Timeline for Picard number and torsion of Neron-Severi group of abelian varieties over a number field
Current License: CC BY-SA 3.0
14 events
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| May 25, 2011 at 12:39 | comment | added | Stefan Keil | @Daniel: Thanks for the answer. How about the boundedness of the rank? And since $NS(\bar A)$ is a fin. gen. torsion-free ab. gr., for $\bar A=A/k \times_k \mathbb{C}$, I can conclude that $NS_A(k)$ and $NS_A(\bar k)$ (with notation from my question) are also fin. gen. torsion-free ab. gr., because they are subgroups? | |
| May 25, 2011 at 12:27 | vote | accept | Stefan Keil | ||
| May 25, 2011 at 11:47 | comment | added | Daniel Loughran | This means that there are injections $F(X) \to F(\overline{X})$, where $\overline{X}$ is the base change of $X$ to $\mathbb{C}$, and $F$ can be the Picard group, Néron-severi group or your favourite cohomology theory. So if there is no torsion in the Néron-severi group over $\mathbb{C}$ then there is certainly none over the ground field. | |
| May 25, 2011 at 11:43 | comment | added | Daniel Loughran | @stefan: I was trying to be careful since if you want to do cohomology and stuff for varieties over arbitrary fields you should probably use something like étale cohomology instead. However, I guess that didnt make clear that almost any functor you care about in algebraic geometry behaves nicely with respect to base change. | |
| May 25, 2011 at 10:13 | comment | added | Stefan Keil | How about the torsion-freeness of $H_1(X,\mathbb{Z})$ for abelian varieties over number fields? | |
| May 24, 2011 at 19:06 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
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| May 24, 2011 at 19:05 | comment | added | Daniel Loughran | @Ulrich. Actually I see now why $NS(X)\to H^2(X,\mathbb{Z})$ is in fact injective. Thanks! | |
| May 24, 2011 at 18:02 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
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| May 24, 2011 at 18:02 | comment | added | Daniel Loughran | Yes I think I need $H_1(X,\mathbb{Z})$ instead, I shall edit my answer accordingly. Although should not numerically trivial divisors also be homologically trivial, so sent to zero in $H^2(X,\mathbb{Z})$? | |
| May 24, 2011 at 17:51 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
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| May 24, 2011 at 17:45 | comment | added | naf | $H^1(X,\mathbb{Z})$ is always torsion free. In fact, $NS(X)$ injects into $H^2(X,\mathbb{Z})$ via the exponential sequence. | |
| May 24, 2011 at 17:40 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
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| May 24, 2011 at 17:34 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
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| May 24, 2011 at 17:26 | history | answered | Daniel Loughran | CC BY-SA 3.0 |