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Dear colleague, I'm wondering how the generalization of the curvature endomorphism for vector field $w \to R(u,v)w$ looks like for tensor fields of higher rank, eg $W \to T(U_1,U_2,...)W$ for some tensor $T$. For a single vector field $w$ this iso is a linear transform of the tangent bundle. What is it gonna be for a tensor field of higher ranks?

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I am not sure that I understand the question. It would help if you could elaborate. At any given point $p$ in the manifold $M$, the curvature defines a linear map $\Lambda^2 T_pM \to \mathrm{End}(T_pM)$. Is your question about other linear maps $\mathcal{T}(T_pM) \to \mathrm{End}(T_pM)$, where $\mathcal{T}$ is some other space of tensors? Which sort of maps do you have in mind? – José Figueroa-O'Farrill Apr 22 2010 at 19:25
Roughly, there is a $R(u,v)$ term for every $TM$ factor of a homogeneous section of $\bigotimes TM$; the naturality of this comes from choosing the connections $\nabla$ on $\bigotimes TM$ to satisfy a Leibniz rule w.r.t $\otimes$: $\nabla_X (W\otimes V) = (\nabla_X W)\otimes V + W\otimes(\nabla_X V) $. It is easy to check that the terms in $[\nabla_X,\nabla_Y]$ with $\nabla_X$ and $\nabla_Y$ on distinct factors will cancel. A similar Leibniz condition describes how to deal with dual factors. – some guy on the street Apr 22 2010 at 20:01
I'm not sure that this is what the OP means, though. You're describing the action of the curvature operator on tensor fields, whereas the OP explicitly talks about an endomorphism of $TM$ of the form $T(U_1,U_2,\dots)$. – José Figueroa-O'Farrill Apr 22 2010 at 21:20
A map of the form $\Lambda^n T_p M\to End(\otimes_1^{n-2} TM)$ which acts as W\toT(U_1,\dots,U_n)W . In this case W should be a tensor of the rank n−1 , I suppose (so for the Riemann tensor it becomes a vector field). – Peter Apr 22 2010 at 21:53
The action of R via Leibnitz rule wrt $\otimes$ seems to be one of the possibilities, but it involves only two vector fields X and Y as noted above. – Peter Apr 22 2010 at 22:00
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