Timeline for Action of PGL(2) on Projective Space
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 5, 2011 at 11:08 | vote | accept | Jacob Lurie | ||
| Jun 5, 2011 at 4:57 | answer | added | user631 | timeline score: 28 | |
| Jun 5, 2011 at 4:47 | comment | added | Allen Knutson | ...In particular, the first step in that theorem is to work with the group of collineations; Desargues' condition says that this is doubly transitive on the points. Crazy idea: can one build $P^2$ from $P^1$, say as $X^2/S_2$, and reduce your 1-dim question to this known 2-dim result? | |
| Jun 5, 2011 at 2:44 | comment | added | Allen Knutson | The question reminds me of the theorem that abstract projective planes satisfying Desargues' "theorem" are of the form $DP^2$ for a division ring $D$, and satisfying Pappus' "theorem" are of the form $FP^2$ for a field $F$. The proofs do rather a lot, in that they need to use incidence geometry to build addition, subtraction, multiplication, and division on a set built from the plane. Presumably a couple of the same ideas could be put into play here. | |
| Jun 4, 2011 at 5:03 | comment | added | Jacob Lurie | Yes, that is what I should have said. | |
| Jun 4, 2011 at 5:01 | history | edited | Jacob Lurie | CC BY-SA 3.0 |
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| Jun 4, 2011 at 3:33 | comment | added | Daniel Litt | Ah, of course that works. | |
| Jun 4, 2011 at 3:19 | comment | added | Tom Goodwillie | And the remedy is to redefine $3$-transitive as meaning that the action of $G$ on the set of ordered triples of distinct elements is transitive, i.e. has exactly one orbit. | |
| Jun 4, 2011 at 3:19 | comment | added | Tom Goodwillie | Daniel, your silliness can be generalized: $X$ could be empty; and if it is empty or a singleton then $G$ can be arbitrary. | |
| Jun 4, 2011 at 3:01 | comment | added | algori | .. and, of course, they do not satisfy condition 2. unless the ring is commutative... | |
| Jun 4, 2011 at 2:50 | comment | added | algori | Apart from projectve lines over fields there are projective lines over (not necessarily commutative) division rings, such as the quaternions. These examples are not there if $\#X<\infty$ since all finite division rings are fields, by Wedderburn's theorem. | |
| Jun 4, 2011 at 2:34 | comment | added | Daniel Litt | Perhaps I am missing something silly here--isn't the one-element set acted on by the trivial group a counterexample? (Both conditions are vacuous here). Only slightly more seriously, what about the two-element set acted on faithfully by $\mathbb{Z}/2\mathbb{Z}$? (1) is vacuous again, and (2) is satisfied... But clearly $X$ is too small in both cases to be a projective space over any field. | |
| Jun 4, 2011 at 2:20 | history | asked | Jacob Lurie | CC BY-SA 3.0 |