There are two given differential forms:

$\omega_1 = 2 dz + e^xdy + t^2dt$

$\omega_2 =t dx \wedge dy + z dt \wedge dy$

I am supposed to calculate $\omega_1 \wedge \omega_2$ and $d(\omega_1 \wedge \omega_2)$.

So, I think that $\omega_1 \wedge \omega_2 = (2 dz + e^xdy + t^2dt) \wedge (t dx \wedge dy + z dt \wedge dy) = $

$= 2t dz \wedge dx \wedge dy + t^3 dt \wedge dx \wedge dy + 2z dz \wedge dt \wedge dy$

But I'm not not sure what $d(\omega_1 \wedge \omega_2)$ should look like. From what I understand, I think it would be:

$d(\omega_1 \wedge \omega_2) = d(2t) \wedge dz \wedge dx \wedge dy + d(t^3)\wedge dt \wedge dx \wedge dy + d(2z) \wedge dz \wedge dt \wedge dy$

$=2 dt \wedge dz \wedge dx \wedge dy + 3t^2 dt \wedge dt \wedge dx \wedge dy + 2 dz \wedge dz \wedge dt \wedge dy$

$= 2 dx \wedge dy \wedge dz \wedge dt$

Since each function standing in front of a "smaller" form depends only on one variable, so all the other partial derivatives turn out to be zero. After that I'm using the properties of the exterior product.

Am I correct?


Seems that it should be $-2dx\wedge dy\wedge dz\wedge dt$ instead because of anti-commutativity.

  • $\begingroup$ Right, I forgot that I "moved" $dt$ as well, thanks! $\endgroup$ – Angie Jun 6 '17 at 18:38

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