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Let $X = P^{n-1}(\Bbb C)$ ($(n-1)$-dimensional projective space) with $n \geq 3$, and let $K \subset X$ denote a compact subset. I have a bijective, continuous map $\phi:K \to K$ which satisfies the condition $$ \hat u + \hat v \supset \hat w \implies \phi(\hat u) + \phi(\hat v) \supset \phi(\hat w) \tag{1} $$ for all $\hat u,\hat v,\hat w \in P^{n-1}(\Bbb C)$. Here, I am interpreting $\hat u \in P^{n-1}(\Bbb C)$ to be a $1$-dimensional subspace of $\Bbb C^n$, so that $\hat u + \hat v$ is the smallest subspace containing both $\hat u$ and $\hat v$.

The question is as follows: can I guarantee that $\phi$ will have an extension $\Phi:X \to X$ which satisfies (1) for all $\hat u,\hat v, \hat w \in X$?

The motivation here is that, by the fundamental theorem of projective geometry, such a map $\Phi$ is necessarily induced by a semilinear map; it would be nice if I could guarantee that $\phi$ is induced by such a map. If anyone could point me in the direction of a theorem or reference that might be useful here, I'd appreciate it.


Note: I've done my best to make this question self-contained and concise, and have therefore left out background that I felt was extraneous. Feel free to comment if you feel that I should provide further context or clarification

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  • $\begingroup$ Do you mean to assume that $\phi$ is injective? Bijective? Any constant map $\Phi:X\to X$ satisfies the condition as written. $\endgroup$ Aug 13, 2018 at 13:18
  • $\begingroup$ @TomGoodwillie well spotted... either seems to be enough, but for now let's suppose that $\phi$ is bijective. $\endgroup$ Aug 13, 2018 at 14:23

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Your condition (1) means: if $\hat{u}$, $\hat{v}$, and $\hat{w}$ are linearly dependent, then so are $\widehat{\varphi(u)}$, $\widehat{\varphi(v)}$, and $\widehat{\varphi(w)}$. So $\varphi$ preserves linear dependence of $3$ vectors. But it doesn't detect linear dependence among $4$ or more vectors.

The other answers have given counterexamples where $K$ is a finite set of points. Basically, it's not hard to construct finite sets of points where no $3$ are linearly dependent, but there is a dependence among some $4$ or more of them.

I'll give a counterexample where $K$ is connected. Let $K$ be a rational normal curve, that is, the curve parametrized by $[s^{n-1} : s^{n-2} t : s^{n-3} t^2 : \dotsb : t^{n-1} ]$ in homogeneous coordinates on $X = \mathbb{P}^{n-1}$. Now $K$ doesn't have any trisecant lines, i.e., there are no $3$ points $\hat{u},\hat{v},\hat{w} \in K$ that are collinear in projective space (equivalently, $u,v,w \in \mathbb{C}^n$ are linearly dependent). So condition (1) is vacuously true. Then take $\phi$ to be any old map on $K$; then $\varphi$ satisfies (1). But if $\phi$ isn't semilinear then it doesn't extend to $X$.

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  • $\begingroup$ Very good point. I appreciate the fact that you've given me a connected example; if my approach is going to work, then I'll have to lean far more on the nice structure of $K$. Thanks for the help. $\endgroup$ Aug 13, 2018 at 14:36
  • $\begingroup$ @Omnomnomnom You're welcome! $\endgroup$ Aug 15, 2018 at 3:32
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Take $n=2$. If $x$, $y$ are two distinct points of $\mathbb{P}^1$, the condition $x+y\supset z$ is satisfied for any $z$, so the condition on $\phi$ holds whenever $\phi$ is injective, while $\phi$ has (in general) no semilinear extension if $K$ is big enough.

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  • $\begingroup$ Well spotted. I edited the question to specify that $n \geq 3$. Otherwise, as you have said, the "fundamental theorem of projective geometry" doesn't apply. $\endgroup$ Aug 13, 2018 at 11:56
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For larger $n$ we can still make a counterexample, can't we?

Let $S$ be a set of $n+1$ points in general position, and choose them to have rational coordinates. Any linear map which fixes these points must be the identity (on the projective space), and any semilinear map which fixes them must therefore be induced by a field automorphism, and must therefore fix every rational point. Now let $K$ consists of $S$ and two more rational points, and let $\phi:K\to K$ be the transposition interchanging these two points.

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  • $\begingroup$ I also should have specified that $\phi$ is continuous. Sorry for the confusion $\endgroup$ Aug 13, 2018 at 14:26
  • $\begingroup$ Oh, never mind, I see now. Well done. $\endgroup$ Aug 13, 2018 at 14:27

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