What does it mean for a complex differential form on a complex manifold to be real?












0












$begingroup$


I am trying to read Kobayashi's "Differential geometry of complex vector bundles". There are many places where a complex differential form is referred to as being real. e.g



Chapter I, Proposition 7.24 p. 28 " A closed real $(p,p)$ form $omega$ on a compact Kähler manifold M is cohomologous to zero if and only if $ omega = id' d'' phi $ for some real $(p-1,p-1)$-form $phi. $ "



Similarly, p. 41 Chapter II, Proposition 2.23 " Given any closed real $(1,1)$-form $phi$ representing $c_{1}(E),$ there is an Hermitian structure $h$ in $E$ such that $phi = c_{1}(E)$ provided $M$ is compact Kähler."



What does it mean? Does it mean that the coefficients are real valued function? Or does it mean that it is the same under complex conjugation?
i.e. $bar{phi} = phi ? $



To be more explicit,



Let $alpha, beta : mathbb{C}^{2} rightarrow mathbb{R}$ be real valued smooth functions and let $$ phi = [alpha(z_{1}, z_{2}) + i beta(z_{1}, z_{2})] dz_{1}wedge dbar{z_{2}} - [alpha(z_{1}, z_{2}) - i beta(z_{1}, z_{2})] dz_{2}wedge dbar{z_{1}} . $$



This is a $(1,1)$-form on $mathbb{C}^{2}$ with the property that $bar{phi} = phi, $ since $ overline{dz_{1}wedge dbar{z_{2}}} = - dz_{2}wedge dbar{z_{1}},$ but this is a real valued form only if $beta equiv 0$ on $mathbb{C}^{2}.$ Or does $phi$ being real mean something else?



Thanks in advance for any help.






differential-geometry differential-forms complex-geometry







share|cite|improve this question











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    0












    $begingroup$


    I am trying to read Kobayashi's "Differential geometry of complex vector bundles". There are many places where a complex differential form is referred to as being real. e.g



    Chapter I, Proposition 7.24 p. 28 " A closed real $(p,p)$ form $omega$ on a compact Kähler manifold M is cohomologous to zero if and only if $ omega = id' d'' phi $ for some real $(p-1,p-1)$-form $phi. $ "



    Similarly, p. 41 Chapter II, Proposition 2.23 " Given any closed real $(1,1)$-form $phi$ representing $c_{1}(E),$ there is an Hermitian structure $h$ in $E$ such that $phi = c_{1}(E)$ provided $M$ is compact Kähler."



    What does it mean? Does it mean that the coefficients are real valued function? Or does it mean that it is the same under complex conjugation?
    i.e. $bar{phi} = phi ? $



    To be more explicit,



    Let $alpha, beta : mathbb{C}^{2} rightarrow mathbb{R}$ be real valued smooth functions and let $$ phi = [alpha(z_{1}, z_{2}) + i beta(z_{1}, z_{2})] dz_{1}wedge dbar{z_{2}} - [alpha(z_{1}, z_{2}) - i beta(z_{1}, z_{2})] dz_{2}wedge dbar{z_{1}} . $$



    This is a $(1,1)$-form on $mathbb{C}^{2}$ with the property that $bar{phi} = phi, $ since $ overline{dz_{1}wedge dbar{z_{2}}} = - dz_{2}wedge dbar{z_{1}},$ but this is a real valued form only if $beta equiv 0$ on $mathbb{C}^{2}.$ Or does $phi$ being real mean something else?



    Thanks in advance for any help.






    differential-geometry differential-forms complex-geometry







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am trying to read Kobayashi's "Differential geometry of complex vector bundles". There are many places where a complex differential form is referred to as being real. e.g



      Chapter I, Proposition 7.24 p. 28 " A closed real $(p,p)$ form $omega$ on a compact Kähler manifold M is cohomologous to zero if and only if $ omega = id' d'' phi $ for some real $(p-1,p-1)$-form $phi. $ "



      Similarly, p. 41 Chapter II, Proposition 2.23 " Given any closed real $(1,1)$-form $phi$ representing $c_{1}(E),$ there is an Hermitian structure $h$ in $E$ such that $phi = c_{1}(E)$ provided $M$ is compact Kähler."



      What does it mean? Does it mean that the coefficients are real valued function? Or does it mean that it is the same under complex conjugation?
      i.e. $bar{phi} = phi ? $



      To be more explicit,



      Let $alpha, beta : mathbb{C}^{2} rightarrow mathbb{R}$ be real valued smooth functions and let $$ phi = [alpha(z_{1}, z_{2}) + i beta(z_{1}, z_{2})] dz_{1}wedge dbar{z_{2}} - [alpha(z_{1}, z_{2}) - i beta(z_{1}, z_{2})] dz_{2}wedge dbar{z_{1}} . $$



      This is a $(1,1)$-form on $mathbb{C}^{2}$ with the property that $bar{phi} = phi, $ since $ overline{dz_{1}wedge dbar{z_{2}}} = - dz_{2}wedge dbar{z_{1}},$ but this is a real valued form only if $beta equiv 0$ on $mathbb{C}^{2}.$ Or does $phi$ being real mean something else?



      Thanks in advance for any help.






      differential-geometry differential-forms complex-geometry







      share|cite|improve this question











      $endgroup$




      I am trying to read Kobayashi's "Differential geometry of complex vector bundles". There are many places where a complex differential form is referred to as being real. e.g



      Chapter I, Proposition 7.24 p. 28 " A closed real $(p,p)$ form $omega$ on a compact Kähler manifold M is cohomologous to zero if and only if $ omega = id' d'' phi $ for some real $(p-1,p-1)$-form $phi. $ "



      Similarly, p. 41 Chapter II, Proposition 2.23 " Given any closed real $(1,1)$-form $phi$ representing $c_{1}(E),$ there is an Hermitian structure $h$ in $E$ such that $phi = c_{1}(E)$ provided $M$ is compact Kähler."



      What does it mean? Does it mean that the coefficients are real valued function? Or does it mean that it is the same under complex conjugation?
      i.e. $bar{phi} = phi ? $



      To be more explicit,



      Let $alpha, beta : mathbb{C}^{2} rightarrow mathbb{R}$ be real valued smooth functions and let $$ phi = [alpha(z_{1}, z_{2}) + i beta(z_{1}, z_{2})] dz_{1}wedge dbar{z_{2}} - [alpha(z_{1}, z_{2}) - i beta(z_{1}, z_{2})] dz_{2}wedge dbar{z_{1}} . $$



      This is a $(1,1)$-form on $mathbb{C}^{2}$ with the property that $bar{phi} = phi, $ since $ overline{dz_{1}wedge dbar{z_{2}}} = - dz_{2}wedge dbar{z_{1}},$ but this is a real valued form only if $beta equiv 0$ on $mathbb{C}^{2}.$ Or does $phi$ being real mean something else?



      Thanks in advance for any help.






      differential-geometry differential-forms complex-geometry




      differential-geometry differential-forms complex-geometry






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      edited Dec 21 '18 at 15:13







      Swarnendu Sil

















      asked Dec 21 '18 at 14:59









      Swarnendu SilSwarnendu Sil

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          $begingroup$

          That means if you write $dz_j=dx_j+sqrt{-1}dy_j$ and expand everything you are left with only real coefficients. For example, a $(1,1)$-form $xi dzwedge d overline{z}$ is real if writing $xi=alpha+sqrt{-1}beta$ and $dz=dx+sqrt{-1}dy$,$$xi dzwedge doverline{z}=(alpha+sqrt{-1}beta)(-2sqrt{-1}dxwedge dy)$$ has real coefficient, i.e., $(alpha+sqrt{-1}beta)(-2sqrt{-1})$ is a real function.






          share|cite|improve this answer









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            $begingroup$
            I am not absolutely sure, but I think what is you said is always equivalent to requiring $bar{phi} = phi .$ At any rate, your answer clarifies the point that being real and being real-valued are distinct notions. My example above is real, since in real co-ordinates, it would be $2 alpha ( dx_{1}wedge dx_{2} + dy_{1}wedge dy_{2}) + 2 beta ( dx_{1}wedge dy_{2} + dx_{2}wedge dy_{1}).$ But it is not real-valued unless $beta = 0.$ Thanks.
            $endgroup$
            – Swarnendu Sil
            Jan 11 at 19:21












          • $begingroup$
            You’re welcome, yes they are equivalent
            $endgroup$
            – Chris Huang
            Jan 11 at 20:15











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          $begingroup$

          That means if you write $dz_j=dx_j+sqrt{-1}dy_j$ and expand everything you are left with only real coefficients. For example, a $(1,1)$-form $xi dzwedge d overline{z}$ is real if writing $xi=alpha+sqrt{-1}beta$ and $dz=dx+sqrt{-1}dy$,$$xi dzwedge doverline{z}=(alpha+sqrt{-1}beta)(-2sqrt{-1}dxwedge dy)$$ has real coefficient, i.e., $(alpha+sqrt{-1}beta)(-2sqrt{-1})$ is a real function.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I am not absolutely sure, but I think what is you said is always equivalent to requiring $bar{phi} = phi .$ At any rate, your answer clarifies the point that being real and being real-valued are distinct notions. My example above is real, since in real co-ordinates, it would be $2 alpha ( dx_{1}wedge dx_{2} + dy_{1}wedge dy_{2}) + 2 beta ( dx_{1}wedge dy_{2} + dx_{2}wedge dy_{1}).$ But it is not real-valued unless $beta = 0.$ Thanks.
            $endgroup$
            – Swarnendu Sil
            Jan 11 at 19:21












          • $begingroup$
            You’re welcome, yes they are equivalent
            $endgroup$
            – Chris Huang
            Jan 11 at 20:15
















          1












          $begingroup$

          That means if you write $dz_j=dx_j+sqrt{-1}dy_j$ and expand everything you are left with only real coefficients. For example, a $(1,1)$-form $xi dzwedge d overline{z}$ is real if writing $xi=alpha+sqrt{-1}beta$ and $dz=dx+sqrt{-1}dy$,$$xi dzwedge doverline{z}=(alpha+sqrt{-1}beta)(-2sqrt{-1}dxwedge dy)$$ has real coefficient, i.e., $(alpha+sqrt{-1}beta)(-2sqrt{-1})$ is a real function.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I am not absolutely sure, but I think what is you said is always equivalent to requiring $bar{phi} = phi .$ At any rate, your answer clarifies the point that being real and being real-valued are distinct notions. My example above is real, since in real co-ordinates, it would be $2 alpha ( dx_{1}wedge dx_{2} + dy_{1}wedge dy_{2}) + 2 beta ( dx_{1}wedge dy_{2} + dx_{2}wedge dy_{1}).$ But it is not real-valued unless $beta = 0.$ Thanks.
            $endgroup$
            – Swarnendu Sil
            Jan 11 at 19:21












          • $begingroup$
            You’re welcome, yes they are equivalent
            $endgroup$
            – Chris Huang
            Jan 11 at 20:15














          1












          1








          1





          $begingroup$

          That means if you write $dz_j=dx_j+sqrt{-1}dy_j$ and expand everything you are left with only real coefficients. For example, a $(1,1)$-form $xi dzwedge d overline{z}$ is real if writing $xi=alpha+sqrt{-1}beta$ and $dz=dx+sqrt{-1}dy$,$$xi dzwedge doverline{z}=(alpha+sqrt{-1}beta)(-2sqrt{-1}dxwedge dy)$$ has real coefficient, i.e., $(alpha+sqrt{-1}beta)(-2sqrt{-1})$ is a real function.






          share|cite|improve this answer









          $endgroup$



          That means if you write $dz_j=dx_j+sqrt{-1}dy_j$ and expand everything you are left with only real coefficients. For example, a $(1,1)$-form $xi dzwedge d overline{z}$ is real if writing $xi=alpha+sqrt{-1}beta$ and $dz=dx+sqrt{-1}dy$,$$xi dzwedge doverline{z}=(alpha+sqrt{-1}beta)(-2sqrt{-1}dxwedge dy)$$ has real coefficient, i.e., $(alpha+sqrt{-1}beta)(-2sqrt{-1})$ is a real function.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 18:33









          Chris HuangChris Huang

          13019




          13019








          • 1




            $begingroup$
            I am not absolutely sure, but I think what is you said is always equivalent to requiring $bar{phi} = phi .$ At any rate, your answer clarifies the point that being real and being real-valued are distinct notions. My example above is real, since in real co-ordinates, it would be $2 alpha ( dx_{1}wedge dx_{2} + dy_{1}wedge dy_{2}) + 2 beta ( dx_{1}wedge dy_{2} + dx_{2}wedge dy_{1}).$ But it is not real-valued unless $beta = 0.$ Thanks.
            $endgroup$
            – Swarnendu Sil
            Jan 11 at 19:21












          • $begingroup$
            You’re welcome, yes they are equivalent
            $endgroup$
            – Chris Huang
            Jan 11 at 20:15














          • 1




            $begingroup$
            I am not absolutely sure, but I think what is you said is always equivalent to requiring $bar{phi} = phi .$ At any rate, your answer clarifies the point that being real and being real-valued are distinct notions. My example above is real, since in real co-ordinates, it would be $2 alpha ( dx_{1}wedge dx_{2} + dy_{1}wedge dy_{2}) + 2 beta ( dx_{1}wedge dy_{2} + dx_{2}wedge dy_{1}).$ But it is not real-valued unless $beta = 0.$ Thanks.
            $endgroup$
            – Swarnendu Sil
            Jan 11 at 19:21












          • $begingroup$
            You’re welcome, yes they are equivalent
            $endgroup$
            – Chris Huang
            Jan 11 at 20:15








          1




          1




          $begingroup$
          I am not absolutely sure, but I think what is you said is always equivalent to requiring $bar{phi} = phi .$ At any rate, your answer clarifies the point that being real and being real-valued are distinct notions. My example above is real, since in real co-ordinates, it would be $2 alpha ( dx_{1}wedge dx_{2} + dy_{1}wedge dy_{2}) + 2 beta ( dx_{1}wedge dy_{2} + dx_{2}wedge dy_{1}).$ But it is not real-valued unless $beta = 0.$ Thanks.
          $endgroup$
          – Swarnendu Sil
          Jan 11 at 19:21






          $begingroup$
          I am not absolutely sure, but I think what is you said is always equivalent to requiring $bar{phi} = phi .$ At any rate, your answer clarifies the point that being real and being real-valued are distinct notions. My example above is real, since in real co-ordinates, it would be $2 alpha ( dx_{1}wedge dx_{2} + dy_{1}wedge dy_{2}) + 2 beta ( dx_{1}wedge dy_{2} + dx_{2}wedge dy_{1}).$ But it is not real-valued unless $beta = 0.$ Thanks.
          $endgroup$
          – Swarnendu Sil
          Jan 11 at 19:21














          $begingroup$
          You’re welcome, yes they are equivalent
          $endgroup$
          – Chris Huang
          Jan 11 at 20:15




          $begingroup$
          You’re welcome, yes they are equivalent
          $endgroup$
          – Chris Huang
          Jan 11 at 20:15


















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