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This is a proposition in 33 page of Foundation in Differential Geometry - KN I need some detail.

Let $D^r(M)$ be a set of $r$-form. Then derivation (resp. skew-derivation) of degree $k$ is a linear mapping from $D^r(M)$ to $D^{r+k}(M)$ s.t. $$ D(\omega \wedge \tau ) = D\omega \wedge \tau + \omega \wedge D\tau $$ (resp. $ D(\omega \wedge \tau ) = D\omega \wedge \tau + (-1)^r \omega \wedge D\tau $) where $\omega \in D^r(M)$

Note that ${\bf exterior\ differentiation}$ is skew of degree $1$.

${\bf Question}$ : I wanto prove that $[D,D']$ is skew of degree $k+k'$ where $D$ is derivation of degree $k$ and $D'$ is skew of degree $k'$

${\bf Try}$ : $$ DD' (\omega \wedge \tau ) = D\{ D'\omega \wedge \tau + (-1)^r\omega \wedge D'\tau \} $$ $$ = DD'\omega \wedge \tau + D'\omega \wedge D\tau +(-1)^rD\omega \wedge D'\tau + (-1)^r\omega \wedge D D'\tau$$

$$D'D(\omega \wedge \tau ) = D'\{ D\omega \wedge \tau + \omega \wedge D\tau \} $$ $$ = D'D\omega\wedge \tau + (-1)^{k+r}D\omega \wedge D'\tau +D'\omega \wedge D\tau + (-1)^r\omega \wedge D'D\tau $$

So $$ [D,D'] (\omega \wedge \tau )=[D,D']\omega \wedge \tau + (-1)^r \omega \wedge [D,D']\tau + [(-1)^r - (-1)^{k+r}]D\omega \wedge D'\tau $$

Am I miss something ? Why do not the last term disappear ?

HK Lee
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    I have the same problem. I posted the same question, but I am not getting any answers. Did you solve the problem in the meanwhile? – Gibbs Jan 22 '19 at 16:01
  • I add my explanation. – HK Lee Jan 22 '19 at 16:37
  • Note that there is also a similar statement right below concerning DD' + D'D for two skew-derivations whose calculation similarly doesn't seem to yield a skew-derivation as claimed. A similar example shows the failure of that statement. – Nuno Hultberg Sep 23 '23 at 17:07

1 Answers1

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1) When $M$ is Riemannian manifold, if $L$ is a derivation of degree $k$ from $\bigwedge^r(M)$ to $\bigwedge^{k+r}(M)$, then

$$ L(fdx_i)=L(f)dx_i +f L(dx_i) $$

$$ L(dx_i f)=fL(dx_i) + dx_i L(f) = f L(dx_i) + (-1)^k L(f) dx_i $$

When $k$ is even, then we are done. If $k$ is odd, then we have $L(f)=0$ for all function $f$.

\begin{align*}L(dx_1dx_2)& =L(dx_1)dx_2+ dx_1L(dx_2) \\&= dx_2L(dx_1)+ L(dx_2) dx_1\\&= L(dx_2dx_1)\end{align*} so that $L=0$ on $\bigwedge^{2m}(M)$ for all $m$.

2) When $M=\mathbb{R}^3$, then assume that $L(f)=0,\ L (dx_1)=dx_1dx_2,\ L(dx_i)=0,\ i>1$.

Hence $ \omega =dx_1,\ \tau=f$ so that $$ L(\omega) df= f_{x_3}dx_1dx_2 dx_3 \neq 0 $$

That is, $[L,d]$ is not skew derivation.

HK Lee
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    I checked again my answer, I deleted because there are some problems with it. As regards your answer: suppose you work on $\mathbb{R}$. Then the exterior derivative $d$ is a derivation of order $1$, because $d(f(x)dx) = 0$ and $d(fg) = df\cdot g + fdg$. Then your argument does not give any information on $k$, which is actually odd. – Gibbs Jan 22 '19 at 21:49
  • The argument only shows that it is killed by any form of degree 1. Choosing it to be 0 was helpful in constructing a derivation with the required properties. It is still necessary to check that this indeed extends to a derivation (which it does) – Nuno Hultberg Sep 23 '23 at 17:13