Timeline for Generating $\mathbf{PGL}_2(\mathbb{Z})$ and $\mathbf{PGL}_2(\mathbb{Q})$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 13, 2023 at 11:37 | answer | added | Robbie Lyman | timeline score: 5 | |
| Jun 13, 2023 at 0:59 | comment | added | Lee Mosher | I'm willing to be that any fast, natural way to see that $\mathbf{PGL}_2(\mathbb{Z})$ is finitely generated without explicit construction of generators will just have an explicit construction of generators hidden in its proof. It's so simple to prove that fact about $\mathbf{PGL}_2(\mathbb{Z})$ that it has been generalized to the hilt, and all those generalizations are proved by, well, generalizing the construction for $\mathbf{PGL}_2(\mathbb{Z})$. | |
| Jun 7, 2023 at 17:24 | comment | added | paul garrett | @YCor, ah, ok, good, thanks! Somehow I wasn't parsing it correctly. :) | |
| Jun 7, 2023 at 17:24 | comment | added | YCor | @paulgarrett I don't think there's a typo. The abelianization argument doesn't work for $\mathrm{SL}_m$. | |
| Jun 7, 2023 at 16:09 | comment | added | paul garrett | @YCor, good comment, but I am confused by your "unlike any abelianization argument"... Is there a typo? Or can you clarify? :) | |
| Jun 7, 2023 at 12:56 | comment | added | YCor | If $\mathrm{PGL}_2(\mathbf{Q})$ were f.g., it would equal $\mathrm{PGL}_2(\mathbf{Z}[1/n])$ for some $n\ge 1$. But this is not the case since the latter group misses the image of the matrix $e_{12}(1/p)$ for every prime $p$ not dividing $n$. This argument also works for an arbitrary infinite field (since no infinite field is a finitely generated ring), and, unlike any abelianization argument, works for $\mathrm{SL}_m$ as well. | |
| Jun 7, 2023 at 12:45 | answer | added | Aurel | timeline score: 11 | |
| Jun 7, 2023 at 12:30 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
deleted 4 characters in body
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| Jun 7, 2023 at 12:23 | history | asked | THC | CC BY-SA 4.0 |