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I'm just beginning to learn about Riemannian geometry and I ran into the following exercise: Show that every Riemannian 1-manifold is flat.

I know that this is supposed to be a very basic exercise, but I am having troubles with it because I missed the part of class where we talked about isometries and I am having trouble understanding. Here are my thoughts so far:

Let $(M^1,g)$ be a Riemannian manifold (dimension 1) and let $(U,\phi)$ be a local chart. That is to say that $\phi$ is a diffeomorphism with its image and $\phi(U)\subset \mathbb{R}$ is an open interval. Let $p\in U$ and consider $u,v\in T_pM$. Since this is a 1-dimensional vector space, $u=u^1\frac{\partial}{\partial x^1}$ and $v=v^1\frac{\partial}{\partial x^1}$. I am not sure how to prove from here that $\phi^* \tilde{g}=g$ where $\tilde{g}$ is the Euclidean inner product on $\mathbb{R}$. It feels obvious but I am not sure exactly how to write it

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    $\begingroup$ Here's a hint: After using a chart to map a piece of your 1-manifold onto an open interval, what is the most general form that the metric $g$ can take? Now we need to compare to the standard metric on the interval, i.e. $g_0=dx^2$. Given a diffeomorphism $\phi$ from the interval to itself, write down $\phi^*g_0$. Now compare this to your general form of the metric $g$, and see if you can write down a $\phi$ that realizes it. $\endgroup$
    – Danu
    Jan 19 '20 at 21:44
  • $\begingroup$ So if $A:=g(\frac{\partial}{\partial x},\frac{\partial}{\partial x})>0$ (here $(U,x)$ is a coordinate domain), can I choose a new coordinate $(U,y)$ where $y=\sqrt{A} x$? Does that work? Because then $\frac{\partial x}{\partial y}=1/\sqrt{A}$ $\endgroup$ Jan 19 '20 at 21:49
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Flat means that the curvature vanishes since the curvature is a $2$-form, it vanishes on a $1$-dimensional manifold.

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Your $\phi$ won't automatically be an isometry with $\mathbb{R}$, but there is the following trick to modify it properly:

First, following the hint of Danu, note that if $\gamma:\mathbb{R}\to M$ is a smooth map, then $\gamma^*g=f\,dt^2$ for a smooth function $f$ (symetric tensors of $\mathbb{R}$ are spanned by $dt^2$). Since $\gamma^*g=g(d\gamma(\bullet),d\gamma(\bullet))$ by definition, we read that

$$f=f\,dt^2\Big(\frac{\partial}{\partial t},\frac{\partial}{\partial t}\Big)=g\Big(d\gamma\Big(\frac{\partial}{\partial t}\Big),d\gamma\Big(\frac{\partial}{\partial t}\Big)\Big)=g(\gamma'(t),\gamma'(t)).$$

So our goal to construct an isometry will be to find a $\gamma$ such that $f=1$, i.e. $g(\gamma',\gamma')=1$.

Note $\psi:=\phi^{-1}:(a,b)\to M$. The map $l:t\mapsto\int_c^t|\psi'(s)|ds$ - where $c\in(a,b)$ and $|\psi'(s)|=\sqrt{g(\psi'(s),\psi'(s))}$ - is differentiable, of differential $l'(t)=|\psi'(t)|>0$. So $l$ has a differentiable inverse, say $h$. Note that

$$h'(t)=\frac{1}{l'(h(t))}=\frac{1}{|\psi'(h(t))|}$$

and so $|(\psi\circ h)'(t)|=|\psi'(h(t))\cdot h'(t)|=\Big|\frac{\psi'(h(t))}{|\psi'(h(t))|}\Big|=1$. Thus $\gamma:=\psi\circ h$ is as required a unit speed parametrization of $M$: since $\gamma^*g=g(\gamma'(t),\gamma'(t))\,dt^2=dt^2$, $\gamma$ is an isometry.

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