Is any k-form $\mathcal{C}^\infty$-multilinear?

Question

I mean, does it hold $$w(fX_1,...,X_k)=fw(X_1,...,X_k)$$ with $w\in\mathcal{\Omega}^k(M)$, $f\in\mathcal{C}^\infty(M)$, $X_i\in\mathcal{T}(M)$ with M smooth manifold?

Answer 1

Yes it is, since a k-form $w\in\Gamma((\mathcal{T}^*(M))^{\otimes k}):=\Gamma(\mathcal{T}^*(M)\otimes...\otimes\mathcal{T}^*(M))$ is a section of the tensor bundle $(\mathcal{T}^*(M))^{\otimes k}$, i. e. a map $w:M\rightarrow(\mathcal{T}^*(M))^{\otimes k}$, for which to any point $p\in M$ it associates a (0,k)-tensor $w_p:=w(p)$ which is by definition $\mathbb{R}-$multilinear. In the same way any vector field is $X_j\in\Gamma(\mathcal{T}(M))$ is a section of the tangent bundle $\mathcal{T}(M)$, i. e. a map $X_j:M\rightarrow\mathcal{T}(M)$, that associates to any point $p\in M$ a vector (or a (1,0)-tensor) $X_j(p)$. Then the vector field $fX_1$ associates to any point $p\in M$ the vector $f(p)X_1(p)$. So the action of $w$ on $(fX_1,...,X_k)$ takes p and gets $w_p(f(p)X_1(p),...,X_k(p))=f(p)w_p(X_1(p),...,X_k(p))$, since $f(p)\in\mathbb{R}$ and $w_p$ is $\mathbb{R}$-multilinear. On the other hand, the action of $fw$ on $(X_1,...,X_k)$ takes $p$ and gets $f(p)w_p(X_1(p),...,X_k(p))$. Since this holds $\forall p\in M$, we get $w(fX_1,...,X_k)=fw(X_1,...,X_k)$.