Timeline for Finite group of units in quaternion orders
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| S Sep 26, 2016 at 23:01 | history | suggested | Kimball |
added qa tag
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| Sep 26, 2016 at 22:52 | review | Suggested edits | |||
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| S Sep 26, 2016 at 22:39 | history | suggested | Aurel |
added a relevant tag
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| Sep 26, 2016 at 22:18 | review | Suggested edits | |||
| S Sep 26, 2016 at 22:39 | |||||
| Sep 26, 2016 at 22:18 | answer | added | Aurel | timeline score: 3 | |
| Sep 26, 2016 at 19:53 | history | edited | SashaP | CC BY-SA 3.0 |
added 15 characters in body
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| Sep 26, 2016 at 18:12 | comment | added | David E Speyer | One more correction: The unit group of $\mathcal{O}_F$ will embed into $\mathcal{O}^{\ast}$ so, unless $F = \mathbb{Q}$, this will always be infinite. You might have wanted to compute $\mathcal{O}^{\ast}/\mathcal{O}^{\ast}_F$? That's what the Magma command Ben Linowitz lists does. | |
| Sep 26, 2016 at 16:04 | comment | added | Henri Johnston | In the case that $\mathcal{O}$ is a maximal order, there's a brief discussion of this problem in section 4.5 of arxiv.org/abs/1006.4381 | |
| Sep 26, 2016 at 15:55 | comment | added | user1073 | Actually, the condition that you need to impose is that $B$ be a definite quaternion algebra (i.e., ramified at all infinite places of $F$). Otherwise the subgroup $\mathcal O^1$ of the unit group which consists of elements with reduced norm $1$ will embed into a product of $\mathrm{SL}_2(\mathbb R)$'s and $\mathrm{SL}_2(\mathbb C)$'s as a cocompact discrete arithmetic subgroup with finite covolume. In the case that $\mathcal O$ is a definite quaternion order, $\mathcal O^*$ will be finite and can be computed by Magma: math.lsu.edu/doc/magma/html/text1025.htm | |
| Sep 26, 2016 at 14:53 | comment | added | SashaP | @Ben Linowitz sure, I changed unramified to ramified, thank you! | |
| Sep 26, 2016 at 14:51 | history | edited | SashaP | CC BY-SA 3.0 |
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| Sep 26, 2016 at 13:52 | comment | added | user1073 | Am I missing something? If $F=\mathbb Q$ and $R=M_2(\mathbb Q)$ then $\mathcal O=M_2(\mathbb Z)$ is an order whose group of units is infinite. | |
| Sep 26, 2016 at 13:12 | history | asked | SashaP | CC BY-SA 3.0 |