4

Setting. Let $V$ be a vector space over some field $F$ with $\dim(V) \geq 3$. A map $\phi : V \to V$ is called flat if the image of any subspace of dimension $2$ is again a subspace of dimension $2$.

Question. Is the implication $$f \text{ flat} \implies f \text{ injective}$$ always true?

Motivation. In the case of a finite field $F = \mathbb F_q$, the answer is "yes": Assume that $\phi(v) = \phi(w)$ for two distinct vectors $v,w\in V$. Let $U$ be a subspace of dimension $2$ containing $v$ and $w$. Then $\#U = q^2$, but $\#\phi(U) \leq q^2 - 1$, which is a contradiction.

Clarifications.

  1. The map $\phi$ is not assumed to be linear! (For linear maps $\phi$, it is not hard to see that the answer is "yes".)
  2. The term "subspace" means "vector subspace" (and not "affine subspace").
azimut
  • 22,696
  • This seems to be related: https://math.stackexchange.com/questions/2035208/is-a-map-that-preserves-lines-and-fixes-the-origin-necessarily-linear also https://math.stackexchange.com/questions/3985327/is-lines-remain-lines-sufficient-to-characterise-linearity-in-mathbb-r2 – Gerry Myerson Feb 21 '23 at 12:39
  • Possibly https://mathoverflow.net/questions/163923/properties-of-vector-spaces-without-ac is related. – Gerry Myerson Feb 21 '23 at 12:45
  • Concerning the 2 links to MSE: of course, I immediately thought of this "fundamental theorem of affine geometry". But the difficulty lies in proving that if planes remain planes then lines remain lines. As for the link to MO: I cannot see how it could be related. – Anne Bauval Feb 21 '23 at 12:46
  • 1
    @GerryMyerson Many thanks for your pointers. Yes, there is some relation to your first pair of links, in the sense that they explore the outskirts of the fundamental theorem of projective geometry (the one which states that every collineation of a projective Desarguesian geometry of dimension $\geq 2$ is induced by a semilinear map). But I don't see how they should help making progress here. Concerning the mathoverflow-Link: Admittedly, that one is a bit scary to me :) But yes, maybe it is related. I'd not be too surprised if in the end, the property depends on the axiom of choice... – azimut Feb 21 '23 at 12:52
  • Right, I humbly retract my comment. +1 for the interesting question. – Vivaan Daga Feb 21 '23 at 13:20
  • 1
    @GerryMyerson I'm not sure I see how the MO post is related. The words flat and injective mean completely different things in that question. – ronno Feb 21 '23 at 15:28

1 Answers1

4

Assuming subspace means vector subspace and not affine subspace, and with $V$ over $\mathbb{R}$, take some $v \ne 0$, map $kv$ for $k \in [2, \infty)$ to $(k - 1) v$ and map everything else by identity. Then the image of every line is itself, so the image of every plane is itself.

ronno
  • 11,390
  • thank you! I think your are right. Seems that I was thinking too complicated... – azimut Feb 21 '23 at 15:31
  • What goes wrong with assuming affine subspace? – Vivaan Daga Feb 21 '23 at 15:59
  • 1
    @Shinrin-Yoku Consider an affine plane through $2v$ which does not contain the origin. This particular map will take such a plane to a punctured plane and an isolated point. – ronno Feb 21 '23 at 17:11