8
$\begingroup$

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!

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – Chris Brav
    Feb 17, 2013 at 12:02
  • $\begingroup$ I'm not familiar with derived schemes, but I'm very interested in your explanation. $\endgroup$ Feb 18, 2013 at 16:07
  • $\begingroup$ chris is probably referring to this people.maths.ox.ac.uk/joyce/BBDJ1.pdf $\endgroup$
    – Jacob Bell
    Feb 18, 2013 at 23:29

1 Answer 1

6
$\begingroup$

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 $E^0 \cong \Omega_{W \times \mathbf{A}^n}$.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Feb 18, 2013 at 16:01
  • $\begingroup$ Hopefully the edit will answer your question. $\endgroup$ Feb 18, 2013 at 22:32
  • $\begingroup$ I'm curious if you have an application in mind for this fact. I've never been able to use it, myself. $\endgroup$ Feb 23, 2013 at 1:33
  • $\begingroup$ Actually I don't have one either, I just want to understand better. $\endgroup$ Feb 23, 2013 at 2:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.