Expressing $*(z_1wedge bar z_2)$ without Hodge star operator












2












$begingroup$


I have been stuck on a computation for hours, and still cannot figure out where is the mistake:



For $2$-dimensional complex vector space, let $z_1,z_2$ be the basis. I want to compute $*(z_1wedge bar z_2)$. I am pretty sure the answer is $-z_2wedgebar z_1$. But if I write $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ then $x_1,y_1,x_2,y_2$ is the standard basis. Then by the formula here,
we have
begin{equation}
*z_1wedge bar z_2\
=*(x_1+iy_1)wedge(x_2-iy_2)\
=*(x_1wedge x_2-ix_1wedge y_2+i y_1wedge x_2+y_1wedge y_2)\
=-y_1wedge y_2-iy_1wedge x_2+ix_1wedge y_2-x_1wedge x_2\
=-(x_1+iy_1)wedge (x_2-iy_2)\
=-z_1wedge bar z_2
end{equation}



which I also think is correct. So where is my mistake?






linear-algebra complex-geometry exterior-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
    $endgroup$
    – md2perpe
    Jul 30 '18 at 22:24










  • $begingroup$
    @md2perpe Because I am pretty sure there is a formula $*(dz_I wedge dz_J)=(text{some coefficient})dz_{I^0}wedge dbar z_{J^0}$ where $I,J$ are any multi-indexes and $I^0$ is the complement of $I$.
    $endgroup$
    – User X
    Jul 31 '18 at 6:51












  • $begingroup$
    @md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
    $endgroup$
    – User X
    Jul 31 '18 at 7:18










  • $begingroup$
    But then you have just move "pretty sure" to the statement $*(dz_I wedge dbar z_J) sim dz_{I^0} wedge dbar z_{J^0}.$ Why are you pretty sure about that?
    $endgroup$
    – md2perpe
    Jul 31 '18 at 7:45






  • 1




    $begingroup$
    If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
    $endgroup$
    – md2perpe
    Aug 1 '18 at 11:52
















2












$begingroup$


I have been stuck on a computation for hours, and still cannot figure out where is the mistake:



For $2$-dimensional complex vector space, let $z_1,z_2$ be the basis. I want to compute $*(z_1wedge bar z_2)$. I am pretty sure the answer is $-z_2wedgebar z_1$. But if I write $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ then $x_1,y_1,x_2,y_2$ is the standard basis. Then by the formula here,
we have
begin{equation}
*z_1wedge bar z_2\
=*(x_1+iy_1)wedge(x_2-iy_2)\
=*(x_1wedge x_2-ix_1wedge y_2+i y_1wedge x_2+y_1wedge y_2)\
=-y_1wedge y_2-iy_1wedge x_2+ix_1wedge y_2-x_1wedge x_2\
=-(x_1+iy_1)wedge (x_2-iy_2)\
=-z_1wedge bar z_2
end{equation}



which I also think is correct. So where is my mistake?






linear-algebra complex-geometry exterior-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
    $endgroup$
    – md2perpe
    Jul 30 '18 at 22:24










  • $begingroup$
    @md2perpe Because I am pretty sure there is a formula $*(dz_I wedge dz_J)=(text{some coefficient})dz_{I^0}wedge dbar z_{J^0}$ where $I,J$ are any multi-indexes and $I^0$ is the complement of $I$.
    $endgroup$
    – User X
    Jul 31 '18 at 6:51












  • $begingroup$
    @md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
    $endgroup$
    – User X
    Jul 31 '18 at 7:18










  • $begingroup$
    But then you have just move "pretty sure" to the statement $*(dz_I wedge dbar z_J) sim dz_{I^0} wedge dbar z_{J^0}.$ Why are you pretty sure about that?
    $endgroup$
    – md2perpe
    Jul 31 '18 at 7:45






  • 1




    $begingroup$
    If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
    $endgroup$
    – md2perpe
    Aug 1 '18 at 11:52














2












2








2





$begingroup$


I have been stuck on a computation for hours, and still cannot figure out where is the mistake:



For $2$-dimensional complex vector space, let $z_1,z_2$ be the basis. I want to compute $*(z_1wedge bar z_2)$. I am pretty sure the answer is $-z_2wedgebar z_1$. But if I write $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ then $x_1,y_1,x_2,y_2$ is the standard basis. Then by the formula here,
we have
begin{equation}
*z_1wedge bar z_2\
=*(x_1+iy_1)wedge(x_2-iy_2)\
=*(x_1wedge x_2-ix_1wedge y_2+i y_1wedge x_2+y_1wedge y_2)\
=-y_1wedge y_2-iy_1wedge x_2+ix_1wedge y_2-x_1wedge x_2\
=-(x_1+iy_1)wedge (x_2-iy_2)\
=-z_1wedge bar z_2
end{equation}



which I also think is correct. So where is my mistake?






linear-algebra complex-geometry exterior-algebra







share|cite|improve this question











$endgroup$




I have been stuck on a computation for hours, and still cannot figure out where is the mistake:



For $2$-dimensional complex vector space, let $z_1,z_2$ be the basis. I want to compute $*(z_1wedge bar z_2)$. I am pretty sure the answer is $-z_2wedgebar z_1$. But if I write $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ then $x_1,y_1,x_2,y_2$ is the standard basis. Then by the formula here,
we have
begin{equation}
*z_1wedge bar z_2\
=*(x_1+iy_1)wedge(x_2-iy_2)\
=*(x_1wedge x_2-ix_1wedge y_2+i y_1wedge x_2+y_1wedge y_2)\
=-y_1wedge y_2-iy_1wedge x_2+ix_1wedge y_2-x_1wedge x_2\
=-(x_1+iy_1)wedge (x_2-iy_2)\
=-z_1wedge bar z_2
end{equation}



which I also think is correct. So where is my mistake?






linear-algebra complex-geometry exterior-algebra




linear-algebra complex-geometry exterior-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 2:19









epimorphic

2,74231533




2,74231533










asked Jul 30 '18 at 19:56









User XUser X

32911




32911












  • $begingroup$
    Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
    $endgroup$
    – md2perpe
    Jul 30 '18 at 22:24










  • $begingroup$
    @md2perpe Because I am pretty sure there is a formula $*(dz_I wedge dz_J)=(text{some coefficient})dz_{I^0}wedge dbar z_{J^0}$ where $I,J$ are any multi-indexes and $I^0$ is the complement of $I$.
    $endgroup$
    – User X
    Jul 31 '18 at 6:51












  • $begingroup$
    @md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
    $endgroup$
    – User X
    Jul 31 '18 at 7:18










  • $begingroup$
    But then you have just move "pretty sure" to the statement $*(dz_I wedge dbar z_J) sim dz_{I^0} wedge dbar z_{J^0}.$ Why are you pretty sure about that?
    $endgroup$
    – md2perpe
    Jul 31 '18 at 7:45






  • 1




    $begingroup$
    If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
    $endgroup$
    – md2perpe
    Aug 1 '18 at 11:52


















  • $begingroup$
    Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
    $endgroup$
    – md2perpe
    Jul 30 '18 at 22:24










  • $begingroup$
    @md2perpe Because I am pretty sure there is a formula $*(dz_I wedge dz_J)=(text{some coefficient})dz_{I^0}wedge dbar z_{J^0}$ where $I,J$ are any multi-indexes and $I^0$ is the complement of $I$.
    $endgroup$
    – User X
    Jul 31 '18 at 6:51












  • $begingroup$
    @md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
    $endgroup$
    – User X
    Jul 31 '18 at 7:18










  • $begingroup$
    But then you have just move "pretty sure" to the statement $*(dz_I wedge dbar z_J) sim dz_{I^0} wedge dbar z_{J^0}.$ Why are you pretty sure about that?
    $endgroup$
    – md2perpe
    Jul 31 '18 at 7:45






  • 1




    $begingroup$
    If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
    $endgroup$
    – md2perpe
    Aug 1 '18 at 11:52
















$begingroup$
Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
$endgroup$
– md2perpe
Jul 30 '18 at 22:24




$begingroup$
Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
$endgroup$
– md2perpe
Jul 30 '18 at 22:24












$begingroup$
@md2perpe Because I am pretty sure there is a formula $*(dz_I wedge dz_J)=(text{some coefficient})dz_{I^0}wedge dbar z_{J^0}$ where $I,J$ are any multi-indexes and $I^0$ is the complement of $I$.
$endgroup$
– User X
Jul 31 '18 at 6:51






$begingroup$
@md2perpe Because I am pretty sure there is a formula $*(dz_I wedge dz_J)=(text{some coefficient})dz_{I^0}wedge dbar z_{J^0}$ where $I,J$ are any multi-indexes and $I^0$ is the complement of $I$.
$endgroup$
– User X
Jul 31 '18 at 6:51














$begingroup$
@md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
$endgroup$
– User X
Jul 31 '18 at 7:18




$begingroup$
@md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
$endgroup$
– User X
Jul 31 '18 at 7:18












$begingroup$
But then you have just move "pretty sure" to the statement $*(dz_I wedge dbar z_J) sim dz_{I^0} wedge dbar z_{J^0}.$ Why are you pretty sure about that?
$endgroup$
– md2perpe
Jul 31 '18 at 7:45




$begingroup$
But then you have just move "pretty sure" to the statement $*(dz_I wedge dbar z_J) sim dz_{I^0} wedge dbar z_{J^0}.$ Why are you pretty sure about that?
$endgroup$
– md2perpe
Jul 31 '18 at 7:45




1




1




$begingroup$
If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
$endgroup$
– md2perpe
Aug 1 '18 at 11:52




$begingroup$
If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
$endgroup$
– md2perpe
Aug 1 '18 at 11:52










1 Answer
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$begingroup$

For the formula $(alpha,alpha)=int alphawedge * alpha$ to be valid for complex $alpha$ we need a complex conjugation in the definition of $*alpha$. Doing that we get
$$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$






share|cite|improve this answer









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    $begingroup$

    For the formula $(alpha,alpha)=int alphawedge * alpha$ to be valid for complex $alpha$ we need a complex conjugation in the definition of $*alpha$. Doing that we get
    $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      For the formula $(alpha,alpha)=int alphawedge * alpha$ to be valid for complex $alpha$ we need a complex conjugation in the definition of $*alpha$. Doing that we get
      $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        For the formula $(alpha,alpha)=int alphawedge * alpha$ to be valid for complex $alpha$ we need a complex conjugation in the definition of $*alpha$. Doing that we get
        $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$






        share|cite|improve this answer









        $endgroup$



        For the formula $(alpha,alpha)=int alphawedge * alpha$ to be valid for complex $alpha$ we need a complex conjugation in the definition of $*alpha$. Doing that we get
        $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 2 '18 at 8:08









        md2perpemd2perpe

        7,80111028




        7,80111028






























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