# De Rham differential is a derivation of wedge product?

Let $M$ be a smooth manifold and $k$ a positive integer. Let us define $$\Omega^k(M)=\left\{\begin{array}{lcl}C^\infty(M) & \textrm{if}& k=0\\ \mathsf{Alt}^k_{C^\infty(M)}(\mathfrak{X}(M), C^\infty(M)) & \textrm{if} & k>0 \end{array}\right.$$ where $\mathfrak{X}(M)$ is the $C^\infty(M)$-module of vector fields on $M$ (derivations of the algebra $C^\infty(M)$) and $\mathsf{Alt}^k_{C^\infty(M)}(\mathfrak{X}(M), C^\infty(M))$ is the $C^\infty(M)$-module of $C^\infty(M)$-multilinear skew-symmetric maps $$\omega:\underbrace{\mathfrak{X}(M)\times \ldots \times \mathfrak{X}(M)}_{k}\longrightarrow C^\infty(M).$$ The De Rham differential is the operator $$d:\Omega^k(M)\longrightarrow \Omega^{k+1}(M)$$ defined by $$(d\omega)(X_0, \ldots, X_k)=\sum_{j=0}^k (-1)^j X_j(\omega(X_0, \ldots, \widehat{X_j}, \ldots, X_k))+\sum_{i<j} (-1)^{i+j} \omega([X_i, X_j], X_0, \ldots, \widehat{X_i}, \ldots, \widehat{X_j}, \ldots, X_k)$$ where $\widehat{}$ means omit. We have a product $$\wedge:\Omega^i(M)\times \Omega^j(M)\longrightarrow \Omega^{i+j}(M)$$ defined by $$(\omega\wedge \tau)(X_1, \ldots, X_{i+j})=\frac{1}{i!j!} \sum_{\sigma\in\mathsf{S}_{i+j}} \mathsf{sgn}(\sigma)\omega(X_{\sigma(1)}, \ldots, X_{\sigma(i)})\tau(X_{\sigma(i+1)}, \ldots, X_{\sigma(i+j)}),$$ where $\mathsf{S}_{i+j}$ is the set of permutations of $i+j$ (that is, bijections of the set $\{1, \ldots, i+j\}$) and where $\mathsf{sgn}(\sigma)$ is the sign of $\sigma$.

Does anyone know to prove or where to find the proof of the identity $$d(\omega\wedge \tau)=d\omega\wedge \tau+(-1)^{|\omega|}\omega\wedge d\tau?$$

I know this is a classical result but I need a proof using the above definitions and the explicit computations (not by induction).

Thanks.

• Are you familiar with local representations of differential forms in coordinates? – Amitai Yuval Jun 2 '17 at 18:53
• Yes, I know how to do locally but, as I described, the problem is purely algebraic and I'm interested in the algebraic solution of this (the only difficulty is the sum indexed by permutations but with some trick we could prove the desired identity forgetting the geometry behind). – PtF Jun 2 '17 at 22:55
• Working with coordinates is still very algebraic. It just makes all the computations easier, as all the coefficients are with respect to a commuting frame. – Amitai Yuval Jun 3 '17 at 1:12
• And when working with coordinates, all the definitions are still the same like your definitions. – Amitai Yuval Jun 3 '17 at 1:13
• The point is that I want to use the idea of the proof in another context in which there is not coordinates available but the computations will be analogous. – PtF Jun 3 '17 at 2:18