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?







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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?







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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?







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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?









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      edited 2 days ago









      barto

      13.3k32581




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      asked Aug 7 at 18:54









      Akatsuki

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      8721623




















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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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            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
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              active

              oldest

              votes








              2 Answers
              2






              active

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              active

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              active

              oldest

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              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 $mathbb C$.






              share|cite|improve this answer

























                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 $mathbb C$.






                share|cite|improve this answer























                  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 $mathbb C$.






                  share|cite|improve this answer













                  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 $mathbb C$.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered 2 days ago









                  Akatsuki

                  8721623




                  8721623




















                      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.)






                      share|cite|improve this answer

























                        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.)






                        share|cite|improve this answer























                          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.)






                          share|cite|improve this answer













                          This is by de Rham's theorem. (Which is used in the proof that the Abel-Jacobi map is an isomorphism.)







                          share|cite|improve this answer













                          share|cite|improve this answer



                          share|cite|improve this answer











                          answered 2 days ago









                          barto

                          13.3k32581




                          13.3k32581






















                               

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