# Isometry between circles

Let $C_{1}$ and $C_{2}$ be two circles, respectively with radius $1$ and $2$ endowed with the riemannian metric induced by the euclidean metric. I need to show that they are not isometric. This seems simple but I cannot get a proof.

Here is what I have tried. I work with a generic circle $C_{r}$ of radius $r>0$. First, we have the local coordinate $\theta$ on the circle such that, locally:

$$\phi(\theta)=(r\cos(\theta),r\sin(\theta)),\quad \theta\in[0,2\pi)$$

Denote by $g_{0}=(dx)^{2}+(dy)^{2}$ the euclidean metric on $\mathbb{R}^{2}$. The induced metric is given by

$$\phi^{\ast}g_{0}=r^{2}(d\theta)^{2}$$

If $\gamma:I\to M$ is a differentiable curve (in the sequel, a curve) from an interval $I\subset\mathbb{R}$ to a riemannian manifold $M,g$, the length of $\gamma$ is defined as

$$\ell(\gamma)=\int_{I}\sqrt{g(\dot\gamma(t),\dot\gamma(t))}$$

The distance between two points $x,y\in M$ relatively to the metric $g$ is defined as

$$d_{g}(x,y) = \inf\{\ell(\gamma)\mid\gamma:[0,1]\to M,\gamma\text{ is a curve},\gamma(0)=x,\gamma(1)=y\}$$

It is easy to see that if

$$i:M,g\to N,h$$

is an isometry (i.e. a diffeomorphism such that $i^{\ast}h=g$), then for any curve $\gamma:I\to M$ we have

$$\ell(\gamma)=\ell(i\circ\gamma).$$

Hence, we have

$$d_{g}(x,y) = d_{h}(i(x),i(y))$$

Therefore, let

$$\gamma:[0,1]\to C_{r}:t\mapsto \gamma(t) = (r\cos(s(t)),r\sin(s(t)))$$

A quick computation shows that

$$\ell(\gamma)=r\int_{0}^{1}\vert \dot s(t)\vert_{1} dt$$

where $\vert v\vert_{k}$ denotes the euclidean norm of $v\in\mathbb{R}^{k}$.

As we want to prove that $C^{1}$ and $C^{2}$ are not isometric, we need to show that there exists no isometry $i:C^{1}\to C^{2}$. If we suppose there exists such an isometry, we must have

$$\ell(\gamma)=\ell(i\circ \gamma)$$

for any curve $\gamma:[0,1]\to C_{1}$.

But I am stuck. How to conclude from this?

The point is that the lenght of a curve does not depends on the parametrization. So $i \circ \gamma_1$ should be a reparametrization of $\gamma_2$, in particular they have the same lenght. So if $C_1, C_2$ are isometric this would implies as @Henning Makholm said that $2\pi = 4\pi$ which is absurd.

• Thanks. To be sure to understand it correctly : what you say is that there is a curve, say, $\gamma_{1}(t)=(\cos(t),\sin(t))$ which is bijective from $[0,2\pi)$ to the small circle and that $(i\circ\gamma_{1})(t)$ is also bijective from $[0,2\pi)$ to the big circle. As $(i\circ\gamma_{1})$ is a reparametrization of $t\mapsto (r\cos(t),r\sin(t))$, they have the same length $2\pi r$ and as length is invariant under isometries, we should have $2\pi = 2\pi r$. – MoebiusCorzer Jun 16 '17 at 14:57
• Exactly. Notice that the square root is very important for being invariant under parametrization, so if you take $e(\gamma) := \int g(\overset{\cdot}{\gamma},\overset{\cdot}{\gamma})dt$ (the "energy") you won't obtain something independant of the parametrization. (But it is still useful to physicists since it contains some "dynamical" informations.) – user171326 Jun 16 '17 at 15:04
• There still is something I am not sure about. In order to have a reparametrization, one needs $\gamma_{1}^{-1}\circ i^{-1}\circ\gamma_{2}:[0,2\pi)\to [0,2\pi)$ to be of positive derivative if I am not mistaken. This is ensured by the fact that they are all $C^{1}$-diffeomorphisms? – MoebiusCorzer Jun 16 '17 at 15:24
• You are right, but reversing the direction doesn't affect the lenght, so it doesn't matter if the derivative is negative (just change it by e.g $\tilde \gamma_1(t) := \gamma_1(2\pi - t)$.) – user171326 Jun 16 '17 at 15:29
• You're welcome, glad I could help ! – user171326 Jun 16 '17 at 21:20

My approach would be to prove that the metric induces a measure on the manifold which is preserved by isometries.

Then the two circles have total measure $2\pi$ and $4\pi$, respectively, and therefore cannot be isometric.

• Thank you. That's an interesting approach, but I am beginning with riemannian geometry and this exercise is given to "beginners". Introducing a (useful) measure on a riemannian manifold can be done naturally, I guess, by using Riesz representation theorem for Borel measures on locally compact spaces and by mimicking the construction of the Lebesgue measure. However, all this seems overly complicated for this problem. – MoebiusCorzer Jun 16 '17 at 14:25

Hint One can save some effort and conceptual difficulty here by specializing the general parameterized curve $\gamma$ to a curve that suitably respects the Riemannian structure, i.e., a unit-speed geodesic. Since the isometry $i$ preserves (parameterized) geodesics, $i \circ \gamma$ is also unit-speed geodesic.

Computing $\gamma$ explicitly using the usual embeddings into $\Bbb R^2$ shows that our unit-speed geodesic $\gamma$ satisfies $\gamma(0) = \gamma(2 \pi)$. Now, apply $i$ to both sides.