Proving h is a Differential Form

[Canavar]

Hello,

I try to understand differential forms. For istance i want to prove that
$$h=e_1\wedge e_2 + e_3\wedge e_4$$
is a differential form, where e_1,..,e_4 are elements of my basis.



Do you have a idea, why this is a differential form?

Regards

 

[HallsofIvy]

First, your 'basis' elements are themselves "differentials". If you think of $e_1= dx$, $e_2= dy$, $e_3= dz$, and $e_4= dt$ then $e_1\wedge e_2+ e_3\wedge e_4= dxdy+ dzdt$. It would convert the function f(x,y,z, t) into
$$\int\int f(x,y,z,t) dxdy+ f(x,y,z,t)dzdt$$

The "wedge", $\wedge$, is there because this product is "anti-symmetric" $dx\wedge dy= -dy\wedge dx$ so the, in particular, such things as "$dx\wedge dx$" will be 0.

 

[Canavar]

Hello,

thank you, but why it is a differential form? We have defined differential form as a smooth section of the projection map.
Therefore i have to show this. But for instance i do not see why it is smooth.

Regards