Flatness of Surfaces of Revolution

Lee Smooth Manifolds

We aim to prove the following theorem:

Main Theorem (Prop. 13.19)) Let $C\subseteq H$ be a connected embedded $1$-dimensional submanifold of the half-plane $H=\{(r,z):r>0\}$, and let $S_C$ be the surface of revolution generated by $C$. The induced metric on $S_C$ is flat if and only if $C$ is part of a straight line.

First, some useful facts: if an injective immersion maps an open subset of ${\mathbb{R}}^n$ into an open subset of an embedded $n$-submanifold, it must be a local parametrization as smooth structures on embedded submanifolds are unique. Furthermore, if $(M,g)$ is a Riemannian manifold, any smooth curve $\gamma :J\to M$ with nonvanishing velocity has a unit-speed reparametrization $\tilde{\gamma }$ such that $|\tilde{\gamma }(t{)}^{\prime }{|}_g=1$ for all $t$ in the domain of $\gamma$. We assume this for now as its proof requires Riemannian distance.

Assume that $C$ is part of a straight line; take a unit-speed parametrization $\gamma (t)=(Pt+K,Qt+L)$ defined on some interval $J\subset \mathbb{R}$ (open, as $C$ doesn’t have boundary). One can observe that, for any $(t,\theta )\in J\times \mathbb{R}$, in some neighborhood of $(t,\theta )$ the function $\tau (t,\theta )=((Pt+K)\cos \theta ,(Pt+K)\sin \theta ,(Qt+L))$ is a smooth parametrization (by our first fact). In these coordinates, the induced Riemannian metric can be computed as

$${\tau }^{\ast }\bar{g}=d{x}^2+d{y}^2+d{z}^2=d((Pt+K)\cos \theta {)}^2+d((Pt+K)\sin \theta {)}^2+d(Qt+L{)}^2$$ $$=(Pt+K{)}^{2}d{\theta }^2+(P^2+Q^2)d{t}^2=(Pt+K{)}^{2}d{\theta }^2+d{t}^2$$

We want an isometry that pushes this into the Euclidean metric; observe above that $d{x}^2+d{y}^2=P^{2}d{t}^2+(Pt+K{)}^{2}d{\theta }^2$, which is close to the final expression. Tweaking it to $\tilde{\tau }(t,\theta )=((Pt+K)\cos P\theta ,(Pt+K)\sin P\theta )/P=(u,v)$ to add a $P^2$ in front of $d{\theta }^2$ and rescaling defines a diffeomorphism $\tilde{\tau }$ where ${\tilde{\tau }}^{\ast }\bar{g}=d{u}^2+d{v}^2=(Pt+K{)}^{2}d{\theta }^2+d{t}^2$, as desired. Thus, the induced metric on $S_C$ is locally isometric to the Euclidean metric, and by definition $S_C$ is flat. Note this reparametrization doesn’t work if $P=0$, but this is simply the trivial cylinder case.

We now show the converse. Take any $p=(p_1,0,p_2)\in S_C$ such that $p_1>0$ and let $\gamma (t)=(a(t),b(t))$ be a local parametrization of $C$ centered at $(p_1,p_2)$, assumed to be unit-speed (by fact #2). Defining $\tau (t,\theta )=(a(t)\cos \theta ,a(t)\sin \theta ,b(t))$, we can see that, restricted to some neighborhood $U$ of $(0,0)$, $\tau$ is a local parametrization of $S_C$ centered at $p$. The induced metric in these coordinates is

$${\tau }^{\ast }\bar{g}(t,\theta )=(a^{\prime }(t{)}^2+b^{\prime }(t{)}^2)d{t}^2+a(t{)}^{2}d{\theta }^2=d{t}^2+a(t{)}^{2}d{\theta }^2$$

which, by hypothesis, is flat. Let $V=\tau (U)$ be a subset of $S_C$. The local sections $E_1$ and $E_2$ on $V$, defined by $E_1=\partial /\partial t$ and $E_2=(1/a)\partial /\partial \theta$ (note $a$ cannot vanish as $C$ is in the positive-$x$ half), are thus an orthonormal local frame on $V$. As the induced metric is flat, there exists some connected neighborhood $W\subset V$ of $p$ on which there is a commuting orthonormal local frame $\{F_1,F_2\}$. They can then be written as $F_1=u_1{E}_1+v_1{E}_2$, $F_2=u_2{E}_1+v_2{E}_2$ where $u_i,v_i$ are smooth functions on $W$. These must satisfy

$$⟨F_1,F_2⟩=0\Rightarrow u_1{u}_2+(v_1{v}_2)=0,⟨F_i,F_i⟩=1\Rightarrow u_i^2+v_i^2=1$$

Furthermore, as they commute,

$$0=[F_1,F_2]=\left[u_1\frac{\partial }{\partial t}+\frac{v_1}{a}\frac{\partial }{\partial \theta },u_2\frac{\partial }{\partial t}+\frac{v_2}{a}\frac{\partial }{\partial \theta }\right]=\left(u_1\frac{\partial u_2}{\partial t}+\frac{v_1}{a}\frac{\partial u_2}{\partial \theta }-u_2\frac{\partial u_1}{\partial t}-\frac{v_2}{a}\frac{\partial u_1}{\partial \theta }\right)\frac{\partial }{\partial t}$$ $$+\left(u_1\frac{\partial (v_2/a)}{\partial t}+\frac{v_1}{a}\frac{\partial (v_2/a)}{\partial \theta }-u_2\frac{\partial (v_1/a)}{\partial t}-\frac{v_2}{a}\frac{\partial (v_1/a)}{\partial \theta }\right)\frac{\partial }{\partial \theta }$$

The first two equations forces, for any $q\in W$, $(u_2(q),v_2(q))=\pm (-v_1(q),u_1(q))$. By continuity and the connectedness of $W$, the sign must be fixed throughout $W$; WLOG assume $(u_2(q),v_2(q))=(-v_1(q),u_1(q))$ for all $q\in W$. We can further simplify by noting $\partial a/\partial \theta =0$. Writing $u=u_1=v_2$, $v=v_1=-u_2$, and $a^{\prime }=\partial a/\partial a_t$, the final equation becomes

$$-u{v}_t-\frac{v}{a}{v}_{\theta }+v{u}_t-\frac{u}{a}{u}_{\theta }=0,\frac{v{u}_{\theta }}{a^2}-\frac{u{v}_{\theta }}{a^2}+\frac{u{u}_t}{a}-\frac{u^2{a}^{\prime }}{a^2}+\frac{v{v}_t}{a}-\frac{v^2{a}^{\prime }}{a^2}=0$$

Using the identity $u^2+v^2=1$ to simplify, we get $u{v}_t=v{u}_t$ and $a^{\prime }=v{u}_{\theta }-u{v}_{\theta }$. Assume, WLOG, that $v$ doesn’t vanish on some regular ball $Y\subset W$ containing $p$ in $S_C$; the first equation implies $(u/v{)}_t=0$, and thus $u=vf(\theta )$ on $Y$ where $f$ is smooth and nonvanishing. The identity $u^2+v^2=1$ forces $v^2(1+f^2(\theta ))=1$, and therefore $v$ is a function of $\theta$ alone, implying $u$ is too. This implies $a^{\prime }$ is independent of $t$, and so $a^{\prime }\equiv c$ for some constant $c$ on $Y$. as $a^{\prime 2}+b^{\prime 2}=1$, $b^{\prime }$ must also be constant, and both $a$ and $b$ are affine functions of $t$ in $Y$. By smoothness and connectedness, $C$ is thus forced to be part of a straight line.