Let $f: M \to N$ be a diffeomorphism between two riemannian Manifolds. Suppose there exist constants $0 < c \leq C$ such that for all $p \in M$, we have $c \leq |df_p| \leq C$. Here, $df$ denotes the differential of $f$ as usual and $|df_p|$ denotes the energy density of $f$ at $p$, i.e, the norm coming from the induced riemannian metric on the bundle $TM \otimes f^*(TN)$. I wonder now if it's true that there are constants $0 < b \leq B$ with $b \leq |df^{-1}_q| \leq B$ for every $q \in N$.
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Lower bound yes. Upper bound no.
If you take orthonormal frames for $M$ and $N$, then the norm $|df_p|$ is the Frobenius norm on the matrix expression in those frames. Note that in any bases $df_p$ and $df^{-1}_{f(p)}$ are inverse matrices. The upper bound on the Frobenius norm of $df_p$ implies that each of the singular values are bounded above by $C$, and hence their inverses are bounded below by $1/C$, which gives the lower bound.
To see that the upper bound need not hold, let $M = N = \mathbb{R}^2$ and take
$$ (u,w) = f(x,y) = (x, \sinh^{-1}(y) ) $$
For $|df|$ we have the bound $ 1 \leq |df| \leq \sqrt{2}$.
For $|df^{-1}|$ you can compute
$$ df^{-1}(u,w) = \begin{pmatrix} 1 & 0 \\ 0 & \cosh w \end{pmatrix} $$
is not bounded.
You can of course assume that $M, N$ are compact, but then the desired results are entirely trivial.
answered Sep 10, 2015 at 20:56
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$\begingroup$ Thanks for your answer. It might still be true, though, that the upper bound holds in my specific setting. Let me elaborate: I have $M$ a non-compact riemannian manifold and some value $0 < \beta < i(M)$ and constants $0 < c \leq C$ as above, such that for each $p \in M$, the differential of the diffeomorphism $exp_p|_{B_\beta(0_p)}$ is bounded as described. $\endgroup$– H1ghfiv3Sep 11, 2015 at 6:51
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$\begingroup$ Clearly, there then are upper and lower bounds for the differential of the inverse map on $U_\beta(p)$ (the geodesic ball about $p$ of radius $\beta$), however, can they be chosen independent of $p$ ? $\endgroup$– H1ghfiv3Sep 11, 2015 at 7:06