2 (minor editing for clarity sake)

This question is motivated by http://mathoverflow.net/questions/8829/what-manifolds-are-bounded-by-rpodd (as well as a question a fellow grad student asked me) but I can't seem to generalize any of the provided answers to this setting.

Allow me to give some background. Take all (co)homology groups with $\mathbb{Z}_2$ coefficients.

Given a smooth compact manifold $M^n$, let $w_i = w_i(M)\in H^i(M)$ denote the Stiefel-Whitney classes of (the tangent bundle of) M. Let $[M]\in H_n(M)$ denote the fundamental class (mod 2). Consider the Stiefel-Whitney numbers of $M$, defined as the set of all outputs of $\langle w_{i_1}...w_{i_k} , [M] \rangle$. Of course this is only interesting when $\sum i_{j} = n$.

Pontrjagin proved that if $M$ is the boundary of some compact n+1 manifold, then all the Steifel-Whitney numbers are 0. Thom proved the converse - that if all Stiefel-Whitney numbers are 0, then $M$ can be realized as a boundary of some compact n+1 manifold.

As a quick aside, the Euler characteristic $\chi(M)$ mod 2 is equal to $w_n$. Hence, we see immediately that if $\chi(M)$ is odd, then $M$ is NOT the boundary of a compact manifold.

As an immediate corollary to this, none of $\mathbb{R}P^{even}$, $\mathbb{C}P^{even}$, nor $\mathbb{H}P^{even}$ are boundaries of compact manifolds.

Conversely, one can show that all Stiefel-Whitney numbers of $\mathbb{R}P^{odd}$, $\mathbb{C}P^{odd}$ and $\mathbb{H}P^{odd}$ are 0, so these manifolds can all be realized as boundaries.

What is an example of a manifold $M$ with $\partial M = \mathbb{H}P^{2n+1}$ (and please assume $n>0$ as $\mathbb{H}P^1 = S^4$ is obviously a boundary)?

The question for $\mathbb{R}P^{odd}$ is answered in the link at the top. The question for $\mathbb{C}P^{odd}$ is similar, but slightly harder:

Consider the (standard) inclusions $Sp(n)\times S^1\rightarrow Sp(n)\times Sp(1)\rightarrow Sp(n+1)$. The associated homogeneous fibration is given as

$$Sp(n)\times Sp(3)/ Sp(n)\times S^1\rightarrow Sp(n+1)/Sp(n)\times S^1\rightarrow Sp(n+1)/Sp(n)\times Sp(1),$$ which is probably better recognized as

$$S^2\rightarrow \mathbb{C}P^{2n+1}\rightarrow \mathbb{H}P^{n}.$$

One can apply the same trick of "filling fill in the fibers" - fill the $S^2$ to $D^3$ to get a compact manifold $M$ with boundary equal to $\mathbb{C}P^{2n+1}$.

I'd love to see $\mathbb{H}P^{odd}$ described in a similar fashion, but I don't know if this is possible.

Assuming it's impossible to describe $\mathbb{H}^{odd}$ as above, I'd still love an answer along the lines of "if you just do this simple process to this often used class of spaces, you get the manifolds you're looking for".

1

# What manifold has $\mathbb{H}P^{odd}$ as a boundary?

This question is motivated by http://mathoverflow.net/questions/8829/what-manifolds-are-bounded-by-rpodd (as well as a question a fellow grad student asked me) but I can't seem to generalize any of the provided answers to this setting.

Allow me to give some background. Take all (co)homology groups with $\mathbb{Z}_2$ coefficients.

Given a smooth compact manifold $M^n$, let $w_i = w_i(M)\in H^i(M)$ denote the Stiefel-Whitney classes of (the tangent bundle of) M. Let $[M]\in H_n(M)$ denote the fundamental class (mod 2). Consider the Stiefel-Whitney numbers of $M$, defined as the set of all outputs of $\langle w_{i_1}...w_{i_k} , [M] \rangle$. Of course this is only interesting when $\sum i_{j} = n$.

Pontrjagin proved that if $M$ is the boundary of some compact n+1 manifold, then all the Steifel-Whitney numbers are 0. Thom proved the converse - that if all Stiefel-Whitney numbers are 0, then $M$ can be realized as a boundary of some compact n+1 manifold.

As a quick aside, the Euler characteristic $\chi(M)$ mod 2 is equal to $w_n$. Hence, we see immediately that if $\chi(M)$ is odd, then $M$ is NOT the boundary of a compact manifold.

As an immediate corollary to this, none of $\mathbb{R}P^{even}$, $\mathbb{C}P^{even}$, nor $\mathbb{H}P^{even}$ are boundaries of compact manifolds.

Conversely, one can show that all Stiefel-Whitney numbers of $\mathbb{R}P^{odd}$, $\mathbb{C}P^{odd}$ and $\mathbb{H}P^{odd}$ are 0, so these manifolds can all be realized as boundaries.

What is an example of a manifold $M$ with $\partial M = \mathbb{H}P^{2n+1}$ (and please assume $n>0$ as $\mathbb{H}P^1 = S^4$ is obviously a boundary)?

The question for $\mathbb{R}P^{odd}$ is answered in the link at the top. The question for $\mathbb{C}P^{odd}$ is similar, but slightly harder:

Consider the (standard) inclusions $Sp(n)\times S^1\rightarrow Sp(n)\times Sp(1)\rightarrow Sp(n+1)$. The associated homogeneous fibration is given as

$$Sp(n)\times Sp(3)/ Sp(n)\times S^1\rightarrow Sp(n+1)/Sp(n)\times S^1\rightarrow Sp(n+1)/Sp(n)\times Sp(1),$$ which is probably better recognized as

$$S^2\rightarrow \mathbb{C}P^{2n+1}\rightarrow \mathbb{H}P^{n}.$$

One can apply the same trick of "filling in the fibers" to get a compact manifold $M$ with boundary equal to $\mathbb{C}P^{2n+1}$.

I'd love to see $\mathbb{H}P^{odd}$ described in a similar fashion, but I don't know if this is possible.

Assuming it's impossible to describe $\mathbb{H}^{odd}$ as above, I'd still love an answer along the lines of "if you just do this simple process to this often used class of spaces, you get the manifolds you're looking for".