Let $N=N_{X/\mathbb P^m}$ be the <normal bundle>, of rank $m-n$. Since $TX$ is trivial, its exact sequence and the Euler-sequence calculation give
$$
c(N)=c(T\mathbb P^m|_X)=(1+H)^{m+1}.
$$
If $m<2n$, put $i=m-n+1$, so $1\leq i\leq n$. Since $i$ exceeds the rank of $N$, $c_i(N)=0$, yet the formula gives
$$
0=c_i(N)=\binom{m+1}{i}H^i.
$$
Intersecting with $H^{n-i}$ and integrating over $X$ gives
$$
0=\binom{m+1}{i}\deg X,
$$
contradicting positivity of the degree. Hence $m\geq2n$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 162 / 3 / iii / Solution

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For a smooth closed embedding $i:X\hookrightarrow Y$, the normal bundle fits into $0\to TX\to i^*TY\to N_{X/Y}\to0$.


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