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When does one use infinitesimal deformation (over the ring of dual numbers $k[t]/(t^2)$) versus local deformation (over $k[t]$ or $k[t_1,\ldots, t_n]$)?

It seems that one works over the ring of dual numbers in order to remove or obtain certain singularities (or to compute for degenerate schemes) while one of the reasons one works over $k[t]$ or $k[t_1,\ldots, t_n]$ is to study and relate fibers over various base points, assuming that the ring of interest is a free $k[t]$ or $k[t_1,\ldots, t_n]$-module.

Would you say that this correct?

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2 Answers 2

up vote 0 down vote accepted

I think what I wrote above is correct since $k[x,y]/(xy)$ has a 1-dimensional space of deformations over the dual numbers.

$$ $$ This link is also helpful.

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The standard example of a deformation is a proper flat map $f:X \longrightarrow S$ between schemes. The terms infinitesimal or local characterize the base $S$.

A local base like $S = Spec \ k[t_1,...,t_n]$ often serves as an affine neighbourhood of a base point. $S$ has many closed points. So you can compare fibres $X_s = f^{-1}(s), s \in S$, over different base points $s \in S$ as well as the variation of the fibre $X_s$ when $s$ moves through $S$.

The other extrem is a base like $S = Spec \ \mathbb C$. Here you study a single fibre of $f$ on its own.

The intermediate steps are infinitesimal neighbourhoods of a single fibre $X_s$: You choose

$$S = Spec \ k[t_1, ..., t_n]/<t_1,...,t_n>^k, k \in \mathbb N,$$

an infinitesimal neighbourhood of the point $s$. The most simple non-trivial case is $S= Spec \ k[t]/<t^2>$, the double point. The family of all infinitesimal neighbourhoods is the formal neighbourhood of $s$. Extending information about $f$ from the formal neighbourhood of a base point to a local neighbourhood starts with comparison theorems for base change.

One of the first examples of building bottom-up a deformation via infinitesimal neighbourhoods is the construction of a versal deformation of a compact complex manifold $X_0$ with $H^2(X_0, \Theta_{X_0}) = 0$ by Kodaira-Nirenberg-Spencer.

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