4 added 62 characters in body

Edit: The statement below is wrong. The signature is 0. However the Pontryagin number argument still holds when $n>1$ (obviously the four sphere bounds a ball which is orientable). I will rewrite the post tomorrow.

Jason, this not an answer, just an observation. Using your formula for $\mathbb{H}P^{2n+1}$ has signature 1 p_1$,$< p_1^{2n+1}, [\mathbb{H}P^{2n+1}]> = (or -1, depending on your choice of orientation!) 2n-2)^{2n+1} < u,[\mathbb{H}P^{2n+1}]> \neq 0$if$n>1$, so it$\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}$(alternatively, one of the Pontryagin numbers, which can be computed from the classes in your comment is not zero). \mathbb{C}P^{2n+1}$. 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.

3 added 5 characters in body

Edit: The statement below is wrong. The signature is 0. However the Pontryagin number argument still holds when $n>1$ (obviously the four sphere bounds a ball which is orientable). I will rewrite the post tomorrow.

Jason, this not an answer, just an observation. $\mathbb{H}P^{2n+1}$ has signature 1 (or -1, depending on your choice of orientation!) so it cannot be the boundary of an oriented manifold, unlike the examples you give for $\mathbb{R}P^{2n+1}$ and $\mathbb{C}P^{2n+1}$ (alternatively, one of the Pontryagin numbers, which can be computed from the classes in your comment , are is not zero). 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!!!

2 added 220 characters in body

Edit: The statement below is wrong. The signature is 0. However the Pontryagin number argument still holds when $n>1$ (obviously the four sphere bounds a ball which is orientable). I will rewrite the post tomorrow.

Jason, this not an answer, just an observation. $\mathbb{H}P^{2n+1}$ has signature 1 (or -1, depending on your choice of orientation!) so it cannot be the boundary of an oriented manifold, unlike the examples you give for $\mathbb{R}P^{2n+1}$ and $\mathbb{C}P^{2n+1}$ (alternatively, the Pontryagin numbers, which can be computed from the classes in your comment, are not zero). 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!!!

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