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Hi, I understand the notion of Infinitesimal and global deformations and the fact that global deformations lead to certain infinitesimal deformations. But I could not find any criterion or idea to understand when an infinitesimal deformation cannot be lifted to a global deformation.

edited Feb 15 2012 at 20:25

Charles Matthews
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asked Feb 15 2012 at 20:17

Naga Venkata
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There is an obstruction living on an $H^2$. You might more useful answers if you ask a more specific question. – Felipe Voloch Feb 15 2012 at 20:24

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Here's a ``germ'' of an idea that might help: let $f:C\to C$ be $f(x)=x^2$. The infinitesimal deformations of $f^{-1}(0)$ is the 1-diml space $\ker df$. But $f^{-1}(0)$ is isolated, hence not deformable. Most of the examples of what you ask are elaborations of this simple example. – Paul Feb 16 2012 at 0:49

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@Paul What is C? There is the tag algebraic geometry, so presumable it is a curve? If so, then I don't think $f(x)=x^2$ makes sense, or "0". Please elaborate. – Matt Feb 16 2012 at 1:18

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Hmm...maybe there is just a translation issue. So you're thinking of $\mathbb{C}$ as the closed points of $Spec (\mathbb{C}[x])$? Then you pick the (scheme-theoretic?) fiber over the ramified point? Are you considering deformations of the map or of this fiber as a scheme? It seems to me there should be no non-trivial infinitesimal deformations as well...but maybe I'm just complicating matters. You could post this example expanded as an answer. I'm apparently just being dense, since someone else thinks it's a good answer as they upvoted it. Thanks. Sorry for the confusion. – Matt Feb 16 2012 at 5:18

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Dear Naga, As a consideration of Felipe's comment and Paul's comment and answer will show, there is uncertainty about your question: Do you want to understand why formal deformation spaces can be singular (or, equivalently, obstructed), i.e. why first order deformations do not always lift to higher order deformations (this is what Felipe's and Paul's comments/answer address)? Or do you want to understand whether formal deformations can be turned into true algebraic deformations? This latter question is address for example by the Grothendieck existence theorem; the basic criterion is ... – Emerton Feb 16 2012 at 16:05

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Paul · Answer 1 · 2012-02-16 14:44:15Z UTC

@Matt. I'm not really an expert here, and maybe I misunderstood the question, so here's my low tech explanation, and someone can add details or correct me. First I think the question really was "infinitesimal" v. "local" deformations, as I'm not sure what a "global" deformation is. I interpret "infinitesimal" deformations as elements of a zariski tangent space. If $f:{\bf C}^a\to {\bf C}^b$ is a map, then the tangent space at $v\in V=f^{-1}(0)$ is $\ker df$. At a smooth (i.e.submersion) point, $V$ is locally isomorphic to $\ker df$, and so every infinitesimal deformation is locally deformable. But at a singular point some formal tangent vectors (i.e. vectors in the zariski tangent space) aren't tangent vectors, e.g. in the example of $z^2$.

Typically one is trying to deform some structure, i.e. move around in some kind of moduli space, and there is a group acting, so that the zariski tangent space to the moduli space at a point is a quotient $\ker df_v/im~ g$, where $g$ is the orbit map. Frequently this quotient is identified with $H^1$ of some complex ($g= d_0, df_v= d_1$), so that the infinitesimal deformations near $v$ are identified with some $H^1$. Kuranishi's method (for example) is a general trick which locally describes the moduli space near $v$ as cut out by a non-linear map $k:H^1\to H^2$, so that $k^{-1}(0)$ describes the local deformations inside the infinitesimal deformations $H^1$. In many cases the quadratic part of $k$ is given algebraically (i.e. by cup products) and in really nice cases (e.g. $H^2=0$) this is enough to completely describe the local deformations (hence Felipe's comment).

All this fits in a larger context (e.g. Artin' paper http://www.ams.org/mathscinet-getitem?mr=232018 and $R[t]/t^n$, etc.)

Naga Venkata · Answer 2 · 2012-02-17 09:33:27Z UTC

@Emerton and Paul: Thanks a lot for the answers. Sorry for being so vague. However the answers include what I was looking for. More specifically, I was interested in the following: As I understand, infinitesimal deformation of a variety $V$ in $k^n$, means considering an ideal $I'(V)$ in $B:=k[X_1,...,X_n,t]/(t^2)$ such that the image of $I'(V)$ in $B/(t)$ is $I(V)$. So I guess if take the image in $B/(t-a)$ I should get ideals isomorphic to $I(V)$, hence with the same Hilbert polynomial. This is what I meant by global deformations coming from local deformations. But the confusion is that if this is true, then what happens when the family containing $V$ is not smooth (eg. non-reduced)? Would I still get an ideal isomorphic to $I(V)$?

When do infinitesimal deformations lift to global deformations?

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