H/wk 10. Due Friday, April 13.
NAME:
1. Let
ω
be an
r
form in a manifold
M
n
such that for some
r
chain
σ
in
M
with
∂σ
= 0 we have
R
σ
ω
6
= 0. Prove that
ω
is not exact.
Solution.
Suppose that
ω
is exact, so that
ω
=
dα
for some (
r

1)form
α
. Then by Stokes’
Theorem
Z
σ
ω
=
Z
σ
dα
=
Z
∂σ
α
=
Z
0
α
= 0
,
contradicting the assumption that
R
σ
ω
6
= 0.
2. Let
T
2
=
{
(
x, y, z, w
)
∈
R
4
:
x
2
+
y
2
=
z
2
+
w
2
= 1
}
.
Consider the following 1forms on
T
2
:
ω
1
:=

y
x
2
+
y
2
dx
+
x
x
2
+
y
2
dy
and
ω
2
:=

w
z
2
+
w
2
dz
+
z
z
2
+
w
2
dw
.
Consider the following 1cubes
γ
1
: [0
,
1]
→
T
2
and
γ
2
: [0
,
1]
→
T
2
in
T
2
:
γ
1
(
t
) = (cos 2
πt,
sin 2
πt,
1
,
0)
,
γ
2
(
t
) = (1
,
0
,
cos 2
πt,
sin 2
πt
)
,
where
t
∈
[0
,
1]
.
(a) Show that
∂γ
1
=
∂γ
2
= 0.
(b) Compute
R
γ
1
ω
1
and
R
γ
2
ω
2
. Conclude that
ω
1
and
ω
2
are not exact on
T
2
.
(c) Consider the chart (
U, φ
= (
θ
1
, θ
2
)) on
T
2
where
φ
(
U
) = (0
,
2
π
)
×
(0
,
2
π
)
and where
φ

1
(
θ
1
, θ
2
) = (cos
θ
1
,
sin
θ
1
,
cos
θ
2
,
sin
θ
2
). Compute
ω
1
and
ω
2
in the
chart (
U, φ
), that is, calculate (
φ

1
)
*
ω
i
for
i
= 1
,
2. Then verify that
dω
1
= 0 and
dω
2
= 0 in this chart hence, by continuity,
dω
1
= 0 and
dω
2
= 0 on
T
2
. Thus
ω
1
and
ω
2
are closed on
T
2
.
(d) Let