Timeline for A question about pairs of lines in 3D projective space
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 2, 2014 at 15:25 | history | bounty ended | Gil Kalai | ||
| May 1, 2014 at 19:50 | comment | added | Vít Tuček | Let $\ell_4$ be represented by a matrix $A$ and let $A_{ij}$ denote the determinant of a submatrix that arises by deleting columns $i$ and $j$. Then the condition $\ell_4 \cap \ell_4' = \emptyset$ is equivalent to $A_{12} \neq 0$, the condition $\ell_4 \cap \ell_6' \neq \emptyset$ is equivalent to $A_{34} = 0$ and $\ell_4 \cap \ell_5'\neq\emptyset$, $\ell_4\cap \ell_7' \neq \emptyset$ are equivalent to $A_{34} + A_{12} + A_{14}-A_{23} = 0$ and $b^2A_{34} + A_{12} + b(A_{14}-A_{23}) = 0$ respectively. This forces $b=1$. | |
| May 1, 2014 at 18:46 | comment | added | David E Speyer | Eight, in this case :) | |
| May 1, 2014 at 18:24 | comment | added | Vít Tuček | Right. Sorry for my silly question. I should really practice counting to ten more often. ;) | |
| May 1, 2014 at 18:13 | comment | added | David E Speyer | @VítTuček That lets me define $(\ell_1, \ell_2, \ell_3, \ell_7, \ell_8)$ and $(\ell'_4, \ell'_5, \ell'_6, \ell'_7, \ell'_8)$ meeting in the required manner. But I don't know whether or not I can fill in $(\ell_4, \ell_5,\ell_6)$ and $(\ell'_1, \ell'_2, \ell'_3)$. | |
| May 1, 2014 at 17:59 | comment | added | Vít Tuček | I am quite a bit confused, probably because OP is asking for smallest $m$, but ... wouldn't $a_1 = i$, $b_1 = j$, $a_2 = j$, $b_2 = i$ give you what you want? | |
| Apr 30, 2014 at 19:04 | vote | accept | Gil Kalai | ||
| Apr 29, 2014 at 9:00 | comment | added | Gil Kalai | Amazing!! Many thanks, David. Very nice result. | |
| Apr 29, 2014 at 1:52 | comment | added | David E Speyer | A less number theoretic example of a division algebra with $2N$ elements obeying $a_i b_j = a_j b_i$ for $i \neq j$ is the fraction field of the quantum torus. The quantum torus is $R:=\mathbb{Q}(q)\langle x_1, \ldots, x_{2n}\rangle$ with relations $x_i x_j = x_j x_i$ for $i \neq 2n+1-j$ and $x_i x_{2n+1-i} = q x_{2n+1-i} x_i$. $R$ is an Ore domain (see the appendix of arxiv.org/abs/math/0404446 ), so it has a skew field of fractions. | |
| Apr 29, 2014 at 1:07 | history | edited | David E Speyer | CC BY-SA 3.0 |
deleted 16 characters in body
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| Apr 29, 2014 at 0:49 | history | answered | David E Speyer | CC BY-SA 3.0 |