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.