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, 2019 at 4:27
  • $\begingroup$ Thank you @SteveD! $\endgroup$ Aug 10, 2019 at 14:58

1 Answer 1


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.

  • $\begingroup$ That's a nice exposition, thanks! $\endgroup$ Sep 16, 2019 at 11:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.