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Consider the following 2-dimensional CW-complex: its 1-skeleton is $K_8$, which we write as an edge-disjoint union $K_5\cup K_{5,3}\cup K_3$. Then for any two edges $ab\in E(K_5)$ and $cd\in E(K_3)$ there are 2-cells attached along $abcda$ and $abdca$.

Question: does this complex embed in $\Bbb R^4$ (say, piecewise linearly)?

The figure shows the 1-skeleton and the red edges are the boundaries of two exemplary 2-cells.

enter image description here

Note that any two 2-cells share a vertex, so the van Kampen obstruction vanishes and yields no information.

If the answer to the question is Yes, does this change if we fill in $K_3$ with another 2-cell or if we replace $K_5$ by some larger complete graph $K_n$?

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  • $\begingroup$ @HenrikRüping I called it "join" for lack of a better name. I can imagine that $K_3*K_5:=K_8$ is called join in graph theory, but I am not sure. I don't think it is the 2-skeleton of the usual join because here there are no triangles filled by 2-cells, only 4-gons. You are right that I should clarify whether my edges are directed (which I clarified now), but I suspect that the answer will not depends on whether you include only $abcda$ or also $abdca$. $\endgroup$
    – M. Winter
    Commented Feb 1, 2023 at 15:40
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    $\begingroup$ Your question looks to be imprecise. In particular, your 2-cell attachments do not appear to be realizable by continuous functions. The union you describe defining $K_8$ is not well-defined. Do you mean disjoint union? $\endgroup$
    – Ryan Budney
    Commented Feb 1, 2023 at 17:06
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    $\begingroup$ I think the idea is to take $K_8$ and then subdivide the vertices in two sets A,B of size 3 and 5. Then any edge either connects two vertices from A in which case it is in $K_3$ or two vertices from B in which case it is in $K_5$ or one vertex from A to one vertex from B. In that case it is in $K_{5,3}$, The rule where to glue 2 cells is precise and it always is a closed loop. $\endgroup$
    – HenrikRüping
    Commented Feb 1, 2023 at 18:49
  • $\begingroup$ I would expect that it matters whether one adds one or both 4-gons. For the smaller example of $K_4$ with divided into sets of size two we either get a square or a moebius strip. The smallest $n$ into which these embed are different. EDIT: wrong it is not a square but a square with two 1-cells attached so that it does not embed anymore into $\mathbb{R}^2$. $\endgroup$
    – HenrikRüping
    Commented Feb 1, 2023 at 18:51
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    $\begingroup$ Did you check the van Kampen obstruction? $\endgroup$
    – Moishe Kohan
    Commented Feb 13, 2023 at 16:33

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