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Let $X$ be a 2-dimensional CW complex and let $c\subset X$ be a 2-cell attached along a simple closed loop $\gamma$ in the 1-skeleton $X^{(1)}$. For a natural number $n\ge 2$ consider the operation of $n$-rewinding $c$, which means to replace $c$ by a new 2-cell $c^n$ whose boundary is attached to $X^{(1)}$ along $\gamma^n=\gamma\circ\dotsb\circ\gamma$, the $n$-fold concatenation of $\gamma$ with itself. In other words, while $c$ winds around $\gamma$ once, $c^n$ winds around $\gamma$ $n$ times.

Question: Suppose that $X$ does not embed in $\Bbb R^4$. Let $X^n$ be obtained from $X$ by $n$-rewinding a 2-cell $c$ for some $n\ge 2$. Is it possible that $X^n$ now does embed in $\Bbb R^4$? Or does rewinding always preserve non-embeddability?

If there is a general answer to this question that would be highly interesting, but I would also be satisfied with an answer to the following special case: $X=\mathcal K_7^2$, the full 2-complex on $K_7$, obtained from the complete graph $K_7$ by attaching a 2-cell along each triangle. This complex does not embed in $\Bbb R^4$, but it becomes embeddable if a single 2-cell is deleted (see this answer to "Almost embedding" the complete 2-dimensional complex $\mathcal K_7^2$ into $\Bbb R^4$). I wonder whether the same result can be achieved by rewinding a 2-cell.

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Here is the answer for the specific example $X = K_7^2$ (also known as the 2-skeleton of the 6-simplex, $\Delta_6^{(2)}$). Let $\tau$ be an arbitrarily chosen triangle of $X$ and let $X^-$ be obtained by removing $\tau$ from $X$. As pointed out in the question, it is well known that $X^-$ embeds into $\mathbb{R}^4$. Let $\Sigma$ be the 2-sphere formed by triangles of $X^-$ not hitting $\tau$. It is also well known that in any embedding $g\colon X^- \to \mathbb{R}^4$, the linking number of $g(\partial \tau)$ and $g(\Sigma)$ is odd. This follows from the fact that the mod 2 van Kampen obstruction of $X$ is nonzero. For a reference, I am aware of Lemma 6 in Van Kampen’s embedding obstruction is incomplete for 2-complexes in $\mathbb{R}^4$ by Freedman, Krushkal and Teichner. (One has to check the proof as the complex considered there is slightly bigger.)

Now assume that there is an embedding $g$ of $X^n$ (in the notation of the question) into $\mathbb{R}^4$. Assume that the boundary of the rewinded cell is $\partial \tau$. Because $X^2$ contains $X^-$, we get that the linking number of $g(\partial \tau)$ and $g(\Sigma)$ is some odd number $k$. Then also the linking number of $n g(\partial \tau)$ and $g(\Sigma)$ is $nk \neq 0$ considering $n g(\partial \tau)$ as an image of $S^1$ winding $n$-times around $g(\partial \tau)$. But this means that we cannot glue a disk to $n g(\partial \tau)$ without hitting $g(\Sigma)$. That is, the embedding of $X^-$ cannot be extended to $X^n$ contradicting our original assumption.

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    $\begingroup$ This does not seem to be an answer to the actual question, even in the case of $K_7^2$ $\endgroup$ Commented Oct 31 at 7:49
  • $\begingroup$ My fault. I have recalled Melikhov's example too quickly and I have overlooked that the construction is slightly different. I will edit the answer providing an argument in this setting. $\endgroup$ Commented Oct 31 at 18:32
  • $\begingroup$ Great answer, thank you! Do I see correctly that this applies more generally to any complex $X$ of non-zero mod 2 van Kampen obstruction for which removing the 2-cell in question makes it embeddable in $\Bbb R^4$? $\endgroup$
    – M. Winter
    Commented Nov 8 at 12:08
  • $\begingroup$ I have to think a little bit about the more general setting. At least in the proof, I use the existence of $\Sigma$. I am not sure whether in every 2-complex with nonzero mod 2 van Kampen obstruction and for which removing the 2-cell makes it embeddable, this happens because of a similar linking argument. $\endgroup$ Commented Nov 12 at 11:44

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