Timeline for Pontryagin square of first Stiefel-Whitney class
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 7, 2019 at 20:02 | vote | accept | user34104 | ||
| Oct 7, 2019 at 12:29 | answer | added | Mark Grant | timeline score: 3 | |
| Oct 5, 2019 at 18:18 | comment | added | Mark Grant | I still don't see how that integral is defined when $M$ is non-orientable. If $M$ is orientable then you can evaluate on the fundamental class and get an integer mod 4, in which case I agree that the exponential of the integral is well-defined. | |
| Oct 5, 2019 at 14:57 | comment | added | user34104 | @MarkGrant This is precisely the problem posted in the question, i.e., when plug in $u=w_1^2$, the well-defined $\exp(i \pi/2 \int_M \mathcal{P}(u))$ becomes not well defined. So it seems to suggest that identifying $u$ with $w_1^2$ is somewhat problematic. But I do not know why. | |
| Oct 5, 2019 at 14:50 | comment | added | user34104 | @MarkGrant Yes, as M.G. explained, $M_4=M$. | |
| Oct 4, 2019 at 20:15 | comment | added | Mark Grant | I'm not sure I understand what the integration means. In particular, if M is non-orientable then there is neither a volume form or an integral fundamental class to evaluate a mod 4 cohomology class against. | |
| Oct 4, 2019 at 16:05 | comment | added | M.G. | @MarkGrant: I suspect $M_4 :=M$ to emphasize that $M$ is a 4-manifold. | |
| Oct 4, 2019 at 8:50 | comment | added | Mark Grant | What does the notation $M_4$ mean? Also, do you have a reference for "It is known that..."? | |
| Oct 3, 2019 at 19:13 | history | asked | user34104 | CC BY-SA 4.0 |