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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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    You can get proper norm bars using `\lVert` and `\rVert`.2012-09-27
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    Try to differentiate $\|\alpha(s+h) - \alpha(s)\|^2$ with respect to $s$.2012-09-27
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    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...2012-09-27
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    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.2012-09-28
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    Any progress yet?2012-10-01
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    Not yet. Even the teacher couldn't solve it through this way, though it seems to be the right way to go. He mentioned something about it implying that the curve is arc-lenght parameterized. I'm going to try again later.2012-10-01
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    In the statement of the problem, do you mean $\alpha(I)$ is a subset of a line or a circle?2012-10-05
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    I don't have time at the moment to try this approach, so I want to throw it out as a comment. I want to try to show that there's a Mobius transformation that carries $I$ to a subset of a circle or a line.2012-10-05
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    Yes @MichaelAlbanese.2012-10-05

2 Answers 2

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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.2012-10-04
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    You are right, i just saw it. Im trying to solve this problem here... Any idea?2012-10-04
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    I've tried a lot of things using this same starting point. None of them resulted in what I want :(2012-10-04
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    Ok, I'm gonna check your answer within the hour, just arrived at home. Looks promising! Thanks!2012-10-05
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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 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, $\alpha(t)=\gamma(t-s)\alpha(s)$, 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.2012-10-04
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    Just say that every differentiable function is continuous, and hence Ewan's proof applies.2012-10-04
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    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).2012-10-05
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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.2012-10-05