Let $D$ be a tensor derivation on a mnaifold $M$. I have to show that if $D(\partial_i)=\sum F_i^j \partial_j$, then $D(dx^j)=-\sum F_i^j dx^i$.

Any help on how to do this? Thanks in advance.

  • $\begingroup$ Welcome to Math.SE! Can you show how far you got yourself? This helps others to provide better answers. $\endgroup$ – Hrodelbert Sep 28 '15 at 13:54
  • $\begingroup$ Hi. Thanks. Not far, I'm afraid. I've been playing around with the product rule, but it has not given me much. Also, the minus sign throws me off. I feel like there is something pretty trivial I'm missing. Forgive me, but I'm completely new to tensors. $\endgroup$ – L. Henry Sep 28 '15 at 15:37
  • $\begingroup$ @L.Henry, I have reverted your edit — which turned the question incomprehensible. Please do not do that. $\endgroup$ – Mariano Suárez-Álvarez Sep 28 '15 at 20:22
  • $\begingroup$ If you find an answer satisfying, accept it. Not accepting any answers may repulse some people to bother with your questions. $\endgroup$ – Fallen Apart Oct 24 '15 at 10:28
  • $\begingroup$ BTW this is the first part of exercise 6 in Barrett O'Neill's book on semi-Riemannian geometry. $\endgroup$ – idontgetoutmuch Feb 18 '16 at 23:10

Let $D$ be a tensor derivation on a manifold $M$

I assume it means $D$ is $\mathbb{R}$-linear and obeys Leibniz rules with respect to tensor multiplication and contraction See axioms 2 and 3 here. Actually here we only need 2 things:

(1) For every vector field $X$ and 1-form $\omega$ we have $$D(\omega(X))=D(\omega)(X)+\omega(D(X)).$$ (2) For function $f$ and vector field $X$ we have $$D(fX)=D(f)X +fD(X)$$ In order to show $$D(dx^j)=-F_i^j dx^i$$ we have to prove that for every vecotr field $X$ the following holds: $$\color{red}{D(dx^j)(X)}=\color{blue}{-F_i^j dx^i(X)}.$$ So let $X$ be a vector field and $X=X^i\partial_i$ (denote it $\star$).

Ladies and Gentelmen! Let's compute $$D(dx^j(X))\stackrel{(1)}{=}\color{red}{D(dx^j)(X)}+dx^j(D(X))\stackrel{\star}{=}\color{red}{D(dx^j)(X)}+dx^j(D(X^i\partial_i))\stackrel{(2)}{=}\\ \color{red}{D(dx^j)(X)}+dx^j(D(X^i)\partial_i+X^iD(\partial_i))=\color{red}{D(dx^j)(X)}+D(X^j)+X^idx^j(D(\partial_i))=\\ \color{red}{D(dx^j)(X)}+D(X^j)+X^idx^j(F^k_i\partial_k)=\color{red}{D(dx^j)(X)}+D(X^j)+X^iF^j_i $$ Compute the same again, but this time $$D(dx^j(X))\stackrel{\star}{=}D(dx^j(X^i\partial_i))=D(X^idx^j(\partial_i))=D(X^j).$$ So if we extract $\color{red}{D(dx^j)(X)},$ we get that $$\color{red}{D(dx^j)(X)}=\color{green}{-X^iF^j_i}$$ To complete the proof we just need to compute $\color{blue}{-F_i^j dx^i(X)}.$ So $$\color{blue}{-F_i^j dx^i(X)}\stackrel{\star}{=}-F^j_i dx^i(X^k\partial_k)=\color{green}{-X^iF^j_i}.$$ As a result $$\color{red}{D(dx^j)(X)}=\color{green}{-X^iF^j_i}=\color{blue}{-F_i^j dx^i(X)}.$$

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  • $\begingroup$ Thanks alot!! That was more hairy than I thought. I see that there's a lot of notation in this topic that I have to decipher, before I can solve problems like this help-free :) $\endgroup$ – L. Henry Sep 28 '15 at 20:36
  • $\begingroup$ @L.Henry Feel free to ask what bothers you $\endgroup$ – Fallen Apart Sep 28 '15 at 20:38
  • $\begingroup$ +1 Nicely done. May I point out a typo: you wrote "excract", where extract was meant. $\endgroup$ – Hrodelbert Sep 29 '15 at 9:41
  • $\begingroup$ @Hrodelbert Fixed, Thx $\endgroup$ – Fallen Apart Sep 29 '15 at 11:33
  • $\begingroup$ @FallenApart You say for instance that $X^i dx^j (\partial_i) = X^j$. Can you explain how you arive at this, because I have difficulty seeing it directly from the definitions? $\endgroup$ – L. Henry Sep 29 '15 at 22:14

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