TITLE: Exact sequence involving the nabla operator
QUESTION [7 upvotes]: Recently I noticed that $$0 \longrightarrow \Bbb R \overset{\text{const.}}\longrightarrow \mathcal{C}^\infty(\Bbb R^3,\Bbb R) \overset{\text{grad}}\longrightarrow \mathcal{C}^\infty(\Bbb R^3,\Bbb R^3) \overset{\text{rot}}\longrightarrow \mathcal{C}^\infty(\Bbb R^3,\Bbb R^3) \overset{\text{div}}\longrightarrow \mathcal{C}^\infty(\Bbb R^3,\Bbb R)\longrightarrow 0$$
is an exact sequence of $\Bbb R$-algebras, where the second arrow is given by $\text{const}:c \mapsto f(\vec x)\equiv c$ and grad, rot ,div are the gradient, rotation and divergence operators.
Is the existence of such an exact sequence a mere curiosity or does it have its origins from deep results in homological algebra. 
If so, are there generelizations to $\Bbb R^n$ with higher $n$ or even to other smooth manifolds?

REPLY [2 votes]: Yes, you can replace the derivatives here with exterior derivatives.  This is the proper generalization of divergence, gradient, and curl.
Each real vector space $\mathbb R^n$ admits a geometric algebra on it called $\mathbb G^n$.  This is a clifford algebra, and its member objects are called multivectors.
These multivectors can be separated by "grades."  Each grade forms its own subspace.  They are as follows.  In $\mathbb G^n$ there is/are...

1 linearly independent scalar
$n$ linearly independent vectors
$n(n-1)/2 = \binom{n}{2}$ linearly independent bivectors
$\binom{n}{3}$ linearly independent trivectors
...
$n$ linearly independent $(n-1)$-vectors, also called pseudovectors
1 linearly independent $n$-vector, also called pseudoscalar

There are $2^n$ linearly independent elements.  You might see how this progression goes according to Pascal's triangle, and that in 3d, it goes 1-3-3-1.
Typically, we interpret the $k$-vectors (for any integer $k$ such that $0 \leq k \leq n$) geometrically.  A vector is an oriented line with a weight (magnitude).  A bivector is an oriented plane with a magnitude.  Trivectors are oriented volumes, and so on.
As vector calculus allows us to talk about vector and scalar fields, we can talk about bivector and trivector fields, arbitrary $k$-vector fields, or even general multivector fields with elements of several grades!
The vector derivative $\nabla$ can be taken to act on such fields.  We say that $\nabla \wedge A$ is a differential operator that acts a $k$-vector field $A$ and returns a $(k+1)$-vector field.  So from the space of scalar fields, we can build up a space of vector fields.  From vector fields, we can build up bivector fields, and so on.
But wait, there's more! When you have a metric, you can also do this in the opposite direction!  There is a "interior" derivative $\nabla \cdot A$ that acts on a $k$-vector field $A$ and returns a $(k-1)$-vector field.  This is also called the coderivative and by some other names.  The existence of this operator is why I consider it a mistake to implicitly identify gradient, divergence, and curl solely with the exterior derivative.  The "gradient" of a pseudoscalar field is not an exterior derivative at all, so such a claim that these three operators are only the exterior derivative in various guises is really incomplete.
What this means, then, is that as long as there is a metric involved, you can make the chain run the other way around.  Make a constant pseudoscalar field (as you made a constant scalar field) and run it backwards.
When dealing with general manifolds, let's consider the embedded case first.  Any embedded manifold has a pseudoscalar, which in an embedding will vary with position on the manifold.  The behavior of the pseudoscalar (how it changes with position) actually characterizes most of the manifold's properties!  But naturally, a $k$-vector field on the manifold cannot exceed the dimension of the pseudoscalar.  If your pseudoscalar is a bivector--a plane--then obviously trivector fields (which correspond to volumes) cannot live in that tangent space.