Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 26I / d

Table of contents
- Solution d

Solution

The answer is no. Let

a_{i} \neq = 0

tend to zero and take

p_{i} = (a_{i}, 0, 0, 0) \in X_{a_{i}}, p_{i} \to p = (0, 0, 0, 0) \in X_{0} .

At a point

x \in X_{a}

with

a \neq = 0

, differentiating the defining equations gives

x_{1} w_{1} + 2 x_{2} w_{2} = 0, w_{3} = a w_{4} .

Thus

w_{i} = (0, 1, 0, 0) \in T_{p_{i}} X_{a_{i}}

for every

i

, and

w_{i} \to w = (0, 1, 0, 0)

. But

X_{0}

is the

x_{4}

-axis, so

T_{p} X_{0} = span {(0, 0, 0, 1)}, w \in / T_{p} X_{0} .

Hence

T_{p} X_{a} does not depend continuously on (p, a) at a = 0.

The obstruction is the dimension jump in a smooth family of manifolds.

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