1. True or false: if ω is a k-form and k is odd, then ω ω = 0. What if k

```1. True or false: if ω is a k-form and k is odd, then ω ∧ ω = 0. What if k is even and k ≥ 2?
P
I claim that this statement is true. Assume that ω = I aI (x)dxI , where I = I1 , ..., In and each Iu
represents a specific iu1 , ..., iuk Then, we have
ω∧ω =
n X
n
X
aIc (x)aId (x)dxIc ∧ dxId
c=1 d=1
=
X
aIc (x)aId (x)dxIc ∧ dxId +
c=d
=
X
X
aIc (x)aId (x)dxIc ∧ dxId
c6=d
aIc (x)aId (x)dxIc ∧ dxId
c6=d
=
X
aIc (x)aId (x)dxIc ∧ dxId +
c&lt;d
=
X
X
aIc (x)aId (x)dxIc ∧ dxId
c&gt;d
aIc (x)aId (x)dxIc ∧ dxId + aIc (x)aId (x)dxId ∧ dxIc
c&lt;d
=
X
aIc (x)aId (x)(dxIc ∧ dxId + dxId ∧ dxIc )
c&lt;d
Claim: dxIc ∧ dxId + dxId ∧ dxIc = 0.
Proof of this claim: Let Ic = i1 , ..., ik and let Id = j1 , ..., jk , where k is odd.
Then dxIc ∧ dxId = dxi1 ∧ dxi2 ∧ ... ∧ dxik ∧ dxj1 ∧ dxj2 ∧ ... ∧ dxjk
2
= (−1)k dxj1 ∧ dxj2 ∧ ... ∧ dxjk ∧ dxi1 ∧ dxi2 ∧ ... ∧ dxik = −dxId ∧ dxIc .
Therefore ω ∧ ω = 0 when k is odd.
The statement is false if k is even and k ≥ 2. Counterexample: let ω = dx1 ∧ dx2 + dx3 ∧ dx4 .
Then
ω ∧ ω = (dx1 ∧ dx2 + dx3 ∧ dx4 ) ∧ (dx1 ∧ dx2 + dx3 ∧ dx4 )
= dx1 ∧ dx2 ∧ dx1 ∧ dx2 + dx1 ∧ dx2 ∧ dx3 ∧ dx4 + dx3 ∧ dx4 ∧ dx1 ∧ dx2 + dx3 ∧ dx4 ∧ dx3 ∧ dx4
= dx1 ∧ dx2 ∧ dx3 ∧ dx4 + dx1 ∧ dx2 ∧ dx3 ∧ dx4
And ω ∧ ω 6= 0.
2. a) Check that for any 2-simplex σ = [p0 , p1 , p2 ], we have ∂(∂σ) = 0.
By definition, ∂σ = [p1 , p2 ] − [p0 , p2 ] + [p0 , p1 ].
Therefore ∂(∂σ) = [p2 ] − [p1 ] − ([p2 ] − [p0 ]) + [p1 ] − [p0 ] = [p2 ] − [p2 ] + [p1 ] − [p1 ] + [p0 ] − [p0 ] = 0.
b) Show that for any k -chain Γ, ∂(∂Γ) = 0.
First I will show that this is true for any k-simplex σ.
If σ = [p0 , p1 , ..., pk ], then by definition
∂σ =
k
X
(−1)i σ−i
i=0
1
Where σ−i = [p0 , p1 , ..., pi−1 , pi+1 , pi+2 , ..., pk ]
Then taking the boundary of this k − 1 chain, where we let {0, 1, 2, ..., i − 1, i + 1, i + 2, ..., k} =
{i0 , i1 , i2 , ..., ii , ..., ik−1 , } and σ−i,−j represents first removing i then j,
∂(∂σ) =
k k−1
X
X
(−1)i (−1)j [pi0 , pi1 , ..., pij−1 , pˆij , pij+1 , ..., pik−1 ]
i=0 j=0
If j &gt; i, ij = j + 1, and if j &lt; i, ij = j. Thus
=
k−1
X
k
X
(−1)i (−1)j−1 [p0 , p1 , ..., pi−1 , pˆi , pi+1 , ..., pj−1 , pˆj , pj+1 , ..., pk ]
i=0 j=i+1
+
k X
i−1
X
(−1)i (−1)j [p0 , p1 , ..., pi−1 , pˆi , pi+1 , ..., pj−1 , pˆj , pj+1 , ..., pk ]
i=1 j=0
=
X
(−1)i (−1)j−1 σi,j +
i&lt;j
X
(−1)i (−1)j σi,j = 0
i&gt;j
Because in the sum, for any a,b such that 0 ≤ a &lt; b ≤ k, since we can either remove pa or pb first, we
(−1)a (−1)b−1 [p0 , p1 , ..., pa−1 , pˆa , pa+1 , ..., pb−1 , pˆb , pb+1 , ..., pk ]
+(−1)a (−1)b [p0 , p1 , ..., pa−1 , pˆa , pa+1 , ..., pb−1 , pˆb , pb+1 , ..., pk ] = 0.
Therefore, ∂(∂)σ = 0 for any k-simplex σ. Therefore, for any k - chain Γ = σ1 + σ2 + &middot; &middot; &middot; + σn ,
∂(∂Γ) = ∂(∂(σ1 + &middot; &middot; &middot; + σn ))
= ∂(∂σ1 + ∂σ2 + &middot; &middot; &middot; + ∂σn )
= ∂(∂σ1 ) + ∂(∂σ2 ) + &middot; &middot; &middot; + ∂(∂σn ) = 0
3. Rudin Chapter 10, # 17,18,24,25,26
17. Put J 2 = τ1 + τ2 , where τ1 = [0, e1 , e1 + e2 ], τ2 = −[0, e2 , e2 + e1 ] . Explain why it is reasonable
to call J 2 the positively oriented unit square in R2 . Show that ∂J 2 is the sum of 4 oriented affine
1-simplexes. Find these. What is ∂(τ1 − τ2 )?
If we draw a diagonal line from (0,0) to (1,1), τ1 is the bottom half of the unit square and τ2 is the top
half of the unit square, by defintion. We want to find the orientation of these simplices. First we note
that τ2 = −[0, e2 , e2 + e1 ] = [0, e2 + e1 , e2 ]
We can write τ1 (u) = 0 + A1 u, τ2 (u) = A2 u as in equation (78) on page 266, where u ∈ R2 . Then
1
A1 =
0
1
1
and A2 =
1
1
0
1
and detA1 = 1 and detA2 = 1. By the definition on page 267 of Rudin, τ1 and τ2 are positively oriented.
Thus J 2 is positively oriented.
2
∂J 2 = ∂(τ1 + τ2 ) = ∂(τ1 ) + ∂(τ2 )
= [e1 , e1 + e2 ] − [0, e1 + e2 ] + [0, e1 ] + [e2 + e1 , e2 ] − [0, e2 ] + [0, e2 + e1 ]
= [e1 , e1 + e2 ] + [0, e1 ] + [e1 + e2 , e2 ] + [e2 , 0]
Thus ∂(J 2 ) is the sum of four oriented affine 1-simplexes.
∂(τ1 − τ2 ) = ∂[0, e1 , e1 + e2 ] + ∂[0, e2 , e1 + e2 ]
= [e1 , e1 + e2 ] − [0, e1 + e2 ] + [0, e1 ] + [e2 , e1 + e2 ] − [0, e1 + e2 ] + [0, e2 ]
= [e1 , e1 + e2 ] + [0, e1 ] + [e2 , e1 + e2 ] + [0, e2 ] − 2[0, e1 + e2 ]
18. Consider the oriented affine 3-simplex
σ1 = [0, e1 , e1 + e2 , e1 + e2 + e3 ]
in R3 . Show that σ1 (regarded as a linear transformation) has determinant 1. Thus σ1 is positively
oriented.
Since the first element (p0 ) of σ1 is the zero vector, the transformation A associated with σ1 must
satisfy for any u ∈ Q3 (the standard 3 simplex) σ1 (u) = Au. Thus

1
A = 0
0

1
1 .
1
1
1
0
A is an upper triangular matrix, so multiplying the elements on the diagonal we get that detA = 1.
Thus σ1 is positively oriented.
P
24. Let ω =
ai (x)dxi be a 1-form of class C 00 in a convex open set E ⊂ Rn . Assume dω = 0 and
prove that ω is exact in E by completing the following outline:
Fix p ∈ E. Define
Z
f (x) =
[p,x]
For affine-oriented 2-simplexes [p, x, y] in E, since p, x, y are in E and E is convex, the line segments
connecting any two of the three points are also contained entirely within E. Therefore if we let
σ = [p, x, y], the boundary of σ is given by ∂σ = [x, y] − [p, y] + [p, x]. Since dω = 0, by Stokes’ theorem
Z
0=
Z
σ
Therefore
Z
dω =
Z
∂σ
Z
ω+
[x,y]
Z
ω−
[p,y]
Z
ω−
ω=
[p,y]
Z
ω=
[p,x]
ω
[p,x]
ω.
[x,y]
Note that the left hand side of this equation is equal to f (y) − f (x), and [x, y] is the line segment from
x to y. We can parametrize this line segment by γ(t) = x + (y − x)t. Since x ∈ Rn , we can rewrite
γ(t) = (x1 + (y1 − x1 )t, x2 + (y2 − x2 )t, ..., (xn + (yn − xn )t)).
3
Then taking the line integral as usual, we get
Z
f (y) − f (x) =
ω
[x,y]
=
n Z
X
=
n Z
X
i=1
n
X
ai (x + (y − x)t)γi0 (t)dt
0
i=1
=
1
1
ai (x + (y − x)t)(yi − xi )dt
0
1
Z
(yi − xi )
ai ((1 − t)x + ty)dt
0
i=1
for x, y ∈ E. Therefore for x ∈ E,
f (x + hei ) − f (x)
h
R1
h 0 ai ((1 − t)x + t(x + hei ))dt
= lim
h→0
h
Z 1
= lim
ai ((1 − t)x + t(x + hei ))
(Di f )(x) = lim
h→0
h→0
Z
=
0
1
ai (x)dt = ai (x)
0
Hence ω is exact.
25. Assume that ω is a 1-form in an open set E ⊂ Rn such that
Z
ω=0
γ
for every closed curve γ in E, of class C 0 . Prove that ω is exact in E.
P
Let ω =
ai (x)dxi . Want to show that there exists f (x)suchthatdf = ω.
Fix p ∈ E. E is open, so ∃δ &gt; 0 such that Nδ (p) ⊂ E. Then for any x, y ∈ Nδ (p), let f (x) =
Let γ be the 1-chain γ = [x, y] − [p, y] + [p, x]. So gamma is a closed curve, and
Z
0=
Z
Z
ω=
γ
Z
ω−
ω=
[x,y]−[p,y]+[p,x]
Therefore we have
[x,y]
[p,x]
ω.
Z
ω+
[p,y]
R
ω
[p,x]
Z
f (y) − f (x) =
ω
[x,y]
when x, y ∈ Nδ (p). By the exact same argument as in 24 (we have the same parametrization of [x, y]),
(Di f )(x) = ai (x) and ω is exact.
26. Assume ω is a 1-form in R3 − {0}, of class C 0 and dω = 0. Prove that ω is exact in R3 − {0}.
Let E = R3 − {0} for simplicity - note that E is open. We know that every closed continuously
differentiable curve in E is the boundary of a 2-surface in E. Let γ be a closed continuously differentiable
curve in E; it is the boundary of a 2-surface S in E. That is, let S be the surface such that ∂S = γ.
Then by Stokes’ theorem,
4
Z
Z
0=
0=
dω =
S
So
Z
S
Z
ω
∂S
Z
ω=
∂S
ω=0
γ
and γ can be any closed curve in E of class C’. By the previous problem, since ω is a 1-form in an open
set E ⊂ Rn such that
Z
ω=0
γ
for any closed curve of class C 0 , ω is exact in E.
5
```