The Unapologetic Mathematician

The Algebra of Differential Forms

We’ve defined the exterior bundle $\Lambda^*_k(M)$ over a manifold $M$. Given any open $U\subseteq M$ we’ve also defined a $k$-form over $U$ to be a section of this bundle: a function $\alpha:U\to\Lambda^*_k(U)$ such that $\pi\circ\alpha=I_U:U\to U$. We write $\Omega^k(U)=\Omega_M^k(U)$ for the collection of all such $k$-forms over $U$. It’s straightforward to see that this defines a sheaf on $M$.

This isn’t just a sheaf of sets; it’s a sheaf of modules over the structure sheaf $\mathcal{O}_M$ of smooth functions on $M$. We define the necessary operations pointwise:

\displaystyle\begin{aligned}\left[\alpha+\beta\right](p)&=\alpha(p)+\beta(p)\\\left[f\alpha\right](p)&=f(p)\alpha(p)\end{aligned}

where the right hand sides are defined by the vector space structures on the respective $\mathcal{T}_pM$.

We can go even further and define the sheaf of differential forms

$\displaystyle\Omega_M(U)=\Omega(U)=\bigoplus\limits_{k=1}^n\Omega^k(U)$

This sheaf $\Omega_M$ is not just a sheaf of modules over $\mathcal{O}_M$, it’s a sheaf of algebras. For an $\alpha\in\Omega^k(U)$ and a $\beta\in\Omega^l(U)$, we define their exterior product pointwise:

$\displaystyle\left[\alpha\wedge\beta\right](p)=\alpha(p)\wedge\beta(p)$

In fact, this is a graded algebra, and the multiplication has degree zero:

$\displaystyle\wedge:\Omega^k(U)\otimes\Omega^l(U)\to\Omega^{k+l}(U)$

Even better, this is a unital algebra. We see this by considering the zero grade, since the unit must live in the zero grade. Indeed, $\Lambda_0^*(U)\cong\mathbb{R}$, so sections of $\Lambda_0^*(U)$ are simply functions on $U$. That is, $\Omega^0(U)\cong\mathcal{O}(U)$. Given a function $f\in\mathcal{O}(U)$ we will just write $f\alpha$ instead of $f\wedge\alpha$.

July 12, 2011 - Posted by | Differential Topology, Topology

1. […] just seen that smooth real-valued functions are differential forms with grade zero. We also know that functions pull back along smooth maps; if is a smooth function […]

Pingback by Pulling Back Forms « The Unapologetic Mathematician | July 13, 2011 | Reply

2. […] looks sort of familiar as a derivative, but we have another sort of derivative on the algebra of differential forms: the “exterior derivative”. But this one doesn’t really look like a derivative at […]

Pingback by The Exterior Derivative « The Unapologetic Mathematician | July 15, 2011 | Reply

3. […] really important thing about the exterior derivative is that it makes the algebra of differential forms into a “differential graded algebra”. We had the structure of a graded algebra before, […]

Pingback by De Rham Cohomology « The Unapologetic Mathematician | July 20, 2011 | Reply

4. […] forms” — top because is the highest degree possible for a differential form on a differential form — has rank over the algebra of smooth functions. That is, if is a top form then we can […]

Pingback by Integration on the Standard Cube « The Unapologetic Mathematician | August 2, 2011 | Reply

5. […] smooth maps, and homotopies form a 2-category, but it’s not the only 2-category around. The algebra of differential forms — together with the exterior derivative — gives us a chain complex. Since pullbacks of […]

Pingback by The Poincaré Lemma (setup) « The Unapologetic Mathematician | December 2, 2011 | Reply

6. […] Armstrong: The algebra of differential forms, Pulling back forms, The Lie derivative on forms, The exterior derivative is a derivative, The […]