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Consider a compact smooth Riemannian manifold $(\mathcal{M}, g)$. We consider a conformal change of the metric. Let $\tilde{g}=\phi g$, where $\phi$ is smooth and positive. Moreover, it satisfies $C_1\leq \|\phi\|_{C^2}\leq C_2$ for some positive constants $C_1$ and $C_2$. The geodesic ball in metric $g$ is denoted as $B_r(p)$. Denote the geodesic ball in the conformal metric $\tilde{g}$ as $\tilde{B}_r(p)$. I am wondering if we have $$ B_{c_1r}(p)\subset \tilde{B}_r(p)\subset B_{c_2r}(p)$$ for some positive constants $c_1$ and $c_2$ in a small neighborhood of $p$ by assuming $c_1$ is small and $c_2$ is a little large. I am also curious about how the conformal change is related to the exponential map.

Thanks a lot in advance for your insights.

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    $\begingroup$ As stated, I doubt it. Why is $\|\phi\|_{C^2}$ the thing that is bounded? You can take $\phi$ very small and highly oscillatory so that the first and second derivatives are huge. Just by measuring distances this would give a counterexample. Do you just want $\|\phi\|_{C^2}$ to be bounded and that $\phi$ itself to have some upper and lower bound? $\endgroup$
    – Willie Wong
    Commented Nov 3, 2018 at 3:50
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    $\begingroup$ If you just postulate upper and lower bounds for $\phi$, then the minimizing geodesic for $\tilde{g}$ of length $\alpha$ is a curve of length at most $C \alpha$ with respect to $g$. So this guarantees the inclusion of balls. // The relation to exponential map is much trickier: play around with $\mathbb{S}^2$; you can rather explicitly write down the conformal changes in that case. $\endgroup$
    – Willie Wong
    Commented Nov 3, 2018 at 3:58
  • $\begingroup$ Thanks, I just want $\|\phi\|_{C^2}$ to be bounded with that lower and upper bound. Could you please show me a little more calculation about that minimizing geodesic for $\tilde{g}$ of length $\alpha$ is a curve of length at most $C\alpha$ with respect to $g$? or just use the definition of minimizing geodesic $min \int^b_a | r'(t)|\,dt$? where $r(t)$ is the curve between $p$ and $x$. Thanks! $\endgroup$
    – mathpde
    Commented Nov 3, 2018 at 4:26

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