The Stacks project

110.70 The stack of proper algebraic spaces is not algebraic

In Quot, Section 99.13 we introduced and studied the stack in groupoids

the stack whose category of sections over a scheme $S$ is the category of flat, proper, finitely presented algebraic spaces over $S$. We proved that this satisfies many of Artin's axioms. In this section we why this stack is not algebraic by showing that formal effectiveness fails in general.

The canonical example uses that the universal deformation space of an abelian variety of dimension $g$ has $g^2$ formal parameters whereas any effective formal deformation can be defined over a complete local ring of dimension $\leq g(g + 1)/2$. Our example will be constructed by writing down a suitable non-effective deformation of a K3 surface. We will only sketch the argument and not give all the details.

Let $k = \mathbf{C}$ be the field of complex numbers. Let $X \subset \mathbf{P}^3_ k$ be a smooth degree $4$ surface over $k$. We have $\omega _ X \cong \Omega ^2_{X/k} \cong \mathcal{O}_ X$. Finally, we have $\dim _ k H^0(X, T_{X/k}) = 0$, $\dim _ k H^1(X, T_{X/k}) = 20$, and $\dim _ k H^2(X, T_{X/k}) = 0$. Since $L_{X/k} = \Omega _{X/k}$ because $X$ is smooth over $k$, and since $\mathop{\mathrm{Ext}}\nolimits ^ i_{\mathcal{O}_ X}(\Omega _{X/k}, \mathcal{O}_ X) = H^ i(X, T_{X/k})$, and because we have Cotangent, Lemma 92.23.1 we find that there is a universal deformation of $X$ over

Suppose that this universal deformation is effective (as in Artin's Axioms, Section 98.9). Then we would get a flat, proper morphism

where $Y$ is an algebraic space recovering the universal deformation. This is impossible for the following reason. Since $Y$ is separated we can find an affine open subscheme $V \subset Y$. Since the special fibre $X$ of $Y$ is smooth, we see that $f$ is smooth. Hence $Y$ is regular being smooth over regular and it follows that the complement $D$ of $V$ in $Y$ is an effective Cartier divisor. Then $\mathcal{O}_ Y(D)$ is a nontrivial element of $\mathop{\mathrm{Pic}}\nolimits (Y)$ (to prove this you show that the complement of a nonempty affine open in a proper smooth algebraic space over a field is always a nontrivial in the Picard group and you apply this to the generic fibre of $f$). Finally, to get a contradiction, we show that $\mathop{\mathrm{Pic}}\nolimits (Y) = 0$. Namely, the map $\mathop{\mathrm{Pic}}\nolimits (Y) \to \mathop{\mathrm{Pic}}\nolimits (X)$ is injective, because $H^1(X, \mathcal{O}_ X) = 0$ (hence all deformations of $\mathcal{O}_ X$ to $Y \times \mathop{\mathrm{Spec}}(k[[x_ i]]/\mathfrak m^ n)$ are trivial) and Grothendieck's existence theorem (which says that coherent modules giving rise to the same sheaves on thickenings are isomorphic). If $X$ is general enough, then $\mathop{\mathrm{Pic}}\nolimits (X) = \mathbf{Z}$ generated by $\mathcal{O}_ X(1)$. Hence it suffices to show that $\mathcal{O}_ X(n)$, $n > 0$ does not deform to the first order neighbourhood¹. Consider the cup-product

This is a nondegenerate pairing by coherent duality. A computation shows that the Chern class $c_1(\mathcal{O}_ X(n)) \in H^1(X, \Omega _{X/k})$ in Hodge cohomology is nonzero. Hence there is a first order deformation whose cup product with $c_1(\mathcal{O}_ X(n))$ is nonzero. Then finally, one shows this cup product is the obstruction class to lifting.