# What is the boundary of a tubular neighbourhood of the projective plane embedded in $\mathbb{R}^4$?

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 $$\mathbb{S}^3 / Q_8$$.

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