Manifolds whose boundaries are real / complex projective spaces

Question

1. What are some simple manifolds $M$ whose boundaries are $\mathbb{RP}^2$ or $\mathbb{RP}^n$ in general?

Namely $\partial M= \mathbb{RP}^n.$

2. What are some simple manifolds $N$ whose boundaries are $\mathbb{CP}^2$ or $\mathbb{CP}^n$ in general?

Namely $\partial N= \mathbb{CP}^n.$

Some trial attempt: I know that we can have $\partial D^2= \mathbb{RP}^1.$ and $\partial D^3= \mathbb{CP}^1.$ I dont know whether there are simple topology that also do the job for the above two questions?

Answer 1

$\Bbb{RP}^{2n}$ do not bound any manifolds (odd Euler characteristic $1$).

There is a circle bundle $\Bbb{RP}^{2n+1} \to \Bbb{CP}^n$ given by sending every real line in $\Bbb C^{n+1}$ to the complex line it is contained in. (There is a circle's worth of real lines in a given complex line.) Filling this circle bundle in with a disc bundle gives an oriented manifold that $\Bbb{RP}^{2n+1}$ bounds; an element of this manifold is a complex line $\ell \subset \Bbb C^{n+1}$ equipped with an equivalence class of vector $v \in \ell$ with $\|v\| \leq 1$, setting $v \sim v'$ if $v' = \pm v$.

$\Bbb{CP}^{2n}$ do not bound any manifolds (odd Euler characteristic $2n+1$).

There is a 2-sphere bundle $\Bbb{CP}^{2n+1} \to \Bbb{HP}^n$ given by sending every complex line in $\Bbb H^{n+1}$ to the quaternionic line it is contained in. There is a 2-sphere's worth of complex lines in each quaternionic line (because of the isomorphism $S^3/S^1 = \Bbb{CP}^1 \cong S^2$). We may fill this in fiberwise with a 3-ball to obtain an oriented manifold $\Bbb{CP}^{2n+1}$ bounds; an element of this is a quaternionic line in $\Bbb H^{2n+1}$ and an equivalence class of $v \in \ell$, where we say $v \sim v'$ if there is a unit complex number $\lambda$ with $\lambda v = v$.

Answer 2

This should really be a comment on Mike's excellent answer, but it won't fit. As Mike twice uses the fact that "$\chi(M)$ odd implies $M$ is not a boundary", I figure I would add a proof of the fact.

Let $N$ be a manifold with $\partial N = M$. The goal is to show that $\chi(M)$ is even. If $N$ is even dimensional, $M$ is an odd dimensional closed manifold, so $\chi(M) = 0$, which is even. Thus, we may assume that $N$ is odd dimensional.

Now, consider the manifold $N'$ obtained by gluing to copies of $N$ together along the commmon boundary. Then $\chi(N') = 0$ since $N'$ is an odd dimensional closed manifold. But we also see that $\chi(N') = \chi(N) + \chi(N) - \chi(M)$, so $ \chi(M) = 2\chi(N)$ is even.