Interpretation of $p$-forms

Question

Let $M$ be a smooth manifold, let $C^{\infty}(M)$ be set of all smooth functions from $M$ to $\mathbb R$ and let $Vec(M)$ denote the set of all vector fields on $M$. A $1$-form on $M$ is a $C^{\infty}(M)$-module homomorphism from $Vec(M)$ to $C^{\infty}(M)$. Let us denote the set of all $1$-forms on $M$ by $\Omega^1(M)$. Then a $p$-form is defined to be an element of $\bigwedge^p \Omega^1(M)$.

What is the correct mathematically interpretation for the elements of $\bigwedge^p \Omega^1(M)$? The elements of $\Omega^1(M)$ are just functions from $Vec(M)$ to $C^{\infty}(M)$. Are elements of $\bigwedge^p \Omega^1(M)$ also funtions? If what is their domain and codomains?

Answer 1

Your definition of a differential $p$-form on an $n$-dimensional manifold $M$ is slightly different from the one I learned.

The definition I'm used to is: given an $n$-dimensional manifold $M$, a differential $p$-form $\omega$ is a function which assigns to each point of $M$ an alternating $p$-tensor whose domain is the tangent space at that point. That is, for each $x\in M$ $\omega(x)\in\Lambda^p[T_x(M)^*]$. Once you know that the elementary $p$-tensors form a basis for $\Lambda^p[T_x(M)^*]$ for each point $x\in M$, the elements of $\Lambda^p[T_x(M)^*]$ look like $C^\infty(M)$ linear combinations of the elementary $p$-tensors. Of course, the elementary $p$-tensors look like $dx_{i_1}\wedge\cdots\wedge dx_{i_p}$ with $1\leq i_1<i_2<\cdots<i_p\leq n$.

So, some examples of differential 1-forms on $S^2$ are $dx+dy$, $x^4dx$, and $\frac{1}{y^2+x^4}dy$. And an example of a differential 2-form is $(x^2+y^2)dx\wedge dy$ (this differential 2-form is the same as $dx\wedge dy$ on $S^2$).

My definition is not strictly different from yours per se. The one I'm used to stresses the local nature of a $p$-form while yours stresses the global nature. Let's use our definitions of differential 1-forms to explore the nuance, since you seem okay with that definition.

Your definition says that a 1-form is a $C^\infty(M)$-module homomorphism from $Vec(M)$ to $C^\infty(M)$. In other words, it takes a vector field on $M$ and spits out a smooth function on $M$. Consider two different vector fields on $M$ which share some velocity vectors (or tangent vectors), say, at $x$ for example. The smoothness restrictions say that the two different smooth functions our 1-form returns will have to agree at $x$. So, instead of thinking about a 1-form as acting on $Vec(M)$ you can instead think of it as acting simply on the tangent vectors at $x$ as $x$ varies (this is my definition).

Now let's try to understand $\Lambda^p\Omega^1(M)$. Let's forget about all the extra structure $\Omega^1(M)$ brings with it and only think about it as $C^\infty(M)$ module. Forgetting the extra structure let's you see what the domain is. The domain of the elements of $\Lambda^p\Omega^1(M)$ are $Vec(M)^p$ (an element of $Vec(M)^p$ is a $p$-vector field; a $p$-vector field is like a vector field but instead of 1 tangent vector at each point there are $p$) and they each map to $C^\infty(M)$. If we instead only think about these elements locally, you see that they assign an alternating $p$-tensor to each point of $x\in M$ which 'eats up' $p$ vectors from the tangent space of $x$ and spit out a scalar. This is because the smoothness restrictions are going to force the resulting smooth functions to agree at $x$ when considering the same set of $p$-vectors which all happen to be in distinct $p$-vector fields.

Answer 2

Tensor products are taken with respect to $C^{\infty}(M)$; (multi)linear means $C^{\infty}(M)$-(multi)linear; $\Gamma(E)$ denotes the space of sections of a vector bundle $E\xrightarrow{\pi}M$; an open set $U\subset M$ is said to trivialize $E$ if the restriction $E\big|_{U}\to U$ of $E$ to $U$ is trivial.

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There is a first general fact: given vector bundles $E_1,\dots,E_n$ over $M$, the space of sections of the tensor product bundle $E_1\otimes\dots\otimes E_n\to M$ is isomorphic to the tensor product of the section spaces: $$\Gamma(E_1\otimes\dots\otimes E_n)\simeq\Gamma(E_1)\otimes\cdots\otimes\Gamma(E_n)$$ There is an obvious map $\Gamma(E_1)\otimes\cdots\otimes\Gamma(E_n)\to\Gamma(E_1\otimes\dots\otimes E_n)$ which is always one to one (look locally). It is easy to see that it is onto if $M$ is compact (partition of unity subordinate to a finite covering of $M$ by open sets trivializing all of $E_1,\dots,E_n$). This is still an isomorphism when $M$ is only assumed second countable (i.e. paracompact), but requires a less obvious argument using the covering dimension of $M$ : if $E\to M$ is a vector bundle and if $M$ has dimension $d$, there is a finite open cover by $d+1$ open sets that trivialize $E$.

The same proof extends symmetric and exterior powers : $$\mathrm{Sym}^n\Gamma(E)\simeq\Gamma(\mathrm{Sym}^nE)\quad\text{and}\quad\Lambda^n\Gamma(E)\simeq\Gamma(\Lambda^nE)\,.$$

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A second general fact is the following : if $E,F$ are two vector bundles over $M$, then the space of linear maps $\Gamma(E)\to\Gamma(F)$ is isomorphic to $\Gamma(E^*\otimes F)$ (again, linear means $C^{\infty}(M)$-linear). There is an obvious inclusion $\Gamma(E^*\otimes F)\subset\mathrm{Hom}_{C^{\infty}(M)}(\Gamma(E),\Gamma(F))$, and conversely, given a linear map $c:\Gamma(E)\to\Gamma(F)$ one can construct (locally at first) a section $\tilde c$ of $E^*\otimes F$. As a special case, taking $F=\Bbb R\times E$ the trivial one dimensional bundle (so that $\Gamma(F)=C^{\infty}(M)$), we see that $\Gamma(E)^*\simeq\Gamma(E^*)$.

The same proof extends to show that symmetric map (one that is unphazed by permuting the entries) $$\mathrm{Sym}^n\Gamma(E)\to\Gamma(F)$$ arises from a section of $\mathrm{Sym}^n(E^*)\otimes F\,\big(\simeq \mathrm{Sym}^n(E)^*\otimes F\big)$, and alternating maps $$\Lambda^n\Gamma(E)\to\Gamma(F)$$ arise from sections of $\Lambda^n(E^*)\otimes F\,\big(\simeq \Lambda^n(E)^*\otimes F\big)$.

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What we have established is that $\Gamma$ commutes with all the usual operations on vector spaces (resp. $C^{\infty}(M)$-modules) when applied to finite dimensional vector bundles : taking duals, taking tensor products, symmetric or exterior powers, Hom-sets...

Combine these facts, you recover the two interpretations of $\Lambda^p\Omega^1(M)$ that I'm aware of : $$\Lambda^p\Omega^1(M)=\Lambda^p\Gamma(T^*M)\simeq\Gamma(\Lambda^pT^*M)$$ (which is the usual definition of $\Omega^p(M)$) expresses it as the space of sections of the bundle $\Lambda^pT^*M$, and $$\Lambda^p\Omega^1(M)\simeq\Gamma(\Lambda^pT^*M)\simeq\Gamma((\Lambda^pTM)^*)\simeq\Gamma(\Lambda^pTM)^*\simeq\big(\Lambda^p\Gamma(TM)\big)^*$$ the space of all skew-symmetric $C^{\infty}(M)$-linear maps $$\mathrm{Vec}(M)\times\cdots\times\mathrm{Vec}(M)\to C^{\infty}(M)$$