Timeline for If $G$ is absolutely simple simply connected, why is G(F_v) quasisimple for almost every valuation v?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 13, 2016 at 15:27 | vote | accept | JGR | ||
| Feb 20, 2016 at 5:09 | answer | added | nfdc23 | timeline score: 4 | |
| Feb 19, 2016 at 8:02 | history | edited | JGR | CC BY-SA 3.0 |
deleted 1 character in body; edited title
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| Feb 19, 2016 at 7:57 | comment | added | JGR | @nfdc23: You are right, I forgot to write the assumption G simply connected. Thanks | |
| Feb 19, 2016 at 7:56 | history | edited | JGR | CC BY-SA 3.0 |
I forgot the assumption G simply connected as noted by @nfdc23.
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| Feb 19, 2016 at 3:26 | comment | added | nfdc23 | The assertion is false for most classical split absolutely simple $G$ of adjoint type (step 3 fails), say realized inside ${\rm{GL}}(\mathfrak{g})$ (as a closed subgroup over $\mathbf{Z}$) via the adjoint representation. Indeed, your definition of the notation $G(\mathbf{F}_v)$ then coincides with the group of $\mathbf{F}_v$-points, and if $f:\widetilde{G} \rightarrow G$ is the simply connected central cover then $G(\mathbf{F}_v)$ has commutator subgroup the image of $\widetilde{G}(\mathbf{F}_v)$ away from a few cases. If $G$ is simply connected what you want to prove is true. | |
| Feb 18, 2016 at 17:01 | comment | added | Mikhail Borovoi | Lemma 4.9 on page 18 of the following paper will partially help you with (1): Springer, T. A. Reductive groups. Automorphic forms, representations and L-functions (Proc. Sympos. Pure Math., Oregon State Univ., Corvallis, Ore., 1977), Part 1, pp. 3–27, Proc. Sympos. Pure Math., XXXIII, Amer. Math. Soc., Providence, R.I., 1979. | |
| Feb 18, 2016 at 15:56 | review | First posts | |||
| Feb 18, 2016 at 16:00 | |||||
| Feb 18, 2016 at 15:55 | history | asked | JGR | CC BY-SA 3.0 |