Vector field and 2-form correspondence - is this explanation correct? why a vector field w(x,y,z)=f_1 hat(e)_x+f_2 hat(e)_y+f_3 hat(e)_z can be associated with the 2-form omega_2=f_1 dy ^^ dz+f_2 dz ^^ dx+f_3 dx ^^ dy.

Deon Moran

Deon Moran

Answered question

2022-10-15

why a vector field
w ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z
can be associated with the 2-form
ω 2 = f 1 d y d z + f 2 d z d x + f 3 d x d y .
Is it OK to say this works because the wedge product of the 1-forms ω = u 1 d x + u 2 d y + u 3 d z and ν = v 1 d x + v 2 d y + v 3 d z :
ω ν = ( u 2 v 3 u 3 v 2 ) d y d z + ( u 3 v 1 u 1 v 3 ) d z d x + ( u 1 v 2 u 2 v 1 ) d x d y
has the same components as the cross product of the vectors u = u 1 e ^ x + u 2 e ^ y + u 3 e ^ z and v = v 1 e ^ x + v 2 e ^ y + v 3 e ^ z ?

Answer & Explanation

Momellaxi

Momellaxi

Beginner2022-10-16Added 14 answers

Yes, the wedge product of two vectors in R 3 corresponds to the cross product. But what's going on here is that to a vector field w = ( f 1 , f 2 , f 3 ) you can associate both a 1-form
ω = f 1 d x + f 2 d y + f 3 d z
and a 2-form
η = ω = f 1 d y d z + f 2 d z d x + f 3 d x d y .
If you have a curve γ, then γ ω gives the work done by w along γ. If you have an oriented surface Σ, then Σ η gives the flux of w across Σ
Without any discussions of Stokes's (or the Divergence) Theorem, the reason for the flux interpretation is as follows. Let u,v span an oriented parallelogram P in R 3 , and let n the unit vector normal to the parallelogram. Then
η ( u , v ) = | f 1 f 2 f 3 u 1 u 2 u 3 v 1 v 2 v 3 | = w ( u × v ) = ( w n ) ( n ( u × v ) ) = ( w n ) u × v = ( w n ) area ( P ) .

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