5
$\begingroup$

The gradient/differential/exterior differential/divergence/curl are all strictly related first order differential operators. As far as I understood, they are the base of (co)homological theories in Euclidean spaces/Riemannian manifolds. As an instance of what I am saying just consider the spaces of $k$-differential forms in $\mathbb R^3$ and the various $d$ operators acting between them (which turn out to be - I guess the magic word is "Hodge dual to" - the usual divergence/curl we learn in Calculus).

I am wondering if symmetrized gradient ($\frac{1}{2}(\nabla + \nabla^T)$) can be given a similar interpretation.

The flavour of the question is of course geometric, the applications I have in mind are rather analytical: the original question stems from the well-known fact that $BV$ functions are $n$-dimensional normal currents in $\mathbb R^n$, while I do not know if such an identification is possible for functions of bounded deformation, i.e. functions whose symmetric gradient is a measure.

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ I know something about that and I will write what I know when I have time. If you do not see my answer within 24 hours, please leave a comment for me with a reminder since I may forget about your question. $\endgroup$
    – Piotr Hajlasz
    Commented Jan 30, 2019 at 18:58
  • $\begingroup$ Note that the gradient operator corresponds to the $d$ operator under the musical isomorphism, so for cohomological purposes it suffices to consider symmetrized $d$. If I'm not confusing myself, $d$ has the same kernel and cokernel as symmetrized $d$, both viewed as graded operators on the graded exterior algebra of the tangent bundle, so they are cohomologically equivalent (though doing computations with explicit representatives of cohomology classes may feel a bit different). $\endgroup$
    – Paul Siegel
    Commented Jan 30, 2019 at 19:28
  • $\begingroup$ I'll also add that $d$ and symmetrized $d$ induce the same class in the K-homology of the underlying manifold, so one can pass freely back and forth between them for the purposes of index theory and such. $\endgroup$
    – Paul Siegel
    Commented Jan 30, 2019 at 19:30
  • $\begingroup$ This operator shows up in linear elasticity, see e.g. calvino.polito.it/~salamon/seminar/srni99.pdf ams.org/journals/bull/2010-47-02/S0273-0979-10-01278-4 and link.springer.com/article/10.1007/s00205-014-0806-1 for some work on the "elasticity complex". $\endgroup$
    – j.c.
    Commented Jan 31, 2019 at 2:40
  • $\begingroup$ @PiotrHajlasz I am sorry to bother you but, if you have some time, I would read with pleasure your comment/answer (especially concerning the link between BD functions and currents). Thanks a lot for your help. $\endgroup$
    – Romeo
    Commented Feb 27, 2019 at 18:23

0

Reset to default

You must log in to answer this question.