Equivalence of Sobolev spaces for different metrics

Question

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.

Answer 1

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.