It's easily shown that the wedge product descends to cohomology. So for a smooth manifold $M$, given a closed 1-form $\alpha \in \Omega^1(M)$, one can define the derivation of degree 1:
$$ \Gamma : \Omega^k(M) \to \Omega^{k+1}(M) \qquad \Gamma(\omega) = \omega \wedge \alpha $$
which preserves exact forms, in the sense that $d\omega = 0 \implies d\Gamma(\omega) = 0$.
However, I cannot seem to find any way to define an anti-derivation of degree $-1$ which preserves exactness in a similar manner.
A tempting construction is to use the interior derivative $\iota_X : \Omega^k(M) \to \Omega^{k-1}(M)$ for some suitable $X$. If we assume $M$ is endowed with a Riemmanian metric, then we could raise the index of a closed 1-form $\alpha$ to produce the vector field $X$. Using the Cartan homotopy formula, one can check that, given the definition
$$ \Gamma' : \Omega^k(M) \to \Omega^{k-1}(M) \qquad \Gamma'(\omega) = \iota_X (\omega) $$
the exterior derivative of $\Gamma'(\omega)$ is given by (where $L_X$ is the Lie derivative)
$$ d(\Gamma'(\omega)) = d ( \iota_X (\omega) = L_X(\omega) – \iota_X \underbrace{d\omega }_{=0} = L_X(\omega) $$
so $\Gamma'(\omega)$ is closed $\iff L_X(\omega) = 0$. This seems to be related to the idea of Killing vector field, however requiring $L_X (\omega) = 0$ for all $\omega$ most likely makes $X$ trivial (I haven't proven this but it seems true). Alternatively, we could have $X$ depend on $\omega$ in a predefined way, such that $L_{X_\omega} \omega = 0$. But it's not clear to me in general how one would construct such an $X_\omega$.
Is there any other construction for $\Gamma'$ that has this property? Or is there some reason why going "down" the chain of forms is easier that going "up" ?
Best Answer
In fact, every derivation of $\Omega^*(M)$ of degree $-1$ is of the form $\iota_X$ for some vector field $X$. However, $X=0$ is the only one that preserves exact forms. Here's one way of showing this:
1. Every antiderivation of degree $-1$ is an interior product.
Let $D:\Omega^k(M)\to\Omega^{k-1}(M)$ be an antiderivation of degree $-1$. Note that $D$ is fully determined by its restriction to $\Omega^1(M)$, which we can refer to as $D_1:\mathfrak{X}^*(M)\to C^\infty(M)$. It is a well known result (sometimes called the tensor characterization lemma) that such a map corresponds to a contraction with a vector field iff it is $C^\infty(M)$-linear. The antiderivation condition states exactly that: for $f\in C^\infty(M)$, $\alpha\in\mathfrak{X}^*M$, we have $$ D_1(f\alpha)=D_1(f\wedge\alpha)=D(f)\wedge\alpha+(-1)^0f\wedge D_1(\alpha)=fD_1(\alpha) $$ and so $D=\iota_X$ for some $X\in\mathfrak{X}M$.
2. $\iota_X$ does not preserve exact forms unless $X=0$
Suppose $X$ is not identically zero (and note that this implies $\dim(M)\ge 1$), so that we can find a point $p\in M$ such that $X(p)\neq 0$ and a neighborhood $U$ of $p$ with local coordiantes $x^1,\cdots,x^n$ such that $X=\partial/\partial x^1$. Choosing a exact $1$-form $\alpha:=d(e^{x_1})=e^{x^1}dx^1$, we have $d(\iota_X\alpha)=d(e^{x^1})\neq 0$. This is a local result, but it can be extended to a global one by instead choosing $\alpha=d(\psi e^{x^1})$ for a suitable bump function $\psi\in C^\infty M$.