Integral along different loops on torus gives a full lattice in $mathbb C$
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Let $S$ be a compact Riemann surface we $g=1$, $alpha,beta$ be the generator of $pi_1(S)$, $omeganeq 0$ be a fixed holomorphic $1$-form. How can we know $int_alpha omega$ and $int_beta omega$ are $mathbb R$-linearly independent?
differential-forms riemann-surfaces
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Let $S$ be a compact Riemann surface we $g=1$, $alpha,beta$ be the generator of $pi_1(S)$, $omeganeq 0$ be a fixed holomorphic $1$-form. How can we know $int_alpha omega$ and $int_beta omega$ are $mathbb R$-linearly independent?
differential-forms riemann-surfaces
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $S$ be a compact Riemann surface we $g=1$, $alpha,beta$ be the generator of $pi_1(S)$, $omeganeq 0$ be a fixed holomorphic $1$-form. How can we know $int_alpha omega$ and $int_beta omega$ are $mathbb R$-linearly independent?
differential-forms riemann-surfaces
Let $S$ be a compact Riemann surface we $g=1$, $alpha,beta$ be the generator of $pi_1(S)$, $omeganeq 0$ be a fixed holomorphic $1$-form. How can we know $int_alpha omega$ and $int_beta omega$ are $mathbb R$-linearly independent?
differential-forms riemann-surfaces
edited 2 days ago
barto
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13.3k32581
asked Aug 7 at 18:54
Akatsuki
8721623
8721623
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2 Answers
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Since $g=1$, $omega$ is a basis of $H^1,0(S)$. Fix a point $x_0$, and let $Lambda$ be the lattice spanned by $int_alpha omega$ and $int_beta omega$. Then the Abelian-Jacobian map
$$S to mathbb C /Lambda \
x mapsto(int_x_0^x omega,int_x_0^x omega)$$
is an isomorphism. In particular $int_alpha omega$ and $int_beta omega$ form a $mathbb R$-basis of $mathbb C$.
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up vote
1
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This is by de Rham's theorem. (Which is used in the proof that the Abel-Jacobi map is an isomorphism.)
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since $g=1$, $omega$ is a basis of $H^1,0(S)$. Fix a point $x_0$, and let $Lambda$ be the lattice spanned by $int_alpha omega$ and $int_beta omega$. Then the Abelian-Jacobian map
$$S to mathbb C /Lambda \
x mapsto(int_x_0^x omega,int_x_0^x omega)$$
is an isomorphism. In particular $int_alpha omega$ and $int_beta omega$ form a $mathbb R$-basis of $mathbb C$.
add a comment |Â
up vote
1
down vote
accepted
Since $g=1$, $omega$ is a basis of $H^1,0(S)$. Fix a point $x_0$, and let $Lambda$ be the lattice spanned by $int_alpha omega$ and $int_beta omega$. Then the Abelian-Jacobian map
$$S to mathbb C /Lambda \
x mapsto(int_x_0^x omega,int_x_0^x omega)$$
is an isomorphism. In particular $int_alpha omega$ and $int_beta omega$ form a $mathbb R$-basis of $mathbb C$.
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since $g=1$, $omega$ is a basis of $H^1,0(S)$. Fix a point $x_0$, and let $Lambda$ be the lattice spanned by $int_alpha omega$ and $int_beta omega$. Then the Abelian-Jacobian map
$$S to mathbb C /Lambda \
x mapsto(int_x_0^x omega,int_x_0^x omega)$$
is an isomorphism. In particular $int_alpha omega$ and $int_beta omega$ form a $mathbb R$-basis of $mathbb C$.
Since $g=1$, $omega$ is a basis of $H^1,0(S)$. Fix a point $x_0$, and let $Lambda$ be the lattice spanned by $int_alpha omega$ and $int_beta omega$. Then the Abelian-Jacobian map
$$S to mathbb C /Lambda \
x mapsto(int_x_0^x omega,int_x_0^x omega)$$
is an isomorphism. In particular $int_alpha omega$ and $int_beta omega$ form a $mathbb R$-basis of $mathbb C$.
answered 2 days ago
Akatsuki
8721623
8721623
add a comment |Â
add a comment |Â
up vote
1
down vote
This is by de Rham's theorem. (Which is used in the proof that the Abel-Jacobi map is an isomorphism.)
add a comment |Â
up vote
1
down vote
This is by de Rham's theorem. (Which is used in the proof that the Abel-Jacobi map is an isomorphism.)
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is by de Rham's theorem. (Which is used in the proof that the Abel-Jacobi map is an isomorphism.)
This is by de Rham's theorem. (Which is used in the proof that the Abel-Jacobi map is an isomorphism.)
answered 2 days ago
barto
13.3k32581
13.3k32581
add a comment |Â
add a comment |Â
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