- #1

dyn

- 747

- 57

( ∂/∂x , dx∧dy ) = dy and ( ∂/∂x , dy∧dx ) = -dy

( ∂/∂z , dx∧dy∧dz ) = dx∧dy and ( ∂/∂z , dx∧dz∧dy ) = -dx∧dy

Have I got this right ? Thanks

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- Thread starter dyn
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- #1

dyn

- 747

- 57

( ∂/∂x , dx∧dy ) = dy and ( ∂/∂x , dy∧dx ) = -dy

( ∂/∂z , dx∧dy∧dz ) = dx∧dy and ( ∂/∂z , dx∧dz∧dy ) = -dx∧dy

Have I got this right ? Thanks

- #2

- 19,874

- 10,411

You can then use the fact that ##dx \wedge dy = dx \otimes dy - dy \otimes dx## to verify your identity. When you have ##\partial/\partial x## you would get

$$

\left(\frac{\partial}{\partial x}, dy \wedge dx\right) = \underbrace{dy(\partial_x)}_{=0} dx - \underbrace{dx(\partial_x)}_{=1} dy = -dy.

$$

Of course, when you have the differential form expressed in the coordinate differentials, you can generally just anti-commute the relevant coordinate differential to the first position. All terms where it is not in the first position will vanish identically and the interior product will just correspond to anti-commuting the relevant coordinate differential to the front and then removing it from the exterior product.

- #3

dyn

- 747

- 57

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