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I'm a little confused about the divergence and the curl for two-dimensional vector fields. In 3d, I understand the curl as $d:\Omega^1(M^3)\to\Omega^2(M^3)$ and the divergence as $d:\Omega^2(M^3) \to \Omega^3(M^3)$.

But what is the analog in 2d? It seems the curl is the operator $d:\Omega^1(M^2)\to \Omega^2(M^2)$, and then what could the divergence be?

I recall using before the divergence theorem for two-dimensional vector fields...that $$\int_D\text{div}\,\sigma=\int_{\partial D}\sigma\cdot \nu, \tag{1}$$

where $\nu$ is the unit normal to the two dimensional region $D$. Where does this fit in with generalized Stokes' theorem, if $d:\Omega^1(M^2)\to \Omega^2(M^2)$ is the curl, not the divergence?

I'm thinking now as I write this that (1) just treats the 2d vector field as a 3d field with third component zero.

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    $\begingroup$ wouldn't be the Green's Theorem? $\endgroup$
    – janmarqz
    Jan 2, 2014 at 4:21
  • $\begingroup$ pick a volume form $\Delta$, then define the divergence of a vector field $X$ as $ d (i_X \Delta) = (div X )\Delta$. $\endgroup$
    – user40276
    Jan 2, 2014 at 4:25
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    $\begingroup$ @janmarqz Stokes theorem is a generalization of Green 's Theorem stated in the language of differential forms over manifolds. $\endgroup$
    – user40276
    Jan 2, 2014 at 4:33

2 Answers 2

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The curl in 2D is sometimes called rot: $\text{rot}(u) = \frac{\partial u_2}{\partial x_1} - \frac{\partial u_1}{\partial x_2}$. You can also get it by thinking of the 2D field embedded into 3D, and then the curl is in $z$ direction, that is, it only has one component. As you rightly say, it is in essence the same as the div: $\text{div}(u) = \text{rot}(u^\perp)$, where $u^\perp = (-u_2,u_1)$.

The divergence is $d^*:(\Omega_1(M^2))^* \to (\Omega_0(M^2))^*$, the adjoint to the grad, which is $d:\Omega_0(M^2)) \to \Omega_1(M^2)$.

Also, in 3D we think of $d:\Omega_2(M^3)) \to \Omega_3(M^3)$ as divergence by identifying $dx_1 \wedge dx_2$ with $dx_3$, $dx_2\wedge dx_3$ with $dx_1$, etc. In 2D we do a similar identification $dx_1$ with $dx_2$ and $dx_2$ with $-dx_1$. In this way, the rot and div are essentially the same. (This is the Hodge duality, where in $n$ dimensions $a = dx_{i_1} \wedge \dots \wedge d_{x_k}$ is identified with $b = \pm dx_{j_1}\wedge\dots\wedge dx_{j_{n-k}}$, where $\{x_{j_1},\dots,x_{j_{n-k}}\} = \{x_1,\dots,x_n\} \setminus \{x_{i_1},\dots,x_{i_k}\}$ the plus/minus is chosen so that $a\wedge b$ is the volume form $dx_1\wedge\dots\wedge dx_n$.)

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  • $\begingroup$ Can you cite a textbook/paper that contains this definition of a two dimensional curl? $\endgroup$
    – Sjoerd222888
    Mar 5, 2015 at 9:03
  • $\begingroup$ @Sjoerd2228888 Idon't have any such reference. Bit googling "curl 2d" gives lots of hits that look relevant. $\endgroup$
    – Stephen Montgomery-Smith
    Mar 5, 2015 at 12:52
  • $\begingroup$ Google yields a lot of results but so far I did not find any reference to a textbook. I`ll continue my search. $\endgroup$
    – Sjoerd222888
    Mar 5, 2015 at 14:24
  • $\begingroup$ One way to think of the 2D curl is like this. Embed the plane into 3 dimensions, and think of the 2D field as actually being a 3D field that has its third component zero, and only depends on the first and second components. Then the 3D curl will have only one non-zero component, which will be parallel to the third axis. And the value of that third component will be exactly the 2D curl. So in that sense, the 2D curl could be considered to be precisely the same as the 3D curl. $\endgroup$
    – Stephen Montgomery-Smith
    Mar 5, 2015 at 17:46
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    $\begingroup$ @timur Yes, you are correct. $\endgroup$
    – Stephen Montgomery-Smith
    Oct 29, 2015 at 1:13
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I am late to the party, but I think the 2-dimensional Stoke's Theorem is called Green's Theorem (https://en.wikipedia.org/wiki/Green%27s_theorem)

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