Timeline for When do infinitesimal deformations lift to global deformations?
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| Feb 17, 2012 at 9:33 | answer | added | Naga Venkata | timeline score: 0 | |
| Feb 16, 2012 at 16:07 | comment | added | Emerton | ... that you should also deform an ample line bundle, so that the universal formal deformation is actually projective, and hence algebraic. (This explains e.g. why the formal deformation space of a K3 is 20-dim'l, while to get actual algebraic families of K3's we have to cut down to 19 dimensions. Similarly, the formal deformation space of an abelian variety of dimension $g$ is $g^2$ dimensional, but to get actual algebraic families, we have to cut down to dimension $g(g+1)/2$.) Regards, Matthew | |
| Feb 16, 2012 at 16:05 | comment | added | Emerton | 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 ... | |
| Feb 16, 2012 at 14:37 | answer | added | Paul | timeline score: 2 | |
| Feb 16, 2012 at 5:18 | comment | added | Matt | 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. | |
| Feb 16, 2012 at 3:52 | comment | added | Paul | @Matt:$C=$Complex numbers, or any field. | |
| Feb 16, 2012 at 1:18 | comment | added | Matt | @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. | |
| Feb 16, 2012 at 0:49 | comment | added | Paul | 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. | |
| Feb 15, 2012 at 20:25 | history | edited | Charles Matthews | CC BY-SA 3.0 |
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| Feb 15, 2012 at 20:24 | comment | added | Felipe Voloch | There is an obstruction living on an $H^2$. You might more useful answers if you ask a more specific question. | |
| Feb 15, 2012 at 20:17 | history | asked | Naga Venkata | CC BY-SA 3.0 |