Timeline for Given a complex vector bundle with rank higher than 1, is there always a line bundle embedded in it?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 14, 2020 at 20:10 | comment | added | Chris Schommer-Pries | @DavidESpeyer I don't believe any such example exists. If there were such an X with non-trivial obstruction, then this obstruction would persist on the 3-skeleton of X. So we might as well assume X is a 3-dimensional CW complex. But then by cellular approximation any map from X to BU(2) factors through the 3-skeleton of BU(2), which happens to also be the 2-skeleton $\mathbb{CP}^1$. However the universal bundle already splits into two lines when restricted to $\mathbb{CP}^1$. | |
| Dec 8, 2020 at 21:25 | history | edited | Denis Nardin | CC BY-SA 4.0 |
Added remark about low dimensions
|
| Dec 8, 2020 at 15:06 | comment | added | David E Speyer | Nice answer! So, presumably, the answer I gave was the obstruction in $H^4(X, \pi_3 \mathbb{CP}^1)$. Sure enough, the nonzero class in $\pi_3(\mathbb{CP}^1)$ comes from the Hopf fibration. This suggests I should be able to make an example where the obstruction comes from $H^3(X, \pi_2(\mathbb{CP}^1))$. This would be fun to try. | |
| Dec 8, 2020 at 12:49 | comment | added | Denis Nardin | @DavidESpeyer Corrected, thank you! | |
| Dec 8, 2020 at 12:48 | history | edited | Denis Nardin | CC BY-SA 4.0 |
typo
|
| Dec 8, 2020 at 12:41 | comment | added | David E Speyer | Note that my answer is precisely the finite skeleton Denis mentions: $BU_2 \cong G(2,\infty)$, where as I work with $G(2,3) \cong \mathbb{CP}^2$. | |
| Dec 8, 2020 at 9:00 | history | edited | Ben McKay | CC BY-SA 4.0 |
typo in tex
|
| Dec 8, 2020 at 8:37 | history | answered | Denis Nardin | CC BY-SA 4.0 |