11.15 Deforming to stable ample bundlesE22565883FFD

Proof. At a point of $X\times \{ t_0\} $, lift a local basis of the fiber to $\mathcal E$. By Nakayama's lemma, the resulting map from a free module of finite rank is surjective near that point. Tensor its kernel sequence with the residue field of $t_0$. Flatness over $T$ makes the first Tor term zero, so the kernel restricts to the kernel of the chosen isomorphism on the fiber, namely zero, and Nakayama's lemma shows that the kernel vanishes near that point. Thus $\mathcal E$ is locally free near every point of the central fiber. The locus where $\mathcal E$ is not locally free is closed. Since the projection $X\times T\to T$ is proper, the image of that locus is closed and misses $t_0$, and we remove this image. $\square$

Proof. Put $Y=\mathbb P_{X\times T}(\mathcal E^*)$, with the line convention, and let $p:Y\to T$ be the projection, a projective flat morphism. Write $\mathcal L=\mathcal O_ Y(1)$. If $\mathcal E_{t_0}$ is ample, then $\mathcal L_{t_0}$ is ample. Choose $k$ so that its $k$th power is very ample and has vanishing higher cohomology. By Serre vanishing, semicontinuity, and cohomology and base change, after shrinking $T$ around $t_0$ the sheaf $\mathcal V=p_*\mathcal L^ k$ is locally free and commutes with restriction to fibers. After a further shrinking, the evaluation map $p^*\mathcal V\to \mathcal L^ k$ is surjective: its cokernel vanishes on the central fiber, and the image of its support under the proper map $p$ is closed and misses $t_0$.

Evaluation gives a proper morphism

whose restriction to the central fiber is a closed immersion. The locus where $g$ is quasi-finite is open; its complement has closed image in $T$ by properness and misses $t_0$, so after a further shrinking $g$ is proper and quasi-finite, hence finite. For a finite morphism, pushforward commutes with base change. The cokernel of

therefore vanishes on the central fiber, and its support projects to a closed subset of $T$ missing $t_0$. After removing that subset, the map is surjective, so $g$ is a closed immersion. Consequently $\mathcal L^ k$ is very ample on every fiber, and $\mathcal E_ t$ is ample there. $\square$

Proof. Fix a stable fiber at $t_0$ and put $\nu =c_1(\mathcal E_ t)\cdot P/2$, which is independent of $t$. Choose $k\geq 1$ such that $\mathcal E_{t_0}^*(kP)$ is globally generated and has vanishing higher cohomology. The argument with Serre vanishing, base change and evaluation used in the previous proof gives, over a neighborhood of $t_0$, maps

that are injective on every fiber. By the double-dual argument for rank-one subsheaves in the proof of proposition 11.14.1, a destabilizing subsheaf of $\mathcal E_ t$ yields a nonzero map from a line bundle $M=\mathcal O_ X(aS+bf)$ with $\deg _ P M\geq \nu $. At least one component of its composite with the injection above is nonzero, and since effective classes have nonnegative coefficients,

Since $\deg _ P M=(8e+3)a+4b$, the inequality $\deg _ PM\geq \nu $ also gives

Only finitely many integral pairs $(a,b)$ satisfy these four bounds. For each such pair with $\deg _ P M\geq \nu $, semicontinuity makes the set

closed, and it does not contain $t_0$. Conversely, a nonzero map from such an $M$ is injective as a map of sheaves and violates stability. Thus the set of parameters with unstable fibers is, on this neighborhood, exactly the finite union of these closed subsets, and removing the union proves the claim. $\square$

Proof. The proper algebraic closed subset $T\setminus U$ has empty interior in the analytic topology. In an analytic coordinate ball $B$ around $t_0$, choose $t_ j\in U\cap B$ with $t_ j\to t_0$. Choose a smooth complex connection on $\mathcal E|_{X\times B}$. Parallel transport along radial paths in $B$, keeping $x\in X$ fixed, gives identifications

depending smoothly on $(x,t)$, where the superscript $\mathrm{sm}$ denotes the underlying smooth complex bundle. These identifications need not be holomorphic. Transport the Dolbeault operators by these maps and write

Since both operators satisfy the same Leibniz rule, $A_ t$ is an endomorphism-valued $(0,1)$-form. Its coefficients are smooth in $(x,t)$, and $A_{t_0}=0$. Fix background metrics and a smooth connection on the central bundle. Compactness of $X$ and continuity of every finite collection of derivatives give

The dual identifications are $(\Phi _ t^{-1})^*$. In transported dual frames the perturbation is $-A_ t^{\mathsf T}$, so the same convergence holds for the dual Dolbeault operators. $\square$

Proof. Let $\mathcal P$ be the prioritary stack with invariants $(2,4H,13H^2)$. It contains $E_0$ by proposition 11.14.2, and Walter's theorem makes it smooth and irreducible. Let $\mathcal A$ and $\mathcal S$ be the loci of ample and of $\mu _ P$-stable vector bundles in $\mathcal P$. They are open by the three preceding family lemmas, which apply on smooth local charts of $\mathcal P$; the first lemma allows restriction to locally free families near a point given by a vector bundle. The ample bundle of proposition 11.13.2 is prioritary by lemma 11.13.3, so $\mathcal A$ is nonempty. A nonsplit extension is stable and prioritary by proposition 11.14.1, proposition 11.14.2, so $\mathcal S$ is nonempty. By irreducibility, $\mathcal U=\mathcal A\cap \mathcal S$ is nonempty and dense.

Choose a smooth local chart $q:T\to \mathcal P$ of finite type over $\mathbb C$, with a point $t_0$ representing $E_0$, and let $\mathcal E$ be the corresponding flat family on $X\times T$. Fix an identification $\mathcal E_{t_0}\simeq E_0$. The variety $T$ is smooth because the stack and the chart are smooth. Shrink $T$ to an irreducible neighborhood of $t_0$ and use lemma 11.15.1 to make $\mathcal E$ a vector bundle on the whole product. The inverse image $q^{-1}(\mathcal U)$ is dense: every nonempty open $T'\subset T$ has nonempty open image $q(T')$ under the smooth map $q$, and this image meets $\mathcal U$. Apply lemma 11.15.4 to this dense open set. Each transported operator defines a bundle isomorphic to the corresponding ample stable fiber. The fixed invariants of the stack give its Chern classes, and the lemma also gives the convergence of the dual operators. $\square$

Proof. By proposition 11.13.4, the two line summands of $E_0^*$ have positive degrees with respect to suitable test forms. The preceding theorem gives an ample stable bundle $E$ whose dual Dolbeault operator is as close to the split dual operator in $C^4$ as required. Apply corollary 11.12.7 to $F=E^*$. $\square$