Riemann bilinear relations and meromorphic abelian differentials
I am getting quite confused with Riemann bilinear relations.
Let $Sigma$ be a compact Riemann surface of genus $g$, with a canonical homology basis $a_1,b_1,dots,a_g,b_g$, with associated normalised holomorphic and anti-holomorphic one-forms $omega_I,bar omega_I$, $I=1dots,g$, such that $oint_{a_I} omega_J = delta_{IJ}$.
Let further $tau_{P,Q}$ be an abelian differential on $Sigma$ with poles at $P,Q$ of residues $pm1$, normalised so that its $a$-cycles periods are zero : $$oint_{a_I} tau_{PQ}=0.$$
It is called an abelian differential of the third kind and the normalisation guarantees that it is unique. Besides, it is closed on $Sigma-{P,Q}$.
For more details, see Bobenko or textbooks on Riemann surfaces like Farkas & Kra.
I've read in a few places the following statement:
$$iint_Sigma tau_{PQ} wedge bar omega_J propto int_Q^P Im omega_J$$
up to some factors of $2,pi,i$ and sign.
I can't get where the imaginary part comes from. Calling $f=overline{int^zomega_J}$, it seems to me that the standard proof of the Riemann identities based on Stokes theorem, $int_{partial Sigma} f tau_{PQ} = iint bar omega_J wedge tau_{PQ}$ will hold ($partial Sigma$ means the union of $a,b,a^{-1}$ and $b^{-1}$ cycles that constitute a canonical dissection of the surface; the standard 4g-gon) and because $tau_{PQ}$ has zero a-periods this is evaluated to
$$iint_Sigma tau_{PQ} wedge bar omega_J = sum_{I=1}^g oint_{a_I} bar omega_J oint_{b_I} omega_{P,Q}, $$which, using some reciprocity formula like $oint_{b_i} omega_{P,Q} = 2ipi int_Q^P omega_i $ will yield
$$iint_Sigma tau_{PQ} wedge bar omega_J = 2ipi int_Q^P omega_J$$ without the imaginary part ?
differential-forms complex-integration riemann-surfaces
asked Jan 4 at 20:59
picoppicop
507
add a comment |
I am getting quite confused with Riemann bilinear relations.
Let $Sigma$ be a compact Riemann surface of genus $g$, with a canonical homology basis $a_1,b_1,dots,a_g,b_g$, with associated normalised holomorphic and anti-holomorphic one-forms $omega_I,bar omega_I$, $I=1dots,g$, such that $oint_{a_I} omega_J = delta_{IJ}$.
Let further $tau_{P,Q}$ be an abelian differential on $Sigma$ with poles at $P,Q$ of residues $pm1$, normalised so that its $a$-cycles periods are zero : $$oint_{a_I} tau_{PQ}=0.$$
It is called an abelian differential of the third kind and the normalisation guarantees that it is unique. Besides, it is closed on $Sigma-{P,Q}$.
For more details, see Bobenko or textbooks on Riemann surfaces like Farkas & Kra.
I've read in a few places the following statement:
$$iint_Sigma tau_{PQ} wedge bar omega_J propto int_Q^P Im omega_J$$
up to some factors of $2,pi,i$ and sign.
I can't get where the imaginary part comes from. Calling $f=overline{int^zomega_J}$, it seems to me that the standard proof of the Riemann identities based on Stokes theorem, $int_{partial Sigma} f tau_{PQ} = iint bar omega_J wedge tau_{PQ}$ will hold ($partial Sigma$ means the union of $a,b,a^{-1}$ and $b^{-1}$ cycles that constitute a canonical dissection of the surface; the standard 4g-gon) and because $tau_{PQ}$ has zero a-periods this is evaluated to
$$iint_Sigma tau_{PQ} wedge bar omega_J = sum_{I=1}^g oint_{a_I} bar omega_J oint_{b_I} omega_{P,Q}, $$which, using some reciprocity formula like $oint_{b_i} omega_{P,Q} = 2ipi int_Q^P omega_i $ will yield
$$iint_Sigma tau_{PQ} wedge bar omega_J = 2ipi int_Q^P omega_J$$ without the imaginary part ?
differential-forms complex-integration riemann-surfaces
asked Jan 4 at 20:59
picoppicop
507
1
It's too hard to figure out what's going on from what little you've put here. For example, what are the $1$-forms $omega_{P,Q}$? Do you have a link to a complete text?
– Ted Shifrin
Jan 5 at 0:31
I've updated a bit the text... Don't know if this helps.
– picop
Jan 5 at 8:35
add a comment |
I am getting quite confused with Riemann bilinear relations.
Let $Sigma$ be a compact Riemann surface of genus $g$, with a canonical homology basis $a_1,b_1,dots,a_g,b_g$, with associated normalised holomorphic and anti-holomorphic one-forms $omega_I,bar omega_I$, $I=1dots,g$, such that $oint_{a_I} omega_J = delta_{IJ}$.
Let further $tau_{P,Q}$ be an abelian differential on $Sigma$ with poles at $P,Q$ of residues $pm1$, normalised so that its $a$-cycles periods are zero : $$oint_{a_I} tau_{PQ}=0.$$
It is called an abelian differential of the third kind and the normalisation guarantees that it is unique. Besides, it is closed on $Sigma-{P,Q}$.
For more details, see Bobenko or textbooks on Riemann surfaces like Farkas & Kra.
I've read in a few places the following statement:
$$iint_Sigma tau_{PQ} wedge bar omega_J propto int_Q^P Im omega_J$$
up to some factors of $2,pi,i$ and sign.
I can't get where the imaginary part comes from. Calling $f=overline{int^zomega_J}$, it seems to me that the standard proof of the Riemann identities based on Stokes theorem, $int_{partial Sigma} f tau_{PQ} = iint bar omega_J wedge tau_{PQ}$ will hold ($partial Sigma$ means the union of $a,b,a^{-1}$ and $b^{-1}$ cycles that constitute a canonical dissection of the surface; the standard 4g-gon) and because $tau_{PQ}$ has zero a-periods this is evaluated to
$$iint_Sigma tau_{PQ} wedge bar omega_J = sum_{I=1}^g oint_{a_I} bar omega_J oint_{b_I} omega_{P,Q}, $$which, using some reciprocity formula like $oint_{b_i} omega_{P,Q} = 2ipi int_Q^P omega_i $ will yield
$$iint_Sigma tau_{PQ} wedge bar omega_J = 2ipi int_Q^P omega_J$$ without the imaginary part ?
differential-forms complex-integration riemann-surfaces
asked Jan 4 at 20:59
picoppicop
507
I am getting quite confused with Riemann bilinear relations.
Let $Sigma$ be a compact Riemann surface of genus $g$, with a canonical homology basis $a_1,b_1,dots,a_g,b_g$, with associated normalised holomorphic and anti-holomorphic one-forms $omega_I,bar omega_I$, $I=1dots,g$, such that $oint_{a_I} omega_J = delta_{IJ}$.
Let further $tau_{P,Q}$ be an abelian differential on $Sigma$ with poles at $P,Q$ of residues $pm1$, normalised so that its $a$-cycles periods are zero : $$oint_{a_I} tau_{PQ}=0.$$
It is called an abelian differential of the third kind and the normalisation guarantees that it is unique. Besides, it is closed on $Sigma-{P,Q}$.
For more details, see Bobenko or textbooks on Riemann surfaces like Farkas & Kra.
I've read in a few places the following statement:
$$iint_Sigma tau_{PQ} wedge bar omega_J propto int_Q^P Im omega_J$$
up to some factors of $2,pi,i$ and sign.
I can't get where the imaginary part comes from. Calling $f=overline{int^zomega_J}$, it seems to me that the standard proof of the Riemann identities based on Stokes theorem, $int_{partial Sigma} f tau_{PQ} = iint bar omega_J wedge tau_{PQ}$ will hold ($partial Sigma$ means the union of $a,b,a^{-1}$ and $b^{-1}$ cycles that constitute a canonical dissection of the surface; the standard 4g-gon) and because $tau_{PQ}$ has zero a-periods this is evaluated to
$$iint_Sigma tau_{PQ} wedge bar omega_J = sum_{I=1}^g oint_{a_I} bar omega_J oint_{b_I} omega_{P,Q}, $$which, using some reciprocity formula like $oint_{b_i} omega_{P,Q} = 2ipi int_Q^P omega_i $ will yield
$$iint_Sigma tau_{PQ} wedge bar omega_J = 2ipi int_Q^P omega_J$$ without the imaginary part ?
differential-forms complex-integration riemann-surfaces
differential-forms complex-integration riemann-surfaces
asked Jan 4 at 20:59
picoppicop
507
asked Jan 4 at 20:59
picoppicop
507
edited Jan 6 at 10:20
picop
asked Jan 4 at 20:59
picoppicop
507
asked Jan 4 at 20:59
picoppicop
507
asked Jan 4 at 20:59
picoppicop
507
507
1
It's too hard to figure out what's going on from what little you've put here. For example, what are the $1$-forms $omega_{P,Q}$? Do you have a link to a complete text?
– Ted Shifrin
Jan 5 at 0:31
I've updated a bit the text... Don't know if this helps.
– picop
Jan 5 at 8:35
add a comment |
1
It's too hard to figure out what's going on from what little you've put here. For example, what are the $1$-forms $omega_{P,Q}$? Do you have a link to a complete text?
– Ted Shifrin
Jan 5 at 0:31
I've updated a bit the text... Don't know if this helps.
– picop
Jan 5 at 8:35
1
1
It's too hard to figure out what's going on from what little you've put here. For example, what are the $1$-forms $omega_{P,Q}$? Do you have a link to a complete text?
– Ted Shifrin
Jan 5 at 0:31
It's too hard to figure out what's going on from what little you've put here. For example, what are the $1$-forms $omega_{P,Q}$? Do you have a link to a complete text?
– Ted Shifrin
Jan 5 at 0:31
I've updated a bit the text... Don't know if this helps.
– picop
Jan 5 at 8:35
I've updated a bit the text... Don't know if this helps.
– picop
Jan 5 at 8:35
add a comment |
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It's too hard to figure out what's going on from what little you've put here. For example, what are the $1$-forms $omega_{P,Q}$? Do you have a link to a complete text?
– Ted Shifrin
Jan 5 at 0:31
I've updated a bit the text... Don't know if this helps.
– picop
Jan 5 at 8:35