# 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. Aug 10 '19 at 4:27
• Thank you @SteveD! Aug 10 '19 at 14:58

It might be interesting to describe the standard embedding in the following way. Consider the irreducible 5-dimensional representation of $$SO(3)$$, given by the harmonic quadratics, and restrict this action to $$S^4$$. Then an explicit computation shows that a generic orbit is $$SO(3)/V_4$$, where $$V_4$$ is the diagonal subgroup of $$SO(3)$$, and if $$\pi: SU(2) \to SO(3)$$ is the projection, $$\pi^{-1}(V_4) = Q_8$$, to use Steve D's notation. Therefore a generic orbit is $$S^3/Q_8$$.

The two exceptional orbits are $$SO(3)/O(2) = \Bbb{RP}^2$$, again this follows from explicit computation. It is altogether a pleasant exercise. One of these orbits is the image of the standard embedding.

Now it follows from general theory (this is in Bredon's book on transformation groups, if I recall) that if $$M$$ is a $$G$$-manifold where the generic orbit is codimension 1, then $$M/G$$ is a 1-manifold, possibly with boundary; the boundary points then correspond to the exceptional orbits, and the neighborhood $$(1-\epsilon, 1]$$ of a boundary point corresponds to a tubular neighborhood of the exceptional orbit.

Because $$S^4$$ is compact and certainly is not a fiber bundle $$SO(3) \to S^4 \to S^1$$, we see that $$S^4/SO(3) \cong [0,1]$$, with two exceptional orbits; the neighborhood of either exceptional orbit has boundary isomorphic to the generic orbit $$S^3/Q_8$$.

To say a little more...

The neighborhoods of these exceptional orbits are diffeomorphic to a rank 2 bundle over $$E \to \Bbb{RP}^2$$ with $$\det(E) \cong \det(T\Bbb{RP}^2)$$, so that the total space of the bundle and its subset of norm 1 are both canonically oriented; these are classified by elements of $$H^2(\Bbb{RP}^2; \Bbb Z_w) \cong H_0(\Bbb{RP}^2;\Bbb Z) = \Bbb Z$$ (cohomology with twisted coefficients, I am following Poincare duality here). Now it happens to be the case that $$k \in \Bbb Z$$ here represents the bundle that pulls back to an oriented bundle $$E' \to S^2$$ with $$e(E') = k$$.

So one (not very explicit...) way to describe this neighborhood is as the bundle with twisted Euler class $$4$$. An alternate way to describe it is as the quotient of the bundle $$E \to S^2$$ with Euler class 4 by a lift of the antipodal action.

That is everything I know about this $$SO(3)$$-space.

• That's a nice exposition, thanks! Sep 16 '19 at 11:35