Timeline for Compactness of the automorphic quotient and genericity
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2019 at 15:51 | vote | accept | Desiderius Severus | ||
| Sep 3, 2019 at 0:18 | |||||
| May 15, 2019 at 22:11 | answer | added | paul garrett | timeline score: 3 | |
| May 15, 2019 at 13:28 | answer | added | Kimball | timeline score: 5 | |
| May 15, 2019 at 13:01 | comment | added | Desiderius Severus | @Aurel Thanks for the clarification! What I want to know could be weaker though, and concerns what can be deduced on the global automorphic representation from the global compactness. | |
| May 15, 2019 at 12:37 | comment | added | Aurel | @DesideriusSeverus It can happen that the automorphic quotient is compact without any $G(F_v)$ being compact. For instance, take $G=SL(D)$ for $D$ a division algebra of degree $6$ over $\mathbb{Q}$ with invariants $(1/2,1/2,1/3,2/3)$ at four different places and $0$ elsewhere. | |
| May 15, 2019 at 3:10 | comment | added | Desiderius Severus | @WillSawin Thanks you both for the discussion. Indeed, $G(F_v)$ compact at every places is too strong. The requirement of $G(F_v)$ compact at only one place is what happens for quaternion algebras, and this is also the case for inner forms of $\mathrm{GSp}(4)$. But is this a more general statement? | |
| May 15, 2019 at 2:32 | comment | added | Will Sawin | @LSpice I think the claim that $G(F_v)$ is compact for all $F_v$ is wrong. Instead, isn't $G(F_v)$ compact for one $F_v$ equivalent to compactness of the automorphic quotient? | |
| May 15, 2019 at 2:13 | comment | added | LSpice | @WillSawin, thanks for reminding me of this. Does that mean that the quotient in the question can never be compact, or that my claim that its compactness implies that each $G(F_v)$ is compact is wrong? (That is, what is 'this' in your "this can't happen"?) | |
| May 15, 2019 at 2:11 | comment | added | Will Sawin | @LSpice Any reductive group over a global field is quasi-split at all but finitely many primes. (The moduli space of Borel subgroups of $G$ is geometrically $G/B$, which is a smooth variety, so it is itself smooth, hence it has a model smooth over all but finitely many primes. The fibers are flag varieties, hence have rational points, and because the variety is smooth must have rational lifts). One can probably see this more concretely using the fact that any division algebra is ramified at only finitely many primes. So this can't happen. | |
| May 15, 2019 at 1:48 | history | edited | LSpice | CC BY-SA 4.0 |
groupe -> group
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| May 15, 2019 at 1:46 | comment | added | LSpice | If the quotient is compact, then so is each $G(F_v)$ for $v$ finite, and this is a very strong condition. An absolutely simple, adjoint group over $F_v$ with this property must be the adjoint quotient of a central division algebra over $F_v$. I don't know enough about the global situation to know whether this answers your question. | |
| May 15, 2019 at 1:42 | history | asked | Desiderius Severus | CC BY-SA 4.0 |