Is there a good measure theoretic definition of curl?

To give an idea of the sort of equation that I'm looking for, here's now I define grad and div. For the gradient, say we are given a Fréchet differentiable function $f:X\rightarrow\mathbb{R}$, then we can define $\nabla f(x)$ to be the element in $X$ such that $$ \langle \nabla f(x),\eta \rangle = f^\prime(x)\eta. $$ Hence, $\nabla f(x)$ is the Riesz representative of the Fréchet derivative (Note, we've assumed a Hilbert space.) For the divergence, we have $f:X\rightarrow X$ and can define $$ \nabla \cdot f(x) = \lim_{\Omega\rightarrow \{x\}}\frac{1}{\mu(\Omega)} \int_{\Omega} \nabla\cdot f(x) d\mu =\lim_{\Omega\rightarrow \{x\}}\frac{1}{\mu(\Omega)} \int_{\partial \Omega} f(x)\cdot n d\mu. $$ Here, $\lim_{\Omega\rightarrow \{x\}} 1/\mu(\Omega)$ is shorthand for $\lim_{k\rightarrow \infty} 1/\mu(\Omega_k)$ where $\Omega_{k+1}\subseteq\Omega_k$ and $\lim_{k\rightarrow\infty}\Omega_k = \{x\}$. In addition, $\mu$ denotes some measure. In any case, the first equality follows from the Lebesgue differentiation theorem and the second follows from integration by parts. In this way, we require that $X$ be finite dimensional.

Now, I'd like to get something along these lines for curl. More specifically, I'm interested in a definition of curl that does not use differential forms or the exterior calculus.

I haven't checked this closely, but if it helps, I believe that for $f:X\rightarrow X$, we have that $$ (\nabla \times f(x))\times \delta x = f^\prime(x)\delta x-f^\prime(x)^*\delta x $$ where $f^\prime(x)^*$ denotes the adjoint of the Fréchet derivative of $f$ at $x$. In other words, the cross product between the curl of $f$ and $\delta x$ is the antisymmetric part of the Fréchet derivative.

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    $\begingroup$ What about the definition on the Wikipedia page? en.wikipedia.org/wiki/Curl_%28mathematics%29 $\endgroup$
    – Thomas
    Nov 26, 2012 at 16:06
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    $\begingroup$ Not sure if this might be useful (probably not), but you should get the curl if you replace $\cdot$ with $\times$ in the definition of divergence. This holds in classical vector calculus (cfr. Schey, Div grad curl and all of that, Ex. III-29), and should extend to general measure-theoretic setting. $\endgroup$ Jan 7, 2013 at 2:16
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    $\begingroup$ Maybe this answer helps: mathoverflow.net/questions/52829/… $\endgroup$ Sep 10, 2013 at 15:44

1 Answer 1


Let $v=(v_i)$ be a vector field. If $c=(c_i)$ is the curl of $v$, as you did with the divergence, you can write:

$$ c_i(x) = \lim_{A_i\to \{x\}}\frac{1}{\mu(A_i)} \int_{A_i}(\nabla\times v)_i\,d\mu = \lim_{A_i\to \{x\}}\frac{1}{\mu(A_i)} \int_{\partial A_i} (v\times dx)_i, $$

where: $$ (v\times dx)_i = \sum_{j,k}\epsilon_{ijk} v_j\,dx_k $$

are the components of the usual cross product, and $A_i$ are flat surfaces with smooth boundary, orthogonal to $dx_i$. For example, $A_1$ is parallel to the $yz$ plane, and oriented in that way.

Note: integration by parts (componentwise) is exactly equivalent to exterior differentiation and Stokes theorem.


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