Constructing diffeomorphisms of moduli spaces of $J$-holomorphic curves












3












$begingroup$


Let $M^{2n}$ be a smooth manifold admitting two almost complex structures $J_0$ and $J_1$.



Suppose that $J_0$ and $J_1$ are both regular in the sense that the moduli space $$
mathcal{M}_i:=mathcal{M}(A; J_i)/PSL(2,mathbb{C})
$$ of $J_i$-holomorphic spheres representing the homology class $Ain H_2(M;mathbb{Z})$ is a smooth, compact manifold for both $i=0,1$. Suppose further that there exists a path $(J_t)_{tin [0,1]}$ of regular almost complex structures on $M$ connecting $J_0$ and $J_1$.



In McDuff and Salamon's book they show that the manifolds
$$
mathcal{M}_t:=mathcal{M}(A;J_t)/PSL(2,mathbb{C})
$$
are oriented cobordant for all $tin [0,1]$.



QUESTION: Are the moduli spaces $mathcal{M}_t$ diffeomorphic for all $tin [0,1]$? If not, how does the diffeomorphism fail?






differential-geometry complex-geometry symplectic-geometry







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let $M^{2n}$ be a smooth manifold admitting two almost complex structures $J_0$ and $J_1$.



    Suppose that $J_0$ and $J_1$ are both regular in the sense that the moduli space $$
    mathcal{M}_i:=mathcal{M}(A; J_i)/PSL(2,mathbb{C})
    $$     of $J_i$-holomorphic spheres representing the homology class $Ain H_2(M;mathbb{Z})$ is a smooth, compact manifold for both $i=0,1$. Suppose further that there exists a path $(J_t)_{tin [0,1]}$ of regular almost complex structures on $M$ connecting $J_0$ and $J_1$.



    In McDuff and Salamon's book they show that the manifolds
    $$
    mathcal{M}_t:=mathcal{M}(A;J_t)/PSL(2,mathbb{C})
    $$
    are oriented cobordant for all $tin [0,1]$.



    QUESTION: Are the moduli spaces $mathcal{M}_t$ diffeomorphic for all $tin [0,1]$? If not, how does the diffeomorphism fail?






    differential-geometry complex-geometry symplectic-geometry







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $M^{2n}$ be a smooth manifold admitting two almost complex structures $J_0$ and $J_1$.



      Suppose that $J_0$ and $J_1$ are both regular in the sense that the moduli space $$
      mathcal{M}_i:=mathcal{M}(A; J_i)/PSL(2,mathbb{C})
      $$       of $J_i$-holomorphic spheres representing the homology class $Ain H_2(M;mathbb{Z})$ is a smooth, compact manifold for both $i=0,1$. Suppose further that there exists a path $(J_t)_{tin [0,1]}$ of regular almost complex structures on $M$ connecting $J_0$ and $J_1$.



      In McDuff and Salamon's book they show that the manifolds
      $$
      mathcal{M}_t:=mathcal{M}(A;J_t)/PSL(2,mathbb{C})
      $$
      are oriented cobordant for all $tin [0,1]$.



      QUESTION: Are the moduli spaces $mathcal{M}_t$ diffeomorphic for all $tin [0,1]$? If not, how does the diffeomorphism fail?






      differential-geometry complex-geometry symplectic-geometry







      share|cite|improve this question









      $endgroup$




      Let $M^{2n}$ be a smooth manifold admitting two almost complex structures $J_0$ and $J_1$.



      Suppose that $J_0$ and $J_1$ are both regular in the sense that the moduli space $$
      mathcal{M}_i:=mathcal{M}(A; J_i)/PSL(2,mathbb{C})
      $$       of $J_i$-holomorphic spheres representing the homology class $Ain H_2(M;mathbb{Z})$ is a smooth, compact manifold for both $i=0,1$. Suppose further that there exists a path $(J_t)_{tin [0,1]}$ of regular almost complex structures on $M$ connecting $J_0$ and $J_1$.



      In McDuff and Salamon's book they show that the manifolds
      $$
      mathcal{M}_t:=mathcal{M}(A;J_t)/PSL(2,mathbb{C})
      $$
      are oriented cobordant for all $tin [0,1]$.



      QUESTION: Are the moduli spaces $mathcal{M}_t$ diffeomorphic for all $tin [0,1]$? If not, how does the diffeomorphism fail?






      differential-geometry complex-geometry symplectic-geometry




      differential-geometry complex-geometry symplectic-geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 22:06









      srpsrp

      2588




      2588






















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          $begingroup$

          The issue here is precisely what you mean when you say "regular path $J_t$ of almost complex structures".



          I am not particularly expert in the theory of holomorphic curves, so I won't comment on that; I expect the situation is precisely analagous to the finite-dimensional setting I will describe below (I'm sure there are more technical issues involving the corner points).



          Let $M$ be a smooth manifold, and $S subset N$ a closed submanifold. Given a map $f: M to N$ transverse to $S$, we have that $mathcal M_f = f^{-1}(S)$ is a smooth manifold (with corners etc if $S$ and $N$ have them).



          The condition "$f pitchfork S$" is the precise analogy to "$J_t$ is regular" in this finite-dimensional setting.



          Suppose you have a path of functions $f$, written as a map $f_t: I times M to N$. Then I would say that this is a regular path of functions if the map $I times M to N$ is transverse to $S$, and the same for ${0,1} times M to N$. Then the "parameterized moduli space" $mathcal M_I$ is a smooth manifold with boundary equal to $mathcal M_0 sqcup mathcal M_1$. This space is equipped with a smooth map $mathcal M_I to I$ which is a submersion near the boundary.



          The point is that each individual $f_t$ is not assumed to be regular; spelling it out, it is entirely possible that for $(t,x)$, we have $f_t(x) in S$, and $text{Im}(df_t) subset N_{f_t(x)} S$ is not the entirety of the normal space, but rather of codimension 1; and so that together with the additional vector $frac{d}{dt} f_t(x) in N_{f_t(x)} S$, we span the whole of the normal space.



          The second point is that an individual $mathcal M_t$ is regular if and only if $t in I$ is a regular value of the projection $mathcal M_I to I$.



          So if, in addition, $f_t$ is regular for each $t$, then by definition the projection $mathcal M_I to I$ is a submersion; as you know well, a proper smooth submersion is a fiber bundle, and this provides a global diffeomorphism $mathcal M_I cong I times mathcal M_0$.



          In almost every interesting application, given any two regular functions / almost complex structures / whatever, it will be essentially impossible to find a path through regular (blahs) between them; but it will be possible to find a "regular path" of (blahs) between them in the sense given above. A simple case to think about the discussion above with is $S$ a hypersurface and $M$ a point! If $S$ disconnects the codomain, then to get from one side to another you will need to cross $S$. In this situation, $mathcal M_I$ will be a finite set of points which project to the interior of $I$.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            The issue here is precisely what you mean when you say "regular path $J_t$ of almost complex structures".



            I am not particularly expert in the theory of holomorphic curves, so I won't comment on that; I expect the situation is precisely analagous to the finite-dimensional setting I will describe below (I'm sure there are more technical issues involving the corner points).



            Let $M$ be a smooth manifold, and $S subset N$ a closed submanifold. Given a map $f: M to N$ transverse to $S$, we have that $mathcal M_f = f^{-1}(S)$ is a smooth manifold (with corners etc if $S$ and $N$ have them).



            The condition "$f pitchfork S$" is the precise analogy to "$J_t$ is regular" in this finite-dimensional setting.



            Suppose you have a path of functions $f$, written as a map $f_t: I times M to N$. Then I would say that this is a regular path of functions if the map $I times M to N$ is transverse to $S$, and the same for ${0,1} times M to N$. Then the "parameterized moduli space" $mathcal M_I$ is a smooth manifold with boundary equal to $mathcal M_0 sqcup mathcal M_1$. This space is equipped with a smooth map $mathcal M_I to I$ which is a submersion near the boundary.



            The point is that each individual $f_t$ is not assumed to be regular; spelling it out, it is entirely possible that for $(t,x)$, we have $f_t(x) in S$, and $text{Im}(df_t) subset N_{f_t(x)} S$ is not the entirety of the normal space, but rather of codimension 1; and so that together with the additional vector $frac{d}{dt} f_t(x) in N_{f_t(x)} S$, we span the whole of the normal space.



            The second point is that an individual $mathcal M_t$ is regular if and only if $t in I$ is a regular value of the projection $mathcal M_I to I$.



            So if, in addition, $f_t$ is regular for each $t$, then by definition the projection $mathcal M_I to I$ is a submersion; as you know well, a proper smooth submersion is a fiber bundle, and this provides a global diffeomorphism $mathcal M_I cong I times mathcal M_0$.



            In almost every interesting application, given any two regular functions / almost complex structures / whatever, it will be essentially impossible to find a path through regular (blahs) between them; but it will be possible to find a "regular path" of (blahs) between them in the sense given above. A simple case to think about the discussion above with is $S$ a hypersurface and $M$ a point! If $S$ disconnects the codomain, then to get from one side to another you will need to cross $S$. In this situation, $mathcal M_I$ will be a finite set of points which project to the interior of $I$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The issue here is precisely what you mean when you say "regular path $J_t$ of almost complex structures".



              I am not particularly expert in the theory of holomorphic curves, so I won't comment on that; I expect the situation is precisely analagous to the finite-dimensional setting I will describe below (I'm sure there are more technical issues involving the corner points).



              Let $M$ be a smooth manifold, and $S subset N$ a closed submanifold. Given a map $f: M to N$ transverse to $S$, we have that $mathcal M_f = f^{-1}(S)$ is a smooth manifold (with corners etc if $S$ and $N$ have them).



              The condition "$f pitchfork S$" is the precise analogy to "$J_t$ is regular" in this finite-dimensional setting.



              Suppose you have a path of functions $f$, written as a map $f_t: I times M to N$. Then I would say that this is a regular path of functions if the map $I times M to N$ is transverse to $S$, and the same for ${0,1} times M to N$. Then the "parameterized moduli space" $mathcal M_I$ is a smooth manifold with boundary equal to $mathcal M_0 sqcup mathcal M_1$. This space is equipped with a smooth map $mathcal M_I to I$ which is a submersion near the boundary.



              The point is that each individual $f_t$ is not assumed to be regular; spelling it out, it is entirely possible that for $(t,x)$, we have $f_t(x) in S$, and $text{Im}(df_t) subset N_{f_t(x)} S$ is not the entirety of the normal space, but rather of codimension 1; and so that together with the additional vector $frac{d}{dt} f_t(x) in N_{f_t(x)} S$, we span the whole of the normal space.



              The second point is that an individual $mathcal M_t$ is regular if and only if $t in I$ is a regular value of the projection $mathcal M_I to I$.



              So if, in addition, $f_t$ is regular for each $t$, then by definition the projection $mathcal M_I to I$ is a submersion; as you know well, a proper smooth submersion is a fiber bundle, and this provides a global diffeomorphism $mathcal M_I cong I times mathcal M_0$.



              In almost every interesting application, given any two regular functions / almost complex structures / whatever, it will be essentially impossible to find a path through regular (blahs) between them; but it will be possible to find a "regular path" of (blahs) between them in the sense given above. A simple case to think about the discussion above with is $S$ a hypersurface and $M$ a point! If $S$ disconnects the codomain, then to get from one side to another you will need to cross $S$. In this situation, $mathcal M_I$ will be a finite set of points which project to the interior of $I$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The issue here is precisely what you mean when you say "regular path $J_t$ of almost complex structures".



                I am not particularly expert in the theory of holomorphic curves, so I won't comment on that; I expect the situation is precisely analagous to the finite-dimensional setting I will describe below (I'm sure there are more technical issues involving the corner points).



                Let $M$ be a smooth manifold, and $S subset N$ a closed submanifold. Given a map $f: M to N$ transverse to $S$, we have that $mathcal M_f = f^{-1}(S)$ is a smooth manifold (with corners etc if $S$ and $N$ have them).



                The condition "$f pitchfork S$" is the precise analogy to "$J_t$ is regular" in this finite-dimensional setting.



                Suppose you have a path of functions $f$, written as a map $f_t: I times M to N$. Then I would say that this is a regular path of functions if the map $I times M to N$ is transverse to $S$, and the same for ${0,1} times M to N$. Then the "parameterized moduli space" $mathcal M_I$ is a smooth manifold with boundary equal to $mathcal M_0 sqcup mathcal M_1$. This space is equipped with a smooth map $mathcal M_I to I$ which is a submersion near the boundary.



                The point is that each individual $f_t$ is not assumed to be regular; spelling it out, it is entirely possible that for $(t,x)$, we have $f_t(x) in S$, and $text{Im}(df_t) subset N_{f_t(x)} S$ is not the entirety of the normal space, but rather of codimension 1; and so that together with the additional vector $frac{d}{dt} f_t(x) in N_{f_t(x)} S$, we span the whole of the normal space.



                The second point is that an individual $mathcal M_t$ is regular if and only if $t in I$ is a regular value of the projection $mathcal M_I to I$.



                So if, in addition, $f_t$ is regular for each $t$, then by definition the projection $mathcal M_I to I$ is a submersion; as you know well, a proper smooth submersion is a fiber bundle, and this provides a global diffeomorphism $mathcal M_I cong I times mathcal M_0$.



                In almost every interesting application, given any two regular functions / almost complex structures / whatever, it will be essentially impossible to find a path through regular (blahs) between them; but it will be possible to find a "regular path" of (blahs) between them in the sense given above. A simple case to think about the discussion above with is $S$ a hypersurface and $M$ a point! If $S$ disconnects the codomain, then to get from one side to another you will need to cross $S$. In this situation, $mathcal M_I$ will be a finite set of points which project to the interior of $I$.






                share|cite|improve this answer









                $endgroup$



                The issue here is precisely what you mean when you say "regular path $J_t$ of almost complex structures".



                I am not particularly expert in the theory of holomorphic curves, so I won't comment on that; I expect the situation is precisely analagous to the finite-dimensional setting I will describe below (I'm sure there are more technical issues involving the corner points).



                Let $M$ be a smooth manifold, and $S subset N$ a closed submanifold. Given a map $f: M to N$ transverse to $S$, we have that $mathcal M_f = f^{-1}(S)$ is a smooth manifold (with corners etc if $S$ and $N$ have them).



                The condition "$f pitchfork S$" is the precise analogy to "$J_t$ is regular" in this finite-dimensional setting.



                Suppose you have a path of functions $f$, written as a map $f_t: I times M to N$. Then I would say that this is a regular path of functions if the map $I times M to N$ is transverse to $S$, and the same for ${0,1} times M to N$. Then the "parameterized moduli space" $mathcal M_I$ is a smooth manifold with boundary equal to $mathcal M_0 sqcup mathcal M_1$. This space is equipped with a smooth map $mathcal M_I to I$ which is a submersion near the boundary.



                The point is that each individual $f_t$ is not assumed to be regular; spelling it out, it is entirely possible that for $(t,x)$, we have $f_t(x) in S$, and $text{Im}(df_t) subset N_{f_t(x)} S$ is not the entirety of the normal space, but rather of codimension 1; and so that together with the additional vector $frac{d}{dt} f_t(x) in N_{f_t(x)} S$, we span the whole of the normal space.



                The second point is that an individual $mathcal M_t$ is regular if and only if $t in I$ is a regular value of the projection $mathcal M_I to I$.



                So if, in addition, $f_t$ is regular for each $t$, then by definition the projection $mathcal M_I to I$ is a submersion; as you know well, a proper smooth submersion is a fiber bundle, and this provides a global diffeomorphism $mathcal M_I cong I times mathcal M_0$.



                In almost every interesting application, given any two regular functions / almost complex structures / whatever, it will be essentially impossible to find a path through regular (blahs) between them; but it will be possible to find a "regular path" of (blahs) between them in the sense given above. A simple case to think about the discussion above with is $S$ a hypersurface and $M$ a point! If $S$ disconnects the codomain, then to get from one side to another you will need to cross $S$. In this situation, $mathcal M_I$ will be a finite set of points which project to the interior of $I$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 11 at 7:11









                Mike MillerMike Miller

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                37.2k472139






























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