I struggle to understand the following question. I expect I'm simply being dense about something.

Let $F$ be a vector field on $U$, open in $\mathbb R^3,$ $F = \sum_1^3 f_j (x) \frac{\partial}{\partial x_j}$. Consider the $1$-form $\varphi=\sum_1^3 f_j(x) dx_j.$ Show that $d \varphi$ and $\operatorname{curl} F$ are related in the following way:

$$\operatorname{curl} F = \sum_1^3 g_j(x) \frac{\partial}{\partial x_j}$$

$$d\varphi =g_1(x) dx_2 \wedge dx_3 + g_2(x) dx_3 \wedge dx_1 + g_3(x)dx_1 \wedge dx_2$$

What are the $g_i$?

For reference, this is Exercise 5 in section 13 of Chapter 1.


They are simply what you get when you compute the curl (and the differential): $g_1=\partial f_3/\partial x_2-\partial f_2/\partial x_3$, and so on.

  • $\begingroup$ Then what is there to show? $\endgroup$ – Alfred Yerger Dec 29 '15 at 7:22
  • 1
    $\begingroup$ Not much, just check that you really get the same thing in both cases. $\endgroup$ – Hans Lundmark Dec 29 '15 at 9:21

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