# Moduli Space of Bundles

Let ${\displaystyle (X,{\mathcal {O}}_{X}(1))}$ be a projective variety (so that ${\displaystyle {\mathcal {O}}_{X}(1)}$ is an ample line bundle). Let ${\displaystyle \xi }$ be a coherent sheaf on ${\displaystyle X}$. Define ${\displaystyle \xi (m)=\xi \otimes {\mathcal {O}}_{X}(m)}$. The Hilbert function of ${\displaystyle \xi }$ is ${\displaystyle P_{\xi }(m)={\text{dim}}H^{0}(X,\xi (m))={\text{dim}}\Gamma (X,\xi (m))}$ (this depends on the line bundle we have fixed). Alternatively, define this as ${\displaystyle P_{\xi }(m)=\chi (\xi (m))}$. The idea is that for a sufficiently large ${\displaystyle m}$, all non-zero cohomology of ${\displaystyle \xi (m)}$ will vanish (because ${\displaystyle {\mathcal {O}}_{X}(1)}$ is ample). Define the Hilbert polynomial to be a polynomial in ${\displaystyle m}$ which agrees with ${\displaystyle P_{\xi }(m)}$ for ${\displaystyle m>>0}$ (this exists). Thanks to Riemann-Roch, the Hilbert polynomial depends only on the Chern classes of ${\displaystyle \xi }$ and of ${\displaystyle {\mathcal {O}}_{X}(1)}$.

If ${\displaystyle \xi =E}$ is a vector bundle, then in fact,

${\displaystyle P_{E}(m)=a_{0}m^{r}+a_{1}m^{r-1}+\cdots ,}$
where ${\displaystyle a_{0}={\text{rank}}E\int _{X}{\frac {\omega ^{n}}{n!}}}$, ${\displaystyle a_{1}=\int _{X}(c_{1}(E)+{\text{rank}}E{\frac {c_{1}(X)}{2}}).{\frac {\omega ^{n-1}}{(n-1)!}}}$. We can consider the monic version of this polynomial.

Fix a coherent sheaf ${\displaystyle {\mathcal {F}}}$ on ${\displaystyle (X,{\mathcal {O}}_{X}(1))}$. A quotient of ${\displaystyle {\mathcal {F}}}$ is an exact sequence

${\displaystyle {\mathcal {F}}\rightarrow {\mathcal {Q}}\rightarrow 0.}$
A family of quotients with base ${\displaystyle S}$ is a quotient of ${\displaystyle \pi ^{*}{\mathcal {F}}}$, where ${\displaystyle \pi :X\times S\rightarrow X}$ is the projection, which is flat over ${\displaystyle S}$ and ${\displaystyle {\mathcal {Q}}|_{X\times \{s\}}}$ has the same Hilbert polynomial ${\displaystyle P}$, for all ${\displaystyle s\in S}$.

We put an equivalence relation on quotients by identifying quotients with the same kernel.

Exercise 10.4

Check that when ${\displaystyle S=\{s\}}$ this is just a quotient.

The Quot functor takes a scheme ${\displaystyle S}$ to the set of all families of quotients ${\displaystyle {\mathcal {Q}}}$ of ${\displaystyle {\mathcal {F}}}$ with base ${\displaystyle S}$. We can also fix a polynomial ${\displaystyle P}$ and restrict to quotients with that Hilbert polynomial.

Example 10.3

If ${\displaystyle {\mathcal {F}}={\mathcal {O}}}$, then a quotient is just the structure sheaf of some closed subscheme ${\displaystyle Z\subset X}$, by considering the support of ${\displaystyle {\mathcal {Q}}}$.

To see this, take an affine open set ${\displaystyle U}$: ${\displaystyle {\mathcal {O}}_{X}(U)\cong {\text{Spec}}(A)}$. A quotient is a surjective map ${\displaystyle A\rightarrow {\mathcal {Q}}(U)}$, so ${\displaystyle {\mathcal {Q}}(U)\cong A/I}$, for some ideal ${\displaystyle I}$. Then ${\displaystyle Z\cap U={\text{Spec}}A/I=Z(I)\subset X}$. Let ${\displaystyle {\mathcal {I}}}$ be the ideal sheaf of ${\displaystyle Z}$, i.e., on an open set ${\displaystyle U}$, ${\displaystyle {\mathcal {I}}(U)={\text{ker}}({\mathcal {F}}(U)\rightarrow {\mathcal {Q}})}$. So in this case, the Quot functor takes a scheme ${\displaystyle S}$ to the set of all families of subschemes of ${\displaystyle X}$ with base ${\displaystyle S}$. This is called the Hilbert functor. If it is representable, it is called the Hilbert scheme.

If we consider Example (MISSING), the Hilbert scheme is ${\displaystyle {\text{Spec }}\mathbb {C} [t]/(t^{2})}$.

Exercise 10.5

Set the Hilbert polynomial to be ${\displaystyle P(n)\equiv 1}$. Show that ${\displaystyle {\text{Quot}}({\mathcal {O}}_{X},P)\cong X}$ (as a scheme/functor of points - not just as a set).

Example 10.4

To show that the Quot functor is representable, we will need the Grassmannian functor. Let ${\displaystyle k}$ be a field, and ${\displaystyle V}$ a finite dimensional vector space over ${\displaystyle k}$. Let ${\displaystyle r}$ be an integer such that ${\displaystyle 0\leq r\leq {\text{dim}}(V)}$. Then ${\displaystyle {\text{Grass}}(V,r)}$ is a functor from the opposite category of ${\displaystyle {\text{Sch}}/k}$ to Sets, such that

${\displaystyle {\text{Grass}}(V,r)(X)=\{{\text{quotients }}{\mathcal {O}}_{X}\otimes _{k}V\rightarrow F\rightarrow 0{\text{ that are locally free and of constant rank }}r\}.}$

I will sketch how to show that this functor is representable. First, fix a ${\displaystyle W\subset V}$ of dimension ${\displaystyle r}$. Consider the subset of ${\displaystyle {\text{Grass}}(V,r)}$ such that the restriction of ${\displaystyle {\mathcal {O}}_{X}\otimes _{k}V\rightarrow F}$ to ${\displaystyle {\mathcal {O}}_{X}\otimes W}$ gives an isomorphism ${\displaystyle {\mathcal {O}}_{X}\otimes W\rightarrow F}$. Let ${\displaystyle G_{W}}$ be the subfunctor defined by these subsets.

${\displaystyle G_{W}}$ is representable. To see this, note that there is an isomorphism ${\displaystyle {\mathcal {O}}_{X}\otimes W\leftarrow F}$. Then we get a composition ${\displaystyle {\mathcal {O}}_{X}\otimes _{k}V\rightarrow F\rightarrow {\mathcal {O}}_{X}\otimes _{k}W}$, which is the right inverse of the inclusion ${\displaystyle {\mathcal {O}}_{X}\otimes _{k}W\rightarrow {\mathcal {O}}_{X}\otimes _{k}V.}$ So ${\displaystyle G_{W}}$ corresponds to morphisms ${\displaystyle V\rightarrow W}$ which splits the inclusion. If ${\displaystyle K\subset W}$ such that ${\displaystyle V=K\oplus W}$, then ${\displaystyle G_{W}}$ corresponds to morphisms ${\displaystyle K\rightarrow W}$. So ${\displaystyle G_{W}}$ is the functor ${\displaystyle {\textbf {Hom}}(K,W)}$, defined by

${\displaystyle {\textbf {Hom}}(K,W)(X)={\text{Hom}}({\mathcal {O}}_{X}\otimes _{k}K,{\mathcal {O}}_{X}\otimes _{k}W).}$

${\displaystyle G_{W}={\textbf {Hom}}(K,W)}$ is representable by ${\displaystyle {\overline {G}}_{W}={\text{Spec}}({\text{Sym}}({\text{Hom}}(K,W)^{*}))^{*}}$. To see this, recall that the universal property of the symmetric algebra is the following: given a ${\displaystyle k}$-vector space ${\displaystyle V}$ and a ${\displaystyle k}$-module ${\displaystyle B}$, then there is a natural isomorphism ${\displaystyle {\text{Hom}}_{k-{\text{v.s.}}}(V,B)={\text{Hom}}_{k-{\text{alg}}}({\text{Sym}}V,B)}$. Thus,

${\displaystyle {\text{Hom}}_{\text{sch/k}}(X,{\text{Spec}}({\text{Sym}}({\text{Hom}}(K,W)^{*}))^{*})\cong {\text{Hom}}_{k-{\text{alg}}}({\text{Sym}}({\text{Hom}}(K,W)^{*}))^{*},{\mathcal {O}}_{X}(X))}$
${\displaystyle \cong {\text{Hom}}_{k-{\text{v.s.}}}({\text{Hom}}(K,W)^{*},{\mathcal {O}}_{X}(X))\cong {\mathcal {O}}_{X}(X)\otimes {\text{Hom}}(K,W).}$

The idea now is to show that the ${\displaystyle G_{W}}$ are open (we know that they are affine) subfunctors, that agree on intersections, and so they can be glued together to get ${\displaystyle {\text{Grass}}(V,r)}$. Fix ${\displaystyle X}$ and ${\displaystyle W}$, and consider a quotient ${\displaystyle \phi :{\mathcal {O}}_{X}\otimes _{k}V\rightarrow F}$. Consider open subschemes ${\displaystyle Y}$ of ${\displaystyle X}$ such that the restriction of ${\displaystyle \phi }$ to ${\displaystyle Y}$ lies in the image of ${\displaystyle G_{W}(Y)}$, and choose a maximal such ${\displaystyle Y}$, call it ${\displaystyle Y_{W}}$. Then this is the definition of ${\displaystyle G_{W}}$ being an open sub-functor. As ${\displaystyle W}$ varies, the ${\displaystyle Y_{W}}$ form an open covering of ${\displaystyle X}$, so the ${\displaystyle G_{W}}$ are an affine open covering of ${\displaystyle G}$. After checking the cocycle condition, we can see that we can glue the ${\displaystyle {\overline {G}}_{W}}$ together, and this scheme represents ${\displaystyle G}$.

Finally, you can use the Plucker embedding to show that ${\displaystyle {\text{Grass}}(V,r)}$ is projective (and indeed a smooth irreducible variety).

Theorem 10.1

Let ${\displaystyle {\mathcal {F}}}$ be a coherent sheaf over a projective variety ${\displaystyle (X,{\mathcal {O}}(1))}$. Then ${\displaystyle {\text{Quot}}({\mathcal {F}},P)}$ is representable.

Proof (Outline:)

First, let's fix a quotient ${\displaystyle 0\rightarrow {\mathcal {K}}\rightarrow {\mathcal {F}}\rightarrow {\mathcal {Q}}\rightarrow 0}$. As ${\displaystyle {\mathcal {O}}(1)}$ is ample, there is an ${\displaystyle m>>0}$ such that for all ${\displaystyle i>0}$, ${\displaystyle H^{i}({\mathcal {K}}(m))=0}$. Moreover, for ${\displaystyle m}$ sufficiently large, ${\displaystyle {\mathcal {K}}(m)}$ is globally generated. This means that there is an exact sequence

${\displaystyle H^{0}({\mathcal {K}}(m))\otimes _{\mathbb {C} }{\mathcal {O}}_{X}\rightarrow {\mathcal {K}}(m)\rightarrow 0.}$
Sections of ${\displaystyle {\mathcal {K}}(m)}$ generate ${\displaystyle {\mathcal {K}}(m)}$ inside ${\displaystyle {\mathcal {F}}(m)}$, and taking the quotient defines ${\displaystyle {\mathcal {Q}}(m)}$. So the quotient is determined by the exact sequence of vector spaces
${\displaystyle 0\rightarrow H^{0}({\mathcal {K}}(m))\rightarrow H^{0}({\mathcal {F}}(m))\rightarrow H^{0}({\mathcal {Q}}(m))\rightarrow 0.}$
But this exact sequence is a point of ${\displaystyle {\text{Gr}}(H^{0}({\mathcal {F}}(m)),P(m))}$ (recall that by definition ${\displaystyle P(m)={\text{dim }}H^{0}({\mathcal {F}}(m))}$. We assume that you can choose ${\displaystyle m}$ universally.

We then need to generalize and do the above over a base ${\displaystyle S}$, in order to get a subfunctor of the Grassmannian. The Grassmannian is representable, so this gives a subscheme of the Grassmannian, called Quot. For the details of the proof, see Huybrechts-Lehn.

Now we can give the outline of Simpson's construction of the moduli space of vector bundles over ${\displaystyle (X,{\mathcal {O}}(1))}$ with given Hilbert polynomial ${\displaystyle P}$.

1. Fix a bundle ${\displaystyle \xi }$ on ${\displaystyle X}$ with Hilbert polynomial ${\displaystyle P}$.
2. Find ${\displaystyle N>0}$ such that all higher cohomology of ${\displaystyle \xi (N)}$ vanishes, and ${\displaystyle \xi (N)}$ is generated by its global sections. Again, this means that

${\displaystyle H^{0}(\xi (N))\otimes {\mathcal {O}}(-N)\rightarrow \xi \rightarrow 0.}$

1. Fix an identification ${\displaystyle H^{0}(\xi (N))\cong \mathbb {C} ^{P(N)}.}$
2. Under this identification, ${\displaystyle \xi }$ is a a quotient sheaf of ${\displaystyle {\mathcal {O}}(-N)^{\oplus P(N)}}$, so ${\displaystyle \xi \in {\text{Quot}}({\mathcal {O}}(-N)^{\oplus P(N)},P).}$
3. Divide by the choice of identification in 3). That is, divide by the action of ${\displaystyle SL(P(N))}$.
4. Find a stability condition to do GIT for this quotient, and let Quot' denote the semi-stable points.
5. Go back to step 2, and show that ${\displaystyle N}$ can be chosen universally for all bundles satisfying the stability condition.
6. Check that ${\displaystyle {\text{Quot'}}//SL(N)}$ is a moduli space of stable bundles/sheaves.

In step 4, we identified ${\displaystyle \xi }$ as an element of ${\displaystyle {\text{Quot}}(H^{0}(\xi (N))\otimes H^{0}({\mathcal {O}}(-N)),P),}$ divided by the action.

In a previous exercise, we showed that the stability condition for a subspace ${\displaystyle A\subset V\otimes W}$, under the ${\displaystyle SL(V)}$ action on ${\displaystyle {\text{Gr}}(V\otimes W,r)}$ was for all proper subspaces ${\displaystyle S\subset V}$,

${\displaystyle {\frac {{\text{dim}}A\cap (S\otimes W)}{{\text{dim}}S}}<{\frac {{\text{dim}}A}{{\text{dim}}V}}.}$
So as an element of ${\displaystyle {\text{Quot}}(H^{0}(\xi (N))\otimes H^{0}({\mathcal {O}}(-N)),P,}$${\displaystyle \xi }$ is stable by checking proper subspaces ${\displaystyle S}$ of ${\displaystyle H^{0}(\xi (N))}$. By unwinding ${\displaystyle \xi (N)}$ and evaluating sections, ${\displaystyle S}$ gives rise to a subsheaf of ${\displaystyle \xi }$. ${\displaystyle \xi }$ is stable if and only if for all proper subsheaves ${\displaystyle {\mathcal {F}}\subset \xi }$, and ${\displaystyle n>>0}$,
${\displaystyle {\frac {P_{\mathcal {F}}(n)}{{\text{rank}}{\mathcal {F}}}}<{\frac {P_{\xi }(n)}{{\text{rank}}\xi }}.}$
This is called Gieseker stability. If we consider the leading coefficient as ${\displaystyle n\rightarrow \infty }$, then we get slope stability,
${\displaystyle {\frac {\int _{X}c_{1}({\mathcal {F}})\omega ^{n-1}}{{\text{rank}}{\mathcal {F}}}}<{\frac {\int _{X}c_{1}(\xi )\omega ^{n-1}}{{\text{rank}}\xi }}.}$
Slope stability implies Gieseker stability, and on a curve, these are equivalent.

Stability is a generic condition.

Exercise 10.6

Show that both Gieseker stability and slope stability (${\displaystyle u(E)}$ is the slope of ${\displaystyle E}$) satisfy the see-saw property. That is, for all exact sequences

${\displaystyle 0\rightarrow A\rightarrow \xi \rightarrow B\rightarrow 0,}$
such that ${\displaystyle A\neq 0}$, ${\displaystyle u(A) and similarly when we consider Gieseker stability instead of slope stability.

Exercise 10.7

Show that a stable vector bundle is simple.

Stability rules out the possibility of the 'jump phenomenon'.

Lemma 10.2

If ${\displaystyle \xi ,{\mathcal {F}}}$ are families of stable sheaves parametrized by ${\displaystyle \mathbb {C} }$, and for all ${\displaystyle t\in \mathbb {C} ^{*}}$, ${\displaystyle \xi _{t}\cong {\mathcal {F}}_{t}}$, then ${\displaystyle \xi _{0}\cong {\mathcal {F}}_{0}}$.

Proof

By definition, ${\displaystyle \xi ,{\mathcal {F}}}$ are sheaves on ${\displaystyle X\times \mathbb {C} }$, flat over ${\displaystyle \mathbb {C} }$ with the same Hilbert polynomial on each fibre, and stable on each fibre.

Let ${\displaystyle \pi :X\times \mathbb {C} \rightarrow \mathbb {C} }$ be the projection. ${\displaystyle {\mathcal {H}}om(\xi ,{\mathcal {F}})}$ is a sheaf on ${\displaystyle X\times \mathbb {C} }$ (defined by ${\displaystyle {\mathcal {H}}om(\xi ,{\mathcal {F}})(U)={\text{Hom}}(\xi |_{U},{\mathcal {F}}|_{U})}$), so we can construct the direct image sheaf under ${\displaystyle \pi }$, ${\displaystyle \pi _{*}{\mathcal {H}}om(\xi ,{\mathcal {F}})}$. By definition, for an open ${\displaystyle U\subset \mathbb {C} }$, ${\displaystyle \pi _{*}{\mathcal {H}}om(\xi ,{\mathcal {F}})(U)={\text{Hom}}(\xi |_{X\times U},{\mathcal {F}}|_{X\times U})}$. The fiber over ${\displaystyle t\in \mathbb {C} }$ is ${\displaystyle {\text{Hom}}(\xi _{t},{\mathcal {F}}_{t})}$, as these are coherent sheaves.

Because ${\displaystyle \xi _{t},{\mathcal {F}}_{t}}$ are simple by the previous exercise, the fiber over ${\displaystyle t\in \mathbb {C} -\{0\}}$ is ${\displaystyle {\text{Hom}}(\xi _{t},{\mathcal {F}}_{t})\cong \mathbb {C} }$.

Exercise 10.8

Show that ${\displaystyle \pi _{*}{\mathcal {H}}om(\xi ,{\mathcal {F}})}$ is torsion free over ${\displaystyle \mathbb {C} }$. Recall that a sheaf ${\displaystyle {\mathcal {G}}}$ over a scheme ${\displaystyle (X,{\mathcal {O}}_{X})}$ is torsion-free if for all ${\displaystyle p\in X}$, ${\displaystyle {\mathcal {G}}_{p}}$ is a torsion-free ${\displaystyle {\mathcal {O}}_{X,p}}$- module.

If ${\displaystyle \xi ,{\mathcal {F}}}$ are families of vector bundles, then ${\displaystyle \pi _{*}{\mathcal {H}}om(\xi ,{\mathcal {F}})}$ is a vector bundle, so as it is a line bundle over ${\displaystyle \mathbb {C} -\{0\}}$, it is a line bundle over ${\displaystyle \mathbb {C} }$. For sheaves, we need the previous exercise to show that it is a a line bundle over ${\displaystyle \mathbb {C} }$.

Pick a nonzero element of the fiber over ${\displaystyle 0}$. By definition, this is a non-zero morphism ${\displaystyle \phi :\xi _{0}\rightarrow {\mathcal {F}}_{0}}$. So we can form two exact sequences:

${\displaystyle 0\rightarrow {\text{ker}}\phi \rightarrow \xi _{0}\rightarrow {\text{im}}\phi \rightarrow 0,}$
${\displaystyle 0\rightarrow {\text{im}}\phi \rightarrow {\mathcal {F}}_{0}\rightarrow {\text{coker}}\phi \rightarrow 0.}$
Using the see-saw property of ${\displaystyle \mu }$ and the stability of ${\displaystyle \xi _{0},{\mathcal {F}}_{0}}$, if ${\displaystyle {\text{im}}\phi \neq {\mathcal {F}}_{0}}$, we have that ${\displaystyle \mu ({\text{im}}\phi )>\mu (\xi _{0})}$, and ${\displaystyle \mu ({\text{im}}(\phi ))<\mu ({\mathcal {F}}_{0})}$. By definition of ${\displaystyle \mu }$, it depends on the Hilbert polynomial of a sheaf, and so because ${\displaystyle \xi }$ and ${\displaystyle {\mathcal {F}}}$ are families of fixed Hilbert polynomials, ${\displaystyle \mu ({\mathcal {F}}_{0})=\mu (\xi _{0})}$. So we have a contradiction, unless ${\displaystyle {\text{im}}\phi ={\mathcal {F}}_{0}}$ and ${\displaystyle {\text{ker}}(\phi )=0}$. So ${\displaystyle \phi }$ is an isomorphism.