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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?

Arctic Char
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ciccio
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  • Does minimal in this context just mean that there are no $(-1)$-curves? Or is there more in the definition? – Matt Jan 23 '13 at 03:49
  • I think you mean to say "every birational morphism from a K3 to any other surface is an isomorphism." – Andrew Jan 23 '13 at 03:57
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    @Andrew: no, it's really correct to say that every birational map is an isomorphism. –  Feb 04 '14 at 09:58

2 Answers2

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What Beauville says is that "[c]learly, $K\equiv 0$ implies that they are minimal."

A surface $S$ is minimal if any birational morphism $S\to S'$ to any other surface $S'$ is an isomorphism. Thus, suppose that some K3, say $S,$ is not minimal. This means, by definition, that there exists a birational morphism $S\to S'$ which is not an isomorphism. By Theorem II.11 of Beauville, such a morphism can be factored as a (finite, nonzero) sequence of blowups at a point, and by Proposition II.3(iv), $K_S$ is a nonzero effective linear combination of pullbacks of the exceptional divisors of the successive blowups plus the strict transform of $K_{S'}$. Since $\operatorname{Pic}(S)=\operatorname{Pic}(S')\oplus \mathbb Z^m,$ where $m$ is the number of blowups, this gives $K_S\neq 0$ (i.e., we cannot cancel by linear equivalence) thus the contradiction, hence the K3 must be minimal.

Andrew
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  • thank you! let's suppose the case of a birational morphism $S\rightarrow S'$ between K3 surfaces. then, stating that $S$ is not minimal, $K_S$ must be a linear combination of $E_1,\cdots E_k$, where $E_i$ is the exceptional curve of the i-th blow up. you mean that this linear combination cannot be equivalent to 0 because $Pic(S)=Pic(S')\oplus \mathbb{Z}^m$,and so this is absurd because by definition $K_S=0$, right? – ciccio Jan 23 '13 at 14:59
  • Dear @ciccio, yes that's exactly what I meant. You may have noticed that I made a small edit to my answer, because I realised I do not want/need to assume that $S'$ is also K3. – Andrew Jan 23 '13 at 16:40
  • I think it is correct to say that every birational map between K3 surfaces is an isomorphism (you prove it thanks to Castelnuovo's rationality criterion). – idioteca Feb 02 '14 at 14:51
  • @idioteca: how would you use Castelnuovo's criterion to prove that? –  Feb 04 '14 at 09:59
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    I just followed Beauville's book "Complex algebraic surfaces", Theorem V.19. The proof is long and it involves also Albanese map, but if you want i can write it. Basically it uses the fact that a birational map can be written as composition of blowing up and downs. Infact if $\phi : S\dashrightarrow S^\prime$ is a birational map you have this commutative diagram – idioteca Feb 04 '14 at 10:09
  • Infact if $\phi : S^\prime\dashrightarrow S$ is a birational map between minimal not ruled surfaces you have $\phi=f\circ g^{-1}$, where both $g : \hat{S}\rightarrow S^\prime$ and $f : \hat{S}\rightarrow S$ are morphism, so compositions of blow-ups. Let's put $g=\epsilon_1\circ\dots\circ\epsilon_n$. Among all the diagrams of this type, you choose the one with $n$ minimal. Your aim is to prove that $n=0$, so $S^\prime$ is isomorphic to $S$. – idioteca Feb 04 '14 at 10:26
  • Considering curves on $S$ and the canonical divisor $K_S$, thanks to castelnuovo theorem and noether-enriques theorem you prove that if $n>0$, you always get $S$ rational or ruled. This is a contradiction since you had supposed $S$ to be not ruled. – idioteca Feb 04 '14 at 10:30
  • @ ciccio - Your question is a duplicate. Minimal means the absence of -1 curves. For a proof, why a K3-surface does not have a -1 curve, see http://math.stackexchange.com/questions/348375/k3-surface-is-not-the-blow-up-of-any-other-smooth-complex-surface/937023#937023 – Jo Wehler Jan 14 '15 at 21:17
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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 $\mathbb{CP}^2\#\overline{\mathbb{CP}^2} = \Sigma_1$.

Michael Albanese
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