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I need to show this result:

Let $\alpha :I\rightarrow \mathbb{R}^2$ a smooth curve, where $I$ is a compact interval of the real line. If $\lVert \alpha (s) - \alpha (t) \rVert$ depends only on $|s-t|$, for all $s$, $t$ in $I$, then $\alpha$ must be a subset of a line or a circle.

I have tried calling $\lVert \alpha (s) - \alpha (t) \rVert=f(|s-t|)$, where $f$ is smooth but this led me nowhere. Also, I had a suggestion to fixing $t$ as $0$ and $\alpha(0)=0$ through reparametrization and a rigid movement, so I would have $\lVert\alpha(s) \rVert=f(|s|)$, $f$ is smooth; this lead to nothing also. I strongly believe I must show that the curvature of this function if constant, either $0$ (then it is a line) or it is a constant $\neq 0$ from where it is in a circle.

Can someone please give me a hint?

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1  
You can get proper norm bars using \lVert and \rVert. –  joriki Sep 27 '12 at 1:45
1  
Try to differentiate $\|\alpha(s+h) - \alpha(s)\|^2$ with respect to $s$. –  user31373 Sep 27 '12 at 3:13
    
I have tried that, writing $\parallel \alpha(s+h)-\alpha (s)\parallel ^2t = < \alpha(s+h)-\alpha (s),\alpha(s+h)-\alpha (s)>$ When I call $g(t,s)=\parallel \alpha(t)-\alpha (s)\parallel$, and differentiate $g^2(t+h,t)$ with respect to $t$, this equals $f(|h|)$. Since $\frac{d}{dt}(f(|h|))=0$ the derivative of $g(t+h,t)^2$ is zero. Now give me a minute... –  Marra Sep 27 '12 at 13:09
    
The reason for differentiating the square of distance is that you can write it as product of vector with itself and use the product rule for derivative. I don't promise that it will solve the problem, but it will give a kind of differential equation. –  user31373 Sep 28 '12 at 1:42
    
Any progress yet? –  user31373 Oct 1 '12 at 2:18

2 Answers 2

up vote 6 down vote accepted
+150

Ok let me try it again.

First of all, suppose that $\alpha$ is parametrized by arc length.

Write $$\alpha(s)-\alpha(t)=\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2}+o(|s-t|)$$

Then

\begin{eqnarray} \|\alpha(s)-\alpha(t)\|^{2} &=& \|\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2}+o(|s-t|)\|^{2} \nonumber \\ &=& \|\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2}\|^{2}+2\langle\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2},o(|s-t|)\rangle \nonumber \\ &&+\|o(|s-t|)\|^{2} \nonumber \\ &=& |s-t|^{2}+k(t)\frac{|s-t|^{4}}{4}+o(|s-t|) \end{eqnarray}

So, as you can see, if $k(t)$ is not constant $\|\alpha(s)-\alpha(t)\|$ depends on $k(t)$. Hence the only possibilities are the circle or the line. But an straightforward calculation shows that for the circle and the line the statement is true.

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This only proofs that the derivative of $\parallel \alpha '(s)\parallel=0$, that is, the tangent vector has a constant lenght and thus the curve is arc-lenght parameterized (just divide it by a constant). It does not prove that it is contained in a line or a curve. –  Marra Oct 4 '12 at 18:37
    
You are right, i just saw it. Im trying to solve this problem here... Any idea? –  Tomás Oct 4 '12 at 18:41
    
I've tried a lot of things using this same starting point. None of them resulted in what I want :( –  Marra Oct 4 '12 at 19:17
    
Ok, I'm gonna check your answer within the hour, just arrived at home. Looks promising! Thanks! –  Marra Oct 5 '12 at 21:24

You do not need differentiability. Below is a proof that works when $\alpha$ is only continuous.

We may assume that $I$ starts with $0$ : $I=[0,M]$ for some $M$. Suppose that the image of $\alpha$ is not contained in a line. Then there are $x_1<x_2<x_3$ such that $\alpha(x_1),\alpha(x_2),\alpha(x_3)$ are not collinear. Let $t\in ]0,M-x_3[$. Let now $x=x_4$ be an arbitrary number in $[0,M-t]$. Consider the two families of four points $A_i=\alpha(x_i) (1 \leq i \leq 4)$ and $B_i=\alpha(x_i+t) (1 \leq i \leq 4)$. They are isometric by hypothesis : $d(A_i,A_j)=d(B_i,B_j)$.

By this older StackOverflow question, there is an isometry $\gamma(x,t)$ of the plane sending $A_i$ to $B_i$. Since $\gamma(t)$ is already uniquely determined by its image on $\alpha(x_1),\alpha(x_2),\alpha(x_3)$, we see that $\gamma(x,t)=\gamma(t)$ is in fact independent of $x$, and we can explicitly write out the coefficients of the matrix of $\gamma(t)$ in terms of the coordinates of $\alpha(x_1),\alpha(x_2),\alpha(x_3)$ and $\alpha(x_1+t),\alpha(x_2+t),\alpha(x_3+t)$. So $\gamma$ will be continuous if $\alpha$ is. Unicity also ensures that $\gamma$ is a homomorphism (where it is defined) : $\gamma(s+t)=\gamma(s)\gamma(t)$ when $s,t$ and $s+t$ are all in $I$.

Now ${\sf det}(\gamma(t))=1$ or $-1$, and by continuity this determinant is always $1$ or always $-1$. Since $\gamma(0)$ is the identity, we see that $\gamma$ is always an (affine) rotation. We have $\gamma(s)\gamma(t)=\gamma(s+t)=\gamma(t)\gamma(s)$. We see that $\gamma(s)$ and $\gamma(t)$ commute, so they must share the same center point. We deduce that the center $\Omega$ of $\gamma(t)$ is independent of $t$. Now for any $s<t$, $\alpha(t)=\gamma(t-s)\alpha(s)$, so $\alpha(t)$ and $\alpha(s)$ are located at the same distance from $\Omega$. So $\alpha$ walks on a circle of cnter $\Omega$, qed.

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Well, I'd like some argument using the differentiability of the functions involved. –  Marra Oct 4 '12 at 19:19
4  
Just say that every differentiable function is continuous, and hence Ewan's proof applies. –  Hagen von Eitzen Oct 4 '12 at 20:55
    
Which also does not solve it in an adequate way (it's a differential geometry problem), and also only proofs only that if it's not in a line then it's in a circle;the second part is missing (if it's not in a circle then it's in a line). –  Marra Oct 5 '12 at 0:07
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@Gustavo: Proving "A or B" is the same as proving "B" under the assumption that "A" is false, so there is no "second part" to worry about. –  Jason DeVito Oct 5 '12 at 2:26

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