What is the definition of curl of $\mathbf{F}(x_1, x_2) = ( F_1 (x_1,x_2) F_2(x_1,x_2))$ in $\mathbb{R}^2?$

Question

What is the definition of curl of $\mathbf{F}(x_1, x_2) = ( F_1 (x_1,x_2) F_2(x_1,x_2))$ in $\mathbb{R}^2?$

Most textbook says only of vector fields in the space $\mathbb{R}^3$...

Answer 1

Technically there isn't one as the cross product is only defined for three (and seven) dimensions but we can easily resolve this problem by considering these vectors to be in $\mathbb{R}^3$ with a $z$-coordinate of 0. Note in this case only one coordinate (the $z$-coordinate) will be non-zero so we often write the result as a scalar.

Answer 2

You use a wedge product instead of a cross product. Wedge products can be used in 3d (and beyond) also, generalizing nicely.

The result of a wedge product is not a vector but something called a bivector, representing the plane spanned by the two vectors. Like the cross product, the wedge product is antisymmetric, so $a \wedge a= 0$ for any $a$ and $a \wedge b = -b \wedge a$.

Thus, we have a differentiation operator denoted $\nabla \wedge F$ and is expressed in the standard basis $e_1, e_2$ as

$$\nabla \wedge F = \left(e_1 \frac{\partial}{\partial x_1} + e_2 \frac{\partial}{\partial x_2} \right) \wedge [F_1 e_1 + F_2 e_2] = (e_1 \wedge e_2 )\left(\frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2} \right)$$

This has the same basic expression as if you had used a cross product, but written in terms of $e_1 \wedge e_2$, we have something entirely intrinsic to the 2d plane, no fudging with embedding the plane in $\mathbb R^3$.