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

This question is motivated by What manifolds are bounded by RP^odd? (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 "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".

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Jason, this not an answer, just an observation. Using your formula for $p_1$, $< p_1^{2n+1}, [\mathbb{H}P^{2n+1}]> = (2n-2)^{2n+1} < u,[\mathbb{H}P^{2n+1}]> \neq 0$ if $n>1$, so $\mathbb{H}P^{2n+1}$ cannot be the boundary of an oriented manifold, unlike the examples you give for $\mathbb{R}P^{2n+1}$ and $\mathbb{C}P^{2n+1}$. The point is that filling spherical fibres in oriented bundles will not work.

By the way, this is my first post in Math Overflow. Yay!!!

Note: this post has been edited because the original was very false. I claimed that $\sigma(\mathbb{H}P^{2n+1})=1$ which is silly because the middle cohomology is $H^{4n+2}(\mathbb{H}P^{2n+1}) = 0$. Also the signature being odd would have contradicted the fact that $\chi (\mathbb{H}P^{2n+1})$ is even, which is stated in the question.

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Welcome to Mathoverflow, Torcuato. The fact about Pontryagin numbers follows immediately from the fact that $p_1(\mathbb{H}P^n)$ is nontrivial for $n\neq 1$, so I should have picked up on this myself following my comment to Ryan's answer - thanks for pointing it out! – Jason DeVito Feb 21 '10 at 18:35

A small note on extending the argument I gave in the previous (linked) thread.

You get a free involution on $\mathbb CP^{2n+1}$ by using the fibrewise antipodal map for your bundle $$S^2 \to \mathbb CP^{2n+1} \to \mathbb HP^n$$ so this also gives you $\mathbb CP^{2n+1}$ as the boundary of a mapping cylinder.

$\mathbb HP^{2n+1}$ I'm not sure how to deal with analogously. I suppose a place to start would be to try and find a somehow more natural free involution on $\mathbb CP^{2n+1}$.

Googling around it's not clear to me whether or not it's known if $\mathbb HP^{2n+1}$ admits a free involution.

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$\mathbb{H}P^n$ has no free involutions unless n = 1. This is because the first Pontrjagin class is given as (2(n+1) - 4)u where u is a specific generator of $H^4(\mathbb{H}P^n)$ and any diffeomorphism f will fix this. Hence, $f$ will fix $u^k$ for all $k$, and then, by the Lefshetz fixed point theorem, $f$ will have a fixed point. – Jason DeVito Feb 9 '10 at 1:47
Ah, okay. So that tells us it's not the boundary of an $I$-bundle. – Ryan Budney Feb 9 '10 at 2:21
I've tried responding in the other thread, but the "add comment" button isn't working. Yes, I-bundle means bundle over a space with fiber an interval. – Ryan Budney Feb 9 '10 at 5:37