4
$\begingroup$

Let $S$ be a complex algebraic (smooth) surface and $\widetilde{S}$ be the blowup of $S$ at a point $p\in S$.

I would like to understand the statement:

As a topological manifold, $\widetilde{S}$ is a connected sum of $S$ and the complex projective plane $\Bbb{P}^2$ with reversed orientation.

Even some sketch of an explanation or a reference would be greatly appreciated.

Is this the most useful description of $\widetilde{S}$ if one wants to show $H^2(\widetilde{S},\Bbb{Z}) = H^2(S,\Bbb{Z})\oplus \Bbb{Z}$ ?

$\endgroup$
1
  • 2
    $\begingroup$ For the first question, you'll find a detailed sketch of proof here. For the second, no, there are much easier ways to compute the cohomology of $\widetilde{S}$ -- see e.g. Griffiths-Harris. $\endgroup$
    – abx
    Commented Mar 18, 2018 at 11:19

2 Answers 2

9
$\begingroup$

Here's a proof of the statement in Michael's answer (repeated below) that conceptually has more steps but hopefully more illuminating than the computation found in Huybrechts - we hope to make clear why the $\mathbb{P}^n$ should have the opposite orientation.

Proposition: Let $x\in X$ be a point in a complex manifold $X$. Then the blow-up $\text{Bl}_x(X)$ is diffeomorphic as an oriented differentiable manifold to $X \# \overline{\mathbb{P}^n}$.

Proof: We work locally, assuming $X$ is $\mathbb{C}^n$.

Claim 1: The blow-up at the origin $\text{Bl}_0(\mathbb{C}^n)$ is biholomorphic to the total space $E$ of the tautological bundle $\mathcal{O}(-1)$ over $\mathbb{P}^{n-1}$. The exceptional fibre of the blow-up is the zero section of the bundle.

Claim 2: For a complex line bundle $L$ over some smooth manifold, $L^* \cong \overline{L}$ as complex bundles, where $L^*$ is the dual bundle, while $\overline{L}$ has the same underlying real bundle as $L$ but with the negative complex structure, i.e. the conjugate bundle. (Follows by choosing some fibrewise Hermitian form)

Claim 3: $\mathbb{P}^n \backslash \{x\}$ is a line bundle over $\mathbb{P}^{n-1}$ by projecting from the point $x$, and is isomorphic as complex line bundles to $\mathcal{O}(1)$.

Proof of claims are left to the reader.

Now let us assemble the pieces. By claim 1, we have orientation-preserving (in fact holomorphic) inclusions $\mathbb{C}^n \hookleftarrow \mathbb{C}^n \backslash \{0\} \cong E \backslash \{\text{zero section}\} \hookrightarrow E$. Let $F$ be the total space of $\mathcal{O}(1)$. Then by claim 2, in particular, $E \cong \overline{F}$ as differential manifolds, and sends the zero section to the zero section. By claim 3, the one-point compactification of $F$ is $\mathbb{P}^n$. All of this amounts to saying that (real) rays going into the origin of $\mathbb{C}^n$ get magically transformed (under the identification in claim 1) into rays going out of $x$ in $\mathbb{P}^n$. This allows us to say that the blow-up $\text{Bl}_0(\mathbb{C}^n)$ is the connect sum of $\mathbb{C}^n$ with $\overline{\mathbb{P}^n}$.

$\endgroup$
3
$\begingroup$

A much stronger statement is true. The following is from Huybrechts' Complex Geometry (note however that the emphasis is mine):

Let $x \in X$ be a point in a complex manifold $X$. Then the blow-up $\operatorname{Bl}_x(X)$ is diffeomorphic as an oriented differentiable manifold to $X\#\overline{\mathbb{P}^n}$.

Here $n = \dim X$.

This is Proposition $2.5.8$ which can be found on page $102$, together with a proof.

$\endgroup$
2
  • $\begingroup$ Is there a similar statement for the blow up of a submanifold $Y \subseteq X$? $\endgroup$ Commented Jun 11, 2018 at 14:47
  • $\begingroup$ @unknownymous: See this question. $\endgroup$ Commented Jun 11, 2018 at 16:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .