29

I'll post an answer to what I think is a reasonable question, hoping that someone more expert will improve on this answer. My apologies if this answer is too chatty.
First, a very simple reason you might hope that there is something like a cotangent complex and that it should be an object in the derived category of quasi-coherent sheaves. Given a morphism $...

25

One of my favourite examples is the following theorem, due to S. Mori:
Theorem A. Let $X$ be a smooth complex projective variety such that $-K_X$ is ample. Then $X$ contains a rational curve. In fact, through any point $x \in X$ there is a rational curve $D$ such that $$ 0 < -(D \cdot K_X )\leq \dim X+1.$$
In other words, smooth Fano varieties over $\...

answered Jun 20 '13 at 7:20

Francesco Polizzi

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23

This approach to deformations is taken, for instance, in all of the original papers of Kodaira-Spencer and Nirenberg. You can have a look at
On the existence of deformations of complex analytic structures, Annals, Vol.68, No.2, 1958
http://www.jstor.org/discover/10.2307/1970256?uid=3737608&uid=2129&uid=2&uid=70&uid=4&sid=...

21

The answer in general is no. Nakamura has constructed here (pp.90, 96-99, solvmanifolds of type III-(3b)) an example of a compact complex (non-Kähler) manifold $M$ with $TM$ holomorphically trivial (so in particular $K_M$ is holomorphically trivial) which has arbitrarily small deformations $M_t$ with negative Kodaira dimension.
On the other hand, if $M$...

18

$\DeclareMathOperator{\Ext}{Ext} \newcommand{\G}{\hat{\mathbb{G}}} \DeclareMathOperator{\Maps}{Maps} \renewcommand{\phi}{\varphi}$ The analysis of the infinitesimal deformation space of the Honda formal groups $H_n$ uses three calculations which govern the existence of square-zero deformations. These can be phrased in terms of certain $\Ext$ groups, and ...

17

I agree. Thanks for discovering the error. And by the way there is another error on the same page, line -1, there is a -2 that should be a -4.
Robin Hartshorne

16

As Igor mentions in the comments, this is really a question about deformations of the multiplication map of the exterior algebra in the space of associative multiplications. Since this is a pretty general story, let me try to sketch how it works in this case.
Let's fix some field $k$ and a $k$-algebra $(A,\mu)$. We want to understand deformations of $(A,\mu)...

16

Example. $\pi: \{(z,w)\in {\mathbb C}^2: |zw|<1\}\to {\mathbb C}$, $\pi(z,w)=z$.
Edit. Similarly, to get a nontrivial deformation of ${\mathbb C}^n$, consider
$$
X=\{(z_0, z_1,...,z_n)\in {\mathbb C}^{n+1}: |z_0 z_1|<1\}
$$
and let $\pi$ be the projection of $X$ to ${\mathbb C}$ which the 1-st coordinate line in ${\mathbb C}^{n+1}$. Then $\pi^{-1}(t)...

15

When $n=3$, this is in Stasheff's H-spaces from a homotopy point of view, Chapter 12. For general $n$, it is in a paper of mine with Lu, Wu, and Zhang, "$A_\infty$-structures in Ext algebras, J. Pure Appl. Alg. 213 (2009), 2017--2037 (Theorem 3.1 and Corollary A.5).

15

If you take, say, set of real points of the group-scheme $O(n)$, i.e., $O(n, {\mathbb R})$, then you recover the usual orthogonal (real Lie) group, which you know as $O(n)$. Same applies to $SL(n)$, etc. There is one case when this does not work well, namely when you deal with character varieties. For instance, take $\pi$, say, the free group on two ...

15

The Quillen-Drinfeld-Deligne-etc. philosopy should not be looked at as something too mysterious.
Namely, it reduces to the fact that if the set of objects one is interesting in the infinitesimal deformations of is not too wild, then it can be described in the form $f(v)+Q(v)=0$, where $f:V\to W$ is a linear function and $Q:V \to W$ is a quadratic function. ...

14

I am just posting my comment above as an answer. Of course the only rigid curve is $\mathbb{P}^1$, and this specializes to $\mathbb{P}^1$. Building on Mori's work, Siu proved that every smooth, projective specialization of $\mathbb{P}^n$ is $\mathbb{P}^n$. The analogous result for quadric hypersurfaces in $\mathbb{P}^n$ with $n\geq 4$ was proved by Jun-...

13

Perhaps one way of reading this question is "Why is it important to think about complexes when doing deformation theory?" One must certainly accept that deformation theory is cohomological in nature: consider the standard Čech construction of classes in $H^1(X, T_X)$ and $H^2(X, T_X)$ corresponding respectively to deformations and obstructions to ...

13

There is indeed a proof along these lines. Suppose one has a polynomial $f$ over $\mathbb{Z}_p$; I'll use $f$ to refer to its reductions mod $p^k$ as well. You have a diagram:
$$\text{Spec}((\mathbb{Z}/p^k)[t]/f(t))\to~\text{Spec}((\mathbb{Z}/p^{k+1})[t]/f(t))~~~~~~~~~~~~~$$
$$\downarrow\uparrow~~~~~~~~~~~~~~~~~~~~~~\downarrow$$
$$\text{Spec}(\mathbb{Z}/p^...

13

No, that is not true. I am sure that somebody else has answered this on MO before. I guess the simplest example is a family of genus 6 curves, where some fibers are embeddable as plane quintics. If memory serves, the canonical image of a general genus 6 curve (non-hyperelliptic) is the intersection of a Pfaffian quintic del Pezzo surface and a quadric ...

13

Unfortunately, there is no real textbook on DQ around. One has Fedosov's book on his construction of star products including a detailed exposition of his index theorem.
There is a chapter on DQ in the recent Poisson geometry book by
Laurent-Gengoux, Pichereau, Vanhaecke.
In the conference proceedings of the PQR2003 by Gutt, Rawnsley, and Sternheimer one ...

answered May 13 '16 at 16:50

Stefan Waldmann

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13

The answer is yes, in fact the following result holds.
Theorem. Let $f \colon X \to T$ be a flat deformation of a Fano variety $X_0:=f^{-1}(0)$ having at most terminal, $\mathbb{Q}$-factorial singularities.
Then $X_t:=f^{-1}(t)$ is a Fano variety with at most terminal, $\mathbb{Q}$-factorial singularities, for all $t$ in a neighborhood of $0$ in $T$.
...

answered Apr 7 '17 at 13:12

Francesco Polizzi

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12

Yes, in the following sense. Pick a trivalent graph $G$ with $v$ vertices and regard it as the dual complex of the stable curve $E$ consisting of one copy of $\mathbb P^1$ for each vertex of $G$ and one node for each edge. The genus $g$ of $E$ is given by $2g-2=v$. The stack of stable curves of genus $g$ is irreducible, so any smooth curve $C$ of genus $g$ ...

12

These curves may actually exist, as the following example shows.
Let $C$ be a smooth curve of genus $g$. Then the diagonal $\Delta \subset C \times C$ is isomorphic to $C$ and has self intersection $\Delta^2= 2-2g$.
On the other hand, the tangent space of the Hilbert scheme $\mathscr{H}$ of $\Delta$ in $C \times C$ at the point $[\Delta]$ has dimension ...

answered Mar 4 '13 at 11:05

Francesco Polizzi

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12

If p is a nonsingular point, then we can define a tangent vector as an equivalence class of (nonsingular) curves, just as in the differentiable case. In fact, being nonsingular is equivalent to every tangent vector being tangent to a curve.
In this case, the intersection of all curves in the equivalence class is a zero-dimensional closed subscheme with a one-...

12

If the base of the family is a quasi projective variety, by taking sections and base change you can reduced to the case where $\mathcal X$ is a surface fibered over a smooth curve $B$.
Then one can normalize $\mathcal X$ and then solve the remaining singularities.
In this way one gets a new suface $\mathcal X'$ fibered over the same base $B$.
The general ...

12

The answer is yes (in char. 0). Indeed, it suffices to show that for an infinitesimal deformation $\mathcal{X}$ of $X$ over an artinian algebra $A$, the cohomology $H^0(\mathcal{X}, \omega_{\mathcal{X}/A})$ is locally free with formation commuting with base change. But this is a Hodge cohomology group, and the required assertion follows from results of ...

11

Since the Weyl algebra is a deformation of the polynomial ring in two variables, there is a short transparent proof using deformation theory, cf. M. Gerstenhaber and A. Giaquinto, On the cohomology of the Weyl algebra, the quantum plane, and the q-Weyl algebra, arXiv:1208.0346.

11

One nice way to say this in the language of schemes is that the tangent space of a (say complex) variety $X$ at a point $p$ is the set of morphisms from the scheme $Spec(\mathbb{C}[\epsilon]/(\epsilon^2))$ to $X$ which send the closed point to $p$. In other words, we are looking at paths in $X$, passing through $p$, but we only consider the paths up to 1st ...

11

The rigidity of quotient singularities in dimension greater or equal than $3$ was established by Schlessinger in his paper Rigidity of quotient singularities, Invent. Math. 14 (1971). Roughly speaking, he proved that if $(X, \,x)$ is a local scheme with an isolated singularity at $x$ and $\dim X \geq 3$, then deforming $X$ is equivalent to deform the ...

answered Jan 30 '14 at 10:23

Francesco Polizzi

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10

The Hilbert scheme is pretty horribly behaved, and positive results of this nature are quite rare.
You probably mean to ask if there exists $C\in L$ such that $C$ is both reduced and irreducible. If you don't demand reducedness, the answer is yes. For suppose $H$ is a component of a Hilbert scheme of curves in $\mathbb{P}^n$, and let $C\in H$ be a general ...

10

I do not know about the general deformation theory (if there is such a thing), so I will talk about the special case I am familiar with, namely, representation varieties $R=Hom(\pi,G)$ of representations of finitely-generated groups $\pi$ to a Lie group $G$. In this case, as in some other cases, each analytical germ $(R,r)$ of $R$, regarded as the "...

10

The classical reference is Illusie's PhD thesis "complexe cotangent et deformations".There the result without the marked points is proven in a very, very general context using the cotangent complex; then one needs to know that for a smoooth variety, or for a nodal curve (or, again, in many more cases) the cotangent complex is isomorphic in the derived ...

10

I think I can prove the following
Theorem: Let $R$ be a $k$-algebra of finite type with a closed point $x\in \mathrm{Spec} R$ such that $\mathrm{Spec} R\setminus \{x\}\cong \mathbb{A}^2_k\setminus \{(0,0)\}$. Let $\tilde{R}$ be a $t$-adically complete flat lift of $R$ to $k[[t]]$. Then $\mathrm{Spf} \tilde{R}\setminus \{x\}\cong \mathbb{A}^2_{\mathrm{Spf} ...

10

It is not true, but something similar is true. If $X$ is a scheme, the functor you describe is actually $\mathsf{Qcoh}(X) \to \mathsf{Ab}(\mathsf{QAlg}(X)/\mathcal{O}_X)$, where $\mathsf{QAlg}(X)$ denotes the category of quasi-coherent algebras on $X$, and $\mathsf{QAlg}(X)/\mathcal{O}_X$ is the slice category consisting of homomorphisms $A \to \mathcal{O}_X$...

answered Jul 8 '13 at 10:34

Martin Brandenburg

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