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.

3There is an obstruction living on an $H^2$. You might more useful answers if you ask a more specific question. – Felipe Voloch Feb 15 '12 at 20:24

1Here'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 1diml 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 '12 at 0:49

1@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 '12 at 1:18

1Hmm...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 (schemetheoretic?) 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 nontrivial 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 '12 at 5:18

3Dear 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 '12 at 16:05
@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 nonlinear 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/mathscinetgetitem?mr=232018 and $R[t]/t^n$, etc.)

Oh I see what you're doing now. I was being dumb. The ker df should have tipped me off. I was very confused at what some ambient space was doing. I usually call what you described an "embedded deformation" (maybe this is not standard?). To me an infinitesimal deformation of $X_0/k$ is a flat map $X\to Spec (k[x]/(x^2))$ with special fiber $X_0$. Hartshorne defines global deformation to be any flat map $X\to T$ where some fiber $X_t\simeq X_0$. My guess is that Naga wants specific conditions on T (integral?), otherwise every infinitesimal is global. – Matt Feb 16 '12 at 18:36

OK, my use of "$x^2$" was confusing, since it also arises in $k[x]/(x^2)$, or equivalently in taking the linearization $df$. I should have used $f(x)=x^3$! – Paul Feb 16 '12 at 20:41
@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 $B/(ta)$ 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. nonreduced)? Would I still get an ideal isomorphic to $I(V)$?

1It's probably better to edit your original question to add these comments. – Donu Arapura Feb 17 '12 at 12:28

Dear Naga, What does ``the image in $B/(ta)$ mean? Since you have already imposed $t^2 = 0$, the only value of $a$ such that $ta$ is not coprime to $t^2$ is $a = 0$. In otherwords, you cannot form the reduction mod $t  a$ unless $a = 0$ ; this is the reason they are called formal, or infinitesimal, deformations. Regards, Matthew – Emerton Feb 18 '12 at 0:49

@Emerton: Thanks a lot for the correction. I am looking into the criterion when formal deformations can be turned into true algebraic deformations. This was in the heart of my question although I got it all wrong while formalizing it. I guess the reference for this is FGA (on Grothendieck's existence theorem in formal geometry). – Naga Venkata Feb 19 '12 at 8:18