Does there exist a Fano variety with torsion in $H^3$?

Question

Let $X$ be a (smooth) Fano variety over $\mathbb{C}$. If $\dim(X)=3$, inspection of the Iskovskikh-Mori-Mukai lists seems to indicate that $H^3(X,\mathbb{Z})$ is torsion free. Is there a theoretical reason for that? What happens in higher dimension?

Answer 1

There are smooth Fano $5$-folds $\widetilde Y$ with non-zero $2$-torsion in the Brauer group (i.e., $2$-torsion in $H^3(\widetilde Y,\mathbb Z)$); the examples that follow are directly inspired by Beauville's exposition of the Artin--Mumford examples in terms of Reye congruences [Springer LNM $997$, pp. $28-30$].

Start with a general $5$-dimensional linear system $L=\mathbb P^5$ of quadrics in $\mathbb P^3$ (Beauville uses a $3$-dimensional system). The discriminant locus in $L$ is a quartic $4$-fold $S$ defined by the vanishing of a symmetric $4\times 4$ determinant. The singular locus of $S$ is the scheme $C$ that is defined by the vanishing of the $3\times 3$ minors. It is known that $C$ is a smooth surface. By construction, $C$ is the scheme-theoretic intersection of $10$ cubics.

Let $\pi:\widetilde{L}\to L$ denote the blow-up $Bl_CL$, $\widetilde S$ the strict transform of $S$, $H$ the class of a hyperplane in $L$ and $E$ the exceptional divisor. Since $S$ has ordinary double points along $C$, $\widetilde S$ is smooth and $\widetilde S\sim \pi^*S-2E\sim 4\pi^*H-2E$.

Now let $Y\to L$ be the double cover branched along $S$ and $f:\widetilde Y\to\widetilde{L}$ the double cover branched along $\widetilde S$. Note that $Y$ has ordinary double points along the inverse image $C_1$ of $C$ while $\widetilde Y$ is smooth. Note also that $C_1\to C$ is an isomorphism and that $\widetilde Y= Bl_{C_1}Y$. Moreover, $K_{\widetilde Y}\sim f^*(-4H+E)$, from the adjunction formula (or Riemann--Hurwitz).

Lemma: $4H-E$ is ample on $\widetilde{L}$ and $\widetilde Y$ is Fano.

Proof: The linear system $\vert H\vert$ defines $\pi:\widetilde{L}\to L$ and $\vert 3H-E\vert$ defines a morphism $\widetilde{L}\to\mathbb P^9$, from the description of $C$. It is then clear that the induced morphism $$\widetilde{L}\to L\times\mathbb P^9$$ contracts no curves, and the lemma is proved.

Now let $G$ denote the Grassmannian of lines in $\mathbb P^3$ and consider the incidence variety $$\widetilde G=\{(l,q)\vert l\in G,\ q\in\Pi,\ l\subset q\}.$$ Then $pr_1:\widetilde G\to G$ is a Zariski $\mathbb P^2$-bundle, so that $\widetilde G$ is a smooth rational $6$-fold, while $pr_2:\widetilde G\to L$ factors through $Y$. In fact, $\widetilde G\stackrel{r}{\to} Y\to L$ is the Stein factorization of $pr_2$.

Notice that the fibres of $r$ are exactly the connected components of the scheme of lines in a member $q$ of $L$, so that $r^{-1}(y)$ is a smooth curve of genus zero (a conic) whenever $y$ does not lie on $C_1$. That is, $r$ is an etale $\mathbb P^1$-bundle over $Y-C_1$. On the other hand $Z:=r^{-1}(C_1)\to C_1$ is an etale fibre bundle whose fibre is the union of two copies of $\mathbb P^2$ meeting transversely in a single point.

It is then an elementary local calculation to check that the induced morphism $$\tilde r:G^*=Bl_Z\widetilde G\to \widetilde Y=Bl_{C_1}Y$$ is an etale $\mathbb P^1$-bundle. (In terms of etale local co-ordinates, $r$ is the morphism $$\mathbb A^6=Spec\ k[x,y,z,t,w_1,w_2]\to Spec\ k[v_1,v_2,v_3,v_4,w_1,w_2]/(v_1v_4-v_2v_3)$$ given by $v_1=xz,\ v_2=xt,\ v_3=yz,\ v_4=yt$. Then we make a verification chart by chart. Maybe the only slightly surprising thing here is that the blow-up of the smooth variety $\widetilde G$ along the singular subscheme $Z$ should itself be smooth.)

At this point we have a smooth Fano $5$-fold $\widetilde Y$ and an etale $\mathbb P^1$-bundle $\tilde r:G^*\to \widetilde Y$ whose total space is a rational $6$-fold.

Lemma: $\tilde r$ does not have a generic section.

Proof: Note that restricting to a general linear space $H$ of codimension $2$ in $L$ leads exactly to the picture described by Beauville; namely, a smooth rational $4$-fold $G'$ that is an etale $\mathbb P^1$-bundle over a smooth $3$-fold $X$ which is one of the Artin--Mumford examples. It follows that $G'\to X$ has no generic section, so the same holds for $\tilde r$.

Hence $G^*$ represents a non-trivial $2$-torsion element of the Brauer group $Br(\widetilde Y)$.

[Clearly this is cumbersome. It would be much better to find a smooth Fano variety $\widetilde Y$ of dimension at least $4$ and a smooth hyperplane section $X'$ with non-trivial Brauer group. Then weak Lefschetz would show that $\widetilde Y$ had the same property.]

"Is there a theoretical reason?" I'm not sure that there is; it's remarkable that such a precise classification of Fano $3$-folds is possible.