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Hi, I'm just starting to learn about deformation theory (via Hartshorne's Deformation theory, as well as Fantechi's section of FGA explained), and I feel like I'm confused about fundamental concepts. So please indulge me even if the question is well-known, or trivial.

So suppose we have a projective variety $Y \subseteq \mathbb{P}_k^n$. Then we can consider two natural objects associated to it, namely, the Hilbert scheme, which parametrizes all the objects with the same Hilbert polynomial as $Y$, and the (versal?) deformation space, which represents the deformation functor $F: (Art)_k \to (Sets)$.

My question is, is there a relationship between these two spaces? My hunch is that the Hilbert scheme is included in some way (perhaps via immersion, although that sounds kind of strong) to the deformation space - I guess this means that any deformation over an Artin ring doesn't change the Hilbert polynomial - but I can't formulate any coherent and believable conjecture right now. (If the question doesn't make sense, could you answer the right question that most closely approximates it?)

Thank you for reading.

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up vote 8 down vote accepted

When you say "deformation functor", you have to be careful to specify exactly which functor you are thinking about. There are two relevant deformation functors at play here.

One is the functor $D$ of abstract deformations of $Y$. If $A$ is an Artin local $k$-algebra, then $D(A)$ is the set of isomorphism classes of flat families $Y_A$ over $A$ whose central fiber is isomorphic to $Y$.

The other functor $D'$ is the functor of deformations of $Y$ inside $\mathbb{P}^n_k$. Here, $D'(A)$ is the set of isomorphism classes of flat families $Y_A$ together with an embedding $Y_A \rightarrow \mathbb{P}^n_k \times Spec A$ such that the central fiber is $Y \rightarrow \mathbb{P}^n_k$ (not isomorphic to $Y$, but exactly $Y$, as we are talking about subschemes of $\mathbb{P}^n_k$).

The functor $D$ does not have any direct relationship to the Hilbert scheme, but the functor D' certainly does.

Given any scheme $X$ and a point $x \in X$, the local ring $\mathcal{O}_x$ defines a functor $F: (Art)_k \rightarrow (Sets)$ by setting $F(A) = Hom(Spec A, Spec \mathcal{O}_x)$. We say that $F$ is pro-represented by $\mathcal{O}_x$.

If you take the local ring $\mathcal{O}_{[Y]}$ of the Hilbert scheme (which you probably know to be representable by an honest scheme by a theorem of Grothendieck) at the point $[Y]$, then the functor it pro-represents is none other than $D'$.

Being pro-representable is stronger than having a versal deformation space; it means that the functor has a universal deformation space, which is the formal completion of the point whose local ring is doing the pro-representing.

So the (uni)versal deformation space of $D'$ is the formal completion of the point $[Y]$ in the Hilbert scheme. As for the versal deformation space of $D$, I am not really sure what it looks like, but there is a morphism $D' \rightarrow D$ of deformation functors given by forgetting the embedding of $Y_A \rightarrow \mathbb{P}^n_k \times Spec A$.

In general, I think of a versal deformation space as an infinitesimal object; it is only keeping track of what happens when you deform $Y$ a little bit. The Hilbert scheme, on the other hand, is global; it is keeping track of all subschemes with the same Hilbert polynomial as $Y$, including ones which might be far away from $Y$ (if $Y$ was a point, for instance). Thus, you would expect the versal deformation space of the appropriate functor to map into the Hilbert scheme, rather than the other way around.

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So I have one more question: You defined a functor F: (Art)_k \to (sets) by F(A) = Hom(Spec A, Spec O_x). Will this still be a deformation functor? I don't see why, if it is so. –  user14449 Apr 20 '11 at 16:54
    
Rex, you should be a little careful - the versal deformation space will not map the Hilbert scheme. As you said, the versal deformation space is only keeping track of deformations over local artin rings - and these deformations need not be embedded in projective space. Formally locally around the point [Y] though, there is a map from the Hilbert scheme to the deformation space - which is what both of our answers point out. –  mdeland Apr 20 '11 at 23:52
    
SL - Rex is saying that the local ring of the Hilbert scheme prorepresents the embedded deformation functor (he calls this D'). The local ring of any point of a scheme prorepresents maps to that point in the scheme - which is the functor he writes down. –  mdeland Apr 20 '11 at 23:55
    
SL - I don't know what your precise definition of a deformation functor is, but the functor F: (Art)_k \to (sets) given by F(A) = Hom(Spec A, Spec O_x) is considered the most canonical example of a deformation functor. If your definition does not admit this example, then it is at odds with the designs of deformation theory. I think the only universal requirement for a deformation functor is that F(k) = {pt}. Some people go further and require that some of the Schlessinger conditions be fulfilled. Here, F is pro-representable by definition, so it satisfies all of the Schlessinger conditions. –  user332 Apr 21 '11 at 4:30
    
mdeland - I was not referring specifically to D or D' when I said "versal deformation space". Rather, I meant to point out that versal deformation spaces tend to look like formal completions of schemes at a point, so they are really small, local objects such as Spf O_x. One would not expect a big scheme like the Hilbert scheme to map into them in an interesting way. –  user332 Apr 21 '11 at 4:35
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Here's one interpretation if you're content with thinking about the tangent spaces of these functors:

Suppose that $Y$ is a smooth (you can get away with less here) subvariety of $\mathbb{P}^n$. You have the short exact sequence

$0 \rightarrow T_Y \rightarrow i^*T_{\mathbb{P}^n} \rightarrow N_{Y/\mathbb{P}^n} \rightarrow 0.$

Infinitesimal embedded deformations of $Y$ inside $\mathbb{P}^n$ are given by global sections $H^0(Y, N_{Y/\mathbb{P}^n})$. Infinitesimal deformations of $Y$ abstractly are parameterized by $H^1(Y, T_Y)$. There is a forgetful map of functors sending an embedded deformation to the abstract deformation (that is, forget the embedding). Once you know that those functors are (pro)representable (or even if you don't), then the differential map for the corresponding morphism of spaces/functors is given by the connecting homomorphism in the long exact sequence of cohomology (from the above short exact sequence):

$\delta: H^0(Y, N_{Y/\mathbb{P}^n}) \rightarrow H^1(Y, T_Y)$

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