# How to construct Enriques surface from Fermat K3

Let $x_1^4+x_2^4+x_3^4+x_4^4=0 \subset \mathbb{P}^4$ be the Fermat K3 surface. Is it possible to start from some involution on $\mathbb{P}^3$, do blow-ups to get rid of fixed points and then quotient with respect to resulting involution to get an Enrique surface inside the quotient of some blow-up of $\mathbb{P}^3$.

Or

can you give an example of an Enrique surface inside a simply-connected 3-fold?

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## 2 Answers

Let me show that the answer to your last question is yes, by explaining how to construct an Enriques surface inside a simply connected threefold.

Let $\psi \colon V \to \mathbb{P}^3$ be a double covering branched on a smooth quartic surface and let $\tau \in \textrm{Aut}(V)$ be an involution fixing eight points. Set $X=V/ \tau$. Then one proves that $X$ is a rational threefold with eight singular points of type $\frac{1}{2}(1,1,1)$. The anti-canonical divisor $-K_X$ is not Cartier but we have $-K_X \equiv_{Q} H$, where $H$ is an ample divisor on $X$ with $H^3=8$ such that the general element of $|H|$ is a smooth Enriques surface.

Now consider any desingularization $\pi \colon Y \to X$. Since $X$ is rational, $Y$ is a smooth rational variety, hence $\pi_1(Y)=0$. Moreover $X$ has only isolated singularities, so the general element of $|\pi^*H|$ is again a smooth Enriques surface.

Example. Let $V \subset \mathbb{P}(1^3, \; 2)$ defined by the equation $$u^2=x^4+y^4+z^4+t^4,$$ where $x,\; y, \; z,\; t$ have weight $1$ and $u$ has weight $2$, and consider the involution $\tau \colon \mathbb{P}(1^3, \; 2) \to \mathbb{P}(1^3, \;2)$ given by $$\tau(x:y:z:t:u)=(x:y:-z:-t:-u).$$ Then $V$ is naturally a double cover of $\mathbb{P}^3$ branched on a Fermat quartic surface and moreover $V$ is $\tau$-invariant. The restricted involution $\tau|_V$ fixes the eight points of $V$ given by $u=z=t=0$ and $u=x=y=0$.

For further details see I. Cheltsov, Rationality of Enriques-Fano threefold of genus five, Isvestiya Math. 68 (2004) and the references given therein.

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nice, but let me know if there are other simpler constructions. –  Mohammad F. Tehrani Oct 10 '12 at 13:18

In a first version of this answer I claimed incorrectly (see Francesco's answer), that there are no examples inside smooth simply connected threefolds. In fact what I had in mind is the well known fact that an Enriques surface cannot be a hyperplane section of a smooth threefold. I leave here the closing remark, because it is perhaps useful.

Concerning the example of the Fermat quartic, notice that an involution of $\mathbb P^3$ has fixed points on any surface $S\subset \mathbb P^3$ so you cannot get a free involution this way. Blowing up does not improve things, because you replace a fixed point wih a curve all made of fixed points.

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How is the proof of that well-known fact? –  Mohammad F. Tehrani Oct 10 '12 at 12:57
I think you can find a reference in the paper mentioned in Francesco's answer. Anyway, I think the proof should be something like this: the map $Pic(X)\to Pic(S)$ is injective by Lefschetz theorem, since $S$ is ample. Then $S=-K_X$ (up to numerical equivalence) and therefore $X$ is Fano. I think Fano manifolds have free Picard group, but I don't know a proof of this, so you should check the literature. –  rita Oct 10 '12 at 16:42