Timeline for Do $\mathbb{HP}^2\#\overline{\mathbb{HP}^2}$ and $\mathbb{OP}^2\#\overline{\mathbb{OP}^2}$ arise as sphere bundles over spheres?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 31, 2018 at 18:42 | comment | added | mme | @მამუკაჯიბლაძე I think my picture is unclear: there are no gluing instructions. The fibers in the glued-up thing are still the lines through the origin (intersected with the spherical shell), just as they were on each "half"; this defines a fiber bundle in its own right without bothering to check compatibility conditions. The charts on $\Bbb{KP}^n$ give us local trivializations of this fiber bundle. I can write an answer. | |
| May 31, 2018 at 18:25 | comment | added | მამუკა ჯიბლაძე | @MikeMiller I see as perfectly clear the trivial case of the trivial bundle. When the bundle is not trivial it is not clear to me anymore whether if I start extending my gluing instructions to nearby fibres along all possible directions and eventually reach back to the same fibre these gluing maps return to their original position... | |
| May 31, 2018 at 16:55 | comment | added | mme | the fibers are still the $\Bbb K$-multiples of a point in the innermost (or, equivalently, outermost) `boundary component' = copy of $\Bbb{KP}^{n-1}$. This picture might be clearest when applied to $\Bbb K = \Bbb R$ and $n = 3$ (note that $\Bbb{RP}^3 \# \Bbb{RP}^3$ is not the 3D Klein bottle, which should be what one calls the mapping torus of reflection $S^2 \to S^2$.) | |
| May 31, 2018 at 16:54 | comment | added | mme | @მამუკაჯიბლაძე The clearest picture in my mind represents $\Bbb{KP}^n$ minus a ball as a solid spherical shell in $\Bbb K^n$ with the equivalence relation on the outer boundary given by identifying points equivalent under the action of $S(\Bbb K)$ (unit norm elements). The discs are the $\Bbb K$-multiples of a point in that 'boundary' $\Bbb{KP}^{n-1}$; their boundary is a copy of $S(\Bbb K)$ in the "inner boundary component". Now one doubles this by adding another spherical shell glued to the inner boundary, and an equivalence relation on the new innermost boundary... | |
| May 31, 2018 at 14:57 | comment | added | მამუკა ჯიბლაძე | Well yes if you already know the double is fibred compatibly to the fibrations of its halves. Do you? | |
| May 31, 2018 at 14:55 | comment | added | Mikhail Katz | That's what it means to take the double. The two halves are mirror images of one another so the fibrations are compatible by definition. | |
| May 31, 2018 at 14:51 | comment | added | მამუკა ჯიბლაძე | Is it obvious that the pairs of boundary spheres can be glued to each other matchingly simultaneously in all fibres? | |
| May 31, 2018 at 14:09 | history | answered | Mikhail Katz | CC BY-SA 4.0 |