# Riemannian 1-manifold is flat

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

• 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.
– Danu
Jan 19 '20 at 21:44
• 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 $\frac{\partial x}{\partial y}=1/\sqrt{A}$ Jan 19 '20 at 21:49

Flat means that the curvature vanishes since the curvature is a $$2$$-form, it vanishes on a $$1$$-dimensional manifold.

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.