most recent 30 from http://mathoverflow.net 2013-05-24T18:37:24Z http://mathoverflow.net/feeds/question/88552 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88552/when-do-infinitesimal-deformations-lift-to-global-deformations Naga Venkata 2012-02-15T20:17:46Z 2012-02-17T09:33:27Z

<p>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. </p>

http://mathoverflow.net/questions/88552/when-do-infinitesimal-deformations-lift-to-global-deformations/88634#88634 Paul 2012-02-16T14:37:13Z 2012-02-16T14:44:15Z

<p>@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 $z^2$. </p> <p>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).</p> <p>All this fits in a larger context (e.g. Artin' paper <a href="http://www.ams.org/mathscinet-getitem?mr=232018" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=232018</a> and $R[t]/t^n$, etc.)</p>

http://mathoverflow.net/questions/88552/when-do-infinitesimal-deformations-lift-to-global-deformations/88702#88702 Naga Venkata 2012-02-17T09:33:27Z 2012-02-17T09:33:27Z

<p>@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 $I(V)$? </p>