Relationship between Hilbert schemes and deformation spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:40:35Z http://mathoverflow.net/feeds/question/62159 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62159/relationship-between-hilbert-schemes-and-deformation-spaces Relationship between Hilbert schemes and deformation spaces SL 2011-04-18T19:12:00Z 2011-04-18T22:34:58Z <p>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.</p> <p>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 $F: (Art)_k \to (Sets)$.</p> <p>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?)</p> <p>Thank you for reading.</p> http://mathoverflow.net/questions/62159/relationship-between-hilbert-schemes-and-deformation-spaces/62165#62165 Answer by AH for Relationship between Hilbert schemes and deformation spaces AH 2011-04-18T20:06:25Z 2011-04-18T20:28:08Z <p>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.</p> <p>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 $Y$.</p> <p>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 $\mathbb{P}^n_k$).</p> <p>The functor $D$ does not have any direct relationship to the Hilbert scheme, but the functor D' certainly does.</p> <p>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 <em>pro-represented</em> by $\mathcal{O}_x$.</p> <p>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 $D'$.</p> <p>Being pro-representable is stronger than having a versal deformation space; it means that the functor has a <em>universal deformation space</em>, which is the formal completion of the point whose local ring is doing the pro-representing.</p> <p>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 $Y_A \rightarrow \mathbb{P}^n_k \times Spec A$.</p> <p>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.</p> http://mathoverflow.net/questions/62159/relationship-between-hilbert-schemes-and-deformation-spaces/62179#62179 Answer by mdeland for Relationship between Hilbert schemes and deformation spaces mdeland 2011-04-18T22:34:58Z 2011-04-18T22:34:58Z <p>Here's one interpretation if you're content with thinking about the tangent spaces of these functors:</p> <p>Suppose that $Y$ is a smooth (you can get away with less here) subvariety of $\mathbb{P}^n$. You have the short exact sequence </p> <p>$0 \rightarrow T_Y \rightarrow i^*T_{\mathbb{P}^n} \rightarrow N_{Y/\mathbb{P}^n} \rightarrow 0.$</p> <p>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): </p> <p>$\delta: H^0(Y, N_{Y/\mathbb{P}^n}) \rightarrow H^1(Y, T_Y)$</p>