3
$\begingroup$

Consider two vector spaces $V$ and $V'$ with the same dimension. Let $f: V\longrightarrow V'$ be a bijection such that it maps straight lines into straight lines; I don't know if the following statement is true:

There exists a unique linear function $L:V\longrightarrow V'$ such that $f(v)=L(v)+v_0$ for any $v\in V$ and for a certain fixed $v_0\in V$.

Maybe this is a stupid question, but I can't find the (probably) easy proof of this fact.

$\endgroup$
3
$\begingroup$

If $\dim V=\dim V'=1$ there are many counterexamples. However, if $\dim V=\dim V'=n>1$ and if $\text{char}\mathbb{K}\neq 2$ (where $V$ and $V'$ are vector spaces over $\mathbb{K}$), any map $f:V\to V'$ which sends affine straight lines into affine straight lines sends also affine subspaces into affine subspaces and you can write $$f(v)=f(0)+a(v)$$ where $a:V\to V'$ is an isomorphism of abelian groups such that $a(\lambda v)=\phi(\lambda)a(v)$ for some automorphism $\phi:\mathbb{K}\to\mathbb{K}$. Moreover, if $f$ also preserves the ratio of distances between any three collinear points, then it is an affine map (i.e. the map $a$ above is linear).

Also, if $\mathbb{K}=\mathbb{R}$, the only field automorphism is the identity, therefore also in this case $a$ is linear, hence $f$ is an affine map (without conditions on the simple ratio of collinear points).

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

Hint: Let $V$ and $V'$ be $1$-dimensional ...

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

(What follows is not quite finished, because the computations didn't go as simply as I had thught they would. Still, it seems better to post than to delete.)

I will assume that $f$ is differentiable, and leave it to others to determine is this assumption is important.

If the dimension of $V$ is $1$, the the claim is obviously false, as mentioned by Amitesh Datta. I will give an elementary proof for the dimension equal to $2$. For larger dimensions, just note that $f$ has to map planes to planes, so the considerations for $\operatorname{dim} V = 2$ still apply, so $f$ restricted to any plane is affine. But a function affine on any plane is affine per se, so the claim in higher dimension follows.

So, let us assume that $\operatorname{dim} V = 2$, so that we are talking about a plane. Let $A,B,C$ be any points on the plane, and let $A' = f(A)$, $B' = f(B)$, $C' = f(C)$. For $\alpha \in [0,\infty]$, let $K(\alpha)$ denote the point on $BC$ with $|BK(\alpha)|:|K(\alpha)C| = \alpha$. Likewise, we denote by $L(\beta), M(\gamma)$ analogously defined points on $CA$ and $AB$ respectively. Because of the condition about the lines being preserved, $f$ maps $K(\alpha)$ to a point $K'(\alpha')$ with $K'$ defined analogously, and $\alpha' = t(\alpha)$ for some function $t:\ [0,\infty] \to [0,\infty]$. Likewise, let $r,s$ be functions such that $f(L(\beta)) = L'(r(\beta))$ and $f(M(\gamma)) = M'(s(\gamma))$.

The key geometric insight which I will use is the Menelaus Theorem, which asserts that the lines $AK(\alpha), BL(\beta), CL(\gamma)$ intersect in a single point iff $\alpha\beta\gamma =1$. Thus, if $\alpha\beta\gamma =1$, then we also have $t(\alpha) r(\beta) s(\gamma) = 1$. Put $\gamma = 1/\alpha\beta$ so that $\alpha,\beta$ are independent. Differentiating this equality, we get for any $\alpha,\beta$: $$t'(\alpha) s(\frac{1}{\alpha\beta}) + \frac{-1}{\alpha^2 \beta}t(\alpha)s'(\frac{1}{\alpha\beta}) = 0.$$ Letting $\beta := \frac{1}{\alpha \gamma}$ for arbitrary $\gamma$, we get: $$ \alpha \frac{t'(\alpha)}{t(\alpha)} = \gamma \frac{s'(\gamma)}{s(\gamma)}. $$ Because each side depends only on a single variable, they have to be constant: $$ \alpha \frac{t'(\alpha)}{t(\alpha)} = c, \quad c \text{ - constant}.$$ This is easily solved: $$ (\ln t)'(\alpha) = \frac{c}{\alpha} $$ $$ (\ln t)'(\alpha) = (c \ln )'(\alpha)$$ $$ (\ln t)(\alpha) = c \ln(\alpha) + d $$ $$ t(\alpha) = \alpha^c e^d $$ Note that $c$ is universal: the same for $t,s,r$, while $d$ depends on the line we are looking at (so, $d = d_{BC}$, say).

Now, it didn't go nearly as smoothly as I had thought at first. There is a way to finish this off, but that requires a little bit more computation. Here is what to do: Pick another point, say $C_2$, at $BC$. For the triangle $ABC_2$, one can redo the above reasoning. This gives two formulas for how points on $BC_2$ transform onto point on $B'C_2'$. One is, roughly speaking $ t(\alpha) = \alpha^c e^d $, the other is (even more roughly) $ t_2(\alpha_2) = \alpha_2^c e^{d_2} $. These two are incompatible, unless $c= 1 d = 0$, I believe.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.