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?

My body leaves; my core can stay

should truth entail possible truth

Would an alien lifeform be able to achieve space travel if lacking in vision?

Identify 80s or 90s comics with ripped creatures (not dwarves)

60's-70's movie: home appliances revolting against the owners

Drawing vertical/oblique lines in Metrical tree (tikz-qtree, tipa)

What was the last x86 CPU that did not have the x87 floating-point unit built in?

Match Roman Numerals

Why doesn't a hydraulic lever violate conservation of energy?

Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?

ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?

Am I ethically obligated to go into work on an off day if the reason is sudden?

Is an up-to-date browser secure on an out-of-date OS?

What information about me do stores get via my credit card?

What to do when moving next to a bird sanctuary with a loosely-domesticated cat?

Word to describe a time interval

What is the role of 'For' here?

How do you keep chess fun when your opponent constantly beats you?

Presidential Pardon

Does Parliament hold absolute power in the UK?

Do ℕ, mathbbN, BbbN, symbbN effectively differ, and is there a "canonical" specification of the naturals?

Loose spokes after only a few rides

Does Parliament need to approve the new Brexit delay to 31 October 2019?

How do spell lists change if the party levels up without taking a long rest?



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, $[M]^textrmvir=c_top(T_M)cap [M]$.






share|cite|improve this answer











$endgroup$













    Your Answer








    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f773527%2fconstruction-of-virtual-class-at-homological-mirror-symmetry%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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, $[M]^textrmvir=c_top(T_M)cap [M]$.






    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



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f773527%2fconstruction-of-virtual-class-at-homological-mirror-symmetry%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Solar Wings Breeze Design and development Specifications (Breeze) References Navigation menu1368-485X"Hang glider: Breeze (Solar Wings)"e

            Kathakali Contents Etymology and nomenclature History Repertoire Songs and musical instruments Traditional plays Styles: Sampradayam Training centers and awards Relationship to other dance forms See also Notes References External links Navigation menueThe Illustrated Encyclopedia of Hinduism: A-MSouth Asian Folklore: An EncyclopediaRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to Play10.1353/atj.2005.0004The Illustrated Encyclopedia of Hinduism: A-MEncyclopedia of HinduismKathakali Dance-drama: Where Gods and Demons Come to PlaySonic Liturgy: Ritual and Music in Hindu Tradition"The Mirror of Gesture"Kathakali Dance-drama: Where Gods and Demons Come to Play"Kathakali"Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceMedieval Indian Literature: An AnthologyThe Oxford Companion to Indian TheatreSouth Asian Folklore: An Encyclopedia : Afghanistan, Bangladesh, India, Nepal, Pakistan, Sri LankaThe Rise of Performance Studies: Rethinking Richard Schechner's Broad SpectrumIndian Theatre: Traditions of PerformanceModern Asian Theatre and Performance 1900-2000Critical Theory and PerformanceBetween Theater and AnthropologyKathakali603847011Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceBetween Theater and AnthropologyBetween Theater and AnthropologyNambeesan Smaraka AwardsArchivedThe Cambridge Guide to TheatreRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeThe Garland Encyclopedia of World Music: South Asia : the Indian subcontinentThe Ethos of Noh: Actors and Their Art10.2307/1145740By Means of Performance: Intercultural Studies of Theatre and Ritual10.1017/s204912550000100xReconceiving the Renaissance: A Critical ReaderPerformance TheoryListening to Theatre: The Aural Dimension of Beijing Opera10.2307/1146013Kathakali: The Art of the Non-WorldlyOn KathakaliKathakali, the dance theatreThe Kathakali Complex: Performance & StructureKathakali Dance-Drama: Where Gods and Demons Come to Play10.1093/obo/9780195399318-0071Drama and Ritual of Early Hinduism"In the Shadow of Hollywood Orientalism: Authentic East Indian Dancing"10.1080/08949460490274013Sanskrit Play Production in Ancient IndiaIndian Music: History and StructureBharata, the Nāṭyaśāstra233639306Table of Contents2238067286469807Dance In Indian Painting10.2307/32047833204783Kathakali Dance-Theatre: A Visual Narrative of Sacred Indian MimeIndian Classical Dance: The Renaissance and BeyondKathakali: an indigenous art-form of Keralaeee

            Method to test if a number is a perfect power? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Detecting perfect squares faster than by extracting square rooteffective way to get the integer sequence A181392 from oeisA rarely mentioned fact about perfect powersHow many numbers such $n$ are there that $n<100,lfloorsqrtn rfloor mid n$Check perfect squareness by modulo division against multiple basesFor what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square.Do there exist any positive integers $n$ such that $lfloore^nrfloor$ is a perfect power? What is the probability that one exists?finding perfect power factors of an integerProve that the sequence contains a perfect square for any natural number $m $ in the domain of $f$ .Counting Perfect Powers