# Is there a lower dimensional vector operator analogous to Curl or Div?

The Curl vector operator can be defined as the limit of an infinitesimal closed line integral over a surface divided by the |area| as it approaches zero. Likewise for Div as the limit of an infinitesimal surface integral over a volume divided by the |volume| as it approaches zero. Taking the analogy to a lower dimension:

Is there a lower dimensional vector operator analogous to Curl and Div?

Is there a theorem associated with it analogous to Stokes' for Curl or Ostrogradsky's for Div?

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By "lower dimensional" do you mean: "for scalar and vector fields in plane domains"? – Christian Blatter Sep 13 '12 at 13:33
– user8268 Sep 13 '12 at 15:01
@ChristianBlatter I see Curl as being a lower dimensional vector operator compared to Div and I'm looking for a lower dimensional operator compared to Curl. I'm open to suggestions on how to better express this in my question. – user10389 Sep 13 '12 at 15:05

1
– Jonathan Sep 13 '12 at 16:36

## 1 Answer

Note that all three operators: ${\rm div}$, ${\rm curl}$, and $\nabla$ are first order operators, only the Laplace operator $\Delta:={\rm div}\circ\nabla$ is of second order. Your question then concerns only the dimension of the domains over which one integrates.

The operator ${\rm div}$ acts on flow fields ${\bf v}$, and is implicitly defined by $$\int_{\partial B} {\bf v}\cdot{\rm d}{\omega}\ \doteq\ {\rm div}\,{\bf v}({\bf p})\ {\rm vol}(B)$$ for small three-dimensional balls $B:=B_\epsilon({\bf p})$ with center ${\bf p}$.

The operator ${\rm curl}$ acts on force fields ${\bf F}$, and is implicitly defined by $$\int_{\partial D} {\bf F}\cdot d{\bf x}\ \doteq\ {\rm curl}\,{\bf F}({\bf p})\cdot {\bf n}\ \ {\rm area}(D)$$ for small oriented two-dimensional disks $D\subset{\mathbb R}^3$ with center ${\bf p}$ and normal ${\bf n}$.

Note that the dimension of the considered "balls" has decreased from $3$ to $2$, and the dimension of their boundaries (spheres) has decreased from $2$ to $1$. Now you want to go one step further down. This is indeed possible, and results in the following sentence, formulated in the same way as the first two:

The operator $\nabla$ acts on scalar fields $f$, and is implicitly defined by $$\int_{\partial S} f\ dp\doteq\ \nabla f({\bf p})\cdot {\bf u}\ \ {\rm length}(S)$$ for small oriented segments $S\subset{\mathbb R}^3$ with center ${\bf p}$ and direction ${\bf u}$.

But what is $\partial S$ in this case? It is the endpoint ${\bf b}$ of $S$ with weight $+1$ and the initial point of $S$ with weight $-1$. Therefore the last formula should be read as $$f({\bf b})-f({\bf a})\ \doteq\ \nabla f({\bf p})\cdot ({\bf b}-{\bf a})\ .$$ This formula can be viewed as "implicit" definition of the gradient at ${\bf p}$.

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