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Consider $M$ a manifold and $g_1, g_2$ two different Riemannian metrics. I want to know how the condition $|\nabla^{g_1,k}(g_1-g_2)|_{g_1}\leq C$ implies that the norms of $|\nabla^{g_1,i}u|_{T^{\otimes i}M, g_1}$ and $|\nabla^{g_2,i}u|_{T^{\otimes i}M, g_2}$ are equivalent for $i=1,...,k+1$, where $\nabla^{g_1,i}=\nabla^{g_1}\dots\nabla^{g_1}$ $i$-times, and therefore the Sobolev spaces up to order $k+1$ defined by $g_1$ and $g_2$ are equivalent. For $k=0$, this is quasi-isometry and for first order Sobolev spaces I know how to show it. But for arbitrary order I haven't found a way. So I post it here if someone can give me a reference. Because every time I saw this statement there was not an indication of a proof.

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  • $\begingroup$ In your question, does $|\nabla^{g_1,i}u|_{T^{\otimes i}M, g_1}$ mean the norm at a point, the sup norm over the manifold, or something else? $\endgroup$
    – Deane Yang
    Nov 29, 2019 at 17:29

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You said you know why the $L^2$ norms are equivalent, so let's look at the gradient term $$\int |\nabla^{g_1}u|^2,$$ where the norm associated to $|\cdot|$ doesn't matter. But since $u$ is a function, the gradient is just $du$ contracted with the metric. So again there are no derivatives of the metric involved.

For the Hessian term $$\int|(\nabla^{g_1})^2u|$$ we have schematically $$(\nabla^{g_1})^2u=\partial^2u+\Gamma^{g_1}\partial u,$$ where $\Gamma^{g_1}\approx \partial g_1$. But since $\partial g_1\approx \partial g_2$, the integral is equivalent to $$\int|(\nabla^{g_2})^2u|+\int |\nabla^{g_2}u|^2.$$ So note that the seminorms are not equivalent, but the actual Sobolev norms are.

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  • $\begingroup$ This works in the second order case (after providing a bit of detail). But I was asking for an approach that would cover every case and would not depend on expressions of local coordinates. $\endgroup$ Mar 5, 2019 at 9:29
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    $\begingroup$ It suffices to do it in local coordinates. $\endgroup$
    – Deane Yang
    Apr 3, 2019 at 12:27
  • $\begingroup$ You could do this without using local coordinates. However, since you'll need to integrate by parts and permute the covariant derivatives, lower order terms involving the curvature and its covariant derivatives will appear. The proof is more or less just as messy as using local coordinates. $\endgroup$
    – Deane Yang
    Nov 29, 2019 at 17:26

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