I have a reference request concerning equivalent norms on Sobolev Spaces on manifolds of bounded geometry. This may be obvious to the experts but I am not working in the field and only want to use this result. Let $(M,g)$ be a smooth Riemannian manifold. The Sobolev Space $W_p^m(M)$ with $m\in\mathbb{N}$ and $1\leq p< \infty$ is defined via completion of the space of smooth functions on $M$ in $L^p(M)$ with respect to the norm

$\left\lVert f\right\rVert_{W_p^m(M)}:=\sum_{l=0}^m \left(\int_M \vert\nabla^lf\vert^p\,\mathrm{d}V_g\right)^{\frac{1}{p}}\,.$

Now it is very frequently used that on manifolds with bounded geometry and positive injectivity radius one can use geodesic normal coordinates and appropriate partitions of unity to define an equivalent norm on $W_p^m(M)$ via looking at the norms of the "pulled backed" function in euclidean space. (see for example: Triebel, Theory of function spaces II 7.4.5., identity (8) on page 320).

I am looking for a reference where the equivalence of the norms is shown explicitely.

Thank you!

  • 2
    $\begingroup$ Do you need a citable reference, or are you just looking for a proof? If the latter, here's a sketch (I can expand if you need): bounded geometry will imply uniform C^k bounds on the normal co-ordinates, treated literally as a function (the exponential map), and hence uniform bounds on the metric tensor and its derivatives in normal co-ordinates. Now expand out the "geometric" Sobolev norm in terms of the metric tensor, Christoffel symbols, etc (some of this is written out in Lemma 1.3 of these old notes of mine math.mit.edu/~macbeth/expository/riemannian-inequalities.pdf ). $\endgroup$ – macbeth Apr 8 '19 at 11:44
  • $\begingroup$ Thank you for your answer! I am not 100% sure about the rigorous proof. The implication that if the norm in geodesic coordinates is finite, then so is the geometric one, is clear to me from the formula in Lemma 1.3 and the bounds on the geometric quantities. Concerning the other implication, I guess one needs an estimate from below on the inverse metric? (What does $Q^{-1}\delta_{ij}\leq g_{ij}\leq Q\delta_{ij}$ as bilinear forms mean precisely?) Further I suppose one needs to proceed via induction to be able to treat the lower order terms appearing (linearly!) in the formula in Lemma 1.3? $\endgroup$ – Julia Apr 8 '19 at 13:22
  • $\begingroup$ 1) You're right, $Q^{-1}\delta_{ij}\leq g_{ij} \leq Q\delta_{ij}$ means (in particular) a bound on this inverse metric. (It literally means that for any nonzero vector $v^i$, one has $Q^{-1}|v|^2\leq g_{ij}v^iv^j\leq Q|v|^2$.) Do you know how to prove theorems like Cartan-Hadamard, i.e. theorems guaranteeing that on a certain geodesic ball (of radius given in terms of an upper curvature bound) there are no conjugate points? A similar argument will give a lower bound on the inverse metric on a certain (smaller) geodesic ball. $\endgroup$ – macbeth Apr 8 '19 at 17:07
  • $\begingroup$ 2) Yes, given the formula in Lemma 1.3, one proves equivalence of the Sobolev $W^{k,p}$ norms by induction on $k$. $\endgroup$ – macbeth Apr 8 '19 at 17:10
  • $\begingroup$ Thank you again! Your answers really help a lot. I know of the theorem but I am not very familiar with those kinds of proofs. Is it not possible to deduce from $Q^{-1}\delta_{ij}\leq g_{ij}\leq Q\delta_{ij}$ an analogous bound on the inverse metric using elementary arguments? $\endgroup$ – Julia Apr 9 '19 at 8:50

I think the argument should be written down in detail in the book

Eichhorn, Global analysis on open manifolds, Nova 2007.

Edit: A more general discussion of coordinate systems that yield equivalent definitions of Sobolev spaces is given in Theorem 3.9 from

Große and Schneider, Sobolev spaces on Riemannian manifolds with bounded geometry: General coordinates and traces, Math. Nachr., 286, 2013. arXiv link

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  • $\begingroup$ Can you say where in the book? $\endgroup$ – Amir Sagiv Apr 8 '19 at 13:47
  • $\begingroup$ I don't have a copy at hand, but it should be in the first chapter, which introduces the setting for analysis of linear PDEs. I edited my answer to provide a reference that may be easier to obtain. $\endgroup$ – J.J.L. Apr 8 '19 at 15:48

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