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?.


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.)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.