MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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 $-1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.