Construction of virtual class at Homological Mirror Symmetry The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)About Homological Mirror SymmetryMathematical terminology about Holomorphic vector bundle over Grassmanian.Tangent Space to Moduli Space of Vector Bundles on CurveDimension of Moduli Space of Bundles on CurvesMirror Symmetry of Elliptic CurveIs this a correct definition for fundamental class of a subvariety?Moduli Space of Degree Zero Stable Maps?First Chern class of toric manifoldsA-branes on the mirror to the projective lineWhy is the quintic in $mathbbCP^4$ simply connected?

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Construction of virtual class at Homological Mirror Symmetry



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)About Homological Mirror SymmetryMathematical terminology about Holomorphic vector bundle over Grassmanian.Tangent Space to Moduli Space of Vector Bundles on CurveDimension of Moduli Space of Bundles on CurvesMirror Symmetry of Elliptic CurveIs this a correct definition for fundamental class of a subvariety?Moduli Space of Degree Zero Stable Maps?First Chern class of toric manifoldsA-branes on the mirror to the projective lineWhy is the quintic in $mathbbCP^4$ simply connected?










3












$begingroup$


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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    I don't know the construction, but I have heard it's rather far from easy.
    $endgroup$
    – Kevin Carlson
    Apr 29 '14 at 1:19















3












$begingroup$


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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    I don't know the construction, but I have heard it's rather far from easy.
    $endgroup$
    – Kevin Carlson
    Apr 29 '14 at 1:19













3












3








3


1



$begingroup$


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?










share|cite|improve this question











$endgroup$




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?







algebraic-geometry moduli-space mirror-symmetry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 24 at 15:34









Andrews

1,2812423




1,2812423










asked Apr 28 '14 at 23:53









Euna KoEuna Ko

475212




475212











  • $begingroup$
    I don't know the construction, but I have heard it's rather far from easy.
    $endgroup$
    – Kevin Carlson
    Apr 29 '14 at 1:19
















  • $begingroup$
    I don't know the construction, but I have heard it's rather far from easy.
    $endgroup$
    – Kevin Carlson
    Apr 29 '14 at 1:19















$begingroup$
I don't know the construction, but I have heard it's rather far from easy.
$endgroup$
– Kevin Carlson
Apr 29 '14 at 1:19




$begingroup$
I don't know the construction, but I have heard it's rather far from easy.
$endgroup$
– Kevin Carlson
Apr 29 '14 at 1:19










1 Answer
1






active

oldest

votes


















4












$begingroup$

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 $[M]^textrmvir$ 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 $[M]^textrmvirin A_ast(M)$ 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 $Mhookrightarrow V$ into a smooth scheme $V$, and let $I$ be its ideal. This "choice" will play no role.



Let us call a complex $mathscr Ein D^b(textrmCoh(M))$ 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 $L_M=[I/I^2to Omega_V|_M]$, viewed in degrees $[-1,0]$.



Definition. A perfect obstruction theory on $M$ is a couple $(mathscr E,phi)$ where $mathscr E=[E_1to E_0]$ is a perfect complex concentrated in degrees $[-1,0]$ and $phi:mathscr Eto L_M$ 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 $textrmrk(mathscr E)=textrmrk(E_0)-textrmrk(E_1)$. The virtual class will live in this dimension.



Fact. The vector bundle $T_V|_M$ acts on $N_M/V=textrmSpec Sym I/I^2$ by leaving $$C_M/V=textrmSpec bigoplus_dgeq 0I^d/I^d+1subset N_M/V$$ 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 $$C_Msubset N_M=[N_M/V/T_V|_M],$$ 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 $N_Msubset E$ of the intrinsic normal sheaf inside a vector bundle $E=[E_1^vee/E_0^vee]$. Therefore we have an embedding $C_Msubset E$, and we can now take $$[M]^textrmvir=0_E^![C_M]in A_textrmrk(E)(M).$$




Example. If $M$ is embedded in a $d$-dimensional scheme $Y$ as the zero section $Z(s)$ of a vector bundle $Eto Y$, then the virtual dimension of $M$ is just $d-textrmrk(E)$, the dimension that $M$ would have if $s$ were a regular section. If $s$ is a regular section, then $$[M]^textrmvir=c_top(E)cap [M]in A_d-textrmrk(E)(M).$$
Otherwise, $$[M]^textrmvir=mathbb Z(s),$$ the localized top Chern class of $E$, which is obtained by intersecting the zero section of $E|_M$ with the cone $C_M/Ysubset E|_M$ (such embedding is obtained by "deforming" the embeddings $lambda s(M)cong Mhookrightarrow E$ to the normal cone, where $mathbb P^1nilambdatoinfty$.



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



Example. When $M$ is nonsingular, $[M]^textrmvir=c_top(T_M)cap [M]$.






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    $begingroup$

    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 $[M]^textrmvir$ 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 $[M]^textrmvirin A_ast(M)$ 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 $Mhookrightarrow V$ into a smooth scheme $V$, and let $I$ be its ideal. This "choice" will play no role.



    Let us call a complex $mathscr Ein D^b(textrmCoh(M))$ 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 $L_M=[I/I^2to Omega_V|_M]$, viewed in degrees $[-1,0]$.



    Definition. A perfect obstruction theory on $M$ is a couple $(mathscr E,phi)$ where $mathscr E=[E_1to E_0]$ is a perfect complex concentrated in degrees $[-1,0]$ and $phi:mathscr Eto L_M$ 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 $textrmrk(mathscr E)=textrmrk(E_0)-textrmrk(E_1)$. The virtual class will live in this dimension.



    Fact. The vector bundle $T_V|_M$ acts on $N_M/V=textrmSpec Sym I/I^2$ by leaving $$C_M/V=textrmSpec bigoplus_dgeq 0I^d/I^d+1subset N_M/V$$ 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 $$C_Msubset N_M=[N_M/V/T_V|_M],$$ 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 $N_Msubset E$ of the intrinsic normal sheaf inside a vector bundle $E=[E_1^vee/E_0^vee]$. Therefore we have an embedding $C_Msubset E$, and we can now take $$[M]^textrmvir=0_E^![C_M]in A_textrmrk(E)(M).$$




    Example. If $M$ is embedded in a $d$-dimensional scheme $Y$ as the zero section $Z(s)$ of a vector bundle $Eto Y$, then the virtual dimension of $M$ is just $d-textrmrk(E)$, the dimension that $M$ would have if $s$ were a regular section. If $s$ is a regular section, then $$[M]^textrmvir=c_top(E)cap [M]in A_d-textrmrk(E)(M).$$
    Otherwise, $$[M]^textrmvir=mathbb Z(s),$$ the localized top Chern class of $E$, which is obtained by intersecting the zero section of $E|_M$ with the cone $C_M/Ysubset E|_M$ (such embedding is obtained by "deforming" the embeddings $lambda s(M)cong Mhookrightarrow E$ to the normal cone, where $mathbb P^1nilambdatoinfty$.



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



    Example. When $M$ is nonsingular, $[M]^textrmvir=c_top(T_M)cap [M]$.






    share|cite|improve this answer











    $endgroup$

















      4












      $begingroup$

      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 $[M]^textrmvir$ 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 $[M]^textrmvirin A_ast(M)$ 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 $Mhookrightarrow V$ into a smooth scheme $V$, and let $I$ be its ideal. This "choice" will play no role.



      Let us call a complex $mathscr Ein D^b(textrmCoh(M))$ 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 $L_M=[I/I^2to Omega_V|_M]$, viewed in degrees $[-1,0]$.



      Definition. A perfect obstruction theory on $M$ is a couple $(mathscr E,phi)$ where $mathscr E=[E_1to E_0]$ is a perfect complex concentrated in degrees $[-1,0]$ and $phi:mathscr Eto L_M$ 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 $textrmrk(mathscr E)=textrmrk(E_0)-textrmrk(E_1)$. The virtual class will live in this dimension.



      Fact. The vector bundle $T_V|_M$ acts on $N_M/V=textrmSpec Sym I/I^2$ by leaving $$C_M/V=textrmSpec bigoplus_dgeq 0I^d/I^d+1subset N_M/V$$ 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 $$C_Msubset N_M=[N_M/V/T_V|_M],$$ 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 $N_Msubset E$ of the intrinsic normal sheaf inside a vector bundle $E=[E_1^vee/E_0^vee]$. Therefore we have an embedding $C_Msubset E$, and we can now take $$[M]^textrmvir=0_E^![C_M]in A_textrmrk(E)(M).$$




      Example. If $M$ is embedded in a $d$-dimensional scheme $Y$ as the zero section $Z(s)$ of a vector bundle $Eto Y$, then the virtual dimension of $M$ is just $d-textrmrk(E)$, the dimension that $M$ would have if $s$ were a regular section. If $s$ is a regular section, then $$[M]^textrmvir=c_top(E)cap [M]in A_d-textrmrk(E)(M).$$
      Otherwise, $$[M]^textrmvir=mathbb Z(s),$$ the localized top Chern class of $E$, which is obtained by intersecting the zero section of $E|_M$ with the cone $C_M/Ysubset E|_M$ (such embedding is obtained by "deforming" the embeddings $lambda s(M)cong Mhookrightarrow E$ to the normal cone, where $mathbb P^1nilambdatoinfty$.



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



      Example. When $M$ is nonsingular, $[M]^textrmvir=c_top(T_M)cap [M]$.






      share|cite|improve this answer











      $endgroup$















        4












        4








        4





        $begingroup$

        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 $[M]^textrmvir$ 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 $[M]^textrmvirin A_ast(M)$ 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 $Mhookrightarrow V$ into a smooth scheme $V$, and let $I$ be its ideal. This "choice" will play no role.



        Let us call a complex $mathscr Ein D^b(textrmCoh(M))$ 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 $L_M=[I/I^2to Omega_V|_M]$, viewed in degrees $[-1,0]$.



        Definition. A perfect obstruction theory on $M$ is a couple $(mathscr E,phi)$ where $mathscr E=[E_1to E_0]$ is a perfect complex concentrated in degrees $[-1,0]$ and $phi:mathscr Eto L_M$ 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 $textrmrk(mathscr E)=textrmrk(E_0)-textrmrk(E_1)$. The virtual class will live in this dimension.



        Fact. The vector bundle $T_V|_M$ acts on $N_M/V=textrmSpec Sym I/I^2$ by leaving $$C_M/V=textrmSpec bigoplus_dgeq 0I^d/I^d+1subset N_M/V$$ 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 $$C_Msubset N_M=[N_M/V/T_V|_M],$$ 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 $N_Msubset E$ of the intrinsic normal sheaf inside a vector bundle $E=[E_1^vee/E_0^vee]$. Therefore we have an embedding $C_Msubset E$, and we can now take $$[M]^textrmvir=0_E^![C_M]in A_textrmrk(E)(M).$$




        Example. If $M$ is embedded in a $d$-dimensional scheme $Y$ as the zero section $Z(s)$ of a vector bundle $Eto Y$, then the virtual dimension of $M$ is just $d-textrmrk(E)$, the dimension that $M$ would have if $s$ were a regular section. If $s$ is a regular section, then $$[M]^textrmvir=c_top(E)cap [M]in A_d-textrmrk(E)(M).$$
        Otherwise, $$[M]^textrmvir=mathbb Z(s),$$ the localized top Chern class of $E$, which is obtained by intersecting the zero section of $E|_M$ with the cone $C_M/Ysubset E|_M$ (such embedding is obtained by "deforming" the embeddings $lambda s(M)cong Mhookrightarrow E$ to the normal cone, where $mathbb P^1nilambdatoinfty$.



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



        Example. When $M$ is nonsingular, $[M]^textrmvir=c_top(T_M)cap [M]$.






        share|cite|improve this answer











        $endgroup$



        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 $[M]^textrmvir$ 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 $[M]^textrmvirin A_ast(M)$ 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 $Mhookrightarrow V$ into a smooth scheme $V$, and let $I$ be its ideal. This "choice" will play no role.



        Let us call a complex $mathscr Ein D^b(textrmCoh(M))$ 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 $L_M=[I/I^2to Omega_V|_M]$, viewed in degrees $[-1,0]$.



        Definition. A perfect obstruction theory on $M$ is a couple $(mathscr E,phi)$ where $mathscr E=[E_1to E_0]$ is a perfect complex concentrated in degrees $[-1,0]$ and $phi:mathscr Eto L_M$ 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 $textrmrk(mathscr E)=textrmrk(E_0)-textrmrk(E_1)$. The virtual class will live in this dimension.



        Fact. The vector bundle $T_V|_M$ acts on $N_M/V=textrmSpec Sym I/I^2$ by leaving $$C_M/V=textrmSpec bigoplus_dgeq 0I^d/I^d+1subset N_M/V$$ 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 $$C_Msubset N_M=[N_M/V/T_V|_M],$$ 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 $N_Msubset E$ of the intrinsic normal sheaf inside a vector bundle $E=[E_1^vee/E_0^vee]$. Therefore we have an embedding $C_Msubset E$, and we can now take $$[M]^textrmvir=0_E^![C_M]in A_textrmrk(E)(M).$$




        Example. If $M$ is embedded in a $d$-dimensional scheme $Y$ as the zero section $Z(s)$ of a vector bundle $Eto Y$, then the virtual dimension of $M$ is just $d-textrmrk(E)$, the dimension that $M$ would have if $s$ were a regular section. If $s$ is a regular section, then $$[M]^textrmvir=c_top(E)cap [M]in A_d-textrmrk(E)(M).$$
        Otherwise, $$[M]^textrmvir=mathbb Z(s),$$ the localized top Chern class of $E$, which is obtained by intersecting the zero section of $E|_M$ with the cone $C_M/Ysubset E|_M$ (such embedding is obtained by "deforming" the embeddings $lambda s(M)cong Mhookrightarrow E$ to the normal cone, where $mathbb P^1nilambdatoinfty$.



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



        Example. When $M$ is nonsingular, $[M]^textrmvir=c_top(T_M)cap [M]$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 29 '14 at 8:56

























        answered Apr 29 '14 at 8:37









        BreninBrenin

        9,10331645




        9,10331645



























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