, one checks that only the triangular prism satisfies the condition
* it is a classical fact that a $n$-dimensional **simple** polytope with $n+2$ facets is a Cartesian product of two simplices. So the answer is positive if $P$ is assumed to be simple.



Polytope where each vertex belongs to all but two facets
There are other polytopes with this property that can be obtained via the *free join* construction.


Given two polytopes $P\_1\subset\Bbb R^{d\_1}$ and $P\_2\subset\Bbb R^{d\_2}$, the free join $P\_1\bowtie P\_2$ is obtained by embedding $P\_1$ and $P\_2$ into skew affine subspaces of $\Bbb R^{d\_1+d\_2+1}$ and taking the convex hull.


The claim is that if $P\_1$ and $P\_2$ have your property, then so does $P\_1\bowtie P\_2$.
To see this, note that the facets of $P\_1\bowtie P\_2$ are exactly of the form $P\_1\bowtie f\_2$ and $f\_1\bowtie P\_2$, where $f\_i$ is a facet of $P\_i$. Now, if $v$ is a vertex in, say, $P\_1\subset P\_1\bowtie P\_2$ that is in all facets of $P\_1$ except for $f,f'\subset P\_1$, then $v$ is in all facets of $P\_1\bowtie P\_2$ except for $f\bowtie P\_2$ and $f'\bowtie P\_2$. Equivalently for vertices in $P\_2$.


**Example.** Take the free join of two squares, which is a 5-dimensional polytope with 8 vertices, 8 facets and vertex degree 6. This polytope is not simple and can thus not be the cartesian product of two simplices.




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This construction yields counterexamples in dimension $d\ge 5$.
Concerning $d=4$, there should not be any counterexamples in this dimension.


Suppose that $P\subset\Bbb R^4$ is a counterexample.
According to [Moritz Firsching's comment](https://mathoverflow.net/questions/402842/polytope-where-each-vertex-belongs-to-all-but-two-facets/403080?noredirect=1#comment1030725_402842) we can assume that it has $n\ge 10$ facets. Each vertex is then contained in exactly $n-2\ge 8$ of them. More generally, each set of $k$ vertices is contained in at least $n-2k$ facets. But let $f\subset P$ be a 2-face of $P$ and $v\_1,v\_2,v\_3\in f$ three affinely independent vertices, then this set of vertices, and thus also $f$, is contained in $n-6\ge 4$ facets. This cannot be, since each 2-face of $P$ is contained in exactly two facets.


I believe this generalizes to show that any polytope with your property must have $\le 2d$ facets.


