For the discussion of holomorphic tubular neighborhoods and some criteria for their existence see this question.

Let $X$ be a smooth quasi-projective variety over $\mathbb{C}$. Hironaka tells us that there exists a smooth projective compactification $X \subset \bar{X}$, such that $D =\bar{X}\backslash X$ is a strictly normal crossing divisor. Let $D=\cup_{i}^N D_i$ be its decomposition into irreducible smooth divisors. What is the obstruction to finding such $\bar{X}$, such that each $D_i$ has a holomorphic tubular neighborhood in $\bar{X}$?

Example 1: Let $S$ be a smooth projective variety, $E$ a vector bundle on $S$. The space $X=\text{Tot}(E)$ is compactified to $\bar{X}=\mathbb{P}_S(E\oplus \mathcal{O}_S)$ with the divisor at infinity $D =\mathbb{P}_S(E)\subset\bar{X}$. Its normal bundle should be $\mathcal{O}_{\mathbb{P}(E)}(1)$ and one should be able to find a holomorphic tubular neighborhood by constructing one in each fiber and gluing together.

Example 2: Using the answer by Joey, I have convinced myself that if the divisor $D_i$ is $\mathbb{P}^n$ for some $n\geq 2$ and $n+1=\text{dim}(X)$, then it also has such a neighborhood. If $n=1$, then one needs additionally that $(D_i)^2<0$.

The above examples make me believe that it could be always achievable, as each blow up at a smooth point introduces a new $\mathbb{P}^{n}$ with normal bundle $\mathcal{O}(-1)$.In particular, blowing up seems to only help with the existence of these neighborhoods.

Edit: What I am asking is: Does there exists such a procedure of blowing up the divisor at infinity, that leaves me at the end with $D_i$ having holomorphic tubular neighborhoods? I don't expect to understand all possible compactifications.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.