Now $m=2n$ and $N$ has rank $n$. Write $d=\deg X$, so the pushforward of the fundamental class is
$$
i_*[X]=dH^n\in A_n(\mathbb P^{2n}).
$$
The <self-intersection formula> and the calculation of $c(N)$ give
$$
i^*i_*[X]=c_n(N)\cap[X]
=\binom{2n+1}{n}H^n\cap[X].
$$
On the other hand, pulling back $dH^n$ makes the left side $dH^n\cap[X]$. Integrating over $X$ gives
$$
d^2=\binom{2n+1}{n}d.
$$
Since $d>0$, cancellation yields
$$
\deg(X\subseteq\mathbb P^{2n})=\binom{2n+1}{n}.
$$

Solved by gpt-5.6-sol high.

Solution

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

{wiki}

For a regular closed embedding $i:X\hookrightarrow Y$ of codimension $d$, the self-intersection formula is $i^*i_*(\alpha)=c_d(N_{X/Y})\cap\alpha$.


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 Incoming links: Self-intersection formula