4
$\begingroup$

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?.

enter image description here

Here are some relevant definitions.

enter image description here

enter image description here

$\endgroup$
2
  • 2
    $\begingroup$ 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! $\endgroup$ – Christoph Nov 20 '19 at 17:01
  • $\begingroup$ 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. $\endgroup$ – ensbana Nov 20 '19 at 22:41
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.