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Dimension jump in a smooth family of manifolds
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Past exam of the mathematics course of the University of Cambridge
/
2018
/
ii
/
Paper 1
/
26I
/
d
/
Solution
Created
2026-09-24
Updated
2026-10-03
View more
The answer is no. Let
a
i
=
0
tend to zero and take
p
i
=
(
a
i
,
0
,
0
,
0
)
∈
X
a
i
,
p
i
→
p
=
(
0
,
0
,
0
,
0
)
∈
X
0
.
(1)
At
a
point
x
∈
X
a
with
a
=
0
, differentiating the defining
equations
gives
x
1
w
1
+
2
x
2
w
2
=
0
,
w
3
=
a
w
4
.
(2)
Thus
w
i
=
(
0
,
1
,
0
,
0
)
∈
T
p
i
X
a
i
(3)
for every
i
, and
w
i
→
w
=
(
0
,
1
,
0
,
0
)
. But
X
0
is the
x
4
-axis, so
T
p
X
0
=
span
{(
0
,
0
,
0
,
1
)}
,
w
∈
/
T
p
X
0
.
(4)
Hence
T
p
X
a
does not depend continuously on
(
p
,
a
)
at
a
=
0.
(5)
The obstruction is the
dimension jump in a smooth family of manifolds
.
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:
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