Picard number and topology of Kähler manifolds
$begingroup$
Let $(X,omega)$ be a compact Kähler manifold.
The line bundles $mathcal{L}$ over $X$ with respect to tensor product $otimes$ determine the Picard group $Pic(X)$ of $X$; let $Pic^0(X)$ be the subgroup of degree $0$ line bundles over $X$, where $displaystyledeg(mathcal{L})=int_Xc_1(mathcal{L})wedgeomega^{dim X-1}$; the Néron-Severi group $NS(X)$ is the quotient group $Pic(X)_{displaystyle/Pic^0(X)}$.
One proves that $NS(X)$ is a finitely generated (Abelian) group, and its rank $rho(X)$ is called Picard number of $X$.
From the topological point of view: $Pic(X)=H^1left(X,mathcal{O}_X^{times}right)$, where $mathcal{O}_X^{times}$ is the sheaf of nowhere zero holomorphic functions on $X$; $NS(X)=Imleft(Pic(X)xrightarrow{c_1}H^2(X,mathbb{Z})right)=H^2(X,mathbb{Z})cap H^{(1,1)}(X)$ by Lefschetz $(1,1)$-part theorem, where $c_1$ is the first connection morphism in the long exact sequence in cohomology of the exponential sheaf sequence, in other words is the first Chern class.
I am not able to connect these two diffent point of view; for exact, when I know some information about $Pic(X)$ or $rho(X)$ I am not able to extract topological informations about $X$.
Can someone suggest me books, lecture notes, "toy examples" and other useful material for this complicated (for me) linkage?
Thanks, in advance.
AddendumExample: if $X$ is a projective variety then $NS(X)$ is a finitely generated (Abelian) group (Severi-Néron theorem of the base); so when $rho(X)=1$ any effective (Weil) divisor $D$ determines an ample line bundle $mathcal{L}$ over $X$ (see Mohan's comment below).
algebraic-geometry reference-request kahler-manifolds
asked Dec 30 '18 at 16:41
Armando j18eosArmando j18eos
2,64511428
$endgroup$
|
show 1 more comment
$begingroup$
Let $(X,omega)$ be a compact Kähler manifold.
The line bundles $mathcal{L}$ over $X$ with respect to tensor product $otimes$ determine the Picard group $Pic(X)$ of $X$; let $Pic^0(X)$ be the subgroup of degree $0$ line bundles over $X$, where $displaystyledeg(mathcal{L})=int_Xc_1(mathcal{L})wedgeomega^{dim X-1}$; the Néron-Severi group $NS(X)$ is the quotient group $Pic(X)_{displaystyle/Pic^0(X)}$.
One proves that $NS(X)$ is a finitely generated (Abelian) group, and its rank $rho(X)$ is called Picard number of $X$.
From the topological point of view: $Pic(X)=H^1left(X,mathcal{O}_X^{times}right)$, where $mathcal{O}_X^{times}$ is the sheaf of nowhere zero holomorphic functions on $X$; $NS(X)=Imleft(Pic(X)xrightarrow{c_1}H^2(X,mathbb{Z})right)=H^2(X,mathbb{Z})cap H^{(1,1)}(X)$ by Lefschetz $(1,1)$-part theorem, where $c_1$ is the first connection morphism in the long exact sequence in cohomology of the exponential sheaf sequence, in other words is the first Chern class.
I am not able to connect these two diffent point of view; for exact, when I know some information about $Pic(X)$ or $rho(X)$ I am not able to extract topological informations about $X$.
Can someone suggest me books, lecture notes, "toy examples" and other useful material for this complicated (for me) linkage?
Thanks, in advance.
AddendumExample: if $X$ is a projective variety then $NS(X)$ is a finitely generated (Abelian) group (Severi-Néron theorem of the base); so when $rho(X)=1$ any effective (Weil) divisor $D$ determines an ample line bundle $mathcal{L}$ over $X$ (see Mohan's comment below).
algebraic-geometry reference-request kahler-manifolds
asked Dec 30 '18 at 16:41
Armando j18eosArmando j18eos
2,64511428
$endgroup$
$begingroup$
What sort of topological information do you want to extract? For example, if $X$ is simply connected, $Pic^0=0$. Similarly, (let me assume $X$ is projective), if $rho=1$ (it is always at least one), there are no non-constant morphisms to a smaller dimensional variety.
$endgroup$
– Mohan
Dec 30 '18 at 21:21
$begingroup$
In particular, I'm interested to the case $rho(X)=1$. For example: if $n=2$, do the curves (as complex analytic subspaces) on $X$ (of dimension $1$) satisfy extra conditions?, if $deg(mathcal{L})=0$, does $c_1(mathcal{L})$ vanish, for whatever $n$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 9:30
1
$begingroup$
If $rho=1$, every effective curve is ample. For the second question, it depends where $c_1$ takes values. Typically, in algebraic geometry, $c_1$ takes values in Picard group, so $c_1(L)=0$ would mean $L$ is trivial, stronger than degree being zero. For example, everything in $Pic^0$ has degree zero.
$endgroup$
– Mohan
Dec 31 '18 at 16:09
$begingroup$
Thank you for the first answer, it is very enlighted also for other (unasked) questions. I clarify the second question: let $mathcal{L}$ be a degree $0$ line bundle over a compact Kähler manifold $X$ with $rho(X)=1$; is $c_1(mathcal{L})=0$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 16:27
1
$begingroup$
For $X$ projective, it is correct if you define $c_1$ as you have. In algebraic geometry, as I said, typically chern classes take values in Chow group, then vanishing of degree is not enough.
$endgroup$
– Mohan
Dec 31 '18 at 18:03
|
show 1 more comment
$begingroup$
Let $(X,omega)$ be a compact Kähler manifold.
The line bundles $mathcal{L}$ over $X$ with respect to tensor product $otimes$ determine the Picard group $Pic(X)$ of $X$; let $Pic^0(X)$ be the subgroup of degree $0$ line bundles over $X$, where $displaystyledeg(mathcal{L})=int_Xc_1(mathcal{L})wedgeomega^{dim X-1}$; the Néron-Severi group $NS(X)$ is the quotient group $Pic(X)_{displaystyle/Pic^0(X)}$.
One proves that $NS(X)$ is a finitely generated (Abelian) group, and its rank $rho(X)$ is called Picard number of $X$.
From the topological point of view: $Pic(X)=H^1left(X,mathcal{O}_X^{times}right)$, where $mathcal{O}_X^{times}$ is the sheaf of nowhere zero holomorphic functions on $X$; $NS(X)=Imleft(Pic(X)xrightarrow{c_1}H^2(X,mathbb{Z})right)=H^2(X,mathbb{Z})cap H^{(1,1)}(X)$ by Lefschetz $(1,1)$-part theorem, where $c_1$ is the first connection morphism in the long exact sequence in cohomology of the exponential sheaf sequence, in other words is the first Chern class.
I am not able to connect these two diffent point of view; for exact, when I know some information about $Pic(X)$ or $rho(X)$ I am not able to extract topological informations about $X$.
Can someone suggest me books, lecture notes, "toy examples" and other useful material for this complicated (for me) linkage?
Thanks, in advance.
AddendumExample: if $X$ is a projective variety then $NS(X)$ is a finitely generated (Abelian) group (Severi-Néron theorem of the base); so when $rho(X)=1$ any effective (Weil) divisor $D$ determines an ample line bundle $mathcal{L}$ over $X$ (see Mohan's comment below).
algebraic-geometry reference-request kahler-manifolds
asked Dec 30 '18 at 16:41
Armando j18eosArmando j18eos
2,64511428
$endgroup$
Let $(X,omega)$ be a compact Kähler manifold.
The line bundles $mathcal{L}$ over $X$ with respect to tensor product $otimes$ determine the Picard group $Pic(X)$ of $X$; let $Pic^0(X)$ be the subgroup of degree $0$ line bundles over $X$, where $displaystyledeg(mathcal{L})=int_Xc_1(mathcal{L})wedgeomega^{dim X-1}$; the Néron-Severi group $NS(X)$ is the quotient group $Pic(X)_{displaystyle/Pic^0(X)}$.
One proves that $NS(X)$ is a finitely generated (Abelian) group, and its rank $rho(X)$ is called Picard number of $X$.
From the topological point of view: $Pic(X)=H^1left(X,mathcal{O}_X^{times}right)$, where $mathcal{O}_X^{times}$ is the sheaf of nowhere zero holomorphic functions on $X$; $NS(X)=Imleft(Pic(X)xrightarrow{c_1}H^2(X,mathbb{Z})right)=H^2(X,mathbb{Z})cap H^{(1,1)}(X)$ by Lefschetz $(1,1)$-part theorem, where $c_1$ is the first connection morphism in the long exact sequence in cohomology of the exponential sheaf sequence, in other words is the first Chern class.
I am not able to connect these two diffent point of view; for exact, when I know some information about $Pic(X)$ or $rho(X)$ I am not able to extract topological informations about $X$.
Can someone suggest me books, lecture notes, "toy examples" and other useful material for this complicated (for me) linkage?
Thanks, in advance.
AddendumExample: if $X$ is a projective variety then $NS(X)$ is a finitely generated (Abelian) group (Severi-Néron theorem of the base); so when $rho(X)=1$ any effective (Weil) divisor $D$ determines an ample line bundle $mathcal{L}$ over $X$ (see Mohan's comment below).
algebraic-geometry reference-request kahler-manifolds
algebraic-geometry reference-request kahler-manifolds
asked Dec 30 '18 at 16:41
Armando j18eosArmando j18eos
2,64511428
asked Dec 30 '18 at 16:41
Armando j18eosArmando j18eos
2,64511428
edited Jan 5 at 16:17
Armando j18eos
asked Dec 30 '18 at 16:41
Armando j18eosArmando j18eos
2,64511428
asked Dec 30 '18 at 16:41
Armando j18eosArmando j18eos
2,64511428
asked Dec 30 '18 at 16:41
Armando j18eosArmando j18eos
2,64511428
2,64511428
$begingroup$
What sort of topological information do you want to extract? For example, if $X$ is simply connected, $Pic^0=0$. Similarly, (let me assume $X$ is projective), if $rho=1$ (it is always at least one), there are no non-constant morphisms to a smaller dimensional variety.
$endgroup$
– Mohan
Dec 30 '18 at 21:21
$begingroup$
In particular, I'm interested to the case $rho(X)=1$. For example: if $n=2$, do the curves (as complex analytic subspaces) on $X$ (of dimension $1$) satisfy extra conditions?, if $deg(mathcal{L})=0$, does $c_1(mathcal{L})$ vanish, for whatever $n$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 9:30
1
$begingroup$
If $rho=1$, every effective curve is ample. For the second question, it depends where $c_1$ takes values. Typically, in algebraic geometry, $c_1$ takes values in Picard group, so $c_1(L)=0$ would mean $L$ is trivial, stronger than degree being zero. For example, everything in $Pic^0$ has degree zero.
$endgroup$
– Mohan
Dec 31 '18 at 16:09
$begingroup$
Thank you for the first answer, it is very enlighted also for other (unasked) questions. I clarify the second question: let $mathcal{L}$ be a degree $0$ line bundle over a compact Kähler manifold $X$ with $rho(X)=1$; is $c_1(mathcal{L})=0$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 16:27
1
$begingroup$
For $X$ projective, it is correct if you define $c_1$ as you have. In algebraic geometry, as I said, typically chern classes take values in Chow group, then vanishing of degree is not enough.
$endgroup$
– Mohan
Dec 31 '18 at 18:03
|
show 1 more comment
$begingroup$
What sort of topological information do you want to extract? For example, if $X$ is simply connected, $Pic^0=0$. Similarly, (let me assume $X$ is projective), if $rho=1$ (it is always at least one), there are no non-constant morphisms to a smaller dimensional variety.
$endgroup$
– Mohan
Dec 30 '18 at 21:21
$begingroup$
In particular, I'm interested to the case $rho(X)=1$. For example: if $n=2$, do the curves (as complex analytic subspaces) on $X$ (of dimension $1$) satisfy extra conditions?, if $deg(mathcal{L})=0$, does $c_1(mathcal{L})$ vanish, for whatever $n$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 9:30
1
$begingroup$
If $rho=1$, every effective curve is ample. For the second question, it depends where $c_1$ takes values. Typically, in algebraic geometry, $c_1$ takes values in Picard group, so $c_1(L)=0$ would mean $L$ is trivial, stronger than degree being zero. For example, everything in $Pic^0$ has degree zero.
$endgroup$
– Mohan
Dec 31 '18 at 16:09
$begingroup$
Thank you for the first answer, it is very enlighted also for other (unasked) questions. I clarify the second question: let $mathcal{L}$ be a degree $0$ line bundle over a compact Kähler manifold $X$ with $rho(X)=1$; is $c_1(mathcal{L})=0$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 16:27
1
$begingroup$
For $X$ projective, it is correct if you define $c_1$ as you have. In algebraic geometry, as I said, typically chern classes take values in Chow group, then vanishing of degree is not enough.
$endgroup$
– Mohan
Dec 31 '18 at 18:03
$begingroup$
What sort of topological information do you want to extract? For example, if $X$ is simply connected, $Pic^0=0$. Similarly, (let me assume $X$ is projective), if $rho=1$ (it is always at least one), there are no non-constant morphisms to a smaller dimensional variety.
$endgroup$
– Mohan
Dec 30 '18 at 21:21
$begingroup$
What sort of topological information do you want to extract? For example, if $X$ is simply connected, $Pic^0=0$. Similarly, (let me assume $X$ is projective), if $rho=1$ (it is always at least one), there are no non-constant morphisms to a smaller dimensional variety.
$endgroup$
– Mohan
Dec 30 '18 at 21:21
$begingroup$
In particular, I'm interested to the case $rho(X)=1$. For example: if $n=2$, do the curves (as complex analytic subspaces) on $X$ (of dimension $1$) satisfy extra conditions?, if $deg(mathcal{L})=0$, does $c_1(mathcal{L})$ vanish, for whatever $n$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 9:30
$begingroup$
In particular, I'm interested to the case $rho(X)=1$. For example: if $n=2$, do the curves (as complex analytic subspaces) on $X$ (of dimension $1$) satisfy extra conditions?, if $deg(mathcal{L})=0$, does $c_1(mathcal{L})$ vanish, for whatever $n$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 9:30
1
1
$begingroup$
If $rho=1$, every effective curve is ample. For the second question, it depends where $c_1$ takes values. Typically, in algebraic geometry, $c_1$ takes values in Picard group, so $c_1(L)=0$ would mean $L$ is trivial, stronger than degree being zero. For example, everything in $Pic^0$ has degree zero.
$endgroup$
– Mohan
Dec 31 '18 at 16:09
$begingroup$
If $rho=1$, every effective curve is ample. For the second question, it depends where $c_1$ takes values. Typically, in algebraic geometry, $c_1$ takes values in Picard group, so $c_1(L)=0$ would mean $L$ is trivial, stronger than degree being zero. For example, everything in $Pic^0$ has degree zero.
$endgroup$
– Mohan
Dec 31 '18 at 16:09
$begingroup$
Thank you for the first answer, it is very enlighted also for other (unasked) questions. I clarify the second question: let $mathcal{L}$ be a degree $0$ line bundle over a compact Kähler manifold $X$ with $rho(X)=1$; is $c_1(mathcal{L})=0$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 16:27
$begingroup$
Thank you for the first answer, it is very enlighted also for other (unasked) questions. I clarify the second question: let $mathcal{L}$ be a degree $0$ line bundle over a compact Kähler manifold $X$ with $rho(X)=1$; is $c_1(mathcal{L})=0$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 16:27
1
1
$begingroup$
For $X$ projective, it is correct if you define $c_1$ as you have. In algebraic geometry, as I said, typically chern classes take values in Chow group, then vanishing of degree is not enough.
$endgroup$
– Mohan
Dec 31 '18 at 18:03
$begingroup$
For $X$ projective, it is correct if you define $c_1$ as you have. In algebraic geometry, as I said, typically chern classes take values in Chow group, then vanishing of degree is not enough.
$endgroup$
– Mohan
Dec 31 '18 at 18:03
|
show 1 more comment
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$begingroup$
What sort of topological information do you want to extract? For example, if $X$ is simply connected, $Pic^0=0$. Similarly, (let me assume $X$ is projective), if $rho=1$ (it is always at least one), there are no non-constant morphisms to a smaller dimensional variety.
$endgroup$
– Mohan
Dec 30 '18 at 21:21
$begingroup$
In particular, I'm interested to the case $rho(X)=1$. For example: if $n=2$, do the curves (as complex analytic subspaces) on $X$ (of dimension $1$) satisfy extra conditions?, if $deg(mathcal{L})=0$, does $c_1(mathcal{L})$ vanish, for whatever $n$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 9:30
1
$begingroup$
If $rho=1$, every effective curve is ample. For the second question, it depends where $c_1$ takes values. Typically, in algebraic geometry, $c_1$ takes values in Picard group, so $c_1(L)=0$ would mean $L$ is trivial, stronger than degree being zero. For example, everything in $Pic^0$ has degree zero.
$endgroup$
– Mohan
Dec 31 '18 at 16:09
$begingroup$
Thank you for the first answer, it is very enlighted also for other (unasked) questions. I clarify the second question: let $mathcal{L}$ be a degree $0$ line bundle over a compact Kähler manifold $X$ with $rho(X)=1$; is $c_1(mathcal{L})=0$?
$endgroup$
– Armando j18eos
Dec 31 '18 at 16:27
1
$begingroup$
For $X$ projective, it is correct if you define $c_1$ as you have. In algebraic geometry, as I said, typically chern classes take values in Chow group, then vanishing of degree is not enough.
$endgroup$
– Mohan
Dec 31 '18 at 18:03