Timeline for Non semi-simple monodromy in an algebraic family
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2016 at 14:43 | comment | added | Will Sawin | @PiotrAchinger By Deligne's Weil II theorem the $i$th cohomology of a smooth proper family is pure of weight $i$, hence any two irreducible factors have the same weight and thus by Weil II again the Ext group between them is the first cohomology of a local system of weight $i-i=0$, hence has weight $1$, which means the Frobenius eigenvalues are not $1$ so no extension is defined over the base field. So one is able to show that the global algebraic geometry is completely different from the local / analytic situation using arithmetic and counting, because these only make sense globally. | |
| Feb 22, 2016 at 14:40 | comment | added | Will Sawin | @PiotrAchinger - the argument for this is beautiful! By spreading-out one reduces to characteristic $p$ and then one measures the eigenvalues of Frobenius on this local system. Recall that a local system is pure of weight $w$ if every eigenvalue of $\operatorname{Frob}_q$ has absolute value $q^{w/2}$. | |
| Feb 22, 2016 at 13:59 | answer | added | Geordie Williamson | timeline score: 4 | |
| Feb 19, 2016 at 21:51 | history | edited | Geordie Williamson | CC BY-SA 3.0 |
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| Feb 19, 2016 at 19:37 | answer | added | passerby | timeline score: 3 | |
| Feb 19, 2016 at 18:08 | history | edited | Geordie Williamson | CC BY-SA 3.0 |
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| Feb 19, 2016 at 18:02 | history | edited | Geordie Williamson | CC BY-SA 3.0 |
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| Feb 19, 2016 at 17:53 | comment | added | Geordie Williamson | The monodromy will be unipotent around 0 but there will be other singular fibres (e.g. the Weiestrass family $y^2 = x(x-1)(x-\lambda)$ over $\lambda \in \mathbb{A}^1$ the monodromy is unipotent around 0 and 1, but in total one gets a semi-simple local system). | |
| Feb 19, 2016 at 17:45 | comment | added | Piotr Achinger | I'm confused. In case $Y=\mathbb{A}^1\setminus\{0\}$ (or just the germ of a smooth curve), $f$ smooth and proper, with semistable reduction at $0$, the monodromy action is unipotent, and hence unlikely to be semi-simple. What am I missing? | |
| Feb 19, 2016 at 16:10 | history | asked | Geordie Williamson | CC BY-SA 3.0 |