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I am having real trouble trying to understand a problem in O'Neil's "Semi-Riemannian Geometry" and I can't find much literature on the subject. I will expose the problem and I will be grateful to accept hints, solutions and bibliographic sources on tensor derivations.

Problem 2.10 (p.53) Prove that a tensor derivation $\mathfrak{D}$ has $\mathfrak{D}_0 ^0 = 0$ if and only if $\mathfrak{D}_0^1$ is $\mathfrak{F}(M)$- linear (here, $\mathfrak{F}(M)$ stands for all smooth real-valued functions on $M$). Then by interpretation, $\mathfrak{D}_0 ^1 = B \in \mathfrak{T}_1^1(M)$ (where $\mathfrak{T}_s^r(M)$ stands for all (r,s)-tensors on $M$), and we write $\mathfrak{D} = \mathfrak{D}_B$.

Here is the definition of a tensor derivation according to the book:

Definition (Tensor derivation). A tensor derivation $\mathfrak{D}$ is a set of $\mathbb{R}$-linear functions on a smooth manifold $M$ $$ \mathfrak{D} = \mathfrak{D}^r_s:\mathfrak{T}_s^r \to \mathfrak{T}_s^r \quad\quad (r\geq 0, s\geq 0) $$ such that for any tensors $A$ and $B$:

(1) $\quad \mathfrak{D}(A \otimes B)= \mathfrak{D}A \otimes B+ A \otimes \mathfrak{D}B $

(2) $\quad \mathfrak{D}(\mathbf{C} A) = \mathbf{C}(\mathfrak{D} A )$ for any contraction $\mathbf{C}$.

My first problem is that I cannot see the intuition behind the definition whose notation is already cumbersome and I cannot find additional literature on the subject. Anyway, I believe a solution to the problem is given by the identity (1), for if $\theta$ is a $(1,0)$-tensor and $f$ a smooth real-valued function then $\mathfrak{D}(f)=\mathfrak{D}_0^0(f)=0$ by hypothesis and (1) becomes (hope this makes some sense): $$ \mathfrak{D}(f \theta) = \mathfrak{D}(f) \otimes \theta + f \mathfrak{D}(\theta) = f \mathfrak{D}(\theta)$$ which I hope is enough to prove the necessity and sufficiency of $\mathfrak{F}(M)$-linearity.

Yet I cannot derive how it can be interpreted that $\mathfrak{D}_0^1$ is a (1,1)-tensor! Can anyone explain it? Does someone know a more comprehensive and introductory reference?.

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Your argument to show that $D^1_0$ is $\mathcal F(M)$-linear if and only if $D_0^0=0$ using the product rule is completely correct. Now consider the properties of $D^1_0$ in this case: it is an $\mathcal F(M)$-linear bundle map $TM \to TM$, and thus $$B : (X,\omega) \mapsto \omega(D^1_0 X)$$ is an $\mathcal F(M)$-linear map $TM \otimes TM^* \to \mathcal F(M)$; i.e. a $(1,1)$-tensor. (Depending on your definition of $(1,1)$ tensor this may require the additional step of showing that $\mathcal F(M)$-linear maps on tensor products of the tangent bundle correspond to smooth tensor fields; but I assume you're aware of this correspondence.)

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  • $\begingroup$ Thank you very much for your answer! Do you happen to know an additional textbook that covers the topic? $\endgroup$ – Mauricio G Tec Sep 3 '13 at 3:04
  • $\begingroup$ @mauriciogtec: O'Neill is the only book in which I've seen general tensor derivations defined; most authors first define connections on vector bundles and then define the extension of a connection to the tensor algebra. I found O'Neill good to learn from; what difficulties are you having? $\endgroup$ – Anthony Carapetis Sep 3 '13 at 3:17

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