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Past exam of the mathematics course of the University of Cambridge
/
2022
/
iii
/
Paper 115
/
4
/
d
/
Solution
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Past exam of the mathematics course of the University of Cambridge
2022
iii
Paper 115
4
d
2026-09-28
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For
f
=
t
cos
θ
+
1
and
g
=
sin
θ
,
∂
t
f
−
∂
θ
g
=
cos
θ
−
cos
θ
=
0.
(1)
The connection is therefore
flat
, so the
Frobenius theorem
gives
horizontal sections
locally.
A
global section has the form
z
=
h
(
θ
,
t
)
and is horizontal exactly when
∂
θ
h
=
t
cos
θ
+
1
,
∂
t
h
=
sin
θ
.
(2)
The
second
equation
gives
h
=
t
sin
θ
+
k
(
θ
)
, and the
first
then
forces
k
′
(
θ
)
=
1
. No such
k
is periodic on
S
1
, so no global horizontal section exists. Equivalently, the
horizontal lift
of one positive circuit in the
θ
direction changes
z
by
∫
0
2
π
(
t
cos
θ
+
1
)
d
θ
=
2
π
,
(3)
which is nontrivial
holonomy
.
Ancestors
(11)
d
4
Paper 115
iii
2022
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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