# Geodesic completeness of manifolds with "warped" metrics

Let $$(M,g)$$ be a geodesically complete Riemannian manifold, and let $$f: M\to \mathbb{R}^+$$ be a smooth, bounded strictly positive function on $$M$$; i.e., there exist $$L,A\in \mathbb{R}^+$$ such that $$0 for all $$m\in M$$. Since $$f$$ is positive, $$f^2g$$ defines a metric on $$M$$. Since $$(M,g)$$ is geodesically complete, is $$(M, f^2g)$$ also geodesically complete?

(Note: This question was inspired as a generalization of another question posed recently about the geodesic completeness of $$\mathbb{R}\times \mathbb{S}^{n-1}$$, given a warped metric $$ds^2=dr^2+\psi^2d\theta^2$$ where $$dr$$ and $$d\theta$$ are the metrics from $$\mathbb{R}$$ and $$\mathbb{S}^{n-1}$$ respectively and $$\psi$$ is a positive function on $$\mathbb{R}\times \mathbb{S}^{n-1}$$.)

Let $$L=\inf_{m\in M}(f(m))$$ and $$A=\sup_{m\in M}(f(m))$$. By assumption $$L, A$$ exist and are positive. We define two metric structures, $$d,d': M\times M\to \mathbb{R}^*$$, $$d(x,y)=\inf \bigg\{\int_{0}^1( g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt: \gamma \text{ is a piecewise differentiable curve from } a \text{ to } b \bigg\}$$ And $$d'(x,y)=\inf \bigg\{\int_{0}^1( f^2g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt: \gamma \text{ is a piecewise differentiable curve from } a \text{ to } b \bigg\}$$

Since $$(M,g)$$ is geodesically complete, $$d$$ makes $$M$$ into a complete metric space by the Hopf-Rinow theorem. It suffices to show that the metrics $$d$$ and $$d'$$ are strongly equivalent, i.e. for all $$x,y$$ in $$M$$, there exist $$\alpha,\beta\in \mathbb{R}$$ such that $$\alpha d(x,y)\leq d'(x,y)\leq \beta d(x,y)$$. We can see that for any piecewise differentiable curve $$\gamma:[0,1]\to M$$ such that $$\gamma(0)=x$$ and $$\gamma(1)=y$$ we have $$L(g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}\leq(f^2g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}\leq A(g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}$$ implying that $$L\int_{0}^1(g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt\leq \int_{0}^1 (f^2g_{ij}\dot{\gamma}^i\dot{\gamma}^j)^{1/2}dt\leq A\int_{0}^1(g_{ij}\dot{\gamma} ^i\dot{\gamma} ^j)^{1/2}dt$$ and hence $$Ld(x,y)\leq d'(x,y)\leq Ad(x,y)$$ So $$d$$ and $$d'$$ are strongly equivalent. Because $$(M,d)$$ is a complete metric space and $$(M,d')$$ is strongly equivalent to $$(M,d)$$, then it is also a complete metric space and $$(M,f^2g)$$ is geodesically complete.

• I don't see in the original question any assumption that implies $L>0$. Apr 22 at 18:30
• @Deane : Presumably $\mathbb{R}^+$ in the question means the set of strictly positive real numbers; if it meant the set of non-negative real numbers, then there'd be no point in mentioning $L$ (since one could always use $L=0$). But it would be good to have the question edited to clarify this. Apr 22 at 23:36
• Actually, the comment on the other answer confirms this, so I'll submit an edit to the question myself. Apr 22 at 23:37
• By ‘the other answer’, I meant the one by Michał Miśkiewicz (I should be clear on this in case more answers are added later). Apr 23 at 0:17

[This is an answer to the original question]

$$\newcommand{\R}{\mathbb{R}}$$ Let $$M := \R$$ with the standard metric $$g$$, and $$f(x) := e^{-x^2}$$. Then the whole line $$M$$ has finite length if given the metric $$f^2 g$$, which shows it cannot be geodesically complete.

• I was perhaps too vague in just saying $f$ was bounded, I intended this to mean that $f$ is bounded above and below by positive numbers, rather than allowing it to be bounded below only by $0$. Apr 22 at 23:04