Timeline for Why is the definition of l-adic sheaves so complicated?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 16, 2011 at 14:20 | vote | accept | Jan Weidner | ||
| Dec 16, 2011 at 14:18 | comment | added | Jan Weidner | Thanks Donu, Moosbrugger and anon these comments are useful to me! | |
| Dec 16, 2011 at 5:07 | answer | added | S. Carnahan♦ | timeline score: 11 | |
| Dec 16, 2011 at 4:54 | comment | added | anon | With the naive definition, $H^{1}(X,\mathbb{Z}_\ell)=\operatorname{Hom}_{\text{continuous}}(\pi_1(X),\mathbb{Z}_\ell)$ with the discrete topology on $\mathbb{Z}_\ell$. This is generally zero, because (for nice schemes) $\pi_1(X)$ is profinite. With the nonnaive definition, it is Hom with the natural $\ell$-adic topology on $\mathbb{Z}_\ell$, which is what you want. | |
| Dec 16, 2011 at 0:29 | comment | added | Moosbrugger | @Donu: Indeed, but then the question really is: why $\ell$-adic sheaves at all, instead of some other coefficients? (And this is more or less the question Ryan has answered). @Jan: A good way to get a feel for this definition is to work out what it means for the spectrum of a field, to see that it's the same (modulo equivalence) as continuous representations of the Galois group of the field into a finite extension of $\mathbb{Q}_{\ell}$. | |
| Dec 15, 2011 at 23:26 | comment | added | Donu Arapura | One possible naive definition might include the constant sheaves $\mathbb{Z}_\ell$ etc. But I believe that $H^1(\mathbb{G}_m,\mathbb{Z}_\ell)=0$ with this choice, which would be the wrong answer: it should be $\mathbb{Z}_\ell$ like the circle. | |
| Dec 15, 2011 at 23:00 | answer | added | Ryan Reich | timeline score: 21 | |
| Dec 15, 2011 at 22:13 | comment | added | Moosbrugger | What is a more naive definition? | |
| Dec 15, 2011 at 22:06 | history | asked | Jan Weidner | CC BY-SA 3.0 |