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Consider the surface of revolution $S$ obtained by rotating the curve $(x,y=e^x,z=0)$ about the $x$-axis, and equip $S$ with the induced Riemannian metric from $\mathbb{R}^3$. Show that $S$ is complete and compute it's injectivity radius.

I've been having a lot of trouble with this problem. I began by parameterizing the surface via the function $\varphi:U\to\mathbb{R}^3$ given by $$\varphi(u,v)=(e,e^u\cos(v),e^u\sin(v)),$$ where $U=\{(u,v)\in\mathbb{R}^2~:~v\in[0,2\pi)\}$. I then computed the the metric in these coordinates as $g_{uu}=1+e^{2u}$ and $g_{vv}=e^{2u}$ (and the diagonal terms are zero). A direct computation shows that the Christoffel symbols are given by $$\Gamma^{v}_{uv}=\Gamma^{v}_{vu}=1,\qquad\Gamma^{u}_{vv}=\frac{-e^{2u}}{1+e^{2u}},\qquad\Gamma^{u}_{uu}=\frac{e^{2u}}{1+e^{2u}}.$$ So we have that the geodesic equations are given by $$\frac{d^2v}{dt^2}+2\frac{du}{dt}\frac{dv}{dt}=0,\qquad\frac{d^2u}{dt^2}-\frac{e^{2u}}{1+e^{2u}}\left(\frac{dv}{dt}\right)^2 + \frac{e^{2u}}{1+e^{2u}}\left(\frac{du}{dt}\right)^2=0.$$ Unfortunately, I have no idea how to solve this system of differential equations, and didn't know how to move forward from here. Instead, I tried to guess what the geodesic equations might be in these coordinates: for $(u_0,v_0)\in U$ and $v\in\mathbb{R}^2\cong T_{(u_0,v_0)}U$ I thought that they might given by $$\gamma(t)=(u_0,v_0)+tv.$$ I tried reparameterizing this curve by arclength and showing that it satisfies the geodesic equations, but also haven't had much luck. I think that I might be approaching about this problem incorrectly. I thought that I might need to compute the geodesics since I am asked to compute the injectivity radius of $S$.

If anyone has any hints or pointers on how I might go about showing this that would be extremely helpful!

pomegranate
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1 Answers1

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Rather than working in coordinates, you can consider $S$ as a submanifold of $\mathbf{R}^3$ and use the completeness of the Euclidean distance to show that $(S,d_S)$ is a complete metric space, where $d_S$ denotes the distance coming from the induced Riemannian metric on $S$.

First note that $S$ is a closed subset of $\mathbf{R}^3$. Let $i:S\hookrightarrow\mathbf{R}^3$ denote the inclusion map; since $S$ is closed $i$ is a proper embedding. Endow $S$ with the Riemannian metric $g=i^*(dx\otimes dx + dy\otimes dy + dz\otimes dz).$ Let $d_S$ denote the Riemannian distance on $S$ coming from $g$ and let $d_{\mathbf{R}^3}$ denote the Euclidean distance on $\mathbf{R}^3$. Now that we have this setup out of the way we can start proving our main result:

Fix $p,q\in S$. Let $\gamma:[a,b]\to S$ be a piecewise smooth curve from $p$ to $q$ in $S$. Note that $i\circ\gamma$ is a piecewise smooth curve from $p$ to $q$ in $\mathbf{R}^3$, and $$L_S(\gamma)=L_{\mathbf{R}^3}(i\circ\gamma)\geq d_{\mathbf{R}^3}(p,q).$$ Now by taking the infimum over all admissible curves $\gamma$ we have that $d_S(p,q)\geq d_{\mathbf{R}^3}(p,q).$

Now consider a Cauchy sequence $\{p_n\}_{n=1}^{\infty}$ in $(S,d_S)$. By the above, we see that $\{p_n\}$ is also a Cauchy sequence in $\mathbf{R}^3$. Hence, there is some point $p\in\mathbf{R}^3$ such that $p_n\to p$ in the Euclidean metric. Since $S$ is closed we have $p\in S$. In particular, we can now show $p_n\to p$ in $(S,d_S)$. In light of the Hopf-Rinow theorem we see that $S$ is geodesically complete, and so $\exp_p$ is well defined on all of $T_pS$ for all $p\in S$.

Now we compute the injectivity radius. Intuitively, since $S$ pinches to a point as $x\to-\infty$ we should expect the injectivity radius to be zero. Indeed, this is the case. Note that for some point $p=(x,e^x,0)\in S$ that in the direction $\vec{w}=(0,0,1)$ that a curve can only extend for length $2\pi e^x$ before returning back on itself. In particular, since $\|v\|=L(\exp_ptv|_{t\in[0,1]})$ we see that the injectvity radius is bounded above by $2\pi e^x$. Now we see that in the limit as $x\to-\infty$ that the injectivity radius tends to zero. So we deduce that $\iota(S)=0$, as desired.

Disclaimer: there are a few small details that I didn't fill in.

Yousuf Soliman
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