Proposition $4.16$ from Ballmann's Lectures on Kähler Manifolds
I'm reading proposition 4.16 from Ballmann's Lectures on Kähler Manifolds:
Let $M$ be a complex manifold with a compatible Riemann metric $g$ and Levi-Civita connection $nabla$, then:
$$domega(X,Y,Z)=g((nabla_XJ)Y,Z)+g((nabla_YJ)Z,X)+g((nabla_ZJ)X,Y)$$
$$2g((nabla_XJ)Y,Z)=domega(X,Y,Z)-domega(X,JY,JZ)$$
He begins the proof by saying "Since $M$ is a complex manifold, we can assume that the vector fields $X,Y,Z,JY$ and $JZ$ commute".
I assume "X,Y commutes" (two variables) means $[X,Y]=0$, but I don't understand what it has to do with $M$ being complex and what does he mean by "$X,Y,Z,JY,JZ$ commute" (five variables).
differential-geometry smooth-manifolds complex-geometry kahler-manifolds
asked Nov 23 at 19:22
rmdmc89
2,0511921
add a comment |
I'm reading proposition 4.16 from Ballmann's Lectures on Kähler Manifolds:
Let $M$ be a complex manifold with a compatible Riemann metric $g$ and Levi-Civita connection $nabla$, then:
$$domega(X,Y,Z)=g((nabla_XJ)Y,Z)+g((nabla_YJ)Z,X)+g((nabla_ZJ)X,Y)$$
$$2g((nabla_XJ)Y,Z)=domega(X,Y,Z)-domega(X,JY,JZ)$$
He begins the proof by saying "Since $M$ is a complex manifold, we can assume that the vector fields $X,Y,Z,JY$ and $JZ$ commute".
I assume "X,Y commutes" (two variables) means $[X,Y]=0$, but I don't understand what it has to do with $M$ being complex and what does he mean by "$X,Y,Z,JY,JZ$ commute" (five variables).
differential-geometry smooth-manifolds complex-geometry kahler-manifolds
asked Nov 23 at 19:22
rmdmc89
2,0511921
1
Is it saying we can find $n$ vector fields $X_1,ldots,X_n$ such that $[X_l,X_m]=0$ and any vector field is of the form $sum_{l=1}^n f_l X_l$ with $f_l:M to mathbb{C}$ ?
– reuns
Nov 23 at 20:28
@reuns I guess that makes sense. But using your construction, can we conclude $[X,JY] =[X,JZ]=...=0$? I guess that's what he does, since all the brackets vanish in his calculations
– rmdmc89
Nov 23 at 23:44
add a comment |
I'm reading proposition 4.16 from Ballmann's Lectures on Kähler Manifolds:
Let $M$ be a complex manifold with a compatible Riemann metric $g$ and Levi-Civita connection $nabla$, then:
$$domega(X,Y,Z)=g((nabla_XJ)Y,Z)+g((nabla_YJ)Z,X)+g((nabla_ZJ)X,Y)$$
$$2g((nabla_XJ)Y,Z)=domega(X,Y,Z)-domega(X,JY,JZ)$$
He begins the proof by saying "Since $M$ is a complex manifold, we can assume that the vector fields $X,Y,Z,JY$ and $JZ$ commute".
I assume "X,Y commutes" (two variables) means $[X,Y]=0$, but I don't understand what it has to do with $M$ being complex and what does he mean by "$X,Y,Z,JY,JZ$ commute" (five variables).
differential-geometry smooth-manifolds complex-geometry kahler-manifolds
asked Nov 23 at 19:22
rmdmc89
2,0511921
I'm reading proposition 4.16 from Ballmann's Lectures on Kähler Manifolds:
Let $M$ be a complex manifold with a compatible Riemann metric $g$ and Levi-Civita connection $nabla$, then:
$$domega(X,Y,Z)=g((nabla_XJ)Y,Z)+g((nabla_YJ)Z,X)+g((nabla_ZJ)X,Y)$$
$$2g((nabla_XJ)Y,Z)=domega(X,Y,Z)-domega(X,JY,JZ)$$
He begins the proof by saying "Since $M$ is a complex manifold, we can assume that the vector fields $X,Y,Z,JY$ and $JZ$ commute".
I assume "X,Y commutes" (two variables) means $[X,Y]=0$, but I don't understand what it has to do with $M$ being complex and what does he mean by "$X,Y,Z,JY,JZ$ commute" (five variables).
differential-geometry smooth-manifolds complex-geometry kahler-manifolds
differential-geometry smooth-manifolds complex-geometry kahler-manifolds
asked Nov 23 at 19:22
rmdmc89
2,0511921
asked Nov 23 at 19:22
rmdmc89
2,0511921
asked Nov 23 at 19:22
rmdmc89
2,0511921
asked Nov 23 at 19:22
rmdmc89
2,0511921
asked Nov 23 at 19:22
rmdmc89
2,0511921
2,0511921
1
Is it saying we can find $n$ vector fields $X_1,ldots,X_n$ such that $[X_l,X_m]=0$ and any vector field is of the form $sum_{l=1}^n f_l X_l$ with $f_l:M to mathbb{C}$ ?
– reuns
Nov 23 at 20:28
@reuns I guess that makes sense. But using your construction, can we conclude $[X,JY] =[X,JZ]=...=0$? I guess that's what he does, since all the brackets vanish in his calculations
– rmdmc89
Nov 23 at 23:44
add a comment |
1
Is it saying we can find $n$ vector fields $X_1,ldots,X_n$ such that $[X_l,X_m]=0$ and any vector field is of the form $sum_{l=1}^n f_l X_l$ with $f_l:M to mathbb{C}$ ?
– reuns
Nov 23 at 20:28
@reuns I guess that makes sense. But using your construction, can we conclude $[X,JY] =[X,JZ]=...=0$? I guess that's what he does, since all the brackets vanish in his calculations
– rmdmc89
Nov 23 at 23:44
1
1
Is it saying we can find $n$ vector fields $X_1,ldots,X_n$ such that $[X_l,X_m]=0$ and any vector field is of the form $sum_{l=1}^n f_l X_l$ with $f_l:M to mathbb{C}$ ?
– reuns
Nov 23 at 20:28
Is it saying we can find $n$ vector fields $X_1,ldots,X_n$ such that $[X_l,X_m]=0$ and any vector field is of the form $sum_{l=1}^n f_l X_l$ with $f_l:M to mathbb{C}$ ?
– reuns
Nov 23 at 20:28
@reuns I guess that makes sense. But using your construction, can we conclude $[X,JY] =[X,JZ]=...=0$? I guess that's what he does, since all the brackets vanish in his calculations
– rmdmc89
Nov 23 at 23:44
@reuns I guess that makes sense. But using your construction, can we conclude $[X,JY] =[X,JZ]=...=0$? I guess that's what he does, since all the brackets vanish in his calculations
– rmdmc89
Nov 23 at 23:44
add a comment |
1 Answer
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The equalities you have to prove are of the form $T(X,Y,Z)=0$, where $T$ is a tensor. A tensor something which satisfies $T(a X+bX', Y,Z)=aT(X, Y, Z)+bT(X',T,Z)$ for $a,b$ to functions (and similarly for, $Y,Z$).
Then, you choose a complex chart $(z_1,..., z_n)$, and you check it for the fields $partial over {partial z_i} $, ${partial over {partial z_i }} = i {partial over {partial z_i}}$. This is enough, because these fields generate all vector fields.
Note that these fields commute, and furthermore that $i {partial over {partial z_i}}$ is the local expression for $Jpartial over {partial z_i} $. So to prove that $T(X,Y,Z)=0$ it is enough to check it for all fields $X,Y,Z$ such that $X,Y,Z, JX,JY,JZ$ commute
answered Nov 24 at 10:47
Thomas
3,779510
I'm confused because $J$ is a function from $TM$ to itself, but $frac{partial}{partial z_j},frac{partial}{partial overline{z}_j}in TMotimes mathbb{C}$, so what does $Jfrac{partial }{partial z_j}$ mean?
– rmdmc89
Nov 24 at 14:21
Also, what does $left[frac{partial }{partial z_j},frac{partial }{partial overline{z}_j}right]$ mean since $[cdot,cdot]$ is defined for elements in $TM$?
– rmdmc89
Nov 24 at 14:31
1
$J_x$ at some point $x$ is an element of $End(T_x(M))$. By abuse of notation, one sometimes denotes also by $J$ its complexification, which is a complex endomorphism of $T_x(M) otimes mathbb{C}$.
– Malkoun
Nov 24 at 17:05
1
Similarly, one can consider the Lie bracket of two smooth complex vector fields, by just complexifying the usual (real) Lie bracket.
– Malkoun
Nov 24 at 17:06
add a comment |
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The equalities you have to prove are of the form $T(X,Y,Z)=0$, where $T$ is a tensor. A tensor something which satisfies $T(a X+bX', Y,Z)=aT(X, Y, Z)+bT(X',T,Z)$ for $a,b$ to functions (and similarly for, $Y,Z$).
Then, you choose a complex chart $(z_1,..., z_n)$, and you check it for the fields $partial over {partial z_i} $, ${partial over {partial z_i }} = i {partial over {partial z_i}}$. This is enough, because these fields generate all vector fields.
Note that these fields commute, and furthermore that $i {partial over {partial z_i}}$ is the local expression for $Jpartial over {partial z_i} $. So to prove that $T(X,Y,Z)=0$ it is enough to check it for all fields $X,Y,Z$ such that $X,Y,Z, JX,JY,JZ$ commute
answered Nov 24 at 10:47
Thomas
3,779510
I'm confused because $J$ is a function from $TM$ to itself, but $frac{partial}{partial z_j},frac{partial}{partial overline{z}_j}in TMotimes mathbb{C}$, so what does $Jfrac{partial }{partial z_j}$ mean?
– rmdmc89
Nov 24 at 14:21
Also, what does $left[frac{partial }{partial z_j},frac{partial }{partial overline{z}_j}right]$ mean since $[cdot,cdot]$ is defined for elements in $TM$?
– rmdmc89
Nov 24 at 14:31
1
$J_x$ at some point $x$ is an element of $End(T_x(M))$. By abuse of notation, one sometimes denotes also by $J$ its complexification, which is a complex endomorphism of $T_x(M) otimes mathbb{C}$.
– Malkoun
Nov 24 at 17:05
1
Similarly, one can consider the Lie bracket of two smooth complex vector fields, by just complexifying the usual (real) Lie bracket.
– Malkoun
Nov 24 at 17:06
add a comment |
The equalities you have to prove are of the form $T(X,Y,Z)=0$, where $T$ is a tensor. A tensor something which satisfies $T(a X+bX', Y,Z)=aT(X, Y, Z)+bT(X',T,Z)$ for $a,b$ to functions (and similarly for, $Y,Z$).
Then, you choose a complex chart $(z_1,..., z_n)$, and you check it for the fields $partial over {partial z_i} $, ${partial over {partial z_i }} = i {partial over {partial z_i}}$. This is enough, because these fields generate all vector fields.
Note that these fields commute, and furthermore that $i {partial over {partial z_i}}$ is the local expression for $Jpartial over {partial z_i} $. So to prove that $T(X,Y,Z)=0$ it is enough to check it for all fields $X,Y,Z$ such that $X,Y,Z, JX,JY,JZ$ commute
answered Nov 24 at 10:47
Thomas
3,779510
I'm confused because $J$ is a function from $TM$ to itself, but $frac{partial}{partial z_j},frac{partial}{partial overline{z}_j}in TMotimes mathbb{C}$, so what does $Jfrac{partial }{partial z_j}$ mean?
– rmdmc89
Nov 24 at 14:21
Also, what does $left[frac{partial }{partial z_j},frac{partial }{partial overline{z}_j}right]$ mean since $[cdot,cdot]$ is defined for elements in $TM$?
– rmdmc89
Nov 24 at 14:31
1
$J_x$ at some point $x$ is an element of $End(T_x(M))$. By abuse of notation, one sometimes denotes also by $J$ its complexification, which is a complex endomorphism of $T_x(M) otimes mathbb{C}$.
– Malkoun
Nov 24 at 17:05
1
Similarly, one can consider the Lie bracket of two smooth complex vector fields, by just complexifying the usual (real) Lie bracket.
– Malkoun
Nov 24 at 17:06
add a comment |
The equalities you have to prove are of the form $T(X,Y,Z)=0$, where $T$ is a tensor. A tensor something which satisfies $T(a X+bX', Y,Z)=aT(X, Y, Z)+bT(X',T,Z)$ for $a,b$ to functions (and similarly for, $Y,Z$).
Then, you choose a complex chart $(z_1,..., z_n)$, and you check it for the fields $partial over {partial z_i} $, ${partial over {partial z_i }} = i {partial over {partial z_i}}$. This is enough, because these fields generate all vector fields.
Note that these fields commute, and furthermore that $i {partial over {partial z_i}}$ is the local expression for $Jpartial over {partial z_i} $. So to prove that $T(X,Y,Z)=0$ it is enough to check it for all fields $X,Y,Z$ such that $X,Y,Z, JX,JY,JZ$ commute
answered Nov 24 at 10:47
Thomas
3,779510
The equalities you have to prove are of the form $T(X,Y,Z)=0$, where $T$ is a tensor. A tensor something which satisfies $T(a X+bX', Y,Z)=aT(X, Y, Z)+bT(X',T,Z)$ for $a,b$ to functions (and similarly for, $Y,Z$).
Then, you choose a complex chart $(z_1,..., z_n)$, and you check it for the fields $partial over {partial z_i} $, ${partial over {partial z_i }} = i {partial over {partial z_i}}$. This is enough, because these fields generate all vector fields.
Note that these fields commute, and furthermore that $i {partial over {partial z_i}}$ is the local expression for $Jpartial over {partial z_i} $. So to prove that $T(X,Y,Z)=0$ it is enough to check it for all fields $X,Y,Z$ such that $X,Y,Z, JX,JY,JZ$ commute
answered Nov 24 at 10:47
Thomas
3,779510
answered Nov 24 at 10:47
Thomas
3,779510
answered Nov 24 at 10:47
Thomas
3,779510
answered Nov 24 at 10:47
Thomas
3,779510
3,779510
I'm confused because $J$ is a function from $TM$ to itself, but $frac{partial}{partial z_j},frac{partial}{partial overline{z}_j}in TMotimes mathbb{C}$, so what does $Jfrac{partial }{partial z_j}$ mean?
– rmdmc89
Nov 24 at 14:21
Also, what does $left[frac{partial }{partial z_j},frac{partial }{partial overline{z}_j}right]$ mean since $[cdot,cdot]$ is defined for elements in $TM$?
– rmdmc89
Nov 24 at 14:31
1
$J_x$ at some point $x$ is an element of $End(T_x(M))$. By abuse of notation, one sometimes denotes also by $J$ its complexification, which is a complex endomorphism of $T_x(M) otimes mathbb{C}$.
– Malkoun
Nov 24 at 17:05
1
Similarly, one can consider the Lie bracket of two smooth complex vector fields, by just complexifying the usual (real) Lie bracket.
– Malkoun
Nov 24 at 17:06
add a comment |
I'm confused because $J$ is a function from $TM$ to itself, but $frac{partial}{partial z_j},frac{partial}{partial overline{z}_j}in TMotimes mathbb{C}$, so what does $Jfrac{partial }{partial z_j}$ mean?
– rmdmc89
Nov 24 at 14:21
Also, what does $left[frac{partial }{partial z_j},frac{partial }{partial overline{z}_j}right]$ mean since $[cdot,cdot]$ is defined for elements in $TM$?
– rmdmc89
Nov 24 at 14:31
1
$J_x$ at some point $x$ is an element of $End(T_x(M))$. By abuse of notation, one sometimes denotes also by $J$ its complexification, which is a complex endomorphism of $T_x(M) otimes mathbb{C}$.
– Malkoun
Nov 24 at 17:05
1
Similarly, one can consider the Lie bracket of two smooth complex vector fields, by just complexifying the usual (real) Lie bracket.
– Malkoun
Nov 24 at 17:06
I'm confused because $J$ is a function from $TM$ to itself, but $frac{partial}{partial z_j},frac{partial}{partial overline{z}_j}in TMotimes mathbb{C}$, so what does $Jfrac{partial }{partial z_j}$ mean?
– rmdmc89
Nov 24 at 14:21
I'm confused because $J$ is a function from $TM$ to itself, but $frac{partial}{partial z_j},frac{partial}{partial overline{z}_j}in TMotimes mathbb{C}$, so what does $Jfrac{partial }{partial z_j}$ mean?
– rmdmc89
Nov 24 at 14:21
Also, what does $left[frac{partial }{partial z_j},frac{partial }{partial overline{z}_j}right]$ mean since $[cdot,cdot]$ is defined for elements in $TM$?
– rmdmc89
Nov 24 at 14:31
Also, what does $left[frac{partial }{partial z_j},frac{partial }{partial overline{z}_j}right]$ mean since $[cdot,cdot]$ is defined for elements in $TM$?
– rmdmc89
Nov 24 at 14:31
1
1
$J_x$ at some point $x$ is an element of $End(T_x(M))$. By abuse of notation, one sometimes denotes also by $J$ its complexification, which is a complex endomorphism of $T_x(M) otimes mathbb{C}$.
– Malkoun
Nov 24 at 17:05
$J_x$ at some point $x$ is an element of $End(T_x(M))$. By abuse of notation, one sometimes denotes also by $J$ its complexification, which is a complex endomorphism of $T_x(M) otimes mathbb{C}$.
– Malkoun
Nov 24 at 17:05
1
1
Similarly, one can consider the Lie bracket of two smooth complex vector fields, by just complexifying the usual (real) Lie bracket.
– Malkoun
Nov 24 at 17:06
Similarly, one can consider the Lie bracket of two smooth complex vector fields, by just complexifying the usual (real) Lie bracket.
– Malkoun
Nov 24 at 17:06
add a comment |
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1
Is it saying we can find $n$ vector fields $X_1,ldots,X_n$ such that $[X_l,X_m]=0$ and any vector field is of the form $sum_{l=1}^n f_l X_l$ with $f_l:M to mathbb{C}$ ?
– reuns
Nov 23 at 20:28
@reuns I guess that makes sense. But using your construction, can we conclude $[X,JY] =[X,JZ]=...=0$? I guess that's what he does, since all the brackets vanish in his calculations
– rmdmc89
Nov 23 at 23:44