5
$\begingroup$

Do there exist differentiable almost-everywhere functions on $\mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $\frac{|\langle x, y \rangle|}{|x||y|} \geq \frac{|\langle f(x), f(y) \rangle|}{|f(x)||f(y)|}$? How does one go about constructing one?

$\endgroup$
  • 1
    $\begingroup$ In $\mathbb C$ this is called a conformal map. I'm unaware of analogues in $\mathbb {R}^n$, but they're studied quite a bit in $\mathbb C$ $\endgroup$ – Mark Oct 4 '16 at 6:10
  • 1
    $\begingroup$ The conditions seem to be very rigid - consider the vertices of a simplex centered at the rigin. - Then again, your inequality does no treally match what you write in the title $\endgroup$ – Hagen von Eitzen Oct 4 '16 at 6:10
5
$\begingroup$

I claim that those maps are precisely those that preserves lines through the origin, followed by an orthogonal movement.

For $n=1$, the condition is void (except that we demand $0\mapsto 0$ perhaps?) and hence the claim holds.

Your inequality demands that image vectors are "at least as orthogonal" as the input vectors. In particular, such a map preserves orthogonality. Thus for any $v\ne0$ with $f(v)\ne 0$, it induces a map with the same properties from $v^\perp$ to $f(v)^\perp$, i.e., $\Bbb R^{n-1}\to\Bbb R$. If $e_1,\ldots, e_n$ is the standard basis of $\Bbb R^n$, then $f(e_1),\ldots, f(e_n)$ is an orthogonal base of $\Bbb R^n$ and by performing an orthogonal movement, we may assume $f(e_i)=c_ie_i$ with $c_i>0$. We may assume that $f|_{e_n^\perp}$ is of the claimed form. For $v\in\Bbb R^n$ write $c=ae_n+w$ with $w\in e_n^\perp$. Assume $c\ne0$. Then $v^\perp$ intersects $e_n^\perp$ in an $\Bbb R^{n-2}$ left invariant under$f$, hence $f(v)$ is confined to the perpendicular space of that, which is a $2$-plane. Thus it remains to show the claim for $n=2$.

Indeed, $v=ae_1+be_2$ with non-zero $a,b$ can map at most to $a'e_1+b'e_2$ with $a':b'=\pm a:b$. IN case of negative sign, add a reflection at one of the axes. Then for all other vectors $w$ in the plane, the angle condition relative to $e_1$, $e_2$, $v$ determine that $f(w)$ is on the same line as $w$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The map on the plane in polar co-ordinates $ (r,\theta)\mapsto (r^2,2\theta)$ is an angle-magnifier.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @IwillnotexistIdonotexist: I mixed up polar and cartesian co-ordinates. The map comes from $z\mapsto z^2$. In general higher powers too. I have corrected it now. $\endgroup$ – P Vanchinathan Oct 4 '16 at 6:30
  • $\begingroup$ But then what if you take two 2D vectors, both of magnitude unity, one with $\theta_1 = 0$ and the other with $\theta_2 > 120^o$? Then $\theta'_1 = 0$ and $\theta'_2 > 240^o$ which is the same as $<120^0$ (because circular angles operate on the modulo principle). So this function still decreases the angle between some vectors. $\endgroup$ – Iwillnotexist Idonotexist Oct 4 '16 at 6:49
  • $\begingroup$ But sense (orientation) is important. Let me elaborate with unit vectors. The angle between u and v is to be understood as how much one has to sweep u in counter-clockwise direction to merge with $v$. After applying the transformation $f$ you have to calculate how much $f(u)$ has to be swept in counterclockwise direction to merge with $f(v)$. $\endgroup$ – P Vanchinathan Oct 4 '16 at 6:55
  • $\begingroup$ If the calculation "post-f" is more than "pre-f" then f is to be understood as angle magnifier. Of course visually things have become closer, I agree. $\endgroup$ – P Vanchinathan Oct 4 '16 at 6:58
  • 1
    $\begingroup$ What you say is to be compared with the optical illusion of car wheels moving backwards in a movie. That is, if in between two frames, the car had moved forward in such a way that the tyre made $350^\circ$ rotation, the viewer will see that tyre has moved backwards by ten degrees. But this car motion is 350 degrees forward per unit time (the time elapsed between two frames by the movie camera). $\endgroup$ – P Vanchinathan Oct 4 '16 at 7:18

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.