Holomorphic forms: a very basic notion
$begingroup$
Let $M$ be a holomorphic (complex) manifold with:
$$(1)quad
dim_{C}(M)=m.
$$
What I understand regarding a holomorphic form is that it is:
$$(2)quad
alpha^{(r,0)}=frac{1}{r!}f_{mu_1,dots,mu_r}(z)space dz^{mu_1}wedgedotswedge dz^{mu_r}inOmega^{(r,0)}(M).
$$
Here the totally antisymmetric component functions are themselves holomorphic, and consequently entail only $z$ dependence.
Similarly an antiholomorphic form is given by:
$$(3)quad
beta^{(0,s)}=frac{1}{s!}g_{bar{nu}_1,dots,bar{nu}_s}(bar{z})space dbar{z}^{nu_1}wedgedotswedge dbar{z}^{nu_s}inOmega^{(0,s)}(M),
$$
of course with the component functions being totally antisymmetric and antiholomorphic, consequently entailing only $bar{z}$ dependence.
Taking the wedge of $(2)$ and $(3)$, we get the generic $(r,s)$ form:
$$(4)quad
omega^{(r,s)}=frac{1}{r!s!}f_{mu_1,dots,mu_r}(z)g_{bar{nu}_1,dots,bar{nu}_s}(bar{z})space dz^{mu_1}wedgedotswedge dz^{mu_r}wedge dbar{z}^{nu_1}wedgedotswedge dbar{z}^{nu_s}inOmega^{(r,s)}(M).
$$
Now a form is holomorphic if and only if $bar{partial}omega=0$.
If I begin with the form in $(2)$, which is holomorphic, then there is no problem seeing that $bar{partial}alpha^{(r,0)}=0$, since this would involve $bar{z}$ derivatives of the component functions which are clearly vanishing.
The problem arises when I begin with $eta^{(p,q)}inOmega^{(p,q)}(M)$, require that it satisfy $bar{partial}eta^{(p,q)}=0$, and try to get a quantity of the form $(2)$.
Question 1: Can someone please guide me along this proof?
Question 2: The following is particularly bothering me: consider $gamma^{(p,m)}in Omega^{(p,m)}(M)$, then clearly $bar{partial}gamma^{(p,m)}=0$ (for the wedge having $m+1$ barred indices is vanishing), but I cannot write it in the form $(2)$.
riemannian-geometry smooth-manifolds differential-forms complex-geometry
asked Dec 9 '18 at 15:36
KongKong
315
$endgroup$
add a comment |
$begingroup$
Let $M$ be a holomorphic (complex) manifold with:
$$(1)quad
dim_{C}(M)=m.
$$
What I understand regarding a holomorphic form is that it is:
$$(2)quad
alpha^{(r,0)}=frac{1}{r!}f_{mu_1,dots,mu_r}(z)space dz^{mu_1}wedgedotswedge dz^{mu_r}inOmega^{(r,0)}(M).
$$
Here the totally antisymmetric component functions are themselves holomorphic, and consequently entail only $z$ dependence.
Similarly an antiholomorphic form is given by:
$$(3)quad
beta^{(0,s)}=frac{1}{s!}g_{bar{nu}_1,dots,bar{nu}_s}(bar{z})space dbar{z}^{nu_1}wedgedotswedge dbar{z}^{nu_s}inOmega^{(0,s)}(M),
$$
of course with the component functions being totally antisymmetric and antiholomorphic, consequently entailing only $bar{z}$ dependence.
Taking the wedge of $(2)$ and $(3)$, we get the generic $(r,s)$ form:
$$(4)quad
omega^{(r,s)}=frac{1}{r!s!}f_{mu_1,dots,mu_r}(z)g_{bar{nu}_1,dots,bar{nu}_s}(bar{z})space dz^{mu_1}wedgedotswedge dz^{mu_r}wedge dbar{z}^{nu_1}wedgedotswedge dbar{z}^{nu_s}inOmega^{(r,s)}(M).
$$
Now a form is holomorphic if and only if $bar{partial}omega=0$.
If I begin with the form in $(2)$, which is holomorphic, then there is no problem seeing that $bar{partial}alpha^{(r,0)}=0$, since this would involve $bar{z}$ derivatives of the component functions which are clearly vanishing.
The problem arises when I begin with $eta^{(p,q)}inOmega^{(p,q)}(M)$, require that it satisfy $bar{partial}eta^{(p,q)}=0$, and try to get a quantity of the form $(2)$.
Question 1: Can someone please guide me along this proof?
Question 2: The following is particularly bothering me: consider $gamma^{(p,m)}in Omega^{(p,m)}(M)$, then clearly $bar{partial}gamma^{(p,m)}=0$ (for the wedge having $m+1$ barred indices is vanishing), but I cannot write it in the form $(2)$.
riemannian-geometry smooth-manifolds differential-forms complex-geometry
asked Dec 9 '18 at 15:36
KongKong
315
$endgroup$
2
$begingroup$
A holomorphic $p$-form $alpha$ is a $(p, 0)$-form such that $bar{partial}alpha = 0$. A $(p, q)$-form which is $bar{partial}$-closed is not usually called holomorphic.
$endgroup$
– Michael Albanese
Dec 9 '18 at 18:24
$begingroup$
Got that, thanks :)
$endgroup$
– Kong
Dec 10 '18 at 15:03
$begingroup$
You're wrong when you claim that a general $(r,s)$ form can be written as a wedge product of holomorphic and antiholomorphic forms. You need a general $C^infty$ function for coefficients, not a sum of products of holomorphic and antiholomorphic functions.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:46
$begingroup$
I don't understand your question. You're trying to see that if you start with a $barpartial$-closed $(p,q)$ form, then it must be of type $(p,0)$? That's just not right, of course. Take the Kähler form or its powers, and you'll get $d$-closed (hence $barpartial$-closed) $(p,p)$ forms on any Kähler manifold.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:54
add a comment |
$begingroup$
Let $M$ be a holomorphic (complex) manifold with:
$$(1)quad
dim_{C}(M)=m.
$$
What I understand regarding a holomorphic form is that it is:
$$(2)quad
alpha^{(r,0)}=frac{1}{r!}f_{mu_1,dots,mu_r}(z)space dz^{mu_1}wedgedotswedge dz^{mu_r}inOmega^{(r,0)}(M).
$$
Here the totally antisymmetric component functions are themselves holomorphic, and consequently entail only $z$ dependence.
Similarly an antiholomorphic form is given by:
$$(3)quad
beta^{(0,s)}=frac{1}{s!}g_{bar{nu}_1,dots,bar{nu}_s}(bar{z})space dbar{z}^{nu_1}wedgedotswedge dbar{z}^{nu_s}inOmega^{(0,s)}(M),
$$
of course with the component functions being totally antisymmetric and antiholomorphic, consequently entailing only $bar{z}$ dependence.
Taking the wedge of $(2)$ and $(3)$, we get the generic $(r,s)$ form:
$$(4)quad
omega^{(r,s)}=frac{1}{r!s!}f_{mu_1,dots,mu_r}(z)g_{bar{nu}_1,dots,bar{nu}_s}(bar{z})space dz^{mu_1}wedgedotswedge dz^{mu_r}wedge dbar{z}^{nu_1}wedgedotswedge dbar{z}^{nu_s}inOmega^{(r,s)}(M).
$$
Now a form is holomorphic if and only if $bar{partial}omega=0$.
If I begin with the form in $(2)$, which is holomorphic, then there is no problem seeing that $bar{partial}alpha^{(r,0)}=0$, since this would involve $bar{z}$ derivatives of the component functions which are clearly vanishing.
The problem arises when I begin with $eta^{(p,q)}inOmega^{(p,q)}(M)$, require that it satisfy $bar{partial}eta^{(p,q)}=0$, and try to get a quantity of the form $(2)$.
Question 1: Can someone please guide me along this proof?
Question 2: The following is particularly bothering me: consider $gamma^{(p,m)}in Omega^{(p,m)}(M)$, then clearly $bar{partial}gamma^{(p,m)}=0$ (for the wedge having $m+1$ barred indices is vanishing), but I cannot write it in the form $(2)$.
riemannian-geometry smooth-manifolds differential-forms complex-geometry
asked Dec 9 '18 at 15:36
KongKong
315
$endgroup$
Let $M$ be a holomorphic (complex) manifold with:
$$(1)quad
dim_{C}(M)=m.
$$
What I understand regarding a holomorphic form is that it is:
$$(2)quad
alpha^{(r,0)}=frac{1}{r!}f_{mu_1,dots,mu_r}(z)space dz^{mu_1}wedgedotswedge dz^{mu_r}inOmega^{(r,0)}(M).
$$
Here the totally antisymmetric component functions are themselves holomorphic, and consequently entail only $z$ dependence.
Similarly an antiholomorphic form is given by:
$$(3)quad
beta^{(0,s)}=frac{1}{s!}g_{bar{nu}_1,dots,bar{nu}_s}(bar{z})space dbar{z}^{nu_1}wedgedotswedge dbar{z}^{nu_s}inOmega^{(0,s)}(M),
$$
of course with the component functions being totally antisymmetric and antiholomorphic, consequently entailing only $bar{z}$ dependence.
Taking the wedge of $(2)$ and $(3)$, we get the generic $(r,s)$ form:
$$(4)quad
omega^{(r,s)}=frac{1}{r!s!}f_{mu_1,dots,mu_r}(z)g_{bar{nu}_1,dots,bar{nu}_s}(bar{z})space dz^{mu_1}wedgedotswedge dz^{mu_r}wedge dbar{z}^{nu_1}wedgedotswedge dbar{z}^{nu_s}inOmega^{(r,s)}(M).
$$
Now a form is holomorphic if and only if $bar{partial}omega=0$.
If I begin with the form in $(2)$, which is holomorphic, then there is no problem seeing that $bar{partial}alpha^{(r,0)}=0$, since this would involve $bar{z}$ derivatives of the component functions which are clearly vanishing.
The problem arises when I begin with $eta^{(p,q)}inOmega^{(p,q)}(M)$, require that it satisfy $bar{partial}eta^{(p,q)}=0$, and try to get a quantity of the form $(2)$.
Question 1: Can someone please guide me along this proof?
Question 2: The following is particularly bothering me: consider $gamma^{(p,m)}in Omega^{(p,m)}(M)$, then clearly $bar{partial}gamma^{(p,m)}=0$ (for the wedge having $m+1$ barred indices is vanishing), but I cannot write it in the form $(2)$.
riemannian-geometry smooth-manifolds differential-forms complex-geometry
riemannian-geometry smooth-manifolds differential-forms complex-geometry
asked Dec 9 '18 at 15:36
KongKong
315
asked Dec 9 '18 at 15:36
KongKong
315
asked Dec 9 '18 at 15:36
KongKong
315
asked Dec 9 '18 at 15:36
KongKong
315
asked Dec 9 '18 at 15:36
KongKong
315
315
2
$begingroup$
A holomorphic $p$-form $alpha$ is a $(p, 0)$-form such that $bar{partial}alpha = 0$. A $(p, q)$-form which is $bar{partial}$-closed is not usually called holomorphic.
$endgroup$
– Michael Albanese
Dec 9 '18 at 18:24
$begingroup$
Got that, thanks :)
$endgroup$
– Kong
Dec 10 '18 at 15:03
$begingroup$
You're wrong when you claim that a general $(r,s)$ form can be written as a wedge product of holomorphic and antiholomorphic forms. You need a general $C^infty$ function for coefficients, not a sum of products of holomorphic and antiholomorphic functions.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:46
$begingroup$
I don't understand your question. You're trying to see that if you start with a $barpartial$-closed $(p,q)$ form, then it must be of type $(p,0)$? That's just not right, of course. Take the Kähler form or its powers, and you'll get $d$-closed (hence $barpartial$-closed) $(p,p)$ forms on any Kähler manifold.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:54
add a comment |
2
$begingroup$
A holomorphic $p$-form $alpha$ is a $(p, 0)$-form such that $bar{partial}alpha = 0$. A $(p, q)$-form which is $bar{partial}$-closed is not usually called holomorphic.
$endgroup$
– Michael Albanese
Dec 9 '18 at 18:24
$begingroup$
Got that, thanks :)
$endgroup$
– Kong
Dec 10 '18 at 15:03
$begingroup$
You're wrong when you claim that a general $(r,s)$ form can be written as a wedge product of holomorphic and antiholomorphic forms. You need a general $C^infty$ function for coefficients, not a sum of products of holomorphic and antiholomorphic functions.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:46
$begingroup$
I don't understand your question. You're trying to see that if you start with a $barpartial$-closed $(p,q)$ form, then it must be of type $(p,0)$? That's just not right, of course. Take the Kähler form or its powers, and you'll get $d$-closed (hence $barpartial$-closed) $(p,p)$ forms on any Kähler manifold.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:54
2
2
$begingroup$
A holomorphic $p$-form $alpha$ is a $(p, 0)$-form such that $bar{partial}alpha = 0$. A $(p, q)$-form which is $bar{partial}$-closed is not usually called holomorphic.
$endgroup$
– Michael Albanese
Dec 9 '18 at 18:24
$begingroup$
A holomorphic $p$-form $alpha$ is a $(p, 0)$-form such that $bar{partial}alpha = 0$. A $(p, q)$-form which is $bar{partial}$-closed is not usually called holomorphic.
$endgroup$
– Michael Albanese
Dec 9 '18 at 18:24
$begingroup$
Got that, thanks :)
$endgroup$
– Kong
Dec 10 '18 at 15:03
$begingroup$
Got that, thanks :)
$endgroup$
– Kong
Dec 10 '18 at 15:03
$begingroup$
You're wrong when you claim that a general $(r,s)$ form can be written as a wedge product of holomorphic and antiholomorphic forms. You need a general $C^infty$ function for coefficients, not a sum of products of holomorphic and antiholomorphic functions.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:46
$begingroup$
You're wrong when you claim that a general $(r,s)$ form can be written as a wedge product of holomorphic and antiholomorphic forms. You need a general $C^infty$ function for coefficients, not a sum of products of holomorphic and antiholomorphic functions.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:46
$begingroup$
I don't understand your question. You're trying to see that if you start with a $barpartial$-closed $(p,q)$ form, then it must be of type $(p,0)$? That's just not right, of course. Take the Kähler form or its powers, and you'll get $d$-closed (hence $barpartial$-closed) $(p,p)$ forms on any Kähler manifold.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:54
$begingroup$
I don't understand your question. You're trying to see that if you start with a $barpartial$-closed $(p,q)$ form, then it must be of type $(p,0)$? That's just not right, of course. Take the Kähler form or its powers, and you'll get $d$-closed (hence $barpartial$-closed) $(p,p)$ forms on any Kähler manifold.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:54
add a comment |
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$begingroup$
A holomorphic $p$-form $alpha$ is a $(p, 0)$-form such that $bar{partial}alpha = 0$. A $(p, q)$-form which is $bar{partial}$-closed is not usually called holomorphic.
$endgroup$
– Michael Albanese
Dec 9 '18 at 18:24
$begingroup$
Got that, thanks :)
$endgroup$
– Kong
Dec 10 '18 at 15:03
$begingroup$
You're wrong when you claim that a general $(r,s)$ form can be written as a wedge product of holomorphic and antiholomorphic forms. You need a general $C^infty$ function for coefficients, not a sum of products of holomorphic and antiholomorphic functions.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:46
$begingroup$
I don't understand your question. You're trying to see that if you start with a $barpartial$-closed $(p,q)$ form, then it must be of type $(p,0)$? That's just not right, of course. Take the Kähler form or its powers, and you'll get $d$-closed (hence $barpartial$-closed) $(p,p)$ forms on any Kähler manifold.
$endgroup$
– Ted Shifrin
Dec 10 '18 at 22:54