# Tight closure/Two-dimensional/Semistable vector bundle/Introduction/Section

In the situation of a forcing algebra of homogeneous elements, this torsor ${\displaystyle {}T}$ can also be obtained as ${\displaystyle {}\operatorname {Proj} {\left(B\right)}}$, where ${\displaystyle {}B}$ is the (not necessarily positively) graded forcing algebra. In particular, it follows that the containment ${\displaystyle {}f\in I^{*}}$ is equivalent to the property that ${\displaystyle {}T}$ is not an affine variety. For this properties, positivity (ampleness) properties of the syzygy bundle are crucial. We need the concept of (Mumford-)- semistability.

## Definition

Let ${\displaystyle {}{\mathcal {S}}}$ be a vector bundle on a smooth projective curve ${\displaystyle {}C}$. It is called semistable, if ${\displaystyle {}\mu ({\mathcal {T}})={\frac {\deg({\mathcal {T}})}{\operatorname {rk} ({\mathcal {T}})}}\leq {\frac {\deg({\mathcal {S}})}{\operatorname {rk} ({\mathcal {S}})}}=\mu ({\mathcal {S}})}$ for all subbundles ${\displaystyle {}{\mathcal {T}}\subseteq {\mathcal {S}}}$.

Suppose that the base field has positive characteristic ${\displaystyle {}p>0}$. Then ${\displaystyle {}{\mathcal {S}}}$ is called strongly semistable, if all (absolute)

Frobenius pull-backs ${\displaystyle {}F^{e*}({\mathcal {S}})}$ are semistable.

An important property of a semistable bundle of negative degree is that it can not have any global section ${\displaystyle {}\neq 0}$. Note that a semistable vector bundle need not be strongly semistable, the following is probably the simplest example.

## Example

Let ${\displaystyle {}C}$ be the smooth Fermat quartic given by ${\displaystyle {}x^{4}+y^{4}+z^{4}}$, and consider on it the syzygy bundle ${\displaystyle {}\operatorname {Syz} {\left(x,y,z\right)}}$ (which is also the restricted cotangent bundle from the projective plane). This bundle is semistable. Suppose that the characteristic is ${\displaystyle {}3}$. Then its Frobenius pull-back is ${\displaystyle {}\operatorname {Syz} {\left(x^{3},y^{3},z^{3}\right)}}$. The curve equation gives a global non-trivial section of this bundle of total degree ${\displaystyle {}4}$. But the degree of ${\displaystyle {}\operatorname {Syz} {\left(x^{3},y^{3},z^{3}\right)}(4)}$ is negative, hence it can not be semistable anymore.

The following example is related to example.

## Example

Let ${\displaystyle {}R=K[x,y,z]/{\left(x^{3}+y^{3}+z^{3}\right)}}$, where ${\displaystyle {}K}$ is a field of positive characteristic ${\displaystyle {}p\neq 3}$, ${\displaystyle {}I={\left(x^{2},y^{2},z^{2}\right)}}$, and

${\displaystyle {}C=\operatorname {Proj} {\left(R\right)}\,.}$

The equation ${\displaystyle {}x^{3}+y^{3}+z^{3}=0}$ yields the short exact sequence

${\displaystyle 0\longrightarrow {\mathcal {O}}_{C}\longrightarrow \operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}(3)\longrightarrow {\mathcal {O}}_{C}\longrightarrow 0.}$

This shows that ${\displaystyle {}\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}}$ is strongly semistable.

For a strongly semistable vector bundle ${\displaystyle {}{\mathcal {S}}}$ on ${\displaystyle {}C}$ and a cohomology class ${\displaystyle {}c\in H^{1}(C,{\mathcal {S}})}$ with corresponding torsor we obtain the following affineness criterion.

## Theorem

Let ${\displaystyle {}C}$ denote a smooth projective curve over an algebraically closed field ${\displaystyle {}K}$ and let ${\displaystyle {}{\mathcal {S}}}$ be a strongly semistable vector bundle over ${\displaystyle {}C}$ together with a cohomology class ${\displaystyle {}c\in H^{1}(C,{\mathcal {S}})}$. Then the torsor ${\displaystyle {}T(c)}$ is an affine scheme if and only if ${\displaystyle {}\operatorname {deg} {\left({\mathcal {S}}\right)}<0}$ and ${\displaystyle {}c\neq 0}$ (${\displaystyle {}F^{e}(c)\neq 0}$ for all ${\displaystyle {}e}$ in positive characteristic[1]).

This result rests on the ampleness of ${\displaystyle {}{\mathcal {S}}'^{\vee }}$ occuring in the dual exact sequence ${\displaystyle {}0\rightarrow {\mathcal {O}}_{C}\rightarrow {\mathcal {S}}'^{\vee }\rightarrow {\mathcal {S}}^{\vee }\rightarrow 0}$ given by ${\displaystyle {}c}$ (this rests on work of Gieseker and Hartshorne). It implies for a strongly semistable syzygy bundle the following degree formula for tight closure.

## Theorem

Suppose that ${\displaystyle {}\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}}$ is strongly semistable. Then

${\displaystyle R_{m}\subseteq I^{*}{\text{ for }}m\geq {\frac {\sum d_{i}}{n-1}}{\text{ and (for almost all prime numbers) }}R_{m}\cap I^{*}\subseteq I{\text{ for }}m<{\frac {\sum d_{i}}{n-1}}.}$

If we take on the right hand side ${\displaystyle {}I^{F}}$, the Frobenius closure of the ideal, instead of ${\displaystyle {}I}$, then this statement is true for all characteristics. As stated, it is true in a relative setting for ${\displaystyle {}p}$ large enough.

We indicate the proof of the inclusion result. The degree condition implies that ${\displaystyle {}c\in \delta (f)=H^{1}(C,{\mathcal {S}})}$ is such that ${\displaystyle {}{\mathcal {S}}=\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}(m)}$ has non-negative degree. Then also all Frobenius pull-backs ${\displaystyle {}F^{*}({\mathcal {S}})}$ have non-negative degree. Let ${\displaystyle {}{\mathcal {L}}={\mathcal {O}}(k)}$ be a twist of the tautological line bundle on ${\displaystyle {}C}$ such that its degree is larger than the degree of ${\displaystyle {}\omega _{C}^{-1}}$, the dual of the canonical sheaf. Let ${\displaystyle {}z\in H^{0}(Y,{\mathcal {L}})}$ be a non-zero element. Then ${\displaystyle {}zF^{e*}(c)\in H^{1}(C,F^{e*}({\mathcal {S}})\otimes {\mathcal {L}})}$, and by Serre duality we have

${\displaystyle {}H^{1}(C,F^{e*}({\mathcal {S}})\otimes {\mathcal {L}})\cong H^{0}(F^{e*}({\mathcal {S}}^{\vee })\otimes {\mathcal {L}}^{-1}\otimes \omega _{C})^{\vee }\,.}$

On the right hand side we have a semistable sheaf of negative degree, which can not have a non-trivial section. Hence

${\displaystyle {}zF^{e*}(c)=0\,}$

and therefore ${\displaystyle {}f}$ belongs to the tight closure.

1. Here one has to check only finitely many ${\displaystyle {}e}$s and there exist good estimates how far one has to go. Also, in a relative situation, this is only an extra condition for finitely many prime numbers.