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Edit: I understand your question correctly, this poset I(S) cannot be was including the empty set as a face poset of any polytopein my earlier answer, by which gives the following argument.

Polytopes are in particular regular CW complexeswrong poset I(S). You cannot I have two non-homeomorphic regular CW complexes with now corrected my answer below so as not to include the same face posetempty set, because the order complex of the face poset of a regular CW complex is significantly changing the barycentric subdivision of that regular CW complex, hence homeomorphic to itanswer.On the other hand, your

The poset I(S) with its top element deleted, denoted $I(S)\setminus$ {1}, should cannot be the face poset of the boundary of your desired any polytope, hence because it will have multiple maximal elements of a sphere, in order the form $[v,P]$ for I(S) to be the face poset various vertices of a polytopeP. But $I(S)\setminus$ {1} is I(S) should be the face poset of a subdivision of your original polytope, hence is a ball rather than a sphere, contradictionP. This subdivision is less refined than the barycentric subdivision of the polytope. Its vertices are the barycenters of faces, but its edges only connect barycenters of faces of consecutive dimensions, in contrast to the barycentric subdivision of the original polytope where they would come from all face inclusion pairs. One can continue upward in dimension, likewise describing the faces of the subdivison by progressively filling in its lower skeleta.

You also ask about the cube specifically. Your subdivision in that case is cubical, breaking each i-dimensional cubical face of the original cube into 2^i cubical faces of dimension i.

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If I understand your question correctly, this poset I(PI(S) cannot be the face poset of any polytope, by the following argument.

Polytopes are in particular regular CW complexes. You cannot have two non-homeomorphic regular CW complexes with the same face poset, because the order complex of the face poset of a regular CW complex is the barycentric subdivision of that regular CW complex, hence homeomorphic to it. On the other hand, your poset I(PI(S) with its top element deleted, denoted $I(P)\setminus I(S)\setminus$ {1}, should be the face poset of the boundary of your desired polytope, hence of a sphere, in order for I(PI(S) to be the face poset of a polytope. But $I(P)\setminus I(S)\setminus$ {1} is the face poset of a subdivision of your original polytope, hence is a ball rather than a sphere, contradiction.

This subdivision is less refined than the barycentric subdivision of the polytope. Its vertices are the barycenters of faces, but its edges only connect faces of consecutive dimensions, in contrast to the barycentric subdivision of the original polytope where they would come from all face inclusion pairs. One can continue upward in dimension, likewise describing the faces of the subdivison by progressively filling in its lower skeleta.

You also ask about the cube specifically. Your subdivision in that case is cubical, breaking each i-dimensional cubical face of the original cube into 2^i cubical faces of dimension i.

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If I understand your question correctly, this poset I(P) cannot be the face poset of any polytope, by the following argument.

Polytopes are in particular regular CW complexes. You cannot have two non-homeomorphic regular CW complexes with the same face poset, because the order complex of the face poset of a regular CW complex is the barycentric subdivision of that regular CW complex, hence homeomorphic to it. On the other hand, your poset I(P) with its top element deleted, denoted $I(P)\setminus$ {1}, should be the face poset of the boundary of your desired polytope, hence of a sphere, in order for I(P) to be the face poset of a polytope. But $I(P)\setminus$ {1} is the face poset of a subdivision of your original polytope, hence is a ball rather than a sphere, contradiction.

This subdivision is less refined than the barycentric subdivision of the polytope. Its vertices are the barycenters of faces, but its edges only connect faces of consecutive dimensions, in contrast to the barycentric subdivision of the original polytope where they would come from all face inclusion pairs. One can continue upward in dimension, likewise describing the faces of the subdivison by progressively filling in its lower skeleta.

You also ask about the cube specifically. Your subdivision in that case is cubical, breaking each i-dimensional cubical face of the original cube into 2^i cubical faces of dimension i.