## The Hilbert Scheme

Now that we have a notion of moduli space, we’re going to look at several concrete examples. I mentioned that the Grassmannian is a fine moduli space for $k$-planes in $\mathbb{C}^n$. We’ll make use of this to construct several more. First up on the list: the Hilbert Scheme. This scheme, and it isn’t a variety because it isn’t reduced in general, parameterizes families of projective varieties subschemes of projective space.

We’re going to follow Kollár’s “Rational Curves on Algebraic Varieties” in this construction, because that’s where I learned how to put the Hilbert Scheme together. As such, we’re going to work in a completely relative setting, and forget that we only care about $\mathbb{C}$ for the moment.

For any scheme $S$, we define a scheme over $S$ to be a scheme $X$ along with a map $X\to S$. We generally just say that $X\to S$ or $X/S$ is a scheme over $S$. We get a category of schemes over $S$ by looking at morphisms that commute with the structure maps, and we can talk about functors representable on this category (that is, a functor from schemes over $S$ to Sets such that there is an $S$-scheme $X/S$ such that $\hom_S(-,X)$ is naturally isomorphic to it).

So now we define the Hilbert functor $Hilb(X/S)$ from $S$-schemes to sets by $Hilb(X/S)(T)=$ the subschemes $V\subset X\times_S Z$ which are proper and flat over $Z$. This is the same as the set of quotient sheaves $\mathscr{F}$ of the structure sheaf of $X\times_S Z$ which are flat and have proper support over $Z$, that is, the subscheme where stalk is nonzero is proper over $Z$.

We can actually extend the definition of the Hilbet Polynomial to this relative setting. Look at $g:Y\to Z$ a projective morphism of $S$-schemes, with $Z$ connected. Let $\mathscr{F}$ be a sheaf on $Y$ which is flat over $Z$ and let $\mathscr{O}(1)$ be an ample sheaf on $Y$ such that $Y\to \mathbb{P}^n\times Z\to Z$ is $g$. Then it is a theorem that $g_*\mathscr{F}\otimes\mathscr{O}(n)=g_*\mathscr{F}(n)$ is locally free for large enough $n$. So then there exists a polynomial such that $P(n)=\mathrm{rank}(g_*\mathscr{F}(n))$ for big enough $n$. So then we define the Hilbert polynomial of $V\subset Y$ a closed subscheme flat over $Z$ to be the Hilbert polynomial of $\mathscr{O}_V$. When we take $Z$ to be a point, $S$ to be a point, and $V$ to be a variety, we get the old Hilbert Polynomial. (This isn’t completely obvious, but I’m not proving it, because it’d sidetrack us.)

So now we define, for every polynomial which takes integer values at integers, $Hilb_P(X/S)$ to take $Z$ to subschemes of $X\times_S Z$ which are proper and flat over $Z$ and have Hilbert polynomial $P$. Now, when $Z$ is connected, we get $Hilb(X/S)(Z)=\bigcup_P Hilb_P(X/S)(Z)$. So we take these to be the functors for the moduli problem we care about: flat families of projective schemes and flat families with a given Hilbert Polynomial. The union on the right above is disjoint, because flat families have constant Hilbert Polynomial.

So the real great thing about these functors is the following: $Hilb_P(X/S)$ is representable and the scheme representing it is projective over $S$ (which means that it embeds in $\mathbb{P}^n\times S$.)

Before getting into the details of the construction, we’ll just look at a couple of quick examples. For any scheme, if we take the polynomial to be constant with value 1, then we have the family of dimension zero and degree one subschemes, which means the points.. So $\mathrm{Hilb}_1(X/S)=X/S$. We’ll denote the Hilbert scheme in non-italics from here on out, as we did just now. So now let $C$ be a curve over $\mathbb{C}$. Then $\mathrm{Hilb}_m(C)$ is the collection of degree $d$ subschemes of dimension zero. This is just the set of collections of $m$ points, counted with multiplicities. So our first thought might be that it should be $C\times\ldots\times C$, $m$ times. But this isn’t quite right, because that’s the set of ordered $m$-tuples. So we quotient by $S_m$ to get the correct space. We call this $\mathrm{Sym}^m(C)$, and it is equal to $\mathrm{Hilb}_m(C)$.

Before we can begin (and en route, we’ll state a few hard lemmas without proof) we need to generalize the construction of the Grassmannian. Before we constructed it for a vector space. We can think of this as the case of vector bundles over a point, however, and ask “is there a Grassmannian for general vector bundles?” Well, the answer is yes. Let $S$ be a scheme and $E$ a vector bundle, and then take $r$ a natural number. Then we define a functor $Grass(r,E)$ which takes $Z$ to the subvector bundles of rank $r$ of $E\times_S Z$. This is represented by $\mathrm{Grass}(r,E)$, a space that parameterizes the subbundles of $E$ of rank $r$.

We start by working as though $S$ were a point and $X$ projective space $\mathbb{P}^n$. Let $Y\subset \mathbb{P}^n$ be a subscheme, and look at its ideal sheaf $I_Y$. Then there exists $N$ such that $I_Y(N)$ is generated by global sections (this is one of those lemmas we’re not proving, it’s in a book by Mumford, “Lectures on Curves on an Algebraic Surface”) so $H^0(\mathbb{P}^n,I_Y(N))\subset H^0(\mathbb{P}^n,\mathscr{O}_{\mathbb{P}^n}(N))$ determines $Y$. The strong part of the lemma we are using says that $N$ can be chosen to depend only on the Hilbert Polynomial of $Y$. So we get an injection from the subschemes of $\mathbb{P}^n$ with Hilbert polynomial $P$ to $\mathrm{Grass}(Q(N),H^0(\mathbb{P}^n,\mathscr{O}_{\mathbb{P}^n}(N)))$, where $Q(N)$ is $\binom{n+N}{n}-P(N)$. The image is even an algebraic subset of this Grassmannian. However, we don’t have the scheme structure from this. However, this is the general idea of the proof.

Now, we can reduce to having to work with $\mathbb{P}^n\times S=\mathbb{P}^n_S$. If $X/S$ is an $S$-scheme, all we have to do is embed it in projective space over $S$, and we get an injection of functors, because any family in $X$ must be a family in $\mathbb{P}^n_S$. Next up, we need a map $Hilb(\mathbb{P}^n_S/S)\to Grass$. For any $p:Z\to S$ we can perform a base change and take $Y\subset \mathbb{P}^n\times_S Z$ a closed subscheme, flat over $Z$ with fixed Hilbert Polynomial $P$.

Let $k$ be a field and $\mathrm{Spec}k\to Z$ a morphism. Pulling things back to $\mathrm{Spec}k$, we get $Y_k\subset\mathbb{P}^n_k$. Let $I_k$ be the ideal sheaf of $Y_k$. Then we gave $\chi(I_k(n))=\chi(\mathbb{P}_k^m,\mathscr{O}(n))-\chi(Y_k,\mathscr{O}(n))=\binom{m+n}{n}-P(n)=Q(n)$.

Now we fix $N$ big enough that $I(N)$ is generated by global sections for all ideal sheaves with Hilbert polynomial $P$. So now from $0\to I_Y\to \mathscr{O}_{\mathbb{P}^m}\to \mathscr{O}_Y\to 0$ we find, from flatness, that $0\to (p^*f)_*I_Y(N)\to (p^*f)_*\mathscr{O}_{\mathbb{P}^m}(N)\to (p^*f)_*\mathscr{O}_Y(N)\to 0$ is exact. Now, this tells us that $(p^*f)_*\mathscr{O}_Y(N)$ is locally free of rank $P(N)$. So we get a map $Hilb(\mathbb{P}^m/S)\to Grass(Q(N),(p^*f)_*\mathscr{O}_{\mathbb{P}^m}(N))$, so we have a map into some Grassmannian.

We will call the Grassmannian here $\mathrm{Gr}$, for simplicity of notation. We’re almost through, now. Look at $U_{Gr}$, the universal bundle on $\mathrm{Gr}$, which over each point has the subbundle corresponding to it. We have a map $U_{Gr}\to \mathrm{Gr}$. Now let $\pi_1:\mathrm{Gr}\times_S \mathbb{P}^m\to\mathrm{Gr}$ and $\pi_2:\mathrm{Gr}\times_S \mathbb{P}^m\to \mathbb{P}^m$. Then we can define a map $\pi^*_1 U_{Gr}\to \pi_1^*f^*f_*\mathscr{O}_{\mathbb{P}^m}(N)\to \pi_2^*\mathscr{O}_{\mathbb{P}^m}(N)$, and take $F$ to be the cokernel of the whole composition. Then we look at the sheaf $F(-N)$.

There is a largest subscheme $i:G_P\to \mathrm{Gr}$ with $i^*F(-N)$ as a sheaf on $G_P\times_S \mathbb{P}^m$ is flat with Hilbert polynomial $P$. By largest, we mean that any other such scheme will factor through it. This scheme turns out to represent the functor $Hilb_P(\mathbb{P}^m/S)$. As we’re only going to care about families over connected bases, we then take the disjoint union of these over integer polynomials in order to get the Hilbert scheme.

So now we have the Hilbert Scheme, and we got it by using the Grassmannian. We can now use it to construct some other moduli spaces. Here’s an important one: Let $X/S,Y/S$ be schemes. Then $\hom(X,Y)$, a functor taking $T$ to $X\times_S T\to Y\times_S T$, that is, to families of morphisms parameterized by $T$. Well, we can represent this now. Look at the Hilbert scheme of $X\times_S Y$. Then look at the collection of subschemes which are the graphs of morphisms. Then this locus will precisely represent the $\hom(X,Y)$ functor. Now, I believe that this means that the category of schemes over $S$ is enriched over itself, though I’ve never seen anyone say that, so I’m not sure if it’s really allowed. John, could you answer that one?

In general, if you want to construct a fine moduli space, the first thing to try is to see if it can be written as a subscheme of the Hilbert Scheme.

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
This entry was posted in AG From the Beginning, Algebraic Geometry, Deformation Theory, Examples, Hilbert Scheme. Bookmark the permalink.

### 17 Responses to The Hilbert Scheme

1. Now, I believe that this means that the category of schemes over is enriched over itself, though I’ve never seen anyone say that, so I’m not sure if it’s really allowed.

Allowed to say that a category is enriched over itself? Sure. It’s one of the fundamental interesting properties of topoi, for instance, and it’s necessary for any category to have an “internal hom”. That is, well-behaved categories of sheaves are enriched over themselves, and it’s precisely because we want that property — that internal hom — that we have to deal with certain weird objects.

Remember: it’s often the case that you can’t have both nice objects and a nice category. To get a nice category you have to allow weird objects. And usually you’re better off for it!

2. Jason Starr says:

Why is there such an N? Why is there a flattening stratification G_P? This is a nice post. But the “work” in proving representability of the Hilbert functor lies in answering these 2 questions. By the way, if you want a fun exercise, try to reduce Theorem I.5.16 in Koll\’ar’s book (representability of the Quot functor) to Theorem I.1.4. According to the comments in his book, this is apparently not how Koll\’ar expects you to prove this theorem. And the most important comment I can make — you misspelled Koll\’ar’s name! I now from painful experience that if Koll\’ar learns of this, he will correct you :)

3. Jason Starr says:

2. “now” in the last sentence should be “know”

4. Todd Trimble says:

Charles, when you asked whether it is “allowed” to say that Sch/S is enriched over itself, it sounds like you were really asking “is it true”? To put it more precisely, it sounds like you were asking whether Sch/S is cartesian closed. The answer to that is surely “no”, but I’m not sure I can argue that convincingly right now.

Suppose we just take S = Spec(Z), so that Sch/S = Sch; the question is whether for any scheme X we have that X x – as a functor Sch –> Sch has a right adjoint. A necessary condition would be that X x – preserves colimits, in particular coequalizers. I’d think that this would fail for something like X = Spec(Z_3), roughly because tensoring with Z_3 in the category of commutative rings does not preserve equalizers. There’s still more work to do before this argument could be considered tight, but an assertion of cartesian closure for Sch should be greeted with severe skepticism, I feel.

5. Charles says:

Jason – I’m not completely done. I am intending to keep on the Hilbert Scheme for a bit, and was giving serious thought to proving the lemmas needed next week, but breaking the flow of this post to prove them didn’t feel right.

Todd – I was just asking if it was acceptable terminology or if I had missed a part of the definition. And the definition I’ve seen doesn’t include being cartesian closed, is that a standard part of being enriched over itself? I had thought it was just about the homs being objects in the category, but I’m certainly open to me having misread a definition.

6. Charles says:

Oh, also, I fixed the Kollár’s name.

7. Well, it’s not just about that. What you need is for the category to be a monoidal category, to have “hom-objects” (instead of sets), to have a composition morphism (which depends in its very context on the monoidal structure of the category), and so on.

I think Todd’s point is that to set up internal homs we assume that the category is symmetric, monoidal, and closed. You seem to be using the categorical product as your monoidal structure (look to your composition morphism!), and so you’re asking for your category to be Cartesian-closed.

You can define a category enriched over an arbitrary monoidal category, but you need to be careful when you want to turn the whole thing in on itself.

8. Todd Trimble says:

Sorry, Charles — it may have been rude of me to put words in your mouth. You are right that a priori, enrichment means less than cartesian closure — I was really looking at your description of hom(X, Y) and inferring that cartesian closure was what really being asserted.

Anyway, I’d not heard of a global construction for the scheme hom(X, Y); do you have a reference for that? Is there some fine print that I’m missing?

9. Charles says:

I don’t have my books with me at the moment, but it’s in Kollar. Next week, I think, will be Hilbert Scheme week, so once I’ve patched up the missing bits of this post, I’m going to more carefully discuss the scheme $\hom_S(X,Y)$. I mostly just wanted this post to end on the high point of “Look! hom sets are really hom schemes for us!” and return to it later to be more precise.

10. Todd Trimble says:

Okay, thanks; I’ll sit tight. I’ve been poking around a little and do find references to this construction, but AFAICT there may be some fine print involved (e.g., X is assumed flat over S). Still, it’s very interesting and my interest has been piqued!

11. Jason Starr says:

Todd — Probably Charles will post about this, but you are correct that typically one requires the domain scheme to be projective and flat over S (and one requires the target scheme to be quasi-projective over S) in order to prove the relative Hom scheme is representable by a scheme whose connected components are quasi-projective over S.

12. Charles says:

And John, something I meant to add, but had forgotten: the same dichotomy seems to hold for moduli spaces. A nice moduli space will contain some not-so-nice objects, but if you just want the nice things (like smooth curves of a given genus) you don’t get a nice moduli space.

In that example, the nice moduli space isn’t complete, but to complete it, we need to throw in reducible curves.

13. Anonymous says:

You wrote:
“the Hilbert Scheme. This scheme, and it isn’t a variety because it isn’t reduced in general, parameterizes families of projective varieties.”

This is wrong. Should be “parametrizes families of sub-schemes” and not varieties. For, some points of a Hilbert scheme may correspond to subschemes and not subvarieties. Eg. some points of P^2 correspond to double points of P^1, when we view P^2 as the Hilbert schemes of two points of P^1 (that is, Hilbert scheme of 0-dim subschemes of lenght 2 of P^1)

Just pointing this out since you stress yourself that a Hilbert scheme is a scheme and not a variety since it can be non-reduced…

14. Ack, and worse, that’s something I know and have made use of. I’m a bit annoyed that slipped by me. Correcting!