Timeline for Cross-ratio and projective transformations
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2017 at 7:11 | history | edited | GNiklasch | CC BY-SA 3.0 |
added a missing edge case
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| Apr 25, 2017 at 6:59 | comment | added | GNiklasch | @TopGatLu concerning the unavoidable repetitions: If you let the symmetric group on four letters act to permute the four arguments of the cross-ratio, you'll get only six distinct cross-ratio values (see Igor's answer) - the action of the Klein four-group leaves the cross-ratio unchanged. | |
| Apr 24, 2017 at 22:44 | comment | added | Puzzled | In both cases we get a contradiction because $p_5$ and $p_6$ are general. Is this what you meant? | |
| Apr 24, 2017 at 22:43 | comment | added | Puzzled | Thanks for the answer. What do you mean by "except for the unavoidable repetitions"? If I got it right either $f$ is the identity or is switches $\infty$ with $0$ and $1$ with $p_4$. If $p_4 = [a:b]$ (homogeneous coordinates) this means that the matrix of $f$ is either the identity or $((0,a),(b,0))$. Let us consider the second case. Now $f$ must either fix $p_5$ or map $p_5$ to $p_6$. If $p_5 = [x:y]$ and $p_6 = [z:w]$ in the first case ($f(p_5)=p_5$) we that $x,y$ must satisfy $bx^2-ay^2 = 0$, in the second case we must have $bxz-ayw=0$. | |
| Apr 24, 2017 at 16:48 | history | answered | GNiklasch | CC BY-SA 3.0 |