2

I know it is defined as $ dP \wedge dx $ or explicitly,

$$ \frac{\partial P }{\partial y } dy \wedge dx . $$

The question is, could it be $dx \wedge d P $? Or

$$ \frac{\partial P }{\partial y } dx \wedge dy ? $$

I know the question must be very naive or even stupid. But I am indeed confused.

pie
  • 243

1 Answers1

0

If $\omega$ is a $k$-form and $\eta$ is an $\ell$-form, then $\omega\wedge \eta = (-1)^{k\ell}\eta\wedge\omega$. $dx$ and $dy$ are $1$-forms, and so $dx\wedge dy = -dy\wedge dx$. So, no, $dx\wedge dP$ differs from $dP\wedge dx$ by a minus sign.

florence
  • 12,819