Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\phi:\Omega\to\Omega^'$ be an invertible mapping between two bounded domains in $\mathbb{R}^{n}$ (typically with $n=2$ or $n=3$), and let $F$ be its derivative (i.e., the Jacobian matrix). Let $M:\Omega\to\mathbb{R}^{n\times n}$ be a matrix field. I am wondering if the following is true: $$ F^{-1}\Delta' (FMF^T)F^{-T} = \Delta M, $$ where $\Delta$ acts elementwise, and $\Delta'$ is the same on $\Omega'$. If it is convenient, you can assume $F$ is orthogonal and $M$ is symmetric.

More generally, if the above is not true, is it possible to correct $\Delta$ to $D$, by adding lower order terms, so that

$$ F^{-1}D' (FMF^T)F^{-T} = D M, $$ is true?

Update: Robert Bryant commented that it is not true in general, but is true if $F$ is orthogonal, because then $F$ would have to be constant. I would like to relax the orthogonality condition as it is too restrictive for my purposes. Now I am wondering what happens if we change the metric on $\Omega'$ by the induced metric coming from $\phi$, and replace $\Delta'$ by some appropriate geometric Laplacian. Or in other words, now supposing that $\Omega$ and $\Omega'$ are Riemannian manifolds, and $\Delta$ and $\Delta'$ are suitable Laplacians defined on them, what would be a condition on $\phi$ to guarantee the above invariance?

share|improve this question
1  
@timur: In general, it is not true, even if you modify to use a $D$ that is $\Delta$ plus lower order terms. If $F$ is orthogonal (i.e., $F$ takes value in $\mathrm{O}(n)$ at every point), then $F$ is a constant matrix and the above is certainly true. –  Robert Bryant Aug 6 '12 at 15:04
    
@Robert: Thanks a lot for your comment! Can you please look at the update? –  timur Aug 6 '12 at 19:22
1  
@timur: When you phrase it in terms of a map $\phi$ between Riemannian manifolds, there are several different ways to interpret your question, and I'm not sure which one you mean. Since the tangent bundles are no longer trivial, there is no obvious way to interpret $\phi'$ as a matrix $F$, so it's not clear what your 'elementwise' condition means. For example, are you asking whether the tangent bundles of the two manifolds can be trivialized in such a way that the metric Laplacians on bundle endomorphisms can be compared and so that these Laplacians actually uncouple into scalar operators? –  Robert Bryant Aug 9 '12 at 12:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.