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The following baby version of virtual fundamental cycle is well known:

Let $M\subset V$ be the zero locus of a section $s$ of a vector bandle $E \to V$, in general $s$ is not transversal to the zero section and $M$ does not have the expected dimension, then one uses excess intersection theory to define the virtual fundamental class $[M]^{vir} $ to be $0_E^{!}[C_{M/V}]\in A_{vdim}(M)$, where $C_{M/V}\subset E|_M$ is the normal cone of $M$ in $V$.

If $V$ is smooth, one can reformulate these into Behrend-Fantachi's language by setting $E^{-1}=E^{\vee}|_M, E^0=\Omega_V|_M$, and get a perfect obstruction theory $E^* \to L_{M}$ associated to the model $(V,E,s)$.

My question is whether the reformulation can be reversed:

Given a perfect obstruction theory $E^* \to L_{M}$, can we return to the baby version locally? I mean whether we can find open covering (in some topology) of $M$, such that for each open subset $U$ in the covering, we can find $U\subset V$,a vector bundle $F \to V$ and $U$ is realized as zero locus of a section $s$. And $[E^* \to L_{M}]|_U$ coincides with the perfect obstruction theory associated to the local model $(V,F,s)$. I will appreciate if you could provide the details.

Thank you!

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These kinds of questions seem to be best dealt with in terms of quasi-smooth derived schemes. The basic idea is that any perfect obstruction theory in nature actually comes from the cotangent complex L_X of a quasi-smooth derived scheme X, and it is in fact true that any quasi-smooth derived scheme is locally the derived intersection of two sections in a vector bundle over a smooth base. If I have time later, and someone else hasn't already done so, I'll give a fuller explanation. –  Chris Brav Feb 17 '13 at 12:02
    
I'm not familiar with derived schemes, but I'm very interested in your explanation. –  Xiaobo Zhuang Feb 18 '13 at 16:07
    
chris is probably referring to this people.maths.ox.ac.uk/joyce/BBDJ1.pdf –  Jacob Bell Feb 18 '13 at 23:29
    
Thank you for the reference, I will take a look at it. –  Xiaobo Zhuang Feb 23 '13 at 3:09

1 Answer 1

up vote 4 down vote accepted

The answer is yes. Let $[E^{-1} \rightarrow E^0]$ be a perfect obstruction theory on $M$. After localizing in $M$ we can assume that the map $E^0 \rightarrow \Omega_M$ is induced as $\mathcal{O}_M \otimes \Omega_{\mathbf{A}^n} \rightarrow \Omega_M$ for some map $M \rightarrow \mathbf{A}^n$.(*) As the map on differentials is surjective the relative differentials vanish and the map is unramified. By EGA IV.18.4.7, we can (at least after localizing further in $M$) factor the map $M \rightarrow \mathbf{A}^n$ as a closed embedding $M \rightarrow V$ followed by an étale map $V \rightarrow \mathbf{A}^n$.

Let $I$ be the ideal of $M$ inside $V$. Then we have an exact sequence

$I/I^2 \rightarrow \mathcal{O}_M \otimes \Omega_V \rightarrow \Omega_M \rightarrow 0$

where the first two terms are $\tau_{\geq -1} L_M$. By the definition of an obstruction theory, we are given a map $E^{-1} \rightarrow I/I^2$ inducing a surjection $H^{-1}(E) \rightarrow H^{-1} L_M$. By the 5-lemma, $E^{-1} \rightarrow I/I^2$ is surjective.

Localizing further in $V$ and $M$, we can assume that the map $E^{-1} \rightarrow I/I^2$ is induced from a map $F \rightarrow I$ of coherent sheaves on $V$ with $F$ locally free. By Nakayama's lemma we can assume, possibly after further localization, that $F \rightarrow I$ is surjective. Then consider the section of the vector bundle $F^\vee$ dual to the composition $F \rightarrow I \subset \mathcal{O}_V$. The vanishing locus of this section is $M$ and the induced complex $F \rightarrow \mathcal{O}_M \otimes \Omega_V$ is the obstruction theory we started with.

(*) We can represent $\tau_{\geq -1} L_M$ as $[ J / J^2 \rightarrow \Omega_W ]$ for some closed embedding $M \subset W$ and we are given a map of complexes $E^\bullet \rightarrow \tau_{\geq -1} L_M$. Replacing $E^\bullet$ with a quasi-isomorphic complex, we can assume that the map $E^0 \rightarrow \Omega_W$ is surjective. But $E^0$ and $\Omega_W$ are vector bundles, so after localizing in $M$ we can assume that $\Omega_W$ is a direct summand of $E^0$. Choosing a basis for the complementary summand, we can identify $E^0 \cong \Omega_{W \times \mathbf{A}^n}$.

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Thank you for your great answer! And I have trouble in only one step: in the second line why can we assume E^0\to \Omega_M is induced from some M\to A^n? Maybe this is standard, but I really don't know how to figure it out. –  Xiaobo Zhuang Feb 18 '13 at 16:01
    
Hopefully the edit will answer your question. –  Jonathan Wise Feb 18 '13 at 22:32
    
Thank you for your help! –  Xiaobo Zhuang Feb 22 '13 at 13:36
    
I'm curious if you have an application in mind for this fact. I've never been able to use it, myself. –  Jonathan Wise Feb 23 '13 at 1:33
    
Actually I don't have one either, I just want to understand better. –  Xiaobo Zhuang Feb 23 '13 at 2:56

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