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

In the situation of a forcing algebra of homogeneous elements, this torsor $T$ can also be obtained as $\operatorname {Proj} {\left(B\right)}$, where $B$ is the (not necessarily positively) graded forcing algebra. In particular, it follows that the containment $f\in I^{*}$ is equivalent to the property that $T$ 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 ${\mathcal {S}}$ be a vector bundle on a smooth projective curve $C$. It is called semistable, if $\mu ({\mathcal {T}})={\frac {\deg({\mathcal {T}})}{\operatorname {rk} ({\mathcal {T}})}}\leq {\frac {\deg({\mathcal {S}})}{\operatorname {rk} ({\mathcal {S}})}}=\mu ({\mathcal {S}})$ for all subbundles ${\mathcal {T}}\subseteq {\mathcal {S}}$.
Suppose that the base field has positive characteristic $p>0$. Then ${\mathcal {S}}$ is called strongly semistable, if all (absolute)
$F^{e*}({\mathcal {S}})$
An important property of a semistable bundle of negative degree is that it can not have any global section $\neq 0$. Note that a semistable vector bundle need not be strongly semistable, the following is probably the simplest example.

## Example

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

## Example

Let $R=K[x,y,z]/{\left(x^{3}+y^{3}+z^{3}\right)}$, where $K$ is a field of positive characteristic $p\neq 3$, $I={\left(x^{2},y^{2},z^{2}\right)}$, and
    $C=\operatorname {Proj} {\left(R\right)}\,.$
The equation $x^{3}+y^{3}+z^{3}=0$ yields the short exact sequence
    $0\longrightarrow {\mathcal {O}}_{C}\longrightarrow \operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}(3)\longrightarrow {\mathcal {O}}_{C}\longrightarrow 0.$
This shows that $\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}$ is strongly semistable.
For a strongly semistable vector bundle ${\mathcal {S}}$ on $C$ and a cohomology class $c\in H^{1}(C,{\mathcal {S}})$ with corresponding torsor we obtain the following affineness criterion.

## Theorem

Let $C$ denote a smooth projective curve over an algebraically closed field $K$ and let ${\mathcal {S}}$ be a strongly semistable vector bundle over $C$ together with a cohomology class $c\in H^{1}(C,{\mathcal {S}})$. Then the torsor $T(c)$ is an affine scheme if and only if $\operatorname {deg} {\left({\mathcal {S}}\right)}<0$ and $c\neq 0$ ($F^{e}(c)\neq 0$
$e$
This result rests on the ampleness of ${\mathcal {S}}'^{\vee }$ occuring in the dual exact sequence $0\rightarrow {\mathcal {O}}_{C}\rightarrow {\mathcal {S}}'^{\vee }\rightarrow {\mathcal {S}}^{\vee }\rightarrow 0$ given by $c$ (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 $\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}$ is strongly semistable. Then
    $R_{m}\subseteq I^{*}{\text{ for }}m\geq {\frac {\sum d_{i}}{n-1}}{\text{ and (for almost all prime numbers) }}R_{m}\cap I^{*}\subseteq I{\text{ for }}m<{\frac {\sum d_{i}}{n-1}}.$
If we take on the right hand side $I^{F}$, the Frobenius closure of the ideal, instead of $I$, then this statement is true for all characteristics. As stated, it is true in a relative setting for $p$ large enough.
We indicate the proof of the inclusion result. The degree condition implies that $c\in \delta (f)=H^{1}(C,{\mathcal {S}})$ is such that ${\mathcal {S}}=\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}(m)$ has non-negative degree. Then also all Frobenius pull-backs $F^{*}({\mathcal {S}})$ have non-negative degree. Let
${\mathcal {L}}={\mathcal {O}}(k)$ be a twist of the tautological line bundle on $C$ such that its degree is larger than the degree of $\omega _{C}^{-1}$, the dual of the canonical sheaf. Let $z\in H^{0}(Y,{\mathcal {L}})$ be a non-zero element. Then $zF^{e*}(c)\in H^{1}(C,F^{e*}({\mathcal {S}})\otimes {\mathcal {L}})$, and by Serre duality we have
    $H^{1}(C,F^{e*}({\mathcal {S}})\otimes {\mathcal {L}})\cong H^{0}(F^{e*}({\mathcal {S}}^{\vee })\otimes {\mathcal {L}}^{-1}\otimes \omega _{C})^{\vee }\,.$
On the right hand side we have a semistable sheaf of negative degree, which can not have a non-trivial section. Hence
    $zF^{e*}(c)=0\,$
and therefore $f$ belongs to the tight closure.
* ↑ Here one has to check only finitely many $e$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.