Killing spinors and symmetric tensor fields. - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T02:53:05Zhttp://mathoverflow.net/feeds/question/81900http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/81900/killing-spinors-and-symmetric-tensor-fieldsKilling spinors and symmetric tensor fields.Klaus Kröncke2011-11-25T16:06:25Z2011-11-28T14:28:13Z
<p>Hi all,</p>
<p>I have a question of the following form: Let $(M,g)$ be a Riemannian spin manifold which admits a Killing spinor $\sigma$ and let $h:T M \to T M$ be a symmetric, trace-free and divergence-free tensor. Consider the following expression:
$\sum_{i=1}^n\nabla_i h(Y)\cdot\nabla_i\sigma$</p>
<p>The question is: can this expression be zero for every $Y$ without $\nabla h$ being zero?</p>
<p>In fact, this is equivalent to the following equation:
$\Delta(h(Y)\cdot\sigma)=(\Delta h(Y)+c^2 n h(Y))\cdot\sigma$</p>
<p>$\Delta$ is the usual Laplace Beltrami operator.</p>
<p>I am thankful for any help.</p>
<p>"Edit": I translated the question in a probably more considerable form: I am looking for trace-free and divergence-free symmetric tensors which are in the kernel of the generalized exterior derivative:
$d h(X,Y)=\nabla_X h(Y)-\nabla_Y h(X)$ </p>