Is it easy to see thay none of the $SU(3)$-invariant metrics in your answer "accidentally" has a larger isometry group? E.g., if one considers the same problem on $SU(3)/SU(2)$, there is a 2-parameter family of invariant metrics, but for a 1-parameter sub-family, the isometry group is $O(6)$.

@mme: Thanks! I didn't check details, but I think the compability on the boundary shouldn't be too hard using a collar. It seems that using this, one should be able to assume that near the boundary, the foliation is the obvious foliation of $\partial M\times [0,1)$. Then, of course, things glue with no problems.

What obstructions are there for extending $\Sigma^n$ to a foliation on all of $M$ (without removing any points)?

@ConnorMalin: Thanks to you, too. By the way, the Weiss article your linked is the same Weiss article Igor linked a few comments above ;-).

@IgorBelegradek: Thanks for the references!

This makes perfect sense, thank you! Do you happen to know how to modify this if we restrict to orientation preserving diffeomorphisms and use $SO(n+1)$? Then your argument, at least, implies isomorphisms on $H^1$ and the torsion part of $H^2$.

The link seems dead. Here's an alternative: sciencedirect.com/science/article/pii/S0926224500000309, or a free one here: uni-muenster.de/imperia/md/content/theoretische_mathematik/…

@R.Rankin: The projection $\pi:\widetilde{M}\rightarrow M$ from the universal cover $\widetilde{M}$ of $M$ to $M$ is a local diffeomorphism, so that, $\pi^\ast(TM) = T\widetilde{M}$. Now compute $w_2$ of both sides and use naturality.

The only place I use "simply connected" is in the paragraph giving the map $g$. However it's necessary there. For example, if $N$ denotes the quotient of $S^2\times S^3$ by the diagonal action of $\mathbb{Z}/2\mathbb{Z}$ acting as the antipodal map on the first factor and via a degree $-1$ reflection on the second factor, then $N$ is an orientable rational homology sphere, but any map $S^5\rightarrow N$ has degree $0$. Indeed, any such map lifts to $S^2\times S^3$, where the cohomology ring structure easily implies the claim.

I'm glad all the topology I computed matches what's already known - that gives me some confidence in my answer. It was a great exercise to go through, though!

@IanGershonTeixeira: The bundle structure is quite explicit. Deleting a ball from $\mathbb{R}P^3$, the resulting space is the total space of the disk bundle in the tautological bundle over $\mathbb{R}P^2$. Taking two copies and gluing, we get a map $\mathbb{R}P^3\sharp \mathbb{R}P^3\rightarrow \mathbb{R}P^2$ with fiber obtained by gluing the two fiber $[0,1]$s to themselves along the boundary, i.e., with fiber $S^1$. (Technically, to make everything glue up nicely, we must use $\mathbb{R}P^3\sharp -\mathbb{R}P^3$, but this is diffeomorphic to $\mathbb{R}P^3\sharp \mathbb{R}P^3$.)