Timeline for How do we compute the even cohomology $H^{2i}(Q)$ of the affine hyperquadric?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 14, 2018 at 1:08 | vote | accept | Sergio Charles | ||
| Aug 14, 2018 at 1:08 | vote | accept | Sergio Charles | ||
| Aug 14, 2018 at 1:08 | |||||
| Aug 14, 2018 at 1:00 | history | bounty ended | CommunityBot | ||
| Aug 6, 2018 at 20:13 | comment | added | Jonny Evans | (Having said that, I understand what you're going through: I remember when I first learned about the Riemann curvature tensor that I desperately wanted to compute it from the Christoffel symbols in spherical coordinates on S^3, partly as a test of endurance, partly through a misplaced hope that it would give me an aid to understanding curvature in dimension greater than 2. I computed it, but was none the wiser.) | |
| Aug 6, 2018 at 20:08 | comment | added | Jonny Evans | That's exactly what I mean (except theta is an immersion, not a submersion). The curvature form is probably not too bad in this case because there is a large symmetry group, but I'm going to abstain from computing it on principle! (There are many useful things you can do with Chern classes, and many useful ways to compute them, but this seems on the face of it to be neither!) | |
| Aug 6, 2018 at 19:23 | comment | added | Sergio Charles | Thanks so much @JonnyEvans. However, I would like to see how we compute the curvature form. Also, regarding your first answer, do you simply mean the Hermitian metric $h=\delta_{\alpha\overline\beta}dz^{\alpha}\otimes d{\overline z}^{\beta}=dz^{\alpha}\otimes d{\overline z}^{\alpha}$ for $\alpha=1,...,n+1$ restricted to $Q$ by the pullback of the submersion $\theta:Q\hookrightarrow\mathbb{C}^{n+1}$, i.e. $\theta^*h$? | |
| Aug 6, 2018 at 18:05 | history | answered | Jonny Evans | CC BY-SA 4.0 |