Back to Moduli Spaces

One begins by finding as many discrete invariants as possible, and then considering the moduli problem for fixed invariants: for example, nonsingular projective curves of a fixed genus. Moreover, we want the moduli space to be natural, in that the geometric structure of the moduli space should reflect the geometry of the problem.

Moduli spaces are the solution to a moduli problem. To pose a moduli problem you need the following ingredients:

• The set of objects $A$$A$ you want to classify, perhaps up to an equivalence (for example, isomorphism).
• A definition of a family of these objects with a base space $S$$S$, which satisfies

• • If $S=\{s\}$$S=\{s\}$, then a family over $S$$S$ is just an object in $A$$A$.
  • Families pull back. In other words, if $\phi :S_{1}\rightarrow S_{2}$$\phi :S_{1}\rightarrow S_{2}$ is a morphism, and $X$$X$ is a family over $S_{2}$$S_{2}$, then we need a family over $S_{2}$$S_{2}$, which we denote $\phi ^{*}(X)$$\phi ^{*}(X)$. Pulling back must be functorial, and respect equivalence.

Notice that this means if $X$$X$ is a family over $S$$S$, and $s\in S$$s\in S$, then the pullback of $X$$X$ over the inclusion $\{s\}\in S$$\{s\}\in S$ is an object in $A$$A$, which we denote $X_{s}$$X_{s}$.

Example 10.2

Right notions of families:

• If we want to consider vector bundles over $X$$X$, the correct notion of a family over a base $S$$S$, where $S$$S$ is a scheme, is a bundle $E$$E$ on $X\times _{k}S$$X\times _{k}S$ such that the pullback bundle $X_{s}:=E|_{X\times \{s\}}$$X_{s}:=E|_{X\times \{s\}}$ is stable over $X$$X$ for all $s\in S$$s\in S$. A deformation of a vector bundle $E$$E$ over $X$$X$ is a family of vector bundles over $X$$X$ with base $S$$S$ such that the fibre $X_{s_{0}}$$X_{s_{0}}$ is $E$$E$ for some $s_{0}\in S$$s_{0}\in S$.
• To generalize the above, for sheaves over $X$$X$, a family is a sheaf ${\mathcal {F}}$${\mathcal {F}}$ on $X\times S$$X\times S$ which is flat over $S$$S$ - it sort of varies smoothly over $S$$S$. Flatness implies that the Hilbert polynomial is constant over the fibres $X_{s}$$X_{s}$, $s\in S$$s\in S$ (the fibres over each $s$$s$ have the same topology), for $X$$X$ reduced.
• A family of complex projective varieties over a base $S$$S$, also a complex variety, is a proper surjective morphism $\pi :T\rightarrow S$$\pi :T\rightarrow S$ such that $\pi$$\pi$ is flat with reduced fibres, and has maximal rank. A deformation of a complex projective variety $M$$M$ is a family $\pi :T\rightarrow S$$\pi :T\rightarrow S$ together with an isomorphism $\pi ^{-1}(s_{0})\cong M$$\pi ^{-1}(s_{0})\cong M$ for some $s_{0}\in S$$s_{0}\in S$.
• For the moduli problem of subschemes of $X$$X$, a family is a subscheme of $X\times S$$X\times S$, which is flat over $S$$S$.

 

As mentioned, we need to know more about $M$$M$ than just the points - we need the whole scheme structure. The notion of a family is what allows us to put an algebraic structure on the moduli space. This is because it gives rise to a natural functor Schemes $\rightarrow$$\rightarrow$ Sets,

$$S\mapsto \{{\text{families with base }}S\},$$

$$S\mapsto \{{\text{families with base }}S\},$$

called the moduli functor.

Exercise 10.1

Let ${\mathcal {M}}$${\mathcal {M}}$ be a scheme, and consider the "functor of points" of ${\overline {\mathcal {M}}}$${\overline {\mathcal {M}}}$, Schemes $\rightarrow$$\rightarrow$ Sets, that takes $S\mapsto {\text{Hom}}(S,{\mathcal {M}})$$S\mapsto {\text{Hom}}(S,{\mathcal {M}})$. Show that ${\mathcal {M}}$${\mathcal {M}}$ is determined by its functor of points.

 

Remark 10.1

If ${\mathcal {M}},S$${\mathcal {M}},S$ are schemes, then the $S$$S$-valued points of $M$$M$ is the set ${\overline {\mathcal {M}}}(S)$${\overline {\mathcal {M}}}(S)$. If $M={\text{Spec}}(k[x_{1},\ldots ,x_{n}]/(f_{1},\ldots ,f_{r})$$M={\text{Spec}}(k[x_{1},\ldots ,x_{n}]/(f_{1},\ldots ,f_{r})$ for a field $k$$k$, and $S={\text{Spec}}(R)$$S={\text{Spec}}(R)$ for a $k$$k$-algebra $R$$R$, then the $S$$S$-valued points of $M$$M$ correspond to ring morphisms $k[x_{1},\ldots ,x_{n}]\rightarrow R$$k[x_{1},\ldots ,x_{n}]\rightarrow R$ whose kernel contains $(f_{1},\ldots ,f_{r})$$(f_{1},\ldots ,f_{r})$. That is, it is a choice $x_{i}\mapsto r_{i}$$x_{i}\mapsto r_{i}$ such that $f_{1}(r_{1},\ldots ,r_{n})=\cdots =f_{r}(r_{1},\ldots ,r_{n})=0.$$f_{1}(r_{1},\ldots ,r_{n})=\cdots =f_{r}(r_{1},\ldots ,r_{n})=0.$ So $S$$S$ points of ${\mathcal {M}}$${\mathcal {M}}$ are just solutions to $f_{1}=\cdots =f_{r}=0$$f_{1}=\cdots =f_{r}=0$ in $R$$R$.

 

Applying the functor of points of a scheme ${\mathcal {M}}$${\mathcal {M}}$ to just a point $\{s\}$$\{s\}$ tells us just about the points of ${\mathcal {M}}$${\mathcal {M}}$.

There is also a relative version of the functor of points. If ${\mathcal {M}}$${\mathcal {M}}$ is a scheme over $S$$S$ (that is, ${\mathcal {M}}$${\mathcal {M}}$ and $S$$S$ are schemes and we have a morphism ${\mathcal {M}}\rightarrow S$${\mathcal {M}}\rightarrow S$; morphisms between two schemes over $S$$S$ must make the usual diagram commute), then ${\overline {\mathcal {M}}}_{S}(T)={\text{Hom}}_{S}(T,{\mathcal {M}})\subset {\overline {\mathcal {M}}}(T)$${\overline {\mathcal {M}}}_{S}(T)={\text{Hom}}_{S}(T,{\mathcal {M}})\subset {\overline {\mathcal {M}}}(T)$. If it is clear, you can suppress the $S$$S$. Another thing that will come up is the notion of a subfunctor. Let $A$$A$ be a category, then $F:A\rightarrow {\text{Set}}$$F:A\rightarrow {\text{Set}}$ is a subfunctor of $G:A\rightarrow {\text{Set}}$$G:A\rightarrow {\text{Set}}$ if for all objects $B\in A$$B\in A$, $F(B)\subset F(A)$$F(B)\subset F(A)$, and if $f:B_{1}\rightarrow B_{2}$$f:B_{1}\rightarrow B_{2}$, then $F(f)$$F(f)$ is the restriction of $G(f)$$G(f)$ from $G(B_{i})$$G(B_{i})$ to $F(B_{i})$$F(B_{i})$.

Definition 10.3

Suppose we are given a moduli problem. A (fine) moduli space is a scheme ${\mathcal {M}}$${\mathcal {M}}$ such that ${\overline {\mathcal {M}}}$${\overline {\mathcal {M}}}$ is the functor given by the moduli problem. That is, ${\mathcal {M}}$${\mathcal {M}}$represents$S\mapsto \{{\text{families with base }}S\}$$S\mapsto \{{\text{families with base }}S\}$: ${\text{Mor}}(S,M)\cong \{{\text{families with base }}S\}$${\text{Mor}}(S,M)\cong \{{\text{families with base }}S\}$.

There is a unique family $U$$U$ with base $M$$M$ corresponding to the identity map $1_{\mathcal {M}}\in {\text{Mor}}({\mathcal {M}},{\mathcal {M}})$$1_{\mathcal {M}}\in {\text{Mor}}({\mathcal {M}},{\mathcal {M}})$, and this is called the universal family, because if $F$$F$ is a family over base $X$$X$, then there is a unique morphism $\phi :S\rightarrow M$$\phi :S\rightarrow M$ corresponding to $F$$F$ in ${\text{Mor}}(S,{\mathcal {M}})$${\text{Mor}}(S,{\mathcal {M}})$, and $\phi ^{*}(U)\cong F$$\phi ^{*}(U)\cong F$.

 

The above definition shows that the notion of family over a base $S$$S$, where $S$$S$ is more than just a point, is what gives us the non-reduced information about ${\mathcal {M}}$${\mathcal {M}}$, because it corresponds to considering ${\overline {\mathcal {M}}}$${\overline {\mathcal {M}}}$ for more that just points.

Consider vector bundles over $X$$X$. Then a moduli space ${\mathcal {M}}$${\mathcal {M}}$ has points vector bundles over $X$$X$. The universal bundle $U$$U$ on ${\mathcal {M}}\times X$${\mathcal {M}}\times X$ should have fibre $E$$E$ over $\{E\}\times X$$\{E\}\times X$.

$${\text{Mor}}(S,{\mathcal {M}})\leftrightarrow \{{\text{bundles on }}X\times S\}$$

$${\text{Mor}}(S,{\mathcal {M}})\leftrightarrow \{{\text{bundles on }}X\times S\}$$

$$f:S\rightarrow {\mathcal {M}}\mapsto (f\times {\text{Id}})^{*}(U)$$

$$f:S\rightarrow {\mathcal {M}}\mapsto (f\times {\text{Id}})^{*}(U)$$

$$\{s\in S\mapsto E|_{X\times \{s\}}\}\mapsto E.$$

$$\{s\in S\mapsto E|_{X\times \{s\}}\}\mapsto E.$$

Universal families, or fine moduli spaces, rarely exist. Thus, there is a weaker notion of a coarse moduli space, where instead of a natural isomorphism from the moduli functor to ${\overline {\mathcal {M}}}$${\overline {\mathcal {M}}}$, there is just a natural transformation between them, which is universal among such natural transformations.

Exercise 10.2

Show that $\mathbb {C} ^{*}$$\mathbb {C} ^{*}$ automorphisms of stable bundles mean the moduli functor is not representable (for example, do this for line bundles on a curve of genus $g$$g$).

 

This exercise is related to the jump phenomenon, when, given a family $F$$F$ over a connected base $S$$S$ for some moduli problem, and $s_{0}\in S$$s_{0}\in S$, for all $s,t\in S,s,t\neq s_{0}$$s,t\in S,s,t\neq s_{0}$, $F_{s}=F_{t}$$F_{s}=F_{t}$, but $F_{s_{0}}\neq F_{s}$$F_{s_{0}}\neq F_{s}$.

Exercise 10.3

Show that there exists a family of vector bundles on $\mathbb {CP} ^{n}$$\mathbb {CP} ^{n}$ with base $\mathbb {C}$$\mathbb {C}$, where for $s\neq 0$$s\neq 0$, $X_{s}\cong {\mathcal {O}}\oplus {\mathcal {O}}$$X_{s}\cong {\mathcal {O}}\oplus {\mathcal {O}}$, and $X_{0}\cong {\mathcal {O}}(1)\oplus {\mathcal {O}}(-1)$$X_{0}\cong {\mathcal {O}}(1)\oplus {\mathcal {O}}(-1)$.

 

Jump phenomena prevent any moduli space from being Hausdorff. Similarly to in GIT, one solution to this is to throw away the 'bad' spaces. In fact, the solution to a moduli problem often comes down to forming an orbit space.