Why is the Albanese map well-defined?
$begingroup$
Let $X$ be a compact Kähler manifold of complex dimension $n$.
$Alb(X):=frac{H^0(X, Omega_X)^*}{rho(H_1(X,mathbb{Z}))}$, where $rho:H_1(X, mathbb{Z}) to H^0(X, Omega_X)^*$ is given by $[r]mapsto ([alpha]mapsto int_ralpha)$.
There exists a holomorphic map from $X$ to Alb($X$): If one chooses a point $p_0$ in $X$, then it is
given by $pmapsto([α]mapstoint_{p_0}^{p}α)$.
But why is this map well-defined, why don't we need to worry about the paths between $p_0$ and $p$?
complex-geometry kahler-manifolds
asked Jan 6 at 4:55
66666666
1,373621
$endgroup$
add a comment |
$begingroup$
Let $X$ be a compact Kähler manifold of complex dimension $n$.
$Alb(X):=frac{H^0(X, Omega_X)^*}{rho(H_1(X,mathbb{Z}))}$, where $rho:H_1(X, mathbb{Z}) to H^0(X, Omega_X)^*$ is given by $[r]mapsto ([alpha]mapsto int_ralpha)$.
There exists a holomorphic map from $X$ to Alb($X$): If one chooses a point $p_0$ in $X$, then it is
given by $pmapsto([α]mapstoint_{p_0}^{p}α)$.
But why is this map well-defined, why don't we need to worry about the paths between $p_0$ and $p$?
complex-geometry kahler-manifolds
asked Jan 6 at 4:55
66666666
1,373621
$endgroup$
$begingroup$
If $X$ is a Riemann surface and $p,p_0$ are both in an open set isomorphic to a disk, then this follows directly from Cauchy's integral theorem. The general case can be reduced to this.
$endgroup$
– klirk
Jan 23 at 20:41
add a comment |
$begingroup$
Let $X$ be a compact Kähler manifold of complex dimension $n$.
$Alb(X):=frac{H^0(X, Omega_X)^*}{rho(H_1(X,mathbb{Z}))}$, where $rho:H_1(X, mathbb{Z}) to H^0(X, Omega_X)^*$ is given by $[r]mapsto ([alpha]mapsto int_ralpha)$.
There exists a holomorphic map from $X$ to Alb($X$): If one chooses a point $p_0$ in $X$, then it is
given by $pmapsto([α]mapstoint_{p_0}^{p}α)$.
But why is this map well-defined, why don't we need to worry about the paths between $p_0$ and $p$?
complex-geometry kahler-manifolds
asked Jan 6 at 4:55
66666666
1,373621
$endgroup$
Let $X$ be a compact Kähler manifold of complex dimension $n$.
$Alb(X):=frac{H^0(X, Omega_X)^*}{rho(H_1(X,mathbb{Z}))}$, where $rho:H_1(X, mathbb{Z}) to H^0(X, Omega_X)^*$ is given by $[r]mapsto ([alpha]mapsto int_ralpha)$.
There exists a holomorphic map from $X$ to Alb($X$): If one chooses a point $p_0$ in $X$, then it is
given by $pmapsto([α]mapstoint_{p_0}^{p}α)$.
But why is this map well-defined, why don't we need to worry about the paths between $p_0$ and $p$?
complex-geometry kahler-manifolds
complex-geometry kahler-manifolds
asked Jan 6 at 4:55
66666666
1,373621
asked Jan 6 at 4:55
66666666
1,373621
asked Jan 6 at 4:55
66666666
1,373621
asked Jan 6 at 4:55
66666666
1,373621
asked Jan 6 at 4:55
66666666
1,373621
1,373621
$begingroup$
If $X$ is a Riemann surface and $p,p_0$ are both in an open set isomorphic to a disk, then this follows directly from Cauchy's integral theorem. The general case can be reduced to this.
$endgroup$
– klirk
Jan 23 at 20:41
add a comment |
$begingroup$
If $X$ is a Riemann surface and $p,p_0$ are both in an open set isomorphic to a disk, then this follows directly from Cauchy's integral theorem. The general case can be reduced to this.
$endgroup$
– klirk
Jan 23 at 20:41
$begingroup$
If $X$ is a Riemann surface and $p,p_0$ are both in an open set isomorphic to a disk, then this follows directly from Cauchy's integral theorem. The general case can be reduced to this.
$endgroup$
– klirk
Jan 23 at 20:41
$begingroup$
If $X$ is a Riemann surface and $p,p_0$ are both in an open set isomorphic to a disk, then this follows directly from Cauchy's integral theorem. The general case can be reduced to this.
$endgroup$
– klirk
Jan 23 at 20:41
add a comment |
1 Answer
1
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$begingroup$
Let $c,d$ be paths between $p_0$ and $p$, let $d'$ be $d$ with the reverse orientation $c.d'$ represents an element of $H_1(X,mathbb{Z})$ so $int_{c.d'}alpha=int_calpha-int_dalpha=0$ since it is in the image of $rho$.
answered Jan 6 at 5:07
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let $c,d$ be paths between $p_0$ and $p$, let $d'$ be $d$ with the reverse orientation $c.d'$ represents an element of $H_1(X,mathbb{Z})$ so $int_{c.d'}alpha=int_calpha-int_dalpha=0$ since it is in the image of $rho$.
answered Jan 6 at 5:07
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
add a comment |
$begingroup$
Let $c,d$ be paths between $p_0$ and $p$, let $d'$ be $d$ with the reverse orientation $c.d'$ represents an element of $H_1(X,mathbb{Z})$ so $int_{c.d'}alpha=int_calpha-int_dalpha=0$ since it is in the image of $rho$.
answered Jan 6 at 5:07
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
add a comment |
$begingroup$
Let $c,d$ be paths between $p_0$ and $p$, let $d'$ be $d$ with the reverse orientation $c.d'$ represents an element of $H_1(X,mathbb{Z})$ so $int_{c.d'}alpha=int_calpha-int_dalpha=0$ since it is in the image of $rho$.
answered Jan 6 at 5:07
Tsemo AristideTsemo Aristide
59.6k11446
$endgroup$
Let $c,d$ be paths between $p_0$ and $p$, let $d'$ be $d$ with the reverse orientation $c.d'$ represents an element of $H_1(X,mathbb{Z})$ so $int_{c.d'}alpha=int_calpha-int_dalpha=0$ since it is in the image of $rho$.
answered Jan 6 at 5:07
Tsemo AristideTsemo Aristide
59.6k11446
answered Jan 6 at 5:07
Tsemo AristideTsemo Aristide
59.6k11446
answered Jan 6 at 5:07
Tsemo AristideTsemo Aristide
59.6k11446
answered Jan 6 at 5:07
Tsemo AristideTsemo Aristide
59.6k11446
59.6k11446
add a comment |
add a comment |
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$begingroup$
If $X$ is a Riemann surface and $p,p_0$ are both in an open set isomorphic to a disk, then this follows directly from Cauchy's integral theorem. The general case can be reduced to this.
$endgroup$
– klirk
Jan 23 at 20:41