# Is an Eulerian lattice shellable?

The notion of Eulerian lattice generalizes the notion of face lattice of a convex polytope.

(Bruggesse-Mani): The boundary complex of a convex polytope is shellable.

(Björner-Wachs): A poset is said to be shellable if its order complex is shellable. A bounded poset is shellable if and only if its proper part is shellable.

(Wachs): The order complex of the face poset of a simplicial complex is its barycentric subdivision.

Pre-question: Is the shellability kept by taking the barycentric subdivision?
If so, the face lattice of a convex polytope is shellable. This leads to:

Question: Is an Eulerian lattice shellable?

• At counter example to you question does not come from the lattices of any convex $3$-dimensional polytopal complex. In those cases the barycentric subdivision is always shellalbe. link.springer.com/article/10.1007/s00493-016-3149-8 . But I expect a counterexample for higher dimensions. – Moritz Firsching Mar 16 '18 at 19:14

The simplest example of a nonshellable (even non-Cohen Macaulay) Eulerian lattice is the disjoint union of two boolean algebras of rank three with their bottom elements identified and their top elements identified. The total number of elements is 14.

• Yes! It is drawn in this answer. About "simplest example": first an Eulerian lattice is graded, and it is known that a graded poset of length $1$ or $2$ is Cohen–Macaulay, and a graded poset of length $3$ is Cohen–Macaulay if and only if its proper part is connected. Now, if an Eulerian lattice has lenght $3$ and a disjoint proper part $P$, then the bounded extension of a connected component of $P$ must be an Eulerian lattice too, but the smallest Eulerian lattice of lenght $3$ is the Boolean lattice of rank $3$. The result follows. – Sebastien Palcoux Mar 17 '18 at 7:56

I think that the answer to your question ought to be "not always".

There are examples of non-shellable balls and spheres. A very readable account of these, together with some of the history, can be found at this blog entry.

Now if you take the face lattice of a sphere, that is an Eulerian lattice. Passing from a complex $\Delta$ to the order complex of the face lattice of $\Delta$ yields the barycentric subdivision of $\Delta$.

So to find a negative answer to your question, find a non-shellable sphere whose barycentric subdivision also fails to be shellable. The construction of such spheres is briefly discussed in this paper. There is likely a source somewhere that constructs such a sphere more explicitly.

Update: Also, the answer to your pre-question is "yes". Indeed, the barycentric subdivision of any shellable complex is vertex-decomposable, a somewhat stronger property.

• Why the existence of a non-shellable sphere does not contradict the above theorem of Bruggesse and Mani? – Sebastien Palcoux Mar 16 '18 at 19:32
• @SebastienPalcoux, the boundary of every polytope is a sphere. But not every sphere is the boundary of a polytope! – Russ Woodroofe Mar 16 '18 at 19:46
• If a filled non-shellable sphere is not a polytope, then what is it? – Sebastien Palcoux Mar 17 '18 at 10:59
• @SebastienPalcoux, I think you're assuming that all topological spheres are convex. Take the boundary of a cube in R^3. Subdivide the top facet by "pushing it in". That is, replace the top facet with the cone over its boundary, where the cone point is below the facet. 'Filling' the resulting figure yields something that is not convex. – Russ Woodroofe Mar 17 '18 at 11:36
• Yes, sure. Then a non-shellable sphere is the boundary of non-convex polytope, isn't it? Now, the face lattice of a convex polytope is Eulerian, so I guess that there is a non-convex polytope whose face lattice is not Eulerian. But you are right, the face lattice of a spherical non-convex polytope is still Eulerian. – Sebastien Palcoux Mar 17 '18 at 12:26