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.

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    $\begingroup$ 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 ? $\endgroup$ Jan 23, 2014 at 10:40


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