What manifold has $\mathbb{H}P^{odd}$ as a boundary? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:54:33Z http://mathoverflow.net/feeds/question/14698 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14698/what-manifold-has-mathbbhpodd-as-a-boundary What manifold has $\mathbb{H}P^{odd}$ as a boundary? Jason DeVito 2010-02-08T21:00:13Z 2010-02-21T21:56:22Z <p>This question is motivated by <a href="http://mathoverflow.net/questions/8829/what-manifolds-are-bounded-by-rpodd" rel="nofollow">http://mathoverflow.net/questions/8829/what-manifolds-are-bounded-by-rpodd</a> (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.</p> <p>Allow me to give some background. Take all (co)homology groups with $\mathbb{Z}_2$ coefficients.</p> <p>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 $\sum i_{j} = n$.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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)?</p> <p>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:</p> <p>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</p> <p>$$Sp(n)\times Sp(3)/ Sp(n)\times S^1\rightarrow Sp(n+1)/Sp(n)\times S^1\rightarrow Sp(n+1)/Sp(n)\times Sp(1),$$ which is probably better recognized as</p> <p>$$S^2\rightarrow \mathbb{C}P^{2n+1}\rightarrow \mathbb{H}P^{n}.$$</p> <p>One can "fill in the fibers" - fill the $S^2$ to $D^3$ to get a compact manifold $M$ with boundary equal to $\mathbb{C}P^{2n+1}$.</p> <p>I'd love to see $\mathbb{H}P^{odd}$ described in a similar fashion, but I don't know if this is possible.</p> <p>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".</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/14698/what-manifold-has-mathbbhpodd-as-a-boundary/14706#14706 Answer by Ryan Budney for What manifold has $\mathbb{H}P^{odd}$ as a boundary? Ryan Budney 2010-02-08T23:08:32Z 2010-02-08T23:29:59Z <p>A small note on extending the argument I gave in the previous (linked) thread. </p> <p>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. </p> <p>$\mathbb HP^{2n+1}$ 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 $\mathbb CP^{2n+1}$. </p> <p>Googling around it's not clear to me whether or not it's known if $\mathbb HP^{2n+1}$ admits a free involution. </p> http://mathoverflow.net/questions/14698/what-manifold-has-mathbbhpodd-as-a-boundary/15962#15962 Answer by Torcuato Battaglia for What manifold has $\mathbb{H}P^{odd}$ as a boundary? Torcuato Battaglia 2010-02-21T04:50:12Z 2010-02-21T21:56:22Z <p>Jason, this not an answer, just an observation. Using your formula for $p_1$, $&lt; p_1^{2n+1}, [\mathbb{H}P^{2n+1}]> = (2n-2)^{2n+1} &lt; u,[\mathbb{H}P^{2n+1}]> \neq 0$ if $n>1$, so $\mathbb{H}P^{2n+1}$ cannot be the boundary of an <i>oriented</i> 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.</p> <p>By the way, this is my first post in Math Overflow. Yay!!!</p> <p>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.</p>