4
$\begingroup$

Given a covariant riemannian metric of certain sobolev class (i.e. with square-integrable weak derivatives up to a large enough integer order) on a (compact, if necessary) finite-dimensional smooth manifold, I would like to know the answers to the following related questions:

(1) does its inverse (i.e. the contra-variant type) belong to the same sobolev class?

(2) If a sequence of (covariant) metrics converges in the given sobolev norm to a limit metric, does it follow that the sequence of corresponding inverse metrics converges in the same sobolev norm to the inverse of the above limit metric?

***One of the things that I am iffy about: the inverse of a non-degenerate matrix involves the reciprocal of its determinant. I can believe that the determinant of a metric is a nowhere vanishing sobolev function but I am not sure whether it holds true for its reciprocal.

Thanks very much!

Improve this question
$\endgroup$
9
  • $\begingroup$ If you have large enough order to get, by Sobolev embedding, that the metric is in $L^\infty$, then the question is an issue of arithmetic. // In general, your question is basically the same as asking "if the function $f\in W^{s,p}$ then is the function $1/f \in W^{s,p}$?" $\endgroup$
    – Willie Wong
    Aug 14, 2018 at 20:05
  • $\begingroup$ yes. my knowledge of analysis is at most epsilon above zero. $\endgroup$
    – X-Naut PhD
    Aug 14, 2018 at 20:18
  • 2
    $\begingroup$ These are mostly discussed in Section 5.6 of L.C. Evans' Partial Differential Equations; the statement that $W^{k,p}$ is an algebra is relegated to an exercise at the end of the chapter. $\endgroup$
    – Willie Wong
    Aug 14, 2018 at 20:31
  • 2
    $\begingroup$ For first derivatve: $(1/f)' = -f' / f^2$ so $\| (1/f)'\|_{L^2} \leq \| 1/f\|_{L^\infty}^2 \|f'\|_{L^2}$. For second derivative $(1/f)'' = - f'' / f^2 + 2 (f')^2 / f^3$. To get $L^2$ integrability you need $f'$ is in $L^4$, but this, when $kp > n$, follows from Sobolev embedding and interpolation. Same for higher derivatives. $\endgroup$
    – Willie Wong
    Aug 14, 2018 at 20:55
  • 2
    $\begingroup$ For some reason, I find it easier to understand the general case, rather than for just the function $t \mapsto 1/t$. Instead, assume that you have a smooth function $\Phi: (a,b) \rightarrow \mathbb{R}$ and you want to prove that if $f$ lies in a Sobolev space and its image lies in $(a,b)$, then $\Phi\circ f$ lies in the same Sobolev space. $\endgroup$
    – Deane Yang
    Aug 14, 2018 at 22:56

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.