# Why is Furch's “knotted hole ball” not shellable?

I'm trying to work through the example below, but I need some explanation as to why K' is not shellable. If we try to shell it, where would we get stuck?. Here are some relevant definitions.  • Note that $K$ has $m^3$ cubes as its facets. Removing some of them you end up with $K'$ which has $m^3-q$ cubes as its facets, where $q$ is the number of removed cubes. The boundary complexes of both $K$ and $K'$ are $2$-dimensional, the one of $K$ consists of $6m^2$ squares, the one of $K'$ has more squares. The sides $S$ and $S'$ are not faces in the complex $K$, but subcomplexes, each consisting of $m^2$ squares. Removing one square from $S'$ you get a subcomplex consisting of $m^2-1$ squares whose support is indeed not a polytope, but that's not an issue since it is not a face! – Christoph Nov 20 '19 at 17:01
• Thanks. I got it now. I edited the question so that others don't have to answer those parts again. I hope you don't mind. – ensbana Nov 20 '19 at 22:41

By analogy with Lemma 5.2, indeed, let's prove that pure 3D complex $$G$$ that contains a knotted (in $$G$$) curve with all edges except one on the boundary of $$G$$ can not be shellable.
Suppose $$G$$ is such a shellable complex with minimal number of facets. Let the curve be $$\gamma=C\cup e$$, where $$C$$ is a path in boundary of $$G$$; let $$A$$ and $$B$$ be the vertices of $$e$$. Then remove the last (in the shelling order) facet $$F_N$$ from $$G$$ and get a smaller shellable complex $$\hat{G}$$. We claim is that $$\hat{G}$$ also contains a a knotted (in $$\hat{G}$$) curve with all edges except one on the boundary of $$\hat{G}$$. Once this is established, we are done since $$G$$ was assumed to be minimal complex with such properties.
Now, why does $$\hat{G}$$ contain a knotted loop with all but one edges on the boundary? We claim $$e$$ could not have been an edge of $$F_N$$, for otherwise we could homotope the path $$C$$, keeping $$A$$ and $$B$$ fixed, to a path in boundary of $$F_N$$ and get a knotted curve in the boundary of $$F_N$$, which is not possible since that boundary is a 2-sphere. So the edge $$e$$ was not an edge of $$F_n$$, and then it is still an interior edge of $$\hat{G}$$. Now we can homotope $$C$$, keeping keeping $$A$$ and $$B$$ fixed, to a path $$\hat{C}$$ in the boundary of $$\hat{G}$$, and $$\hat{\gamma}=\hat{C}\cup e$$ is a knotted loop in $$\hat{C}$$ with exactly one edge not on the boundary.