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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 $d \omega$?

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It is the curl of the evident vector field, written as a 2-form. –  Will Jagy May 2 '12 at 2:17
    
I have added more in my answer, hope it will help. –  Shuhao Cao May 2 '12 at 2:54
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up vote 3 down vote accepted

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.

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