Timeline for How do we compute the even cohomology $H^{2i}(Q)$ of the affine hyperquadric?
Current License: CC BY-SA 4.0
36 events
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| 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 | |||||
| S Aug 14, 2018 at 1:00 | history | bounty ended | CommunityBot | ||
| S Aug 14, 2018 at 1:00 | history | notice removed | CommunityBot | ||
| Aug 10, 2018 at 15:39 | comment | added | Todd Trimble | To add a bit to LSpice's comment: it's usually considered okay to ask "why the downvote?" in a comment below the question, as the purpose of comments (from a StackExchange perspective) should be to improve or clarify the question. But the question itself should stick to the math; thus, avoid meta discussion there. | |
| Aug 10, 2018 at 15:27 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 10, 2018 at 12:13 | comment | added | LSpice | As suggested by my rejected edit, I really think that "If you downvote"-type pleading doesn't belong here. If you want to talk about the mechanics of the site rather than mathematics, then that belongs on Meta. | |
| Aug 8, 2018 at 14:34 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 8, 2018 at 5:32 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 7, 2018 at 19:30 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 6, 2018 at 23:30 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 6, 2018 at 18:12 | comment | added | mme | Yes, everything there is a closed manifold. | |
| Aug 6, 2018 at 18:09 | comment | added | Sergio Charles | I came across this here @MikeMIller | |
| Aug 6, 2018 at 18:05 | answer | added | Jonny Evans | timeline score: 4 | |
| Aug 6, 2018 at 17:54 | comment | added | mme | This formula doesn't make sense. $\text{ch}_{n-i}$ is a cohomology class in degree $2n-2i$, the Todd polynomial is a cohomology class in dimension $2i$. On a compact oriented manifold $H^{2n}(X;\Bbb Z)$ is $\Bbb Z$, so this is a number. On a noncompact manifold this group is zero. Note that $\chi(M)$ on something noncompact can be undefined, though here it is the same as $\chi(S^n)$. | |
| Aug 6, 2018 at 17:35 | comment | added | Sergio Charles | @MikeMiller For a compact almost complex manifold $M$ of dimension $2n$ we have $$\chi(M)=\sum_{p=0}^n (-1)^p\sum_{i=0}^{n\choose{p}}\mathrm{ch}_{n-i}(\Omega^p)\frac{T_i}{i!}$$ where $\chi$ is the topological Euler characteristic, $\Omega^p$ is the $p$-th complex exterior power of the cotangent bundle (i.e., the complex dual of the tangent bundle), $\mathrm{ch}$ is the Chern character and $T_i$ is the $i$-th Todd polynomial of $M$. Is this also true for a non-compact complex manifold? | |
| Aug 6, 2018 at 17:13 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 6, 2018 at 16:51 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 6, 2018 at 14:18 | comment | added | Sergio Charles | @ToddTrimble Thanks so much! I think I will try to be more careful when asking next time. | |
| Aug 6, 2018 at 14:13 | comment | added | Todd Trimble | This is simply a guess, since I didn't downvote myself, but my guess is that the downvotes are on the basis of "does not show any research effort". Partly it's that the first question has a trivial answer, noted by Arun. And partly that there are problems with the second part, noted by Mike Miller, that may suggest possibly not thinking things through carefully before asking. To ameliorate this criticism, let me say that I, not being knowledgeable in complex algebraic geometry, wasn't aware of the isomorphism in your display line, so at least I got something out of your question. :-) | |
| Aug 6, 2018 at 13:50 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 6, 2018 at 1:07 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 6, 2018 at 0:59 | comment | added | mme | $Q$ is a complex manifold, but not compact. If anything makes sense, it's the left side of the equality, but I think as $Q$ is a Stein manifold (see here) this is the same as the dimension of $H^0(Q;\mathcal O_Q)$, which is the space of holomorphic vector fields on $Q$. I do not know what this space is; it's possible to be infinite-dimensional when the manifold is not compact (as it is in the case of $\Bbb C^n$). | |
| Aug 6, 2018 at 0:59 | history | edited | LSpice | CC BY-SA 4.0 |
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| Aug 6, 2018 at 0:58 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 6, 2018 at 0:56 | comment | added | Sergio Charles | Thanks for clarifying, so do I mean $\chi(Q,\mathcal{O}_Q)$? @MikeMiller | |
| Aug 6, 2018 at 0:44 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 6, 2018 at 0:38 | comment | added | mme | The second part of your question doesn't make sense to me; it looks like you're trying to state the Hirzebruch-Riemann-Roch theorem, but this is for compact complex manifolds. $S^n$ is not a compact complex manifold and $Q$ is not a holomorphic vector bundle over it unless I guess $n=2$. Even then, what you would be asking for is a Hermitian metric and the curvature of the corresponding connection (if it was a compact complex manifold). A Kahler metric induces these structures on the tangent bundle, and usually one writes $\chi(X, \mathcal O_X)$ to mean the holomorphic Euler characteristic. | |
| Aug 6, 2018 at 0:28 | comment | added | Arun Debray | Since cohomology is a homotopy invariant, $H^*(Q) \cong H^*(S^n)$, with a $\mathbb Z$ in dimensions $0$ and $n$, and $0$ elsewhere. | |
| Aug 6, 2018 at 0:18 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| S Aug 5, 2018 at 23:58 | history | bounty started | Sergio Charles | ||
| S Aug 5, 2018 at 23:58 | history | notice added | Sergio Charles | Canonical answer required | |
| Aug 5, 2018 at 23:56 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Aug 5, 2018 at 3:05 | review | Close votes | |||
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| Jul 30, 2018 at 2:05 | review | Close votes | |||
| Jul 30, 2018 at 10:49 | |||||
| Jul 30, 2018 at 0:28 | history | asked | Sergio Charles | CC BY-SA 4.0 |