Timeline for Fano plane drawings: embedding PG(2,2) into the real plane
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 24, 2021 at 18:44 | answer | added | Anton Petrunin | timeline score: 10 | |
| Jun 13, 2018 at 21:15 | answer | added | Tom | timeline score: 3 | |
| S Sep 30, 2017 at 2:11 | history | suggested | jeq | CC BY-SA 3.0 |
Copied images to imgur.com, as they were not being displayed because of new https rule. Added links to original image sources.
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| Sep 30, 2017 at 1:57 | comment | added | Alexey Ustinov | @Dustin G. Mixon Another way to see it is Sylvester–Gallai theorem en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem | |
| Sep 30, 2017 at 1:32 | review | Suggested edits | |||
| S Sep 30, 2017 at 2:11 | |||||
| May 17, 2013 at 18:22 | comment | added | Seva | Hm-m-m... I'd say you do use this - at least, for the real case. Let $A,B,C,O,A',B',C'$ be as in your comment. How many of the points $A',B',C'$ lie on the edges of the triangle $ABC$? An odd number, on the one hand (they are points of intersection of $OA$, $OB$, $OB$ with the edges), and an even number, on the other hand (they are points of intersection of the straight line through $A',B'$ and $C'$ with the edges) - a contradiction. | |
| May 17, 2013 at 16:16 | comment | added | Noam D. Elkies | Actually I'm not using anything lke that (certainly not for an arbitrary field). Another way to say this is to choose projective coordinates so $A$, $B$, and $C$ are at the unit vectors $(1:0:0)$, $(0:1:0)$, and $(0:0:1)$, and then scale those coordinates so that $O$, which must have all three coordinates nonzero (else it's on one of the lines $AB$, $AC$, $BC$) is on $(1:1:1)$; then $OA$ is the line $y=z$, so $A'=OA \cap BC$ is $(0:1:1)$, and likewise $B = (1:0:1)$ and $C = (0:1:1)$. Now calculate that the determinant of $A,B,C$ is $2$, so $ABC$ are collinear iff we're in characteristic 2. | |
| May 16, 2013 at 8:57 | comment | added | Seva | @Noam: I see, the basic idea is that (1) any line not passing thorough a vertex of a triangle intersects an even number of its edges, while (2) for any triangle $ABC$, and any point $O$ not on its boundary, the three lines $OA$, $OB$, and $OC$ intersect an odd number of the edges. | |
| May 15, 2013 at 6:06 | vote | accept | Seva | ||
| May 15, 2013 at 3:00 | comment | added | Noam D. Elkies | ...and conversely, if $k$ does have characteristic $2$ then $A',B',C'$ are always collinear... | |
| May 15, 2013 at 2:59 | comment | added | Noam D. Elkies | Yes, a line drawing is impossible, over ${\bf R}$ or any field $k$ not of characteristic $2$. Let $A,B,C,O$ be non-collinear points of the Fano plane, and $A',B',C'$ the intersections of $AO,BO,CO$ with $BC,CA,AB$ respectively. By Ceva's theorem (actually proved by Al-Mutaman centuries earlier, and extended algebraically to the case where $O$ is outside the triangle, and indeed to arbitrary $k$), points $A',B',C'$ divide segments $BC,CA,AB$ in signed ratios whose product is $1$. But by Menealus' theorem, $A',B',C'$ are collinear iff that product is $-1$. Since $1 \neq -1$ we're done. | |
| May 15, 2013 at 1:32 | comment | added | Dustin G. Mixon | Is it obvious that you can't draw the Fano plane with lines in $\mathbb{R}^2$? | |
| May 14, 2013 at 23:28 | answer | added | pinaki | timeline score: 26 | |
| May 14, 2013 at 19:20 | history | asked | Seva | CC BY-SA 3.0 |