$\begingroup$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.
$\endgroup$61 Answer
$\begingroup$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 $$\color{red}{D(dx^j)(X)}=\color{green}{-X^iF^j_i}=\color{blue}{-F_i^j dx^i(X)}.$$
$\endgroup$7
