Here is the construction.

Start with $U$ a normal variety of dimension 3 with a unique singular point locally isomorphic to the quotient of $(x, y, z) \to (-x, -y, -z)$. Also assume we have a small contraction which contracts a smooth rational curve $C$ through the singular point.

It is known that we can blow up the singular point in $U$ and get a smooth variety $V$ whose exceptional divisor is $P^2$ with normal bundle $\mathcal{O}(-2)$. Now my question is what is the normal bundle of the strict transform of $C$ in $V$?

It is easy to see the intersection number with $K_V$ should be $0$. In the book Geometry of Higer dimensional Algebraic Variety by Miyaoka and Peternell, the author claims it is easy to see the normal bundle is $\mathcal{O}(-1)\oplus\mathcal{O}(-1)$ (p.184 Example 7.10). But I do not know why.


1 Answer 1


Let the strict transform of $C$ on $V$ be denoted by $\widetilde C$. Obviously, $\widetilde C\simeq \mathbb P^1$. Let $\mathcal O_{\widetilde C}(1):=\mathcal O_{\mathbb P^1}(1)$ via this isomorphism. Writing down the short exact sequence corresponding to the restriction of the cotangent bundle of $V$ to $\widetilde C$, $$ 0 \to \mathcal N_{\widetilde C|V} \to \Omega_V\otimes \mathcal O_{\widetilde C} \to \Omega_{\widetilde C} \to 0 $$ combined with (what you already computed) $K_V\cdot \widetilde C=0$ implies that ${\rm det}\ \mathcal N_{\widetilde C|V}\simeq \mathcal O_{\widetilde C}(-2)$ (I guess I should have just said that the "adjunction formula" does that). Since $\widetilde C\simeq \mathbb P^1$, it follows that $\mathcal N_{\widetilde C|V}\simeq \mathcal O_{\widetilde C}(a)\oplus \mathcal O_{\widetilde C}(b)$ for some $a,b \in\mathbb Z$ such that $a+b=-2$. Since $C$ is contractible, both $a$ and $b$ has to be negative, but the only way that can happen is if they are both equal to $-1$.


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