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Hi all,

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: $\sum_{i=1}^n\nabla_i h(Y)\cdot\nabla_i\sigma$

The question is: can this expression be zero for every $Y$ without $\nabla h$ being zero?

In fact, this is equivalent to the following equation: $\Delta(h(Y)\cdot\sigma)=(\Delta h(Y)+c^2 n h(Y))\cdot\sigma$

$\Delta$ is the usual Laplace Beltrami operator.

I am thankful for any help.

"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: $d h(X,Y)=\nabla_X h(Y)-\nabla_Y h(X)$

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I don't understand. You are saying that $\sum_{i=1}^n \nabla_i h(Y) \cdot \nabla_i \sigma = 0$ for every $Y$ without $\nabla h = 0$ is equivalent to what, exactly? That there exists a solution to the Laplace-type equation? If that's not what you mean (or even if it is), we might be able to help more easily if you show exactly why these two statements (or some two statements) are equivalent. – Spiro Karigiannis Nov 28 '11 at 18:38
Hi, this probably isn't much help. Tensors (of rank two) satisfying $\nabla_Xh(Y)-\nabla_Yh(X)=0$ are also called Codazzi tensors. There are many of them (e.g $g$ or the second fundamental form) and many people find them interesting. For example: According to Berger-Ebin (…) proved that a constant trace Codazzi tensor on a compact manifold with non-negative sectional curvature must be parallel ($\nabla h=0$). – kangdon Dec 6 '11 at 4:13

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