4
$\begingroup$

If we have a form, say, $\omega = f(x,y,z) \, dx + g(x,y,z) \, dy + h(x,y,z) \, dz$, what is the formula for the exterior derivative $d \omega$?

$\endgroup$
  • $\begingroup$ It is the curl of the evident vector field, written as a 2-form. $\endgroup$ – Will Jagy May 2 '12 at 2:17
  • $\begingroup$ I have added more in my answer, hope it will help. $\endgroup$ – Shuhao Cao May 2 '12 at 2:54
5
$\begingroup$

Roughly speaking, it is the anti-symmetric part of the full derivative due to the "odd permutation = sign change" trait of the differential forms, for a 1-form in $\mathbb{R}^3$ you gave if we write it as $\displaystyle\omega = \sum^{3}_{j=1} f_i dx_i$, it is : $$ \begin{aligned} d\omega &= \sum^{3}_{i=1} \sum^{3}_{j=1} \frac{\partial f_i}{\partial x_j}dx_j\wedge dx_i \\ &= \sum^{3}_{j=1} \frac{\partial f_1}{\partial x_j}dx_j\wedge dx_1 + \sum^{3}_{j=1} \frac{\partial f_2}{\partial x_j}dx_j\wedge dx_2 + \sum^{3}_{j=1} \frac{\partial f_3}{\partial x_j}dx_j\wedge dx_3 \\ &= \frac{\partial f_1}{\partial x_2}dx_2\wedge dx_1 + \frac{\partial f_1}{\partial x_3}dx_3\wedge dx_1 \\ &\quad+ \frac{\partial f_2}{\partial x_1}dx_1\wedge dx_2 + \frac{\partial f_2}{\partial x_3}dx_3\wedge dx_2 \\ &\qquad + \frac{\partial f_3}{\partial x_1}dx_1\wedge dx_3 + \frac{\partial f_3}{\partial x_2}dx_2\wedge dx_3 \\ &= (\frac{\partial f_2}{\partial x_1} - \frac{\partial f_1}{\partial x_2})dx_1\wedge dx_2 + (\frac{\partial f_3}{\partial x_2}-\frac{\partial f_2}{\partial x_3})dx_2\wedge dx_3 + (\frac{\partial f_1}{\partial x_3} - \frac{\partial f_3}{\partial x_1})dx_3\wedge dx_1 \\ &= (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y})dx\wedge dy + (\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z})dy\wedge dz + (\frac{\partial f}{\partial z} - \frac{\partial h}{\partial x})dz\wedge dx \end{aligned} $$ We could see it behaves the $\nabla \!\!\times$ operator and you get a 2-form, since we are talking in $\mathbb{R}^3$, by using the Hodge dual operator $\star$, you could get a formula looking more like the old curl formula we learned in vector calculus in the following way: $$ \nabla \times\omega = \star d\omega = \begin{pmatrix} dx & dy & dz \end{pmatrix} \begin{pmatrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0 \end{pmatrix} \begin{pmatrix} f\\g\\h \end{pmatrix} $$ here we could see pretty clear that the anti-symmetricity comes from the anti-commutativity of the wedge product of the exterior algebra.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.