Mixing
Kahler metric and $(1,1)$- forms
$begingroup$
I am following this set of lecture notes. On page 3, the author is considering a metric on a complex Kahler manifold of dimension $n$, which is denoted by $g_{alpha, bar{beta}} = partial_{alpha}partial_{bar{beta}} K(vec{z}, vec{bar{z}})$. Immediately afterwards, the author writes down a (1,1)-form given by $J = g_{alpha, bar{beta}} dz^{alpha} wedge dbar{z}^{bar{beta}}$, which he shows is a $(1,1)$ form belonging to $H^{(1,1)}$ cohomology class, and subsequently defines a Kahler class.
My question is this, the metric is symmetric in its indices, so then how can it be used to construct a (1,1)-form, which is antisymmetric in its indices? What am I missing here? Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
differential-geometry kahler-manifolds complex-manifolds
asked Jan 24 at 6:30
Bruce LeeBruce Lee
187
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add a comment |
$begingroup$
I am following this set of lecture notes. On page 3, the author is considering a metric on a complex Kahler manifold of dimension $n$, which is denoted by $g_{alpha, bar{beta}} = partial_{alpha}partial_{bar{beta}} K(vec{z}, vec{bar{z}})$. Immediately afterwards, the author writes down a (1,1)-form given by $J = g_{alpha, bar{beta}} dz^{alpha} wedge dbar{z}^{bar{beta}}$, which he shows is a $(1,1)$ form belonging to $H^{(1,1)}$ cohomology class, and subsequently defines a Kahler class.
My question is this, the metric is symmetric in its indices, so then how can it be used to construct a (1,1)-form, which is antisymmetric in its indices? What am I missing here? Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
differential-geometry kahler-manifolds complex-manifolds
asked Jan 24 at 6:30
Bruce LeeBruce Lee
187
$endgroup$
$begingroup$
Is the author missing a $wedge$ sign? Some authors just don't bother with them...
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:35
$begingroup$
@LordSharktheUnknown Hi, the author wasn't missing it but I made a typo, thanks for that. My question revolves around the wedge product only, the metric is symmetric in its indices, so in that case how are we using it in a wedge product?
$endgroup$
– Bruce Lee
Jan 24 at 6:40
$begingroup$
@LordSharktheUnknown Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
$endgroup$
– Bruce Lee
Jan 24 at 6:48
add a comment |
$begingroup$
I am following this set of lecture notes. On page 3, the author is considering a metric on a complex Kahler manifold of dimension $n$, which is denoted by $g_{alpha, bar{beta}} = partial_{alpha}partial_{bar{beta}} K(vec{z}, vec{bar{z}})$. Immediately afterwards, the author writes down a (1,1)-form given by $J = g_{alpha, bar{beta}} dz^{alpha} wedge dbar{z}^{bar{beta}}$, which he shows is a $(1,1)$ form belonging to $H^{(1,1)}$ cohomology class, and subsequently defines a Kahler class.
My question is this, the metric is symmetric in its indices, so then how can it be used to construct a (1,1)-form, which is antisymmetric in its indices? What am I missing here? Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
differential-geometry kahler-manifolds complex-manifolds
asked Jan 24 at 6:30
Bruce LeeBruce Lee
187
$endgroup$
I am following this set of lecture notes. On page 3, the author is considering a metric on a complex Kahler manifold of dimension $n$, which is denoted by $g_{alpha, bar{beta}} = partial_{alpha}partial_{bar{beta}} K(vec{z}, vec{bar{z}})$. Immediately afterwards, the author writes down a (1,1)-form given by $J = g_{alpha, bar{beta}} dz^{alpha} wedge dbar{z}^{bar{beta}}$, which he shows is a $(1,1)$ form belonging to $H^{(1,1)}$ cohomology class, and subsequently defines a Kahler class.
My question is this, the metric is symmetric in its indices, so then how can it be used to construct a (1,1)-form, which is antisymmetric in its indices? What am I missing here? Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
differential-geometry kahler-manifolds complex-manifolds
differential-geometry kahler-manifolds complex-manifolds
asked Jan 24 at 6:30
Bruce LeeBruce Lee
187
asked Jan 24 at 6:30
Bruce LeeBruce Lee
187
edited Jan 24 at 6:50
Bruce Lee
asked Jan 24 at 6:30
Bruce LeeBruce Lee
187
asked Jan 24 at 6:30
Bruce LeeBruce Lee
187
asked Jan 24 at 6:30
Bruce LeeBruce Lee
187
187
$begingroup$
Is the author missing a $wedge$ sign? Some authors just don't bother with them...
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:35
$begingroup$
@LordSharktheUnknown Hi, the author wasn't missing it but I made a typo, thanks for that. My question revolves around the wedge product only, the metric is symmetric in its indices, so in that case how are we using it in a wedge product?
$endgroup$
– Bruce Lee
Jan 24 at 6:40
$begingroup$
@LordSharktheUnknown Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
$endgroup$
– Bruce Lee
Jan 24 at 6:48
add a comment |
$begingroup$
Is the author missing a $wedge$ sign? Some authors just don't bother with them...
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:35
$begingroup$
@LordSharktheUnknown Hi, the author wasn't missing it but I made a typo, thanks for that. My question revolves around the wedge product only, the metric is symmetric in its indices, so in that case how are we using it in a wedge product?
$endgroup$
– Bruce Lee
Jan 24 at 6:40
$begingroup$
@LordSharktheUnknown Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
$endgroup$
– Bruce Lee
Jan 24 at 6:48
$begingroup$
Is the author missing a $wedge$ sign? Some authors just don't bother with them...
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:35
$begingroup$
Is the author missing a $wedge$ sign? Some authors just don't bother with them...
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:35
$begingroup$
@LordSharktheUnknown Hi, the author wasn't missing it but I made a typo, thanks for that. My question revolves around the wedge product only, the metric is symmetric in its indices, so in that case how are we using it in a wedge product?
$endgroup$
– Bruce Lee
Jan 24 at 6:40
$begingroup$
@LordSharktheUnknown Hi, the author wasn't missing it but I made a typo, thanks for that. My question revolves around the wedge product only, the metric is symmetric in its indices, so in that case how are we using it in a wedge product?
$endgroup$
– Bruce Lee
Jan 24 at 6:40
$begingroup$
@LordSharktheUnknown Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
$endgroup$
– Bruce Lee
Jan 24 at 6:48
$begingroup$
@LordSharktheUnknown Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
$endgroup$
– Bruce Lee
Jan 24 at 6:48
add a comment |
1 Answer
1
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$begingroup$
Keep in mind that your $J$ is a product of the symmetric metric by the antisymmetric (1,1)-forms $dz^alphawedge dbar{z}^beta$. A symmetric thing times an antisymmetric thing is antisymmetric. So, for example,
begin{align*}
J(X,Y)=left(g_{alphabar{beta}},dz^alphawedge dbar{z}^betaright)(X,Y) = g_{alphabar{beta}}left( dz^alpha(X)dbar{z}^beta(Y) - dz^alpha(Y)dbar{z}^beta(X)right)
end{align*}
is clearly antisymmetric in $(X,Y)$. To see the indices of $J$ explicitly, compute:
begin{align*}
J_{ibar{j}} = J(partial_i,partial_{bar{j}})
= g_{kbar{ell}}(delta_{ik}delta_{jell}-delta_{iell}delta_{jk}),
end{align*}
where $delta$ denotes the Kronecker delta. Now it's again explicit that $J_{ibar{j}}=-J_{jbar{i}}$.
Just to note, your $J$ is usually called the Kahler form and denoted $omega$. People usually reserve $J$ for the (almost-) complex structure.
answered Jan 28 at 3:05
Quinton WestrichQuinton Westrich
1445
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Keep in mind that your $J$ is a product of the symmetric metric by the antisymmetric (1,1)-forms $dz^alphawedge dbar{z}^beta$. A symmetric thing times an antisymmetric thing is antisymmetric. So, for example,
begin{align*}
J(X,Y)=left(g_{alphabar{beta}},dz^alphawedge dbar{z}^betaright)(X,Y) = g_{alphabar{beta}}left( dz^alpha(X)dbar{z}^beta(Y) - dz^alpha(Y)dbar{z}^beta(X)right)
end{align*}
is clearly antisymmetric in $(X,Y)$. To see the indices of $J$ explicitly, compute:
begin{align*}
J_{ibar{j}} = J(partial_i,partial_{bar{j}})
= g_{kbar{ell}}(delta_{ik}delta_{jell}-delta_{iell}delta_{jk}),
end{align*}
where $delta$ denotes the Kronecker delta. Now it's again explicit that $J_{ibar{j}}=-J_{jbar{i}}$.
Just to note, your $J$ is usually called the Kahler form and denoted $omega$. People usually reserve $J$ for the (almost-) complex structure.
answered Jan 28 at 3:05
Quinton WestrichQuinton Westrich
1445
$endgroup$
add a comment |
$begingroup$
Keep in mind that your $J$ is a product of the symmetric metric by the antisymmetric (1,1)-forms $dz^alphawedge dbar{z}^beta$. A symmetric thing times an antisymmetric thing is antisymmetric. So, for example,
begin{align*}
J(X,Y)=left(g_{alphabar{beta}},dz^alphawedge dbar{z}^betaright)(X,Y) = g_{alphabar{beta}}left( dz^alpha(X)dbar{z}^beta(Y) - dz^alpha(Y)dbar{z}^beta(X)right)
end{align*}
is clearly antisymmetric in $(X,Y)$. To see the indices of $J$ explicitly, compute:
begin{align*}
J_{ibar{j}} = J(partial_i,partial_{bar{j}})
= g_{kbar{ell}}(delta_{ik}delta_{jell}-delta_{iell}delta_{jk}),
end{align*}
where $delta$ denotes the Kronecker delta. Now it's again explicit that $J_{ibar{j}}=-J_{jbar{i}}$.
Just to note, your $J$ is usually called the Kahler form and denoted $omega$. People usually reserve $J$ for the (almost-) complex structure.
answered Jan 28 at 3:05
Quinton WestrichQuinton Westrich
1445
$endgroup$
add a comment |
$begingroup$
Keep in mind that your $J$ is a product of the symmetric metric by the antisymmetric (1,1)-forms $dz^alphawedge dbar{z}^beta$. A symmetric thing times an antisymmetric thing is antisymmetric. So, for example,
begin{align*}
J(X,Y)=left(g_{alphabar{beta}},dz^alphawedge dbar{z}^betaright)(X,Y) = g_{alphabar{beta}}left( dz^alpha(X)dbar{z}^beta(Y) - dz^alpha(Y)dbar{z}^beta(X)right)
end{align*}
is clearly antisymmetric in $(X,Y)$. To see the indices of $J$ explicitly, compute:
begin{align*}
J_{ibar{j}} = J(partial_i,partial_{bar{j}})
= g_{kbar{ell}}(delta_{ik}delta_{jell}-delta_{iell}delta_{jk}),
end{align*}
where $delta$ denotes the Kronecker delta. Now it's again explicit that $J_{ibar{j}}=-J_{jbar{i}}$.
Just to note, your $J$ is usually called the Kahler form and denoted $omega$. People usually reserve $J$ for the (almost-) complex structure.
answered Jan 28 at 3:05
Quinton WestrichQuinton Westrich
1445
$endgroup$
Keep in mind that your $J$ is a product of the symmetric metric by the antisymmetric (1,1)-forms $dz^alphawedge dbar{z}^beta$. A symmetric thing times an antisymmetric thing is antisymmetric. So, for example,
begin{align*}
J(X,Y)=left(g_{alphabar{beta}},dz^alphawedge dbar{z}^betaright)(X,Y) = g_{alphabar{beta}}left( dz^alpha(X)dbar{z}^beta(Y) - dz^alpha(Y)dbar{z}^beta(X)right)
end{align*}
is clearly antisymmetric in $(X,Y)$. To see the indices of $J$ explicitly, compute:
begin{align*}
J_{ibar{j}} = J(partial_i,partial_{bar{j}})
= g_{kbar{ell}}(delta_{ik}delta_{jell}-delta_{iell}delta_{jk}),
end{align*}
where $delta$ denotes the Kronecker delta. Now it's again explicit that $J_{ibar{j}}=-J_{jbar{i}}$.
Just to note, your $J$ is usually called the Kahler form and denoted $omega$. People usually reserve $J$ for the (almost-) complex structure.
answered Jan 28 at 3:05
Quinton WestrichQuinton Westrich
1445
answered Jan 28 at 3:05
Quinton WestrichQuinton Westrich
1445
answered Jan 28 at 3:05
Quinton WestrichQuinton Westrich
1445
answered Jan 28 at 3:05
Quinton WestrichQuinton Westrich
1445
1445
add a comment |
add a comment |
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$begingroup$
Is the author missing a $wedge$ sign? Some authors just don't bother with them...
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:35
$begingroup$
@LordSharktheUnknown Hi, the author wasn't missing it but I made a typo, thanks for that. My question revolves around the wedge product only, the metric is symmetric in its indices, so in that case how are we using it in a wedge product?
$endgroup$
– Bruce Lee
Jan 24 at 6:40
$begingroup$
@LordSharktheUnknown Is the antisymmetrization independently in the $z$'s and the $bar{z}$'s and hence this is alright?
$endgroup$
– Bruce Lee
Jan 24 at 6:48