Exact sequence involving the nabla operator 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?  
 A: This is a special case of the deRham complex on $\Bbb R^3$.
Let $M$ be a smooth manifold. Then we get the cotangent bundle $T^\ast M$ of $M$ by letting the cotangent space $T_x^\ast M$ at $x \in M$ be the dual vector space to the tangent space $T_x M$. Recall that given a vector space $V$, we can form the exterior algebra $\Lambda^\ast V$, and we let $\Lambda^p V$ denote the degree $p$ part of $\Lambda^\ast V$ (so if $\{e_1, \dots, e_n\}$ is a basis for $V$, $\Lambda^p V$ is generated by products of the form $e_{i_1} \wedge \cdots \wedge e_{i_p}$). Now we can form the bundle $\Lambda^\ast T^\ast M$ by taking the exterior algebra $\Lambda^\ast T^\ast_x M$ of each cotangent space, and similarly we get bundles $\Lambda^p T^\ast M$. Finally, we define the space of differential $p$-forms on $M$ by
$$\Omega^p(M) = C^\infty(\Lambda^p T^\ast M),$$
i.e. the space of smooth sections of $\Lambda^p T^\ast M$. What this means is the following. We can consider $\Lambda^p T^\ast M$ as
$$\Lambda^p T^\ast M = \coprod_{p \in M} \Lambda^p T_x^\ast M,$$
topologized appropriately. Hence we have a natural projection map $\pi: \Lambda^p T^\ast M \longrightarrow M$ which is given by $\pi(x, v) = x$. Then
$$\Omega^p(M) = \{\alpha: M \longrightarrow \Lambda^p T^\ast M \mid \pi \circ \alpha = \mathrm{Id}_M\}.$$
Note that $\Omega^0(M)$ is just the space of smooth real-valued functions on $M$. We can define the exterior derivative $df$ of $f \in \Omega^0(M)$ by defining $df$ to be the differential of $f$, i.e.
$$df(X) = Xf$$
for any vector field $X$ on $M$. If we impose the Leibniz rule
$$d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\deg(\alpha)} \alpha \wedge d\beta,$$
then the exterior derivative extends uniquely to a map
$$d: \Omega^\ast(M) \longrightarrow \Omega^{\ast + 1}(M).$$
Now one can show that $d^2 = 0$, so that
$$0 \to \Omega^0(M) \xrightarrow{~d~} \Omega^1(M) \xrightarrow{~d~} \Omega^2(M) \xrightarrow{~d~} \cdots$$
is a cochain complex. The cohomology
$$H^\ast_{dR}(M) = H^\ast(\Omega^\ast(M), d)$$
of this complex is called the deRham cohomology of $M$. DeRham's theorem states that deRham cohomology is isomorphic to singular cohomology:
$$H^\ast_{dR}(M) \cong H^\ast_{\text{sing}}(M; \Bbb R).$$
Now let's see why your example is a special case of the deRham complex. When $M = \Bbb R^3$, we have
$$\Omega^0(\Bbb R^3) \cong \Omega^3(\Bbb R^3) \cong C^\infty(\Bbb R^3, \Bbb R), \quad \Omega^1(\Bbb R^3) \cong \Omega^2(\Bbb R^3) \cong C^\infty(\Bbb R^3, \Bbb R^3).$$
All other spaces of $p$-forms on $\Bbb R^3$ are trivial. Now for $f \in \Omega^0(\Bbb R^3)$,
$$df = \sum_{i = 1}^3 \frac{\partial f}{\partial x_i} dx_i ~\leftrightarrow~ \operatorname{grad}(f) = \left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \frac{\partial f}{\partial x_3} \right).$$
For
$$\alpha = \sum_{i = 1}^3 \alpha_{i}(x_1, x_2, x_3) ~dx_i \in \Omega^1(\Bbb R^3) ~\leftrightarrow~ v = (\alpha_1, \alpha_2, \alpha_3) \in C^\infty(\Bbb R^3, \Bbb R^3),$$
we have
$$d\alpha = \sum_{i = 1}^3 \sum_{j = 1}^3 \frac{\partial \alpha_i}{\partial x_j} ~dx_i \wedge dx_j = \left(\frac{\partial \alpha_3}{\partial x_2} - \frac{\partial \alpha_2}{\partial x_3}\right) ~dx_2 \wedge dx_3 - \left(\frac{\partial \alpha_3}{\partial x_1} - \frac{\partial \alpha_1}{\partial x_3}\right) ~dx_1 \wedge dx_3 + \left(\frac{\partial \alpha_2}{\partial x_1} - \frac{\partial \alpha_1}{\partial x_2}\right) ~dx_1 \wedge dx_2 ~\leftrightarrow~ \operatorname{rot}(v).$$
Finally, for
$$\beta = \sum_{i = 1}^3 \sum_{j = 1}^3 \beta_{ij}(x_1, x_2, x_3) ~dx_i \wedge dx_j \in \Omega^2(\Bbb R^3) ~\leftrightarrow~ w = (\beta_{23}, -\beta_{13}, \beta_{12}) \in C^\infty(\Bbb R^3, \Bbb R^3),$$
we have
$$d\beta = \sum_{i,j,k = 1}^3 \frac{\partial \beta_{ij}}{\partial x_k} ~dx_i \wedge dx_j \wedge dx_k = \frac{\partial \beta_{23}}{\partial x_1}~dx_2 \wedge dx_3 \wedge dx_1 + \frac{\partial \beta_{13}}{\partial x_2}~dx_1 \wedge dx_3 \wedge dx_2 + \frac{\partial \beta_{12}}{\partial x_3}~dx_1 \wedge dx_2 \wedge dx_3 ~\leftrightarrow~ \operatorname{div}(w) = \frac{\partial \beta_{23}}{\partial x_1} - \frac{\partial \beta_{13}}{\partial x_2} + \frac{\partial \beta_{12}}{\partial x_3}.$$
Hence we see the correspondence between the exterior derivatives and the vector derivatives. Now deRham's theorem tells us that the cohomology of
$$0 \longrightarrow C^\infty(\Bbb R^3,\Bbb R) \overset{\text{grad}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R^3) \overset{\text{rot}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R^3) \overset{\text{div}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R)\longrightarrow 0$$
is trivial except in degree zero. Hence we augment the cochain complex as you did to get an exact sequence:
$$0 \longrightarrow \Bbb R \overset{\text{const.}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R) \overset{\text{grad}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R^3) \overset{\text{rot}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R^3) \overset{\text{div}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R)\longrightarrow 0.$$
A: 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.
A: Depending on your definition of curiosity, this is not a coincidence. It does in fact generalise into higher dimensions, even into manifolds. For more information I advise you to look up some theory on De Rham Cohomology.
