What is the relationship between k-forms and l-forms on an m-manifold?

  • Level: Graduate 
  • Thread starter Thread starter blendecho
  • Start date Start date
  • Tags Tags
    Manifold Relationship
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 3K views
blendecho
Messages
5
Reaction score
0
So I was wondering about this... if [tex]\omega[/tex] is a [tex]k[/tex]-form and [tex]\eta[/tex] is a [tex]l[/tex]-form, and [tex]m[/tex] is a [tex]k+l+1[/tex] manifold in [tex]\mathbb{R}^n[/tex], what's the relationship between [tex]\int_M \omega\wedge d\eta[/tex] and [tex]\int_M d\omega\wedge \eta[/tex]
given the usual niceness of things being defined where they should be, etc. etc. The manifold has no boundary, so am I correct in writing [tex]\int_{\partial M}\omega\wedge\eta=0[/tex]?
 
Last edited:
Physics news on Phys.org
blendecho said:
So I was wondering about this... if [tex]\omega[/tex] is a [tex]k[/tex]-form and [tex]\eta[/tex] is a [tex]l[/tex]-form, and [tex]m[/tex] is a [tex]k+l+1[/tex] manifold in [tex]\mathbb{R}^n[/tex], what's the relationship between [tex]\int_M \omega\wedge d\eta[/tex] and [tex]\int_M d\omega\wedge \eta[/tex]
given the usual niceness of things being defined where they should be, etc. etc. The manifold has no boundary, so am I correct in writing [tex]\int_{\partial M}\omega\wedge\eta=0[/tex]?

I think it's basically integration by parts. You start with the identity
[tex]\textrm{d}(\omega\wedge \eta) = \textrm{d}\omega\wedge \eta+(-1)^{k}\omega\wedge\textrm{d}\eta[/tex]. Then you integrate both sides over M, taking into account that [tex]\int_M \textrm{d}(...)=0[/tex] since [tex]\partial M=0[/tex].