TITLE: Why are K3 surfaces minimal?
QUESTION [11 upvotes]: I need to prove that all K3 surfaces are minimal surfaces, so that every birational map between K3 surfaces is an isomorphism. I've started to read Beauville's book on complex algebraic surfaces: there it says that the fact that K3 surfaces are minimal comes from the definition ($K=0$ and $H^{1,0}(X)=0$), but I can't see why. Do you know the proof or where I can find it?

REPLY [2 votes]: Here's a topological proof. Note, I'm sure this is not the argument Beauville had in mind.
Let $M$ be a four-dimensional closed, oriented spin manifold (so $w_1(M) = w_2(M) = 0$), then $M$ has even intersection form. This follows from Wu's formula which states that on a closed four-manifold, $x \cdot x = \nu_2\cdot x$ for all $x \in H^2(M; \mathbb{Z}_2)$ where $\nu_2 = w_1(M)^2 + w_2(M)$ is the second Wu class. In particular, there is no element $y \in H^2(M; \mathbb{Z})$ with $y\cdot y = -1$.
Therefore, we have the following:

If $M$ is a compact complex surface which is spin, then $M$ is minimal.

If $M$ is a $K3$ surface, then $c_1(M) = -c_1(K) = 0$, so $w_2(M) = 0$. As $K3$ surfaces are spin, they are minimal.
Note that the converse of the above statement is not true; there are minimal compact complex surfaces which are not spin, e.g. $\mathbb{CP}^2$. There are even examples of minimal compact complex surfaces $M$ with an element $y \in H^2(M; \mathbb{Z})$ such that $y\cdot y = -1$. For example, the Hirzebruch surfaces $\Sigma_{2n+1}$ for $n > 0$ are minimal surfaces which are diffeomorphic to (but not biholomorphic to) the blow-up of $\mathbb{CP}^2$ at a point, namely $\mathbb{CP}^2\#\overline{\mathbb{CP}^2} = \Sigma_1$.