Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
You can find de Rham theory in Warner, Introduction to Differentiable Manifolds and Lie Groups – Neal Jan 31 '13 at 17:58
up vote 5 down vote accepted

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: $$0 \longrightarrow \Bbb R \overset{\text{const.}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R) \overset{\text{grad}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R^3) \overset{\text{rot}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R^3) \overset{\text{div}}\longrightarrow C^\infty(\Bbb R^3,\Bbb R)\longrightarrow 0.$$

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.