The first question is

Is the hemisphere of $S^4$ the unique compact 4-dimensional manifold with $\partial K = S^3$?

In other words, is it obvious?

The question stems from a theoretical physics paper by Micha Berkooz (on the arXiv) on the supergravity dual of a (1,0) field theory in six dimensions. The paper is here. However, my questions really have to do with mathematics.

The second related question is on orbifolds. Specifically,

What is the general relation of $S^{n}/\mathbb{Z}_2$ to some segement (hemisphere?) of $S^{n}$?

For instance, $S^{1}/\mathbb{Z}_2$ is a line interval (simply half the circumference of the circle) with two fixed points. The paper referred to above seems to orbifold $S^{4}$ and replace it by a hemisphere of $S^{4}$.

Suppose I parametrize $S^{n}$ by $n$-dimensional spherical coordinates $(r, \phi_1, \phi_2, \ldots, \phi_{n-2}, \phi_{n-1})$ where $r = \text{constant}$, and $\phi_{i} \in [0, \pi]$ for $1 \leq i \leq n-2$ while $\phi_{n-1} \in [0, 2\pi)$.

Is it correct to think of $S^{n}/\mathbb{Z}_2$ as the $\mathbb{Z}_2$ modding out a particular $\phi$ (which parametrizes the great arc of $S^4$, and ranges from $0$ to $\pi$) to give a hemisphere (by restricting its range to $[0, \pi/2]$)?

EDIT: I changed the $\phi_{n-1}$ in the second question to "a particular $\phi$..."

EDIT #2: A closely related construction of this kind of orbifold is described here: https://books.google.com/books?id=OG0GBwAAQBAJ&lpg=PP1&pg=PA275#v=onepage&q&f=false

  • 2
    $\begingroup$ No, of course not, e.g. you can take connect sum with $\mathbb{CP}^2$. What you mean by $S^n/\mathbb{Z}_2$ depends on what action of $\mathbb{Z}_2$ you have in mind; do you mean a reflection? $\endgroup$ – Qiaochu Yuan May 3 '16 at 19:33
  • $\begingroup$ By $\mathbb{Z}_2$ on a particular coordinate $x$, I mean the identification of $x$ with $-x$. In the physics setting, there's a theory on $AdS_7 \times K$ where $K$ is a compact manifold that realizes an $SO(4)$ global symmetry. The $SO(4)$ symmetry (and the knowledge that $S^3 = SO(4)/SO(3)$) tell me that there's a 3-sphere involved somehow. It turns out that $K$ is a 4-dimensional manifold. The paper says "K is made of a bundle of a 3-sphere over a 1-dimensional manifold." (K is 4D as it is described by $U=f(x^6,..,x^9, x^10) = \text{const.}$) $\endgroup$ – leastaction May 3 '16 at 20:52
  • $\begingroup$ I slightly edited my question. Perhaps it makes more sense now? $\endgroup$ – leastaction May 4 '16 at 0:26

For the sake of having an answer: no, the disk $D^4$ (mathematicians' notation for the hemisphere in $S^4$) is not the unique compact $4$-manifold with boundary $S^3$. Starting from any closed $4$-manifold $M$, you can cut a hole out of it to get a compact $4$-manifold with boundary $S^3$, and most of these are different (they can be distinguished e.g. by their homology); in the comments I effectively suggested that you do this with $M = \mathbb{CP}^2$.

More generally, given any manifold $M$ with boundary $\partial M$, you can take the connected sum with a closed manifold of the same dimension, and the result is another manifold (usually different, and again you can usually tell the difference using e.g. homology) with the same boundary.

With $\mathbb{Z}_2$ acting by reflection, the quotient $S^n/\mathbb{Z}_2$ can be identified as a set with the disk $D^n$, and as an orbifold every point on the boundary $S^{n-1}$ has stabilizer $\mathbb{Z}_2$.

  • $\begingroup$ Thanks for your answer @Qiaochu Yuan! The physics description of the orbifold is at books.google.com/…, if you're interested. $\endgroup$ – leastaction May 4 '16 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.