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.
Best Answer
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 $dx_1\wedge\dots\wedge dx_n$.)