10
$\begingroup$

The Bochner–Lichnerowicz–Weitzenböck formula can be written the following way $$ \Delta \phi-\nabla^*\nabla \phi= R(\phi),$$ here $\phi$ is a section in a Dirac bundle and $R$ the something which can be expressed in terms of its curvature.

One can get something nice applying this formula for different bundles associated to a given Riemannian manifols if $\langle R(\phi),\phi\rangle \ge 0$ for any $\phi$.

For the right choice of bundle, one gets $R$ to be Ricci curvature or scalar curvature and yet number of less popular curvatures. Each of these conditions corresponds to a convex $O(n)$-invariant cone of curvature operators, let us call it Weitzenböck cone.

Question. Is it well understood which cones are Weitzenböck and which are not?

For example assume I have a specific cone (say the cone $sec\ge 0$) is there someone who can tell that it is not a Weitzenböck cone.

$\endgroup$
  • 2
    $\begingroup$ When working on the bundle of $k$-forms, all of these cones contain the cone of nonnegative curvature operators (as operators on $2$-vectors). Can this be proved in general ? $\endgroup$ – Thomas Richard Jan 23 '14 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.