proof of general Stokes theorem

We divide the proof in several steps.

Step One.

Suppose M=(0,1]×(0,1)n-1 and


(i.e. the term dxj is missing). Hence we have

dω(x1,,xn) = (fx1dx1++fxndxn)dx1dxj^dxn
= (-1)j-1fxjdx1dxn

and from the definition of integral on a manifold we get


From the fundamental theorem of CalculusMathworldPlanetmathPlanetmath we get


Since ω and hence f have compact support in M we obtain


On the other hand we notice that Mω is to be understood as Mi*ω where i:MM is the inclusion mapMathworldPlanetmath. Hence it is trivial to verify that when j1 then i*ω=0 while if j=1 it holds


and hence, as wanted


Step Two.

Suppose now that M=(0,1]×(0,1)n-1 and let ω be any differential formMathworldPlanetmath. We can always write


and by the additivity of the integral we can reduce ourself to the previous case.

Step Three.

When M=(0,1)n we could follow the proof as in the first case and end up with M𝑑ω=0 while, in fact, M=.

Step Four.

Consider now the general case.

First of all we consider an oriented atlas (Ui,ϕi) such that either Ui is the cube (0,1]×(0,1)n-1 or Ui=(0,1)n. This is always possible. In fact given any open set U in [0,+)×n-1 and a point xU up to translations and rescaling it is possible to find a “cubic” neighbourhood of x contained in U.

Then consider a partition of unity αi for this atlas.

From the properties of the integral on manifolds we have

M𝑑ω = iUiαiϕ*𝑑ω=iUiαid(ϕ*ω)
= iUid(αiϕ*ω)-iUi(dαi)(ϕ*ω).

The second integral in the last equality is zero since idαi=diαi=0, while applying the previous steps to the first integral we have


On the other hand, being (Ui,ϕ|Ui) an oriented atlas for M and being αi|Ui a partition of unity, we have


and the theorem is proved.

Title proof of general Stokes theorem
Canonical name ProofOfGeneralStokesTheorem
Date of creation 2013-03-22 13:41:43
Last modified on 2013-03-22 13:41:43
Owner paolini (1187)
Last modified by paolini (1187)
Numerical id 9
Author paolini (1187)
Entry type Proof
Classification msc 58C35