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

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

2010-02-08T21:00:13Z

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 "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".

Thanks in advance!

Answer by Ryan Budney for What manifold has $\mathbb{H}P^{odd}$ as a boundary?

2010-02-08T23:08:32Z

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.

Answer by Torcuato Battaglia for What manifold has $\mathbb{H}P^{odd}$ as a boundary?

2010-02-21T04:50:12Z

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.