# Differential forms as fields

I have two question related to differential forms on manifolds.

Question 1: I have seen two definitions of forms: as sections of exterior algebra of cotangent bundle $a\in\Gamma(\Lambda(T^*M))$ and as elements of exterior algebra of sections of cotangent bundle $a\in\Lambda(\Gamma(T^*M))$. Are these two definitions equivalent or one of these is not correct?

I prefer second one because what does it mean by exterior algebra $\Lambda(T^*M)$ if $T^*M$ is not even a vector/module but manifold? Second definition seems to be more appropriate:

We have $C^{\infty}$-module of sections and can build exterior algebra of it consisting only of alternating sections of various ranks with appropriate wedge product of fields and all seems fine.

Question 2: Given some manifold we can work with differential forms as fields or as a particular tensors of tangent spaces at each point. As these objects are different in principle, however in literature it is often used the same wedge product both for fields and for tensors. Although operations are similar they work in different spaces.

Wouldn't it be fair if we distinguish between wedge $\wedge$ in exterior algebra at particular $T_pM$ and wedge $\tilde{\wedge}$ in exterior algebra of sections of a bundle?

Only the first Definition as $\Gamma(\Lambda T^*M)$ is correct, the second leads to something different. This is connected to the answer to Question 2, so let me answer that one first. The right setting here is the one of vector bundles. Basically, you associate to each point $x\in M$ a (finite dimensional) vector space $E_x$ which depends continuously or smoothly on $x$ (this is expressed by local triviality and in particular implies that all $E_x$ have the same dimension). Now it turns out that there is a general scheme to extend functorial constructions with finite dimensional vector spaces to vector bundles. The basic examples here are the direct sum leading to the Whitney-sum and the tensor product and exterior powers or the extrior algebra, which is needed here. The main property of this extension is that for a vector bundle $E$ over $M$ and a point $x\in M$, the fiber of $\Lambda E$ at $x$ is $\Lambda(E_x)$, so basically you apply your functor to each fiber. (I'll continue to work in this general setting, you can specialize everything to $E=T^*M$ when needed.)
Now a section of $E$ associates to each $x\in M$ a vector in $E_x$ such that the dependence is continuous or smooth in an appropriate sense. If you pass to $\Lambda E$ and take two sections $\sigma,\tau\in\Gamma(\Lambda E)$ then you can apply the wedge poroduct in each point, and define $\sigma\wedge\tau\in \Gamma(\Lambda E)$ by $(\sigma\wedge\tau)(x)=\sigma(x)\wedge\tau(x)$. Of course, you could use a separate symbol for this operation, but this would be rather cumbersome. (It's like using different notation for the point-wise product of smooth functions and for the multiplication of real numbers.)
To get back to Question 1, you see that the point-wise definition of the wedge product has consequences. Namely, sections of a vector bundle always form a module over the algebra of smooth function using point wise multiplication. But then bilinearity of the point-wise wedge product implies that for a smooth function $f$ we get $(f\sigma)\wedge\tau=f(\sigma\wedge\tau)=\sigma\wedge(f\tau)$, so the wedge of sections is bilinear over $C^\infty(M,\mathbb R)$. Thus $\Gamma(\Lambda E)$ is the exterior algebra of the (finitely generated) module $\Gamma(E)$ over the algebra $C^\infty(M,\mathbb R)$, and not the exterior algebra of the vector space $\Gamma(E)$ (over $\mathbb R$). The latter still is a module over $C^\infty(M,\mathbb R)$ but certainly not finitely generated.
• So do I understand correctly that the two definitions $\Gamma(\Lambda E)$ and $\Lambda(\Gamma (E))$ are equivalent, as long as we interpret the second as the exterior algebra of $\Gamma(E)$ as a module over $C^\infty(M, \mathbb{R})$? – Phillip Andreae Apr 25 '17 at 12:18
• Oh, great answer, I see. However, if we consider only forms of k-th power $\Gamma(\Lambda^kE)$. How do we prove that we can decompose $\Gamma(\Lambda E)=\bigoplus\limits_{k=0}^{dimE}\Gamma(\Lambda^kE)$ ? Also small question: Is $\Gamma(\Lambda E)$ the same as what's denoted $\Omega(E)$ in the literature? – Sergey Dylda Apr 25 '17 at 15:18
• It easily follows from the definition of the direct sum of vector bundles that $\Gamma(E\oplus F)\cong\Gamma(E)\oplus\Gamma(F)$. Since as a vector bundle $\Lambda E=\oplus_k\Lambda^kE$ this gives the decomposition you are looking for. It may be that sometimes $\Gamma(\Lambda E)$ is denoted as $\Omega(E)$, but since there also is a concept of differential forms with values in a vector bundle, I don't like that notation. – Andreas Cap Apr 25 '17 at 18:11