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Let $K$ be a finte 2-dimensional simplicial complex that admits an embedding into $\mathbb{R}^3$. Further, assume that $H_2(K, \mathbb{R})$ is trivial, so there are no 2-dimensonal cycles in $K$. Let $\gamma$ be a 1-dimensional null-homologous cycle which is homeomorphic to a circle; by the assumption on $H_2$ we know there exists a unique 2-chain whose boundary is $\gamma$. We also assume that $\gamma$ is a formal sum of edges $\gamma = \sum c_i e_i$ such that $c_i \in \{-1, 0, 1\}$ for each $i$.

Now consider the complex $K \cup D$ obtained by taking a complex $D$ that is homeomorphic to a disk and identifying its boundary with $\gamma$. We can assume that $D$ has a nice triangulation such that the edges on its boundary are in bijection with the edges of $\gamma$. Does $K \cup D$ admit an embedding into $\mathbb{R}^3$?

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    $\begingroup$ The boundary of a disk is a circle, so the boundary of D must be homeomorphic to a circle. I think we might be able to provide a counterexample by constructing $K$ such that $\gamma$ is not homeomorphic to a circle. Doodling pictures on my whiteboard suggests that we can pretty easily find a case where $\gamma$ is a figure-8. $\endgroup$ Jul 10 at 5:13
  • $\begingroup$ Good observation. I should have specified that I assume $\gamma$ to be homeomorphic to a circle. I will edit my question to include the assumption. $\endgroup$
    – Will
    Jul 10 at 5:17
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Start with the complete graph on 5 vertices, $K_5$. Let $X$ denote the graph obtained by removing one edge $[v,w]$ from $K_5$. Let $K$ denote the cone over $X$ with the tip $p$. It is easy to see that $X$ is a planar graph, which implies that the cone $K$ embeds in $R^3$. Moreover, $K$ is clearly contractible.

Now, to construct a simple loop $\gamma$ in the 1-skeleton of $K$, let $c$ denote a simple arc in $X$ connecting $v, w$. Add to this arc the edges $[p,v], [p,w]$. The result is a simple loop $\gamma$.

I will leave it to you to prove that $Y:=K\cup_\gamma D^2$ does not embed in $R^3$. (Use the fact that the link of $p$ in $Y$ is $K_5$, which is not a planar graph.)

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