Mixing
Chern class of holomorphic symplectic manifold
$begingroup$
Let $M$ a complex surface and $omegain H^0(Omega_M^2,M)$ a non degenerate holomomorphic form.
I've read somewhere (without proof), that then the first chern class of the symplecitc manifold $(M,Re~ omega)$ vanishes.
Why is this true? As far as I know, the Chern class of $(M, Re~ omega)$ is the chern class of any complex vector bundle with almost complex structure compatible with $Re ~omega$. I would suspect that the original complex structure is of this kind. But doing a local calculation, this seems to be only correct up to a sign.
Update: Now posted on mathoverflow
EDIT: (The calculation)
We need that $Re ~omega(Ju,Jv)=Re~omega(u,v)$ for all tangent vectors $u,v$. By linearity of $omega$ it suffices to check this for a basis of each tangent space. So I check this for $frac partial {partial z_i}$ and $frac partial {partial bar z_i}$.
As $omega$ is of type $(2,0)$, it is locally of the form $sum a_{kl}~dz_kwedge dz_l$.
Hence the only interesting case to check is
$omega(frac partial {partial z_i}, frac partial {partial z_j})$.
We have
$$omega(J frac partial {partial z_i}, J frac partial {partial z_j})
= omega(i frac partial {partial z_i}, i frac partial {partial z_j})
=-omega( frac partial {partial z_i}, frac partial {partial z_j}), $$
$$baromega(J frac partial {partial z_i}, J frac partial {partial z_j})= omega(i frac partial {partial z_i}, i frac partial {partial z_j})=-baromega( frac partial {partial z_i}, frac partial {partial z_j})$$
and hence
$$Re ~omega(J frac partial {partial z_i}, J frac partial {partial z_j})
=frac 1 2[omega(J frac partial {partial z_i}, J frac partial {partial z_j})+bar omega(J frac partial {partial z_i}, J frac partial {partial z_j})]
=-frac 1 2[omega( frac partial {partial z_i}, frac partial {partial z_j})+omega( frac partial {partial z_i}, frac partial {partial z_j})]
= -Re~ omega( frac partial {partial z_i}, frac partial {partial z_j}).$$
Now that I think about it, this just seems to be the proof that Kähler forms are of type $(1,1)$.
differential-geometry complex-geometry symplectic-geometry characteristic-classes
$endgroup$
|
show 7 more comments
$begingroup$
Let $M$ a complex surface and $omegain H^0(Omega_M^2,M)$ a non degenerate holomomorphic form.
I've read somewhere (without proof), that then the first chern class of the symplecitc manifold $(M,Re~ omega)$ vanishes.
Why is this true? As far as I know, the Chern class of $(M, Re~ omega)$ is the chern class of any complex vector bundle with almost complex structure compatible with $Re ~omega$. I would suspect that the original complex structure is of this kind. But doing a local calculation, this seems to be only correct up to a sign.
Update: Now posted on mathoverflow
EDIT: (The calculation)
We need that $Re ~omega(Ju,Jv)=Re~omega(u,v)$ for all tangent vectors $u,v$. By linearity of $omega$ it suffices to check this for a basis of each tangent space. So I check this for $frac partial {partial z_i}$ and $frac partial {partial bar z_i}$.
As $omega$ is of type $(2,0)$, it is locally of the form $sum a_{kl}~dz_kwedge dz_l$.
Hence the only interesting case to check is
$omega(frac partial {partial z_i}, frac partial {partial z_j})$.
We have
$$omega(J frac partial {partial z_i}, J frac partial {partial z_j})
= omega(i frac partial {partial z_i}, i frac partial {partial z_j})
=-omega( frac partial {partial z_i}, frac partial {partial z_j}), $$
$$baromega(J frac partial {partial z_i}, J frac partial {partial z_j})= omega(i frac partial {partial z_i}, i frac partial {partial z_j})=-baromega( frac partial {partial z_i}, frac partial {partial z_j})$$
and hence
$$Re ~omega(J frac partial {partial z_i}, J frac partial {partial z_j})
=frac 1 2[omega(J frac partial {partial z_i}, J frac partial {partial z_j})+bar omega(J frac partial {partial z_i}, J frac partial {partial z_j})]
=-frac 1 2[omega( frac partial {partial z_i}, frac partial {partial z_j})+omega( frac partial {partial z_i}, frac partial {partial z_j})]
= -Re~ omega( frac partial {partial z_i}, frac partial {partial z_j}).$$
Now that I think about it, this just seems to be the proof that Kähler forms are of type $(1,1)$.
differential-geometry complex-geometry symplectic-geometry characteristic-classes
$endgroup$
1
$begingroup$
Note that $omega$ gives you a trivialization of the canonical bundle, and so $c_1$ of the canonical bundle, hence the holomorphic cotangent bundle, hence the holomorphic tangent bundle, vanishes; then recall that the ordinary tangent bundle is isomorphic as a complex vector bundle to the holomorphic tangent bundle.
$endgroup$
– Aleksandar Milivojevic
Jan 28 at 18:11
$begingroup$
@AleksandarMilivojevic: Sorry, but this is not clear to me. I get that the chern class of the holomorphic tangent bundle vanishes. I know that thhe holomorphic tangent bundle is isomorphic to the$(1,0)$-part of the complexified real tangent bundle. What do you mean with the "ordinary tangent bundle"? Further, where in your argument does it enter, that we we want to calculate the chern class of a specific symplectic structure on $M$?
$endgroup$
– user639051
Jan 30 at 16:38
$begingroup$
First note that symplectic does not enter since the Chern class only depends on the complex structure (and you’re starting with a complex surface). By “ordinary” tangent bundle I mean the tangent bundle of the underlying smooth manifold (and this has the structure of a complex bundle because of the complex structure).
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 16:41
$begingroup$
@AleksandarMilivojevic: But why does the complex structure induced by the symplecitc form coincide with the original complex structure?
$endgroup$
– user639051
Jan 30 at 18:11
$begingroup$
I see, sorry for not reading carefully. You said you did a computation to show that the two complex structures coincide “up to sign” though? Can you elaborate on that; maybe then we’ll see how one vanishing Chern class implies the other.
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 21:20
|
show 7 more comments
$begingroup$
Let $M$ a complex surface and $omegain H^0(Omega_M^2,M)$ a non degenerate holomomorphic form.
I've read somewhere (without proof), that then the first chern class of the symplecitc manifold $(M,Re~ omega)$ vanishes.
Why is this true? As far as I know, the Chern class of $(M, Re~ omega)$ is the chern class of any complex vector bundle with almost complex structure compatible with $Re ~omega$. I would suspect that the original complex structure is of this kind. But doing a local calculation, this seems to be only correct up to a sign.
Update: Now posted on mathoverflow
EDIT: (The calculation)
We need that $Re ~omega(Ju,Jv)=Re~omega(u,v)$ for all tangent vectors $u,v$. By linearity of $omega$ it suffices to check this for a basis of each tangent space. So I check this for $frac partial {partial z_i}$ and $frac partial {partial bar z_i}$.
As $omega$ is of type $(2,0)$, it is locally of the form $sum a_{kl}~dz_kwedge dz_l$.
Hence the only interesting case to check is
$omega(frac partial {partial z_i}, frac partial {partial z_j})$.
We have
$$omega(J frac partial {partial z_i}, J frac partial {partial z_j})
= omega(i frac partial {partial z_i}, i frac partial {partial z_j})
=-omega( frac partial {partial z_i}, frac partial {partial z_j}), $$
$$baromega(J frac partial {partial z_i}, J frac partial {partial z_j})= omega(i frac partial {partial z_i}, i frac partial {partial z_j})=-baromega( frac partial {partial z_i}, frac partial {partial z_j})$$
and hence
$$Re ~omega(J frac partial {partial z_i}, J frac partial {partial z_j})
=frac 1 2[omega(J frac partial {partial z_i}, J frac partial {partial z_j})+bar omega(J frac partial {partial z_i}, J frac partial {partial z_j})]
=-frac 1 2[omega( frac partial {partial z_i}, frac partial {partial z_j})+omega( frac partial {partial z_i}, frac partial {partial z_j})]
= -Re~ omega( frac partial {partial z_i}, frac partial {partial z_j}).$$
Now that I think about it, this just seems to be the proof that Kähler forms are of type $(1,1)$.
differential-geometry complex-geometry symplectic-geometry characteristic-classes
$endgroup$
Let $M$ a complex surface and $omegain H^0(Omega_M^2,M)$ a non degenerate holomomorphic form.
I've read somewhere (without proof), that then the first chern class of the symplecitc manifold $(M,Re~ omega)$ vanishes.
Why is this true? As far as I know, the Chern class of $(M, Re~ omega)$ is the chern class of any complex vector bundle with almost complex structure compatible with $Re ~omega$. I would suspect that the original complex structure is of this kind. But doing a local calculation, this seems to be only correct up to a sign.
Update: Now posted on mathoverflow
EDIT: (The calculation)
We need that $Re ~omega(Ju,Jv)=Re~omega(u,v)$ for all tangent vectors $u,v$. By linearity of $omega$ it suffices to check this for a basis of each tangent space. So I check this for $frac partial {partial z_i}$ and $frac partial {partial bar z_i}$.
As $omega$ is of type $(2,0)$, it is locally of the form $sum a_{kl}~dz_kwedge dz_l$.
Hence the only interesting case to check is
$omega(frac partial {partial z_i}, frac partial {partial z_j})$.
We have
$$omega(J frac partial {partial z_i}, J frac partial {partial z_j})
= omega(i frac partial {partial z_i}, i frac partial {partial z_j})
=-omega( frac partial {partial z_i}, frac partial {partial z_j}), $$
$$baromega(J frac partial {partial z_i}, J frac partial {partial z_j})= omega(i frac partial {partial z_i}, i frac partial {partial z_j})=-baromega( frac partial {partial z_i}, frac partial {partial z_j})$$
and hence
$$Re ~omega(J frac partial {partial z_i}, J frac partial {partial z_j})
=frac 1 2[omega(J frac partial {partial z_i}, J frac partial {partial z_j})+bar omega(J frac partial {partial z_i}, J frac partial {partial z_j})]
=-frac 1 2[omega( frac partial {partial z_i}, frac partial {partial z_j})+omega( frac partial {partial z_i}, frac partial {partial z_j})]
= -Re~ omega( frac partial {partial z_i}, frac partial {partial z_j}).$$
Now that I think about it, this just seems to be the proof that Kähler forms are of type $(1,1)$.
differential-geometry complex-geometry symplectic-geometry characteristic-classes
differential-geometry complex-geometry symplectic-geometry characteristic-classes
edited Feb 22 at 13:52
asked Jan 28 at 10:20
user639051
1
$begingroup$
Note that $omega$ gives you a trivialization of the canonical bundle, and so $c_1$ of the canonical bundle, hence the holomorphic cotangent bundle, hence the holomorphic tangent bundle, vanishes; then recall that the ordinary tangent bundle is isomorphic as a complex vector bundle to the holomorphic tangent bundle.
$endgroup$
– Aleksandar Milivojevic
Jan 28 at 18:11
$begingroup$
@AleksandarMilivojevic: Sorry, but this is not clear to me. I get that the chern class of the holomorphic tangent bundle vanishes. I know that thhe holomorphic tangent bundle is isomorphic to the$(1,0)$-part of the complexified real tangent bundle. What do you mean with the "ordinary tangent bundle"? Further, where in your argument does it enter, that we we want to calculate the chern class of a specific symplectic structure on $M$?
$endgroup$
– user639051
Jan 30 at 16:38
$begingroup$
First note that symplectic does not enter since the Chern class only depends on the complex structure (and you’re starting with a complex surface). By “ordinary” tangent bundle I mean the tangent bundle of the underlying smooth manifold (and this has the structure of a complex bundle because of the complex structure).
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 16:41
$begingroup$
@AleksandarMilivojevic: But why does the complex structure induced by the symplecitc form coincide with the original complex structure?
$endgroup$
– user639051
Jan 30 at 18:11
$begingroup$
I see, sorry for not reading carefully. You said you did a computation to show that the two complex structures coincide “up to sign” though? Can you elaborate on that; maybe then we’ll see how one vanishing Chern class implies the other.
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 21:20
|
show 7 more comments
1
$begingroup$
Note that $omega$ gives you a trivialization of the canonical bundle, and so $c_1$ of the canonical bundle, hence the holomorphic cotangent bundle, hence the holomorphic tangent bundle, vanishes; then recall that the ordinary tangent bundle is isomorphic as a complex vector bundle to the holomorphic tangent bundle.
$endgroup$
– Aleksandar Milivojevic
Jan 28 at 18:11
$begingroup$
@AleksandarMilivojevic: Sorry, but this is not clear to me. I get that the chern class of the holomorphic tangent bundle vanishes. I know that thhe holomorphic tangent bundle is isomorphic to the$(1,0)$-part of the complexified real tangent bundle. What do you mean with the "ordinary tangent bundle"? Further, where in your argument does it enter, that we we want to calculate the chern class of a specific symplectic structure on $M$?
$endgroup$
– user639051
Jan 30 at 16:38
$begingroup$
First note that symplectic does not enter since the Chern class only depends on the complex structure (and you’re starting with a complex surface). By “ordinary” tangent bundle I mean the tangent bundle of the underlying smooth manifold (and this has the structure of a complex bundle because of the complex structure).
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 16:41
$begingroup$
@AleksandarMilivojevic: But why does the complex structure induced by the symplecitc form coincide with the original complex structure?
$endgroup$
– user639051
Jan 30 at 18:11
$begingroup$
I see, sorry for not reading carefully. You said you did a computation to show that the two complex structures coincide “up to sign” though? Can you elaborate on that; maybe then we’ll see how one vanishing Chern class implies the other.
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 21:20
1
1
$begingroup$
Note that $omega$ gives you a trivialization of the canonical bundle, and so $c_1$ of the canonical bundle, hence the holomorphic cotangent bundle, hence the holomorphic tangent bundle, vanishes; then recall that the ordinary tangent bundle is isomorphic as a complex vector bundle to the holomorphic tangent bundle.
$endgroup$
– Aleksandar Milivojevic
Jan 28 at 18:11
$begingroup$
Note that $omega$ gives you a trivialization of the canonical bundle, and so $c_1$ of the canonical bundle, hence the holomorphic cotangent bundle, hence the holomorphic tangent bundle, vanishes; then recall that the ordinary tangent bundle is isomorphic as a complex vector bundle to the holomorphic tangent bundle.
$endgroup$
– Aleksandar Milivojevic
Jan 28 at 18:11
$begingroup$
@AleksandarMilivojevic: Sorry, but this is not clear to me. I get that the chern class of the holomorphic tangent bundle vanishes. I know that thhe holomorphic tangent bundle is isomorphic to the$(1,0)$-part of the complexified real tangent bundle. What do you mean with the "ordinary tangent bundle"? Further, where in your argument does it enter, that we we want to calculate the chern class of a specific symplectic structure on $M$?
$endgroup$
– user639051
Jan 30 at 16:38
$begingroup$
@AleksandarMilivojevic: Sorry, but this is not clear to me. I get that the chern class of the holomorphic tangent bundle vanishes. I know that thhe holomorphic tangent bundle is isomorphic to the$(1,0)$-part of the complexified real tangent bundle. What do you mean with the "ordinary tangent bundle"? Further, where in your argument does it enter, that we we want to calculate the chern class of a specific symplectic structure on $M$?
$endgroup$
– user639051
Jan 30 at 16:38
$begingroup$
First note that symplectic does not enter since the Chern class only depends on the complex structure (and you’re starting with a complex surface). By “ordinary” tangent bundle I mean the tangent bundle of the underlying smooth manifold (and this has the structure of a complex bundle because of the complex structure).
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 16:41
$begingroup$
First note that symplectic does not enter since the Chern class only depends on the complex structure (and you’re starting with a complex surface). By “ordinary” tangent bundle I mean the tangent bundle of the underlying smooth manifold (and this has the structure of a complex bundle because of the complex structure).
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 16:41
$begingroup$
@AleksandarMilivojevic: But why does the complex structure induced by the symplecitc form coincide with the original complex structure?
$endgroup$
– user639051
Jan 30 at 18:11
$begingroup$
@AleksandarMilivojevic: But why does the complex structure induced by the symplecitc form coincide with the original complex structure?
$endgroup$
– user639051
Jan 30 at 18:11
$begingroup$
I see, sorry for not reading carefully. You said you did a computation to show that the two complex structures coincide “up to sign” though? Can you elaborate on that; maybe then we’ll see how one vanishing Chern class implies the other.
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 21:20
$begingroup$
I see, sorry for not reading carefully. You said you did a computation to show that the two complex structures coincide “up to sign” though? Can you elaborate on that; maybe then we’ll see how one vanishing Chern class implies the other.
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 21:20
|
show 7 more comments
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$begingroup$
Note that $omega$ gives you a trivialization of the canonical bundle, and so $c_1$ of the canonical bundle, hence the holomorphic cotangent bundle, hence the holomorphic tangent bundle, vanishes; then recall that the ordinary tangent bundle is isomorphic as a complex vector bundle to the holomorphic tangent bundle.
$endgroup$
– Aleksandar Milivojevic
Jan 28 at 18:11
$begingroup$
@AleksandarMilivojevic: Sorry, but this is not clear to me. I get that the chern class of the holomorphic tangent bundle vanishes. I know that thhe holomorphic tangent bundle is isomorphic to the$(1,0)$-part of the complexified real tangent bundle. What do you mean with the "ordinary tangent bundle"? Further, where in your argument does it enter, that we we want to calculate the chern class of a specific symplectic structure on $M$?
$endgroup$
– user639051
Jan 30 at 16:38
$begingroup$
First note that symplectic does not enter since the Chern class only depends on the complex structure (and you’re starting with a complex surface). By “ordinary” tangent bundle I mean the tangent bundle of the underlying smooth manifold (and this has the structure of a complex bundle because of the complex structure).
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 16:41
$begingroup$
@AleksandarMilivojevic: But why does the complex structure induced by the symplecitc form coincide with the original complex structure?
$endgroup$
– user639051
Jan 30 at 18:11
$begingroup$
I see, sorry for not reading carefully. You said you did a computation to show that the two complex structures coincide “up to sign” though? Can you elaborate on that; maybe then we’ll see how one vanishing Chern class implies the other.
$endgroup$
– Aleksandar Milivojevic
Jan 30 at 21:20