Timeline for Does the closure of a smooth algebraic always define a homology class?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2013 at 11:22 | history | bounty ended | Ritwik | ||
| Jun 12, 2013 at 18:52 | comment | added | Allen Knutson | No, I was only commenting on part (3). | |
| Jun 12, 2013 at 3:35 | comment | added | Ritwik | Just to be sure I understand Allen Knutson's comment: You are saying if look up what "cycle map from Chow groups to homology" is I should be able to understand why $\bar{X}$ is the variety $\Phi_1 =0$ minus all irreducible components contained in $\Phi_2 =0$? | |
| Jun 11, 2013 at 12:11 | comment | added | Allen Knutson | This is called the "cycle map" from Chow groups to homology groups. One good way to think of (3) (though terrible way to prove it) is to push the fundamental class forward from a resolution $\widetilde X$ of $\overline X$. | |
| Jun 11, 2013 at 7:37 | comment | added | Ritwik | Regarding Brett Parker's comment: Can you explain why this is true; ``You can think of $\bar{X}$ as the variety defined by $\Phi_1 =0 $ minus all irreducible components contained in $\Phi_2 =0 $." ? And I am also not seeing how the next statement follows from this fact: "It follows that every irreducible component of $\bar{X}$ has complex dimension $k$" | |
| Jun 11, 2013 at 7:25 | history | edited | Ritwik | CC BY-SA 3.0 |
edited title
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| Jun 11, 2013 at 4:54 | comment | added | Mariano Suárez-Álvarez | Your title somehow got cut off? | |
| Jun 11, 2013 at 4:51 | comment | added | Brett Parker | If $\Phi_1$ and $\Phi_2$ are algebraic, then the answer to all your questions is yes, as Serge Lvovski pointed out above. You can think of $\bar X$ as the variety defined by $\{\Phi_1=0\}$ minus all irreducible components contained in $\{\Phi_2=0\}$. It follows that every irreducible component of $\bar X$ has complex dimension $k$. One way to see that $\bar X$ defines a homology class is to use the existence of a Whitney stratification with all strata being complex varieties. | |
| Jun 11, 2013 at 4:07 | comment | added | Ritwik | Actually that was a typo. I meant to ask if $\bar{X}$ is an algebraic variety. | |
| Jun 11, 2013 at 4:03 | history | edited | Ritwik | CC BY-SA 3.0 |
Changed the first question
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| Jun 10, 2013 at 15:11 | comment | added | J.C. Ottem | $\overline{X}-X$ may not be irreducible: Consider a section $\phi_1\in H^0(\mathbb{P}^2,O(2))$ defining a smooth conic $C$, and let $\phi_2\in H^0(\mathbb{P}^2,O(1))$ define a line $L$ intersecting $C$ transversely. Then $X$ is $C$ with two points $p,q$ removed, but $\overline{X}-X=${p,q}, which is not an algebraic variety. | |
| Jun 10, 2013 at 11:19 | history | bounty started | Ritwik | ||
| Jan 25, 2013 at 17:30 | history | edited | Ritwik | CC BY-SA 3.0 |
Explicitly said what sort of a manifold X is.
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| Jan 25, 2013 at 17:24 | comment | added | Ritwik | Thank you for the counter example, in my case I am interested only in algebraic (locally closed) sub manifold. Is there a reference for this fact if X is algebaric, ie locally closed? You are saying in that case the answer is yes. | |
| Jan 25, 2013 at 16:27 | comment | added | Serge Lvovski | Sorry, did you mean smooth algebraic (locally closed) submanifold? Then the answer to all the questions is yes. | |
| Jan 25, 2013 at 16:13 | answer | added | Serge Lvovski | timeline score: 2 | |
| Jan 25, 2013 at 16:03 | history | asked | Ritwik | CC BY-SA 3.0 |