In Homological Mirror Symmetry it is necessary to integrate cohomology class at stable moduli.

For this, we can define virtual dimension that stable moduli space should have, and at moduli defined at cohomology of this degree we can integrate by taking "virtual fundamental class".

So, if moduli is unobstructed, there is no need to take virtual class. However, if it is obstructed, virtual class is needed. Can someone explain construction of this virtual class easily?

  • $\begingroup$ I don't know the construction, but I have heard it's rather far from easy. $\endgroup$ Apr 29, 2014 at 1:19

1 Answer 1


Strategy. Let $M$ be your moduli space. In order to construct a virtual fundamental class, you need a cone (this will be the intrinsic normal cone $C_M$, which is a certain stack quotient) embedded in a vector bundle stack (this embedding "is" the datum of a perfect obstruction theory on $M$). Then $[M]^\textrm{vir}$ will be the intersection of this cone with the zero section of the vector bundle.

Remark. The attached virtual class will in general be dependent on the chosen obstruction theory. The virtual class $[M]^\textrm{vir}\in A_\ast(M)$ lives in the virtual dimension of $M$, which is the rank of the perfect obstruction theory (defined below).

Let us assume $M$ is proper and embeddable. Let us then choose an embedding $M\hookrightarrow V$ into a smooth scheme $V$, and let $I$ be its ideal. This "choice" will play no role.

Let us call a complex $\mathscr E\in D^b(\textrm{Coh}(M))$ perfect in case it is quasi-isomorphic to a (bounded) complex of vector bundles on $M$.

The truncated cotangent complex of $M$, relative to the chosen embedding, is the complex $L_M=[I/I^2\to \Omega_V|_M]$, viewed in degrees $[-1,0]$.

Definition. A perfect obstruction theory on $M$ is a couple $(\mathscr E,\phi)$ where $\mathscr E=[E_1\to E_0]$ is a perfect complex concentrated in degrees $[-1,0]$ and $\phi:\mathscr E\to L_M$ is a morphism (in the derived category) such that, in cohomology,

  1. $h^0(\phi):h^0(\mathscr E)\to h^0(L_M)=\Omega_M$ is an isomorphism, and
  2. $h^{-1}(\phi):h^{-1}(\mathscr E)\to h^{-1}(L_M)$ is an epimorphism.

The rank of the perfect obstruction theory is $\textrm{rk}(\mathscr E)=\textrm{rk}(E_0)-\textrm{rk}(E_1)$. The virtual class will live in this dimension.

Fact. The vector bundle $T_V|_M$ acts on $N_{M/V}=\textrm{Spec Sym }I/I^2$ by leaving $$C_{M/V}=\textrm{Spec }\bigoplus_{d\geq 0}I^d/I^{d+1}\subset N_{M/V}$$ invariant, so that the stack quotient $C_M=[C_{M/V}/T_V|_M]$ makes sense. Moreover, $C_M$ does not depend on the chosen embedding, and therefore deserves the name intrinsic normal cone.

There is an embedding $$C_M\subset N_M=[N_{M/V}/T_V|_M],$$ but the latter (the intrinsic normal sheaf) may not be a vector bundle. However, the datum of an obstruction theory as above provides an embedding $N_M\subset E$ of the intrinsic normal sheaf inside a vector bundle $E=[E_1^\vee/E_0^\vee]$. Therefore we have an embedding $C_M\subset E$, and we can now take $$[M]^\textrm{vir}=0_E^![C_M]\in A_{\textrm{rk}(E)}(M).$$

Example. If $M$ is embedded in a $d$-dimensional scheme $Y$ as the zero section $Z(s)$ of a vector bundle $E\to Y$, then the virtual dimension of $M$ is just $d-\textrm{rk}(E)$, the dimension that $M$ would have if $s$ were a regular section. If $s$ is a regular section, then $$[M]^\textrm{vir}=c_{top}(E)\cap [M]\in A_{d-\textrm{rk}(E)}(M).$$ Otherwise, $$[M]^\textrm{vir}=\mathbb Z(s),$$ the localized top Chern class of $E$, which is obtained by intersecting the zero section of $E|_M$ with the cone $C_{M/Y}\subset E|_M$ (such embedding is obtained by "deforming" the embeddings $\lambda s(M)\cong M\hookrightarrow E$ to the normal cone, where $\mathbb P^1\ni\lambda\to\infty$.

Example. As you said, if the obstruction sheaf $\textrm{ob}:=h^1(\mathscr E^\vee)=0$ then $[M]^\textrm{vir}=[M]$. This happens when $M$ is nonsingular of dimension the virtual dimension.

Example. When $M$ is nonsingular, $[M]^\textrm{vir}=c_{top}(T_M)\cap [M]$.


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