484 CHAPTER 18. DIFFERENTIAL FORMS

For a particular l there are two faces in the boundary term for the Stokes formula.Consider the one where the lth component is bl . Recall that J was the set of increasing listsof k−1 indices. ∫

[a,b]l∑I∈J

((α I ◦r)AI

1l)(ul (bl))dul

Here AI1l = (−1)1+l ∂

(xi1 ,··· ,xik−1

)∂ (u1,··· ,ûl ··· ,uk)

and I = (i1, ..., ik−1) . Letting

J∗ (ul) =√

∑I∈J

(AI

1l

)2(ul (bl)),

this term is of the form∫[a,b]l

∑I∈J

((α I ◦r)

AI1l

J∗ (ul)

)(ul (bl))J∗ (ul)dul (18.8)

Define AI1l

J∗(ul)= 0 if J∗ (ul) = 0 on Zl . By Lemma 17.3.1, H k−1 (rl (Zl)) = 0 so the con-

siderations presented here hold off a set of H k−1 measure zero in rl ([a,b]l). Also we canignore the set where the derivative does not exist thanks to Lemma 17.1.2 which says Lip-schitz mapse of sets of measure zero have measure zero. Using the Binet Cauchy theoremto identify J∗ (ul) with

(det(Drbl (ul)

∗Drbl (ul)))1/2

, 18.8 reduces to∫rbl

([a,b]l)#(x)∑

I∈Jα I (x)NI

bl(x)dH k−1 (x)

where NIbl

(rbl (ul)

)=

AI1l

J∗(ul)(ul (bl)) =

1J∗(ul)

(−1)1+l ∂

(xi1 ,··· ,xik−1

)∂ (u1,··· ,ûl ··· ,uk)

is a component of a

unit vector in RC(p,k−1) at least H k−1 a.e. Assume that rbl is one to one or is one to oneoff a set S which has H k−1

(rbl (S)

)= 0. That way we can eliminate #(x) the number of

times x is hit by rbl replacing it with 1. Thus, generalizing the notation, the boundary termin Stokes theorem is of the form

k

∑l=1

∫rbl

([a,b]l)∑I∈J

α I (x)NIbl(x)dH k−1 (x)

−k

∑l=1

∫ral ([a,b]l)

∑I∈J

α I (x)NIal(x)dH k−1 (x)

where ∑I∈J

(NI

bl(x))2

= 1. Letting NI = NIbl

on rbl ([a,b]l) and −NIal

on ral ([a,b]l) , theabove is of the form ∫

r(∂ [a,b])∑I∈J

α I (x)NI (x)dH k−1

Similarly ∫r

dω ≡ ∑I∈J

∫[a,b]

p

∑j=1

∂α I

∂x j(r (u))

∂(x j,xi1 · · ·xik−1

)∂ (u1, · · · ,uk)

du

=∫[a,b]

p

∑j=1

∑I∈J

∂α I

∂x j(r (u))

∂(x j,xi1 · · ·xik−1

)∂ (u1, · · · ,uk)

du