There are various ways to embed the projective plane into $\mathbb{R}^4$ very nicely, see e.g. Wikipedia. Suppose now that I take such an embedded projective plane $P \subset \mathbb{R}^4$ and fix a tubular neighbourhood $U$ of $P$. Now $\overline{U}$ is a $4$-manifold with a boundary embedded in $\mathbb{R}^4$, and so we know that its boumdary is orientable. Thus $\partial U$ is a closed orientable $3$-manifold, but which one is it?

Edit: As kindly pointed out by @SteveD, this has been answered here. The answer is $\mathbb{S}^3 / Q_8$.

  • 2
    $\begingroup$ It's $S^3/Q_8$; I answered this here. $\endgroup$ – Steve D Aug 10 at 4:27
  • $\begingroup$ Thank you @SteveD! $\endgroup$ – Rami Luisto Aug 10 at 14:58

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