- For every polytope \(P\), the face poset \(L(P)\) is a graded lattice of length \(\dim(P)+1\), with rank function \(r(F)=\dim(F)+1.\)
- Every interval \([G,F]\) of \(L(P)\) is the face lattice of a convex polytope of dimension \(r(F)-r(G)-1.\)
- ("Diamond property") Every interval of length \(2\) has exactly four elements. That is, if \(G \subseteq F\) with \(r(F)-r(G)=2,\) then there are exactly two faces \(H\) with \( G \subset H \subset F, \) and the interval \([G,F]\) looks like
- The opposite poset \(L(P)^{\mathrm{op}}\) is also the face poset of a convex polytope.
- The face lattice \(L(P)\) is both atomic and coatomic.
Proof (i):
To show that \(L(P)\) is a lattice, let \(F\) and \(G\) be any two faces in \(P\). we want to show that there is a unique meet \(F \wedge G\) and a unique join \(F \vee G\). By proposition 2.3(ii), every intersection of faces of \(P\) is a face of \(P\). Hence \(F \cap G\) is a face. We claim that \(F \cap G\) is a meet.To see this, suppose \(H\) was any lower bound, then \(H \subseteq F\) and \(H \subseteq G\) so \(H \subseteq F \cap G\). Therefore, \(F \cap G\) is the greatest lower bound so it’s a unique meet.
For the join, observe that \(P\) is the unique maximal element of \(P\) and \(\emptyset\) is the unique minimal element of\(P\). Let \(U\) be the set of all the common upper bounds of the faces \(F\) and \(G\). We know \(P \in U\) so \(U\) is not empty. We also know that any element in \(U\) is a face. Let \(H\) be the intersection of all elements in \(U\):
\(H\) is a face by proposition 2.3(ii). Since \(F \subseteq K\) and \(G \subseteq K\) for all \(K \in U\), then \(F \subseteq H\) and \(G \subseteq H\). By construction \(H\) is the smallest element contained in \(U\). Hence \(H\) is the least upper bound of the faces \(F\) and \(G\). So it’s a join \(F \vee G\). Thus, \(L(P)\) is a lattice.
Next, we need to show that \(L(P)\) is graded. To do this, we must show that it is bounded and every maximal chain has the same length. We have shown that it is bounded and every maximal chain in a bounded poset must start at \(\emptyset\) and end at \(P\). To show that every maximal chain has the same length, first if we have two \(F\) and \(G\) such that \(G \subseteq F\), then if \(G \neq F\), \(\dim(G) < \dim(F)\). To see this, suppose it wasn’t true and that \(\dim(F) = \dim(G)\). Since \(G \subseteq F\), we have
\(\operatorname{aff}(G)\subseteq \operatorname{aff}(F)\).
But affine spaces having the same dimension imply that
\(\operatorname{aff}(G)=\operatorname{aff}(F)\). Hence by Proposition 2.3(iv)
So \(F = G\) which is a contradiction. Therefore, if \(G \neq F\), then \(\dim(G) < \dim(F)\).
Second, we claim that if \(\dim(F) - \dim(G) \geq 2\), then there exists a face \(H\) such that \(G \subset H \subset F\). By (ii), the interval \([G,F]\) is the face lattice of a polytope of dimension at least \(1\). This implies that we have some nontrivial face other than the minimal and maximal elements. In \([G,F]\), the maximal element is \(F\) and the minimal element is \(G\). So this implies that there must exists some face \(H\) such that \(G \subset H \subset F\).
Now suppose we have a cover relation \(G \prec F\). Then we know we don’t have any faces in between \(G\) and \(F\). By the earlier two arguments, since \(G \neq F\) and since \(\dim(F) - \dim(G) \geq 2\) would imply having an intermediate face \(H\), then we must have \(\dim(F) - \dim(G) = 1\).
Finally, consider any maximal chain
Since every step is a cover relation, the previous argument shows that \(\dim(F_i)-\dim(F_{i-1})=1\) for every \(i\). Since \(\dim(\emptyset)=-1\), then \(\dim(F_i)=i\). Hence every maximal chain contains exactly one face of each dimension: \(-1,0,1,\ldots,\dim(P)\). Therefore, every maximal chain has length \(\dim(P)+1\). Thus every maximal chain has the same length, so \(L(P)\) is graded. Moreover, the rank of a face \(F\) is