Differential forms - order of the differentials?

Question

I understand that when using differential forms the order of the differentials matters. For example, $$dx\wedge dy\wedge dz=-dy\wedge dx\wedge dz.$$So, practically, if I get the order wrong, I can end up with the answer having the wrong sign. Is that because when, again for example, $dx\wedge dy\wedge dz$ acts on the standard $\mathbb{R}^{3}$ ordered basis $\left(e_{1},e_{2},e_{3}\right)$ the answer comes out as positive rather than negative, ie$$\left(dx\wedge dy\wedge dz\right)\left(e_{1},e_{2},e_{3}\right)=\left|\begin{array}{ccc} dx\left(e_{1}\right) & dx\left(e_{2}\right) & dx\left(e_{3}\right)\\ dy\left(e_{1}\right) & dy\left(e_{2}\right) & dy\left(e_{3}\right)\\ dz\left(e_{1}\right) & dz\left(e_{2}\right) & dz\left(e_{3}\right) \end{array}\right|=\left|\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right|=1,$$ie plus one, not negative one? I'm guessing that in some way the $\mathbb{R}^{3}$ orientations of the differential form and the ordered basis have to match up?

Answer 1

One of the way to see why the exterior product needs to be antisymmetric is to rotate the coordinate system in the plane $xy$ by 90°: $x\to y$, $y\to-x$. However, an elementary area in that plane shouldn't depend on the local coordinates: $$ dx\wedge dy = (dy)\wedge(−dx) = -dy\wedge dx $$