Consider the cylinder $\mathbb{R}_t\times X$ where $X$ is a compact manifold without boundary. Consider the cylindrical metric $g_{cyl}=g_X+dt^2$. Clearly $(\mathbb{R}_t\times X, g_X+dt^2)$ is a complete Riemanian manifold and the geodesics is just the pairs of geodesics in $\mathbb{R}$ and $X$. Now consider a conformal deformation of the metric $e^{2u(t,x)}(g_X+dr^2)$ where $u:\mathbb{R}\times X \longrightarrow \mathbb{R}$ is a smooth function.

My questions are the following:

i) for which functions $u$ we obtain again a complete metric $e^{2u}(g_X+dr^2)$ on the cylinder?

ii) is there a simple description of the geodesics in terms of the factor $e^{2u}$?

ii) Observe that when $u(t,x)=t$ the cylinder $(\mathbb{R}_t\times X, e^{2t}(g_X+dt^2))$ is isometric to the cone $(\mathbb{R}^{+}_{r}\times X, r^{2}g_X+dr^2)$ by the change $r=e^t$. In this case is the metric $e^{2t}(g_X+dt^2)$ complete? i.e. the cone with conical metric is complete?

Thanks folks!!

  • 1
    $\begingroup$ The answer to iii) is NO in general: take $X=S^{1}$ then the cone $\mathbb{R}^{+}_{r}\times S^{1}$ is just $\mathbb{R}^{2}\setminus \{0\}$ and the cone metric $r^{2}g_{S^{1}}+dr^{2}$ is the standard flat metric on $\mathbb{R}^{2}$, which is clearly non-complete on $\mathbb{R}^{2}\setminus \{0\}$ Therefore the cylinder $\mathbb{R}\times S^{1}$ is not complete with the conformal deformation $(e^{2t})(g_{S^{1}}+dt^{2})$. $\endgroup$ – Coffee Mar 23 '15 at 21:18

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