Expressing $*(z_1wedge bar z_2)$ without Hodge star operator
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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
edited Dec 3 at 2:19
epimorphic
2,72131533
asked Jul 30 at 19:56
User X
17411
|
show 4 more comments
up vote
2
down vote
favorite
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
edited Dec 3 at 2:19
epimorphic
2,72131533
asked Jul 30 at 19:56
User X
17411
Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
– md2perpe
Jul 30 at 22:24
@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 $I$.
– User X
Jul 31 at 6:51
@md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
– User X
Jul 31 at 7:18
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?
– md2perpe
Jul 31 at 7:45
1
If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
– md2perpe
Aug 1 at 11:52
|
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
edited Dec 3 at 2:19
epimorphic
2,72131533
asked Jul 30 at 19:56
User X
17411
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
edited Dec 3 at 2:19
epimorphic
2,72131533
asked Jul 30 at 19:56
User X
17411
edited Dec 3 at 2:19
epimorphic
2,72131533
asked Jul 30 at 19:56
User X
17411
edited Dec 3 at 2:19
epimorphic
2,72131533
edited Dec 3 at 2:19
epimorphic
2,72131533
edited Dec 3 at 2:19
epimorphic
2,72131533
2,72131533
asked Jul 30 at 19:56
User X
17411
asked Jul 30 at 19:56
User X
17411
asked Jul 30 at 19:56
User X
17411
17411
Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
– md2perpe
Jul 30 at 22:24
@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 $I$.
– User X
Jul 31 at 6:51
@md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
– User X
Jul 31 at 7:18
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?
– md2perpe
Jul 31 at 7:45
1
If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
– md2perpe
Aug 1 at 11:52
|
show 4 more comments
Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
– md2perpe
Jul 30 at 22:24
@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 $I$.
– User X
Jul 31 at 6:51
@md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
– User X
Jul 31 at 7:18
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?
– md2perpe
Jul 31 at 7:45
1
If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
– md2perpe
Aug 1 at 11:52
Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
– md2perpe
Jul 30 at 22:24
Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
– md2perpe
Jul 30 at 22:24
@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 $I$.
– User X
Jul 31 at 6:51
@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 $I$.
– User X
Jul 31 at 6:51
@md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
– User X
Jul 31 at 7:18
@md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
– User X
Jul 31 at 7:18
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?
– md2perpe
Jul 31 at 7:45
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?
– md2perpe
Jul 31 at 7:45
1
1
If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
– md2perpe
Aug 1 at 11:52
If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
– md2perpe
Aug 1 at 11:52
|
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
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
$$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
answered Aug 2 at 8:08
md2perpe
7,28811027
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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
$$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
answered Aug 2 at 8:08
md2perpe
7,28811027
add a comment |
up vote
1
down vote
accepted
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
$$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
answered Aug 2 at 8:08
md2perpe
7,28811027
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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
$$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
answered Aug 2 at 8:08
md2perpe
7,28811027
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
$$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
answered Aug 2 at 8:08
md2perpe
7,28811027
answered Aug 2 at 8:08
md2perpe
7,28811027
answered Aug 2 at 8:08
md2perpe
7,28811027
answered Aug 2 at 8:08
md2perpe
7,28811027
7,28811027
add a comment |
add a comment |
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Why are you "pretty sure" that the answer is $-z_2 wedge bar z_1$?
– md2perpe
Jul 30 at 22:24
@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 $I$.
– User X
Jul 31 at 6:51
@md2perpe there is a obvious typo in my last comment: the first $*(dz_Iwedge dz_J)$ should be $*(dz_Iwedge dbar z_J)$.
– User X
Jul 31 at 7:18
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?
– md2perpe
Jul 31 at 7:45
1
If $*alpha$ takes the complex conjugate of the coefficients, then I get $$*(dz_1 wedge dbar z_2) = - dbar z_1 wedge dz_2 = dz_2 wedge dbar z_1.$$
– md2perpe
Aug 1 at 11:52