Timeline for Cycle class map in non-smooth family of projective varieties
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 1, 2014 at 22:58 | comment | added | Allen Knutson | I don't really see where $\chi$ smooth is relevant. For example, you could base-change using a resolution of $\chi$. | |
| Mar 1, 2014 at 22:03 | comment | added | Vladimir Baranovsky | For a smooth morphism $\pi$, and any flat sheaf over $T$, such as $\mathcal{O}_Z$, there will be a finite resolution by vector bundles (if $T$ is really nasty, it might exits only locally on $T$). Then at least over $\mathbb{Q}$ one can take the Chern character (alternating sum over the terms of the resolution), which agrees nicely with restriction to fibers. | |
| Mar 1, 2014 at 21:55 | comment | added | Vladimir Baranovsky | If $\pi$ is smooth and $Z$ is flat over $T$ then such a class should exist, at least with complex coefficients. | |
| Mar 1, 2014 at 19:02 | history | edited | user46578 | CC BY-SA 3.0 |
edited title
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| Mar 1, 2014 at 13:20 | comment | added | user46578 | @Degtyarev: $\mathcal{Z}$ need not be smooth. Can you still use Poincare duality? | |
| Mar 1, 2014 at 13:20 | comment | added | abx | Alex, I have difficulties to follow. Are you ware that $\mathcal{X}$ is not smooth? | |
| Mar 1, 2014 at 12:05 | comment | added | Alex Degtyarev | I mean, this is a purely topological fact. If $Z, F$ are two submanifolds transversal in $X$ and $i\colon F\to X$ is the inclusion, then $i^![Z]=[Z\cap F]$, where $i^!=D\circ i^*\circ D^{-1}$ is the inverse Hopf homomorphism (composition of the cohomological restriction and Poincare dualities). | |
| Mar 1, 2014 at 12:01 | comment | added | Alex Degtyarev | This is too obvious to give a reference. The keywords are inverse Hopf homomorphism (and Thom isomorphism) and their geometric interpretation. (Thom isomorphism should be applied to the normal bundle of the fiber.) Any good textbook in algebraic topology would do, but personally I learned it long ago and in Russian, so I cannot recommend any English text. (In fact, I just don't know any satisfactory English textbook :) | |
| Mar 1, 2014 at 11:36 | comment | added | user46578 | @Degtyarev: The examples I have in mind it is the case. Could you please suggest some reference for this fact. | |
| Mar 1, 2014 at 9:26 | comment | added | Alex Degtyarev | Of course, first thing coming to one's mind is the class $[\mathcal{Z}]$. A bullet-proof sufficient condition would be the transversality of $\mathcal{Z}$ to the fibers. | |
| Mar 1, 2014 at 1:52 | history | asked | user46578 | CC BY-SA 3.0 |