A computation with derivations and skew-derivations

Question

On a smooth manifold $M$ a derivation $D$ of degree $k$ is an endomorphism of the algebra of forms defined on $M$, which is denoted by $\Omega(M) = \bigoplus_{k \geq 0} \Omega^k(M)$, such that

1. $D\colon \Omega^r(M) \to \Omega^{r+k}(M)$ for all $r$.
2. $D$ satisfies $D(\alpha \wedge \beta) = D\alpha \wedge \beta + \alpha \wedge D\beta.$

It is called skew-derivation if 2. is replaced by $D(\alpha \wedge \beta) = D\alpha \wedge \beta + (-1)^{l}\alpha \wedge D\beta$, where $l$ is the degree of $\alpha$, and means that $\alpha \in \Omega^l(M)$.

Prove that if $D$ is a derivation of degree $k$ and $D'$ is a skew-derivation of degree $k'$, then $[D,D']:= DD'-D'D$ is a skew-derivation of degree $k+k'$.

I cannot see where my computations go wrong:

\begin{align} [D,D'](\alpha \wedge \beta) & = DD'(\alpha \wedge \beta) - D'D(\alpha \wedge \beta) \\ & = D(D'\alpha \wedge \beta + (-1)^{\deg \alpha}\alpha \wedge D'\beta)-D'(D\alpha \wedge \beta + \alpha \wedge D\beta) \\ & = DD'\alpha \wedge \beta + D'\alpha \wedge D\beta + (-1)^{\deg \alpha}D\alpha \wedge D'\beta + (-1)^{\deg \alpha}\alpha \wedge DD'\beta \\ & \quad -D'D\alpha \wedge \beta - (-1)^{\deg \alpha+k}D\alpha \wedge D'\beta-D'\alpha \wedge D\beta-(-1)^{\deg \alpha}\alpha \wedge D'D\beta \\ & = [D,D']\alpha \wedge \beta + (-1)^{\deg \alpha}\alpha \wedge [D,D']\beta+(-1)^{\deg \alpha}(1-(-1)^k)D\alpha \wedge D'\beta. \end{align}

So the statement seems to hold only when $k$ is even. Where is the problem?

Answer 1

Assume $\alpha, \beta$ are differential forms and let $D$ be a derivation of order $k$. We have two relevant cases:

1. $\alpha$ of even degree and $\beta$ of odd degree.
2. Both $\alpha,\beta$ of odd degree.

In the first case we have \begin{align} D(\alpha \wedge \beta) & = D\alpha \wedge \beta + \alpha \wedge D\beta,\\ D(\beta \wedge \alpha) & = D\beta \wedge \alpha + \beta \wedge D\alpha = \alpha \wedge D\beta +(-1)^kD\alpha \wedge \beta. \end{align} First and second line must now agree, which implies that either $k$ is even or $D$ (of odd order) vanishes on even degree forms.

Second case: \begin{align} D(\alpha \wedge \beta) & = D\alpha \wedge \beta + \alpha \wedge D\beta,\\ -D(\beta \wedge \alpha) & = -D\beta \wedge \alpha - \beta \wedge D\alpha = (-1)^k(\alpha \wedge D\beta +D\alpha \wedge \beta). \end{align} Again, the two lines agree. Now, either $k$ is even, or it is odd and then $D$ vanishes on odd degree forms.

Summing up, either $k$ is even, or $D$ vanishes. This should prove that derivations of odd order vanish and that non-trivial derivations have even order.

Thus, there is no loss of generality in assuming $k$ even. In this case the term giving problems vanishes and the statement holds true.