A singular 0-cube in Rn is a function c : {0} → Rn and a (piecewise smooth) singular k-cube in Rn is a function
c : [0, 1]k→ Rn
that can be extended to a smooth function φ : V → Rn defined on an open subset V of Rk containing [0, 1]k. (We will use c for φ for simplicity). A singular 1-cube in Rn is called a singular curve in Rn and a singular two cube in Rn is called a singular surface in Rn. The restriction of the identity function I : Rn → Rn to [0, 1]n is a singular n-cube called the standard n-cube in Rn and denoted by In.
Definition 1.1. A (piecewise smooth) singular k-chain in Rn is a (formal) Z-linear com- bination of (piecewise smooth) singular k-cubes in Rn, i.e. a (piecewise smooth) singular k-chain is of the form
n1γ1+ · · · + nsγs
where n1, · · · , ns∈ Z and γ1, · · · , γs: [0, 1]k→ Rn.
A (piecewise smooth) singular k-chain in Rn can be rewritten as c =X
γ
nγγ
where γ runs through the set of all (piecewise smooth) singular k-cubes and nγ= 0 for all but finitely many (piecewise smooth) singular k-cubes γ in Rn. The set of all (piecewise smooth) singular k-chains in Rn is denoted by Ck(Rn).
We define the sum of two (piecewise smooth) singular k-chains c = P
γnγγ and c0 = P
γn0γγ by
c + c0 =X
γ
(nγ+ n0γ)γ.
Then Ck(Rn) forms an abelian group.
Let Ik : [0, 1]k → Rk be the standard k-cubes in Rk. For each 1 ≤ i ≤ k and 0 ≤ a ≤ 1, we defines functions I(i,a)k : [0, 1]k−1 → Rk by
I(i,0)k (t1, · · · , tk−1) = (t1, · · · , ti−1, 0, ti, · · · , tk) I(i,1)k (t1, · · · , tk−1) = (t1, · · · , ti−1, 1, ti, · · · , tk).
We call I(i,a)k the (i, a)-th face of Ik. The (algebraic) boundary of Ik is defined to be
∂kIk=
k
X
i=1 1
X
a=0
(−1)i+aI(i,a)k . Then ∂kIk is a (piecewise smooth) singular k − 1-chain in Rn.
If γ : [0, 1]k → Rn is a (piecewise smooth) singular k-cube in Rn, we define the (i, a)-th face of γ to be
γ(i,a) = γ ◦ I(i,a)k
for any 1 ≤ i ≤ k and for 0 ≤ a ≤ 1. The algebraic boundary of γ is defined to be
∂kγ =
k
X
i=1 1
X
a=0
(−1)i+aγ(i,a).
1
In general, we define the (algebraic) boundary of a (piecewise smooth) singular k-chain c to be the following singular k − 1-chain
∂kc =X
γ
nγ(∂kγ).
Lemma 1.1. The function ∂k : Ck(Rn) → Ck−1(Rn) is an abelian group homomorphism such that
∂k−1◦ ∂k= 0 for any k ≥ 1.
We denote ker ∂k by Zk(Rn) and Im ∂k+1 by Bk(Rn). Elements of Zk(Rn) are called (piecewise smooth) singular k-cycles in Rn and elements of Bk(Rn) are called (piecewise smooth) singular k-boundaries in Rn. Since ∂k−1◦ ∂k = 0, Bk(Rn) is an abelian subgroup of Zk(Rn). We define the k-th (piecewise smooth) singular homology group of Rn to be the quotient group
Hk(Rn) = Zk(Rn)/Bk(Rn).
Remark. One can prove that
Hk(Rn) =
(0 if k > 0 Z if k = 0.
Now let us state the Stoke’s Theorem. For convenience, all the chains and cubes men- tioned below are assumed to be piecewise smooth.
Let ω = f (x)dx1∧ · · · ∧ dxn be any n-form on an open subset U of Rn. We define the integral of ω over a Jordan measurable subset S of Rn contained in U to be
Z
S
ω = Z
S
f (x)dµ whereR
Sf (x)dµ is the Riemann integral of the function f over S.
Remark. It you are not familiar with the notion of Jordan measurable sets in Rn, you take S to be any n-dimensional compact interval S =Qn
i=1[ai, bi].
If ω is a k-form on Rn and γ : [0, 1]k→ Rn is a singular k-cube, we define Z
γ
ω = Z
[0,1]k
γ∗ω.
In general, if c =P
γnγγ is a singular k-chain in Rn, we define the integral of ω over c by Z
c
ω =X
γ
nγ Z
γ
ω.
Theorem 1.1. (Stoke’s Theorem) Let ω be any k − 1 form on Rn and c be any k-chain in
Rn. Then Z
c
dω = Z
∂c
ω.
Let us prove the case when ω = Q(x, y)dy is a one form on R2 and c is any singular two chain in R2. At first, we prove that
Z
∂I2
ω = Z
I2
dω
where I2 : [0, 1] × [0, 1] → R2 is the standard 2-cube. Since I2 is the restriction of the identity function I : R2 → R2,
(I2)∗dω = dω = Qxdx ∧ dy.
By definition.
Z
I2
dω = Z
[0,1]2
(I2)∗dω = Z
[0,1]2
Qx(x, y)dx ∧ dy
= Z Z
[0,1]×[0,1]
Qx(x, y)dA = Z 1
0
Z 1 0
Qx(x, y)dx
dy
= Z 1
0
(Q(1, y) − Q(0, y))dy.
Since ∂I2= I(2,0)2 + I(1,1)2 − I(2,1)2 − I(1,0)2 , Z
∂I2
ω = Z
I(2,0)2
ω + Z
I(1,1)2
ω − Z
I(2,1)2
ω − Z
I2(1,0)
ω.
One can show that
(I(2,0)2 )∗ω = (I(2,1)2 )∗ω = 0, (I(1,1)2 )∗ω = Q(1, t)dt, (I(1,0)2 )∗ω = Q(0, t)dt.
As a consequence, Z
I2
dω = Z 1
0
(I(1,1)2 )∗ω − Z 1
0
(I(1,0)2 )∗ω = Z 1
0
(Q(1, t) − Q(0, t))dt which coincides with R
I2dω. Now let us prove that the statement is true for any singular 2-cube γ in R2. To do this, we need the following Lemma and its Corollary.
Lemma 1.2. Let f : Rn→ Rm and g : Rm→ Rp be any smooth functions. For any r form ω on Rp,
(g ◦ f )∗ω = f∗(g∗ω).
Proof. When ω = h is a zero form,
(g ◦ f )∗h = h ◦ g ◦ f = f∗(h ◦ g) = f∗(g∗h).
Assume that ω is a r form on Rk. For each p ∈ Rn, and each (v1)p, · · · , (vr)p in Tp(Rn), (g ◦ f )∗ω(p)((v1)p, · · · , (vr)p) = ω((g ◦ f )(p))(d(g ◦ f )p(v1)p, · · · , d(g ◦ f )p(vr)p)
= ω((g ◦ f )(p))(dgf (p)(dfp(v1)p), · · · , dgf (p)(dfp(vr)p))
= (g∗ω)(f (p))(dfp((v1)p), · · · , dfp((v1)p))
= f∗(g∗ω)(p)((v1)p, · · · , (vr)p).
Thus (g ◦ f )∗ω(p) = f∗(g∗ω)(p) for any p ∈ Rn. Hence the statement is true. Corollary 1.1. Let γ be any singular k-cube on Rn and ω be any k-form on Rn. Then
Z
γ
f∗ω = Z
f ◦γ
ω for any smooth function f : Rn→ Rn.
Proof. Using the previous lemma, we find
(f ◦ γ)∗ω = γ∗(f∗ω).
By definition,
Z
f ◦γ
ω = Z
[0,1]k
(f ◦ γ)∗ω = Z
[0,1]k
γ∗(f∗ω) = Z
γ
f∗ω.
Let γ be any 2-cube in R2. We consider γ as a smooth function from an open subset of R2 containing [0, 1] × [0, 1] to R2. By definition, the boundary of γ is
∂γ = γ ◦ I(2,0)2 + γ ◦ I(1,1)2 − γ ◦ I(2,1)2 − γ ◦ I(1,0)2 . By Corollary 1.1,
Z
∂γ
ω = Z
I(2,0)2
γ∗ω + Z
I(1,1)2
γ∗ω − Z
I(2,1)2
γ∗ω − Z
I(1,0)2
γ∗ω = Z
∂I2
γ∗ω.
By Stoke’s Theorem for standard 2-cube, Z
∂I2
γ∗ω = Z
I2
d(γ∗ω).
Since d(γ∗ω) = γ∗(dω), we find Z
I2
d(γ∗ω) = Z
I2
γ∗(dω) = Z
γ◦I2
dω = Z
γ
dω.
We find that the statement is true for any singular 2-cube γ in R2. In general, if c =P
γnγγ is a 2-chain, then
Z
∂c
ω =X
γ
nγ Z
∂γ
ω =X
γ
nγ Z
γ
dω = Z
c
dω.
We prove that the statement is true for any 2-cubes in R2 for ω = Q(x, y)dy. When ω = P (x, y)dx, the proof is similar. When ω = P (x, y)dx + Q(x, y)dy, we let ω1 = P (x, y)dx and ω2 = Q(x, y)dy. Using Stoke’s theorem for ω1 and for ω2 respectively, we obtain
Z
∂c
ω = Z
∂c
ω1+ Z
∂c
ω2 = Z
c
dω1+ Z
c
dω2= Z
c
dω.
Here we use the fact that dω = dω1+ dω2. The idea of the above proof can be applied to the proof of Stoke’s Theorem for general cases. Let us prove that
Z
∂Ik
ω = Z
Ik
dω
holds for k − 1 form of the form ω = f (x)dx2∧ · · · ∧ dxk. Then dω = fx1dx1∧ · · · ∧ dxk. By definition and the Fubini’s Theorem,
Z
Ik
dω = Z
[0,1]k
fx1dx1∧ · · · ∧ dxk
= Z
[0,1]k−1
(f (1, x2, · · · , xn) − f (0, x2, · · · , xn))dµk−1.
Here dµk−1 is the Jordan measure on Rk−1. On the other hand, (I(i,a)k )∗ω = 0 for 2 ≤ i ≤ k and
(I(1,0)k )∗ω = f (0, t1, · · · , tk−1)dt1∧ · · · ∧ dtk−1 (I(1,1)k )∗ω = f (1, t1, · · · , tk−1)dt1∧ · · · ∧ dtk−1. By definition, Ik =Pk
i=1
P1
a=0(−1)i+aI(i,a)k , and hence Z
∂Ik
ω =
k
X
i=1 1
X
a=0
(−1)i+a Z
I(i,a)k
ω.
By the previous observation, Z
I(i,a)k
ω = Z
[0,1]k−1
(I(i,a)k )∗ω = 0 for 2 ≤ i ≤ k and
Z
I(1,0)k
ω = Z
[0,1]k−1
(I(1,0)k )∗ω = Z
[0,1]k−1
f (0, t1, · · · , tk−1)dµn−1
Z
I(1,1)k
ω = Z
[0,1]k−1
(I(1,1)k )∗ω = Z
[0,1]k−1
f (1, t1, · · · , tk−1)dµn−1. We see that
Z
∂Ik
ω = Z
[0,1]k−1
(f (1, t1, · · · , tk−1) − f (0, t1, · · · , tk−1))dµn−1 which coincides withR
Ikdω. For the case when ω is a k − 1 form of the form ω = f (x)dx1∧ · · · dxi−1∧ dxi+1∧ · · · ∧ · · · dxk,
the proof is similar. If ω is a k − 1 form of the form ω =
k
X
i=1
fi(x)dx1∧ · · · dxi−1∧ dxi+1∧ · · · ∧ · · · dxk, we write ωi = fi(x)dx1∧ · · · dxi−1∧ dxi+1∧ · · · ∧ · · · dxk. Then
Z
∂Ik
ωi = Z
Ik
dωi. Since dω =Pk
i=1dωi, we find Z
∂Ik
ω =
k
X
i=1
Z
∂Ik
ωi=
k
X
i=1
Z
Ik
dωi= Z
Ik k
X
i=1
dωi
!
= Z
Ik
dω.
If γ : [0, 1]k→ Rn is a k-cube on Rn and ω is a k-form on Rn, then Z
∂γ
ω = Z
∂Ik
γ∗ω = Z
Ik
d(γ∗ω) = Z
Ik
γ∗(dω) = Z
γ
dω.
We prove that the theorem holds for any k-cubes and any k-forms. One can show that the Stoke’s theorem holds for any k-form and for any k-chains on Rn.
