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Let $P=\{p_1,\ldots,p_6\}\subset\mathbb{P}^1$ be a set of six general points of the projective line. In particular there are no two different subsets $\{p_{i_1},\ldots,p_{i_4}\}$ and $\{p_{j_1},\ldots,p_{j_4}\}$ of four elements of $P$ such that the $p_{i_k}$'s and the $p_{j_k}$'s have the same cross-ratio.

Assume that there exists a projective transformation $f:\mathbb{P}^1\rightarrow\mathbb{P}^1$ such that $f(P) = P$; that is $f$ acts on $P$ by permuting the $p_i$'s.

Do we have then that $f = \operatorname{Id}_{\mathbb{P}^1}$ ?

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  • $\begingroup$ Is the question asking for whether $f(P) = P$ for a particular $P$ in general position or for ALL such $P$'s? $\endgroup$ – Stanley Yao Xiao Apr 24 '17 at 15:29
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As Igor points out, the cross-ratio of a subset of four distinct points is not well-defined. If you meant to stipulate that all tuples of four distinct points from $P$ should have different cross-ratios, except for the unavoidable repetitions, then the answer is yes, you can conclude that $f$ is the identity transformation:

Since $f$ is a projective linear transformation, it preserves cross-ratios. In particular, the cross-ratio of $(f(p_1), f(p_2), f(p_3), f(p_4))$ is the same as that of $(p_1, p_2, p_3, p_4)$. Thus, by the hypothesis, $f$ in fact must permute already the set consisting of the first four points, in a manner which leaves the cross-ratio unchanged. If $f$ isn't the identity, then it must permute them in two pairs. Without loss of generality (renumbering points if necessary and conjugating everything with a suitable projective linear transformation to pin down the first three points) the only possibility is swapping $p_1=\infty$ with $p_2=0$ and $p_3=1$ with some $p_4$; thus $f\colon x\mapsto p_4/x$ and (edit if $f$ moves $p_5$) $p_6=p_4/p_5$; but then the cross-ratio of $(\infty,0,1,p_5)$ would equal that of its $f$-image $(0,\infty,p_4,p_6)$, a tuple supported on a different subset of $P$.

Edit - Or else, this $f$ fixes both $p_5$ and $p_6$, which must then be two distinct square roots of $p_4$ in the field we're working with. But then the cross-ratio of $(\infty,0,1,p_5)$ would equal that of its $f$-image $(0,\infty,p_4,p_5)$; again violating the hypothesis.

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  • $\begingroup$ 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 $bxz-ayw=0$. $\endgroup$ – TopGatLu Apr 24 '17 at 22:43
  • $\begingroup$ In both cases we get a contradiction because $p_5$ and $p_6$ are general. Is this what you meant? $\endgroup$ – TopGatLu Apr 24 '17 at 22:44
  • $\begingroup$ @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. $\endgroup$ – GNiklasch Apr 25 '17 at 6:59
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I think the answer depends on your definition of general position. Consider four points, whose cross ratio is $y.$ Then permutations will have such diverse cross ratios as $ \frac{1}{y}, 1-y, 1-\frac{1}{y}, \frac{1}{1-y}, \frac{y}{y-1}$. Setting one of these quantities equal to $y$ will give you a finite order projective transformation reducing to a permutation on the four points. It is not too easy to conjure up two other points which are permuted by this transformation.

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