Deforming antiholomorphic involutions












3














Let $(M,J)$ be a compact smooth almost complex manifold. We can "deform" $J$ as follows: if$A$ is a smooth section of the endomorphism bundle $mathrm{End}(TM)to M$ satisfying
$
AJ=-JA,
$

it follows that
$
Je^A=e^{-A}J,
$

where $e^A$ is the matrix exponential of $A$. It follows immediately that
$$
J':=Je^A
$$

is another almost complex structure on $M$.



Suppose $phi:Mto M$ is a diffeomorphism which is an anti-$J$-holomorphic involution, i.e. $$
Jcirc dphi=-dphi circ J qquad text{and} qquad phicirc phi =mathrm{id}.
$$



Question.
Can we deform $phi$ into an anti-$J'$-holomorphic involution?



I think this will be true if there exist a vector field $etain Gamma(TM)$ such that the diffeomorphism $phi':Mto M$ defined by
$$
phi'(x)=(phicirc mathrm{exp}circ eta)(x)
$$

is an anti-$J'$-holomorphic involution (here $exp$ is defined with respect to some Riemannian metric on $M$). How can I show the existence/nonexistence of $eta$?










share|cite|improve this question





























    3














    Let $(M,J)$ be a compact smooth almost complex manifold. We can "deform" $J$ as follows: if$A$ is a smooth section of the endomorphism bundle $mathrm{End}(TM)to M$ satisfying
    $
    AJ=-JA,
    $

    it follows that
    $
    Je^A=e^{-A}J,
    $

    where $e^A$ is the matrix exponential of $A$. It follows immediately that
    $$
    J':=Je^A
    $$

    is another almost complex structure on $M$.



    Suppose $phi:Mto M$ is a diffeomorphism which is an anti-$J$-holomorphic involution, i.e. $$
    Jcirc dphi=-dphi circ J qquad text{and} qquad phicirc phi =mathrm{id}.
    $$



    Question.
    Can we deform $phi$ into an anti-$J'$-holomorphic involution?



    I think this will be true if there exist a vector field $etain Gamma(TM)$ such that the diffeomorphism $phi':Mto M$ defined by
    $$
    phi'(x)=(phicirc mathrm{exp}circ eta)(x)
    $$

    is an anti-$J'$-holomorphic involution (here $exp$ is defined with respect to some Riemannian metric on $M$). How can I show the existence/nonexistence of $eta$?










    share|cite|improve this question



























      3












      3








      3


      1





      Let $(M,J)$ be a compact smooth almost complex manifold. We can "deform" $J$ as follows: if$A$ is a smooth section of the endomorphism bundle $mathrm{End}(TM)to M$ satisfying
      $
      AJ=-JA,
      $

      it follows that
      $
      Je^A=e^{-A}J,
      $

      where $e^A$ is the matrix exponential of $A$. It follows immediately that
      $$
      J':=Je^A
      $$

      is another almost complex structure on $M$.



      Suppose $phi:Mto M$ is a diffeomorphism which is an anti-$J$-holomorphic involution, i.e. $$
      Jcirc dphi=-dphi circ J qquad text{and} qquad phicirc phi =mathrm{id}.
      $$



      Question.
      Can we deform $phi$ into an anti-$J'$-holomorphic involution?



      I think this will be true if there exist a vector field $etain Gamma(TM)$ such that the diffeomorphism $phi':Mto M$ defined by
      $$
      phi'(x)=(phicirc mathrm{exp}circ eta)(x)
      $$

      is an anti-$J'$-holomorphic involution (here $exp$ is defined with respect to some Riemannian metric on $M$). How can I show the existence/nonexistence of $eta$?










      share|cite|improve this question















      Let $(M,J)$ be a compact smooth almost complex manifold. We can "deform" $J$ as follows: if$A$ is a smooth section of the endomorphism bundle $mathrm{End}(TM)to M$ satisfying
      $
      AJ=-JA,
      $

      it follows that
      $
      Je^A=e^{-A}J,
      $

      where $e^A$ is the matrix exponential of $A$. It follows immediately that
      $$
      J':=Je^A
      $$

      is another almost complex structure on $M$.



      Suppose $phi:Mto M$ is a diffeomorphism which is an anti-$J$-holomorphic involution, i.e. $$
      Jcirc dphi=-dphi circ J qquad text{and} qquad phicirc phi =mathrm{id}.
      $$



      Question.
      Can we deform $phi$ into an anti-$J'$-holomorphic involution?



      I think this will be true if there exist a vector field $etain Gamma(TM)$ such that the diffeomorphism $phi':Mto M$ defined by
      $$
      phi'(x)=(phicirc mathrm{exp}circ eta)(x)
      $$

      is an anti-$J'$-holomorphic involution (here $exp$ is defined with respect to some Riemannian metric on $M$). How can I show the existence/nonexistence of $eta$?







      differential-geometry riemannian-geometry complex-geometry symplectic-geometry






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 22 at 17:55

























      asked Dec 22 at 17:35









      srp

      2098




      2098






















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          At least in the 2-dimensional case, your deformations of a given complex structure yield all complex structures on the given oriented surface. (A proof of this is a linear algebra exercise where you analyze endomorphisms of a 2-dimensional real vector space.) However, with few exceptions (among compact surfaces, the only exception is the sphere), a generic complex structure on a surface admits no anti-holomorphic automorphisms (see below). Hence, your $phi'$ generically does not exist.



          Edit. Let us check that a generic smooth elliptic curve (aka a real flat 2-torus) has no antiholomorphic automorphisms. Let $T^2$ be a flat torus of unit area. An antiholomorphic involution $h$ of $T^2$ is an orientation-reversing isometry of the flat metric. If $h$ has nonempty fixed point set then it is a union of two parallel geodesic circles on $T^2$ which divide $T^2$ in two isometric annuli $A_1, A_2$ (swapped by $h$). The conformal type of $T^2$ is then uniquely determined by the modulus of $A_1$ which is a single real number. Hence, the space of flat tori of unit area which admit such an involution is real 1-dimensional. The second possibility is that $h$ acts freely on $T^2$ and the quotient is the Klein bottle. The moduli space of flat Klein bottles of fixed area is again real 1-dimensional. (Proving this is requires a tiny bit of work.) On the other hand, the moduli space of elliptic curves is complex one-dimensional. Hence, a generic elliptic curve admits no antiholomorphic involutions. (One can actually do better: The moduli space of $T^2$ is the complex line with two marked points corresponding to elliptic curves with groups extra holomorphic symmetries. Then the set of of elliptic curves which admit an antiholomorphic involution is the unique real line passing through these two marked points.)



          The proof in the higher genus case is a similar dimension count. The moduli space of compact Riemann surfaces of genus $g$ has dimension $3g-3$. On the other hand, if $h: Sto S$ is an antiholomorphic involution of a genus $g$ surface with nonempty fixed point set $F$, then $S$ is determined by the conformal type of one component $C$ of $S-F$. You have that $chi(S)=2chi(C)$. Then you compute the dimension of the moduli space of $C$, it equals $3g'-3+2p$ where $g'$ is the genus of $C$ and $p$ is the number of components of $F$. By the Euler characteristic formula above, you get: $2- 2g = 2(2g'-p+2)$.
          Then make a computation and conclude that $3g'-3+2p$ is strictly less than $3g-3$. The proof when $h$ has no fixed points is a similar dimension count.






          share|cite|improve this answer























          • Thank you for your answer. Why can a generic genus $g>0$ compact complex surface not admit any antiholomorphic automorphisms? I'd appreciate any references.
            – srp
            2 days ago













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          active

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          1














          At least in the 2-dimensional case, your deformations of a given complex structure yield all complex structures on the given oriented surface. (A proof of this is a linear algebra exercise where you analyze endomorphisms of a 2-dimensional real vector space.) However, with few exceptions (among compact surfaces, the only exception is the sphere), a generic complex structure on a surface admits no anti-holomorphic automorphisms (see below). Hence, your $phi'$ generically does not exist.



          Edit. Let us check that a generic smooth elliptic curve (aka a real flat 2-torus) has no antiholomorphic automorphisms. Let $T^2$ be a flat torus of unit area. An antiholomorphic involution $h$ of $T^2$ is an orientation-reversing isometry of the flat metric. If $h$ has nonempty fixed point set then it is a union of two parallel geodesic circles on $T^2$ which divide $T^2$ in two isometric annuli $A_1, A_2$ (swapped by $h$). The conformal type of $T^2$ is then uniquely determined by the modulus of $A_1$ which is a single real number. Hence, the space of flat tori of unit area which admit such an involution is real 1-dimensional. The second possibility is that $h$ acts freely on $T^2$ and the quotient is the Klein bottle. The moduli space of flat Klein bottles of fixed area is again real 1-dimensional. (Proving this is requires a tiny bit of work.) On the other hand, the moduli space of elliptic curves is complex one-dimensional. Hence, a generic elliptic curve admits no antiholomorphic involutions. (One can actually do better: The moduli space of $T^2$ is the complex line with two marked points corresponding to elliptic curves with groups extra holomorphic symmetries. Then the set of of elliptic curves which admit an antiholomorphic involution is the unique real line passing through these two marked points.)



          The proof in the higher genus case is a similar dimension count. The moduli space of compact Riemann surfaces of genus $g$ has dimension $3g-3$. On the other hand, if $h: Sto S$ is an antiholomorphic involution of a genus $g$ surface with nonempty fixed point set $F$, then $S$ is determined by the conformal type of one component $C$ of $S-F$. You have that $chi(S)=2chi(C)$. Then you compute the dimension of the moduli space of $C$, it equals $3g'-3+2p$ where $g'$ is the genus of $C$ and $p$ is the number of components of $F$. By the Euler characteristic formula above, you get: $2- 2g = 2(2g'-p+2)$.
          Then make a computation and conclude that $3g'-3+2p$ is strictly less than $3g-3$. The proof when $h$ has no fixed points is a similar dimension count.






          share|cite|improve this answer























          • Thank you for your answer. Why can a generic genus $g>0$ compact complex surface not admit any antiholomorphic automorphisms? I'd appreciate any references.
            – srp
            2 days ago


















          1














          At least in the 2-dimensional case, your deformations of a given complex structure yield all complex structures on the given oriented surface. (A proof of this is a linear algebra exercise where you analyze endomorphisms of a 2-dimensional real vector space.) However, with few exceptions (among compact surfaces, the only exception is the sphere), a generic complex structure on a surface admits no anti-holomorphic automorphisms (see below). Hence, your $phi'$ generically does not exist.



          Edit. Let us check that a generic smooth elliptic curve (aka a real flat 2-torus) has no antiholomorphic automorphisms. Let $T^2$ be a flat torus of unit area. An antiholomorphic involution $h$ of $T^2$ is an orientation-reversing isometry of the flat metric. If $h$ has nonempty fixed point set then it is a union of two parallel geodesic circles on $T^2$ which divide $T^2$ in two isometric annuli $A_1, A_2$ (swapped by $h$). The conformal type of $T^2$ is then uniquely determined by the modulus of $A_1$ which is a single real number. Hence, the space of flat tori of unit area which admit such an involution is real 1-dimensional. The second possibility is that $h$ acts freely on $T^2$ and the quotient is the Klein bottle. The moduli space of flat Klein bottles of fixed area is again real 1-dimensional. (Proving this is requires a tiny bit of work.) On the other hand, the moduli space of elliptic curves is complex one-dimensional. Hence, a generic elliptic curve admits no antiholomorphic involutions. (One can actually do better: The moduli space of $T^2$ is the complex line with two marked points corresponding to elliptic curves with groups extra holomorphic symmetries. Then the set of of elliptic curves which admit an antiholomorphic involution is the unique real line passing through these two marked points.)



          The proof in the higher genus case is a similar dimension count. The moduli space of compact Riemann surfaces of genus $g$ has dimension $3g-3$. On the other hand, if $h: Sto S$ is an antiholomorphic involution of a genus $g$ surface with nonempty fixed point set $F$, then $S$ is determined by the conformal type of one component $C$ of $S-F$. You have that $chi(S)=2chi(C)$. Then you compute the dimension of the moduli space of $C$, it equals $3g'-3+2p$ where $g'$ is the genus of $C$ and $p$ is the number of components of $F$. By the Euler characteristic formula above, you get: $2- 2g = 2(2g'-p+2)$.
          Then make a computation and conclude that $3g'-3+2p$ is strictly less than $3g-3$. The proof when $h$ has no fixed points is a similar dimension count.






          share|cite|improve this answer























          • Thank you for your answer. Why can a generic genus $g>0$ compact complex surface not admit any antiholomorphic automorphisms? I'd appreciate any references.
            – srp
            2 days ago
















          1












          1








          1






          At least in the 2-dimensional case, your deformations of a given complex structure yield all complex structures on the given oriented surface. (A proof of this is a linear algebra exercise where you analyze endomorphisms of a 2-dimensional real vector space.) However, with few exceptions (among compact surfaces, the only exception is the sphere), a generic complex structure on a surface admits no anti-holomorphic automorphisms (see below). Hence, your $phi'$ generically does not exist.



          Edit. Let us check that a generic smooth elliptic curve (aka a real flat 2-torus) has no antiholomorphic automorphisms. Let $T^2$ be a flat torus of unit area. An antiholomorphic involution $h$ of $T^2$ is an orientation-reversing isometry of the flat metric. If $h$ has nonempty fixed point set then it is a union of two parallel geodesic circles on $T^2$ which divide $T^2$ in two isometric annuli $A_1, A_2$ (swapped by $h$). The conformal type of $T^2$ is then uniquely determined by the modulus of $A_1$ which is a single real number. Hence, the space of flat tori of unit area which admit such an involution is real 1-dimensional. The second possibility is that $h$ acts freely on $T^2$ and the quotient is the Klein bottle. The moduli space of flat Klein bottles of fixed area is again real 1-dimensional. (Proving this is requires a tiny bit of work.) On the other hand, the moduli space of elliptic curves is complex one-dimensional. Hence, a generic elliptic curve admits no antiholomorphic involutions. (One can actually do better: The moduli space of $T^2$ is the complex line with two marked points corresponding to elliptic curves with groups extra holomorphic symmetries. Then the set of of elliptic curves which admit an antiholomorphic involution is the unique real line passing through these two marked points.)



          The proof in the higher genus case is a similar dimension count. The moduli space of compact Riemann surfaces of genus $g$ has dimension $3g-3$. On the other hand, if $h: Sto S$ is an antiholomorphic involution of a genus $g$ surface with nonempty fixed point set $F$, then $S$ is determined by the conformal type of one component $C$ of $S-F$. You have that $chi(S)=2chi(C)$. Then you compute the dimension of the moduli space of $C$, it equals $3g'-3+2p$ where $g'$ is the genus of $C$ and $p$ is the number of components of $F$. By the Euler characteristic formula above, you get: $2- 2g = 2(2g'-p+2)$.
          Then make a computation and conclude that $3g'-3+2p$ is strictly less than $3g-3$. The proof when $h$ has no fixed points is a similar dimension count.






          share|cite|improve this answer














          At least in the 2-dimensional case, your deformations of a given complex structure yield all complex structures on the given oriented surface. (A proof of this is a linear algebra exercise where you analyze endomorphisms of a 2-dimensional real vector space.) However, with few exceptions (among compact surfaces, the only exception is the sphere), a generic complex structure on a surface admits no anti-holomorphic automorphisms (see below). Hence, your $phi'$ generically does not exist.



          Edit. Let us check that a generic smooth elliptic curve (aka a real flat 2-torus) has no antiholomorphic automorphisms. Let $T^2$ be a flat torus of unit area. An antiholomorphic involution $h$ of $T^2$ is an orientation-reversing isometry of the flat metric. If $h$ has nonempty fixed point set then it is a union of two parallel geodesic circles on $T^2$ which divide $T^2$ in two isometric annuli $A_1, A_2$ (swapped by $h$). The conformal type of $T^2$ is then uniquely determined by the modulus of $A_1$ which is a single real number. Hence, the space of flat tori of unit area which admit such an involution is real 1-dimensional. The second possibility is that $h$ acts freely on $T^2$ and the quotient is the Klein bottle. The moduli space of flat Klein bottles of fixed area is again real 1-dimensional. (Proving this is requires a tiny bit of work.) On the other hand, the moduli space of elliptic curves is complex one-dimensional. Hence, a generic elliptic curve admits no antiholomorphic involutions. (One can actually do better: The moduli space of $T^2$ is the complex line with two marked points corresponding to elliptic curves with groups extra holomorphic symmetries. Then the set of of elliptic curves which admit an antiholomorphic involution is the unique real line passing through these two marked points.)



          The proof in the higher genus case is a similar dimension count. The moduli space of compact Riemann surfaces of genus $g$ has dimension $3g-3$. On the other hand, if $h: Sto S$ is an antiholomorphic involution of a genus $g$ surface with nonempty fixed point set $F$, then $S$ is determined by the conformal type of one component $C$ of $S-F$. You have that $chi(S)=2chi(C)$. Then you compute the dimension of the moduli space of $C$, it equals $3g'-3+2p$ where $g'$ is the genus of $C$ and $p$ is the number of components of $F$. By the Euler characteristic formula above, you get: $2- 2g = 2(2g'-p+2)$.
          Then make a computation and conclude that $3g'-3+2p$ is strictly less than $3g-3$. The proof when $h$ has no fixed points is a similar dimension count.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Moishe Cohen

          45.4k342102




          45.4k342102












          • Thank you for your answer. Why can a generic genus $g>0$ compact complex surface not admit any antiholomorphic automorphisms? I'd appreciate any references.
            – srp
            2 days ago




















          • Thank you for your answer. Why can a generic genus $g>0$ compact complex surface not admit any antiholomorphic automorphisms? I'd appreciate any references.
            – srp
            2 days ago


















          Thank you for your answer. Why can a generic genus $g>0$ compact complex surface not admit any antiholomorphic automorphisms? I'd appreciate any references.
          – srp
          2 days ago






          Thank you for your answer. Why can a generic genus $g>0$ compact complex surface not admit any antiholomorphic automorphisms? I'd appreciate any references.
          – srp
          2 days ago




















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