Kenneth Kuttler

18.5. WHAT DOES IT MEAN? 485

Now that determinant is only nonzero if j is none of the is. By the Binet Cauchy theorem,

det(Dr (u)∗Dr (u)

)= J∗ (u)

2 = ∑xi1<···<xik

(∂(xi1 · · · ,xik

)∂ (u1, · · · ,uk)

and so it follows from the area formula that there exists NIj for each I an increasing list of

k−1 indices such that for J all such increasing lists, ∑pj=1 ∑I∈J

(NI

)2= 1 and

∫r

dω =∫[a,b]

∑j=1

∑I∈J

∂α I

∂x j(r (u))

∂

(x j ,xi1 ···xik−1

)∂ (u1,··· ,uk)

J∗ (u)J∗ (u)du (18.9)

=∫r([a,b])

#(x)∑I∈J

∑j=1

∂α I

∂x j(x)NI

j (x)dH k

where #(x) is the number of times x is hit by r. Thus if r is one to one off Ŝ where r(Ŝ)

has H k measure zero, it follows that we can remove #(x) from the formula. As before,we can ignore the set where J∗ (u) = 0 thanks to Lemma 17.3.1. Also we can ignore theset of u where the function is not differentiable by Lemma 17.1.2.

Observation 18.5.1 Stokes theorem may be thought of as a statement about r ([a,b])and r (∂ [a,b]) which involves geometrical concepts dependent on these sets. This holdswhenever Lipschitz r restricted to each face of [a,b] is one to one off a set S where r (S)has H k−1 measure zero, and r is one to one off Ŝ where r

(Ŝ)

has H k measure zero.

I think that the most important case is where k = p and in this case we have the diver-gence theorem. Here there are exactly p−1 increasing lists of indices I and these are of theform

(1, · · · , ĵ, · · · , p

). We let α I be denoted as a j (−1) j+1 where j is the index missing

in I. Therefore, the formula for∫r dω reduces to

±∫r([a,b])

∑j=1

∂a j

∂x j(x)dH k

assuming that r is one to one off a set S for which r (S) has H p measure zero. This isbecause from 18.9

∂α I

∂x j(r (u))

∂

(x j ,xi1 ···xik−1

)∂ (u1,··· ,uk)

J∗ (u)J∗ (u) =

((−1) j+1

)2ai (r (u))

∂(x1,··· ,x j ··· ,xp)∂ (u1,··· ,uk)

J∗ (u)=±1

The boundary terms reduce to

∑l=1

∫rbl

([a,b]l)

∑j=1

a j (x)N j (x)dH k−1 (x)−k

∑l=1

∫ral ([a,b]l)

∑j=1

a j (x)N j (x)dH k−1 (x)

where for x= r (u) ,N j (x) = (−1) j+l ∂(x1,··· ,x̂ j ··· ,xp)∂(u1,··· ,ûl ,··· ,up)

. This can be written in the form

±∫r([a,b])

∑j=1

∂a j

∂x j(x)dH k =

∫r(∂ [a,b])

∑j=1

a j (x)N j (x)dH k−1 (x)

18.5. WHAT DOES IT MEAN? 485Now that determinant is only nonzero if j is none of the i;. By the Binet Cauchy theorem,oe Uk)0 (x; _)\?det (Dr (u)* Dr (u)) = J (u)? = y (Si)jp <i <Xiy U1,"and so it follows from the area formula that there exists Ni for each J an increasing list of2k —1 indices such that for J all such increasing lists, Li Yes (v1) = | anddo = / COT Cp (qy)) —2 (an) du (18.9)| yey ay OD) ta=f ROEY SA wwny(ayaretr((a, 2) IéJ j=1where #(a:) is the number of times & is hit by r. Thus if r is one to one off § where r (S)has #* measure zero, it follows that we can remove #(a) from the formula. As before,we can ignore the set where J, (w) = 0 thanks to Lemma 17.3.1. Also we can ignore theset of wu where the function is not differentiable by Lemma 17.1.2.Observation 18.5.1 Stokes theorem may be thought of as a statement about r ([a, b])and r (0 [a,b]) which involves geometrical concepts dependent on these sets. This holdswhenever Lipschitz r restricted to each face of [a,b] is one to one off a set S where r (S)has #2*—| measure zero, and r is one to one off 8 where r (S) has #7* measure zero.I think that the most important case is where k = p and in this case we have the diver-gence theorem. Here there are exactly p — 1 increasing lists of indices J and these are of theform (1, oo ah ooo 7) . We let a be denoted as a; (-1)/"! where / is the index missingin J. Therefore, the formula for [,,d@ reduces toy 24+ (x) dC"[ ((a,b)) f » ox, j (®)assuming that r is one to one off a set S for which r(S) has #? measure zero. This isbecause from 18.9day (emia) 2 2G a0)Fu, (0 (4) Ede) = (19) ai (r (u)) SE =The boundary terms reduce tokk1 le) ni leva vel aLT ass Yaj(e w)d Ie yl (as) 4" a; (a) N! (x) d 7" (a)where for a = r(u),N/ (a) = (—1)/! a This can be written in the formy 241 ( k k-1= Loom Y oe (watt J Lae a) dH (a)