Why does a holomorphic differential has $2g-2$ zeros? The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraThe sum of orders of zeros of a holomorphic function in a genus $g$ Riemann surface is equal to $2g-2$Application of Riemann-RochOn the original Riemann-Roch theoremHolomorphic line bundle with degree zeroMeromorphic function with bounded order of zeros and polesHow to prove the number of poles minus the number of zeros is $2-2g$?Is there an integral formula for the degree of a holomorphic embedding of the Riemann sphere into projective space?Problem underestanding holomorphic quadratic diferentialsElementary integrals and Riemann surfacesConstruct the meromorphic function on Riemann surface of genus $3$The sum of orders of zeros of a holomorphic function in a genus $g$ Riemann surface is equal to $2g-2$
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Why does a holomorphic differential has $2g-2$ zeros?
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraThe sum of orders of zeros of a holomorphic function in a genus $g$ Riemann surface is equal to $2g-2$Application of Riemann-RochOn the original Riemann-Roch theoremHolomorphic line bundle with degree zeroMeromorphic function with bounded order of zeros and polesHow to prove the number of poles minus the number of zeros is $2-2g$?Is there an integral formula for the degree of a holomorphic embedding of the Riemann sphere into projective space?Problem underestanding holomorphic quadratic diferentialsElementary integrals and Riemann surfacesConstruct the meromorphic function on Riemann surface of genus $3$The sum of orders of zeros of a holomorphic function in a genus $g$ Riemann surface is equal to $2g-2$
$begingroup$
If $X$ is a compact Riemann surface, then any holomorphic differential on $X$ has $2g-2$ zeros.
I would like to know how to prove this. If possible, without some "heavy machinery" like divisors and the Riemann-Roch theorem, which I don't understand very well yet. I really think that we can use the Riemann-Hurwitz to do this, but I am not sure how.
complex-analysis riemann-surfaces
asked Apr 8 at 6:39
GabrielGabriel
1,520623
$endgroup$
|
show 2 more comments
$begingroup$
If $X$ is a compact Riemann surface, then any holomorphic differential on $X$ has $2g-2$ zeros.
I would like to know how to prove this. If possible, without some "heavy machinery" like divisors and the Riemann-Roch theorem, which I don't understand very well yet. I really think that we can use the Riemann-Hurwitz to do this, but I am not sure how.
complex-analysis riemann-surfaces
asked Apr 8 at 6:39
GabrielGabriel
1,520623
$endgroup$
$begingroup$
Is it a joke, I answered to that question two days ago for nothing ? This is not heavy machinery but the strict minimum to define the different objects and obtain the result. math.stackexchange.com/questions/3176319/…
$endgroup$
– reuns
Apr 8 at 6:57
$begingroup$
Before Riemann-Hurwitz you need to show a meromorphic function $u$ has the same number of zeros/poles, showing $int_gamma fracduu=0$ for $gamma$ enclosing the whole of $X$ and applying the argument principle (look first at $X = BbbCP^1$ and $X=BbbC/(Z+iZ)$)
$endgroup$
– reuns
Apr 8 at 7:06
$begingroup$
@reuns they are not the same question. It first I thought this result would be valid for holomorphic functions but your answer helped clarify that.
$endgroup$
– Gabriel
Apr 8 at 7:15
$begingroup$
I'll do some thinking here and try to understand what you mean. Thanks a lot for your help
$endgroup$
– Gabriel
Apr 8 at 7:16
1
$begingroup$
For $X=mathbbC/(mathbbZ+imathbbZ)$ then $D = (0,1)+i(0,1)$ is a fundamental domain that is an open set such that every point of $X$ is contained in $overlineD$ and $D$ doesn't contain any element twice. For $u$ having no pole/zero in $partial D$ then look at $gamma = partial D$ and show the edges of $gamma$ cancel so $int_gamma fracduu=0$. For more complicated $X$ you'll have $gamma= bigcup_j=1^J Gamma_jbigcup_j=1^J Gamma_sigma(j)^-$ for some curves $ Gamma_j$ and some permutation $sigma$
$endgroup$
– reuns
Apr 8 at 7:48
|
show 2 more comments
$begingroup$
If $X$ is a compact Riemann surface, then any holomorphic differential on $X$ has $2g-2$ zeros.
I would like to know how to prove this. If possible, without some "heavy machinery" like divisors and the Riemann-Roch theorem, which I don't understand very well yet. I really think that we can use the Riemann-Hurwitz to do this, but I am not sure how.
complex-analysis riemann-surfaces
asked Apr 8 at 6:39
GabrielGabriel
1,520623
$endgroup$
If $X$ is a compact Riemann surface, then any holomorphic differential on $X$ has $2g-2$ zeros.
I would like to know how to prove this. If possible, without some "heavy machinery" like divisors and the Riemann-Roch theorem, which I don't understand very well yet. I really think that we can use the Riemann-Hurwitz to do this, but I am not sure how.
complex-analysis riemann-surfaces
complex-analysis riemann-surfaces
asked Apr 8 at 6:39
GabrielGabriel
1,520623
asked Apr 8 at 6:39
GabrielGabriel
1,520623
asked Apr 8 at 6:39
GabrielGabriel
1,520623
asked Apr 8 at 6:39
GabrielGabriel
1,520623
asked Apr 8 at 6:39
GabrielGabriel
1,520623
1,520623
$begingroup$
Is it a joke, I answered to that question two days ago for nothing ? This is not heavy machinery but the strict minimum to define the different objects and obtain the result. math.stackexchange.com/questions/3176319/…
$endgroup$
– reuns
Apr 8 at 6:57
$begingroup$
Before Riemann-Hurwitz you need to show a meromorphic function $u$ has the same number of zeros/poles, showing $int_gamma fracduu=0$ for $gamma$ enclosing the whole of $X$ and applying the argument principle (look first at $X = BbbCP^1$ and $X=BbbC/(Z+iZ)$)
$endgroup$
– reuns
Apr 8 at 7:06
$begingroup$
@reuns they are not the same question. It first I thought this result would be valid for holomorphic functions but your answer helped clarify that.
$endgroup$
– Gabriel
Apr 8 at 7:15
$begingroup$
I'll do some thinking here and try to understand what you mean. Thanks a lot for your help
$endgroup$
– Gabriel
Apr 8 at 7:16
1
$begingroup$
For $X=mathbbC/(mathbbZ+imathbbZ)$ then $D = (0,1)+i(0,1)$ is a fundamental domain that is an open set such that every point of $X$ is contained in $overlineD$ and $D$ doesn't contain any element twice. For $u$ having no pole/zero in $partial D$ then look at $gamma = partial D$ and show the edges of $gamma$ cancel so $int_gamma fracduu=0$. For more complicated $X$ you'll have $gamma= bigcup_j=1^J Gamma_jbigcup_j=1^J Gamma_sigma(j)^-$ for some curves $ Gamma_j$ and some permutation $sigma$
$endgroup$
– reuns
Apr 8 at 7:48
|
show 2 more comments
$begingroup$
Is it a joke, I answered to that question two days ago for nothing ? This is not heavy machinery but the strict minimum to define the different objects and obtain the result. math.stackexchange.com/questions/3176319/…
$endgroup$
– reuns
Apr 8 at 6:57
$begingroup$
Before Riemann-Hurwitz you need to show a meromorphic function $u$ has the same number of zeros/poles, showing $int_gamma fracduu=0$ for $gamma$ enclosing the whole of $X$ and applying the argument principle (look first at $X = BbbCP^1$ and $X=BbbC/(Z+iZ)$)
$endgroup$
– reuns
Apr 8 at 7:06
$begingroup$
@reuns they are not the same question. It first I thought this result would be valid for holomorphic functions but your answer helped clarify that.
$endgroup$
– Gabriel
Apr 8 at 7:15
$begingroup$
I'll do some thinking here and try to understand what you mean. Thanks a lot for your help
$endgroup$
– Gabriel
Apr 8 at 7:16
1
$begingroup$
For $X=mathbbC/(mathbbZ+imathbbZ)$ then $D = (0,1)+i(0,1)$ is a fundamental domain that is an open set such that every point of $X$ is contained in $overlineD$ and $D$ doesn't contain any element twice. For $u$ having no pole/zero in $partial D$ then look at $gamma = partial D$ and show the edges of $gamma$ cancel so $int_gamma fracduu=0$. For more complicated $X$ you'll have $gamma= bigcup_j=1^J Gamma_jbigcup_j=1^J Gamma_sigma(j)^-$ for some curves $ Gamma_j$ and some permutation $sigma$
$endgroup$
– reuns
Apr 8 at 7:48
$begingroup$
Is it a joke, I answered to that question two days ago for nothing ? This is not heavy machinery but the strict minimum to define the different objects and obtain the result. math.stackexchange.com/questions/3176319/…
$endgroup$
– reuns
Apr 8 at 6:57
$begingroup$
Is it a joke, I answered to that question two days ago for nothing ? This is not heavy machinery but the strict minimum to define the different objects and obtain the result. math.stackexchange.com/questions/3176319/…
$endgroup$
– reuns
Apr 8 at 6:57
$begingroup$
Before Riemann-Hurwitz you need to show a meromorphic function $u$ has the same number of zeros/poles, showing $int_gamma fracduu=0$ for $gamma$ enclosing the whole of $X$ and applying the argument principle (look first at $X = BbbCP^1$ and $X=BbbC/(Z+iZ)$)
$endgroup$
– reuns
Apr 8 at 7:06
$begingroup$
Before Riemann-Hurwitz you need to show a meromorphic function $u$ has the same number of zeros/poles, showing $int_gamma fracduu=0$ for $gamma$ enclosing the whole of $X$ and applying the argument principle (look first at $X = BbbCP^1$ and $X=BbbC/(Z+iZ)$)
$endgroup$
– reuns
Apr 8 at 7:06
$begingroup$
@reuns they are not the same question. It first I thought this result would be valid for holomorphic functions but your answer helped clarify that.
$endgroup$
– Gabriel
Apr 8 at 7:15
$begingroup$
@reuns they are not the same question. It first I thought this result would be valid for holomorphic functions but your answer helped clarify that.
$endgroup$
– Gabriel
Apr 8 at 7:15
$begingroup$
I'll do some thinking here and try to understand what you mean. Thanks a lot for your help
$endgroup$
– Gabriel
Apr 8 at 7:16
$begingroup$
I'll do some thinking here and try to understand what you mean. Thanks a lot for your help
$endgroup$
– Gabriel
Apr 8 at 7:16
1
1
$begingroup$
For $X=mathbbC/(mathbbZ+imathbbZ)$ then $D = (0,1)+i(0,1)$ is a fundamental domain that is an open set such that every point of $X$ is contained in $overlineD$ and $D$ doesn't contain any element twice. For $u$ having no pole/zero in $partial D$ then look at $gamma = partial D$ and show the edges of $gamma$ cancel so $int_gamma fracduu=0$. For more complicated $X$ you'll have $gamma= bigcup_j=1^J Gamma_jbigcup_j=1^J Gamma_sigma(j)^-$ for some curves $ Gamma_j$ and some permutation $sigma$
$endgroup$
– reuns
Apr 8 at 7:48
$begingroup$
For $X=mathbbC/(mathbbZ+imathbbZ)$ then $D = (0,1)+i(0,1)$ is a fundamental domain that is an open set such that every point of $X$ is contained in $overlineD$ and $D$ doesn't contain any element twice. For $u$ having no pole/zero in $partial D$ then look at $gamma = partial D$ and show the edges of $gamma$ cancel so $int_gamma fracduu=0$. For more complicated $X$ you'll have $gamma= bigcup_j=1^J Gamma_jbigcup_j=1^J Gamma_sigma(j)^-$ for some curves $ Gamma_j$ and some permutation $sigma$
$endgroup$
– reuns
Apr 8 at 7:48
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If you know what is the Euler characteristic, this is easy. A holomorphic form is a section of the cotangent bundle, whose Euler characteristic is $-(2-2g)$ (this bundle is the dual of the tangent bundle), and all zeroes of this section have to be counted with a $+1$, as we choose an holomorphic form.
answered Apr 8 at 6:46
ThomasThomas
4,182510
$endgroup$
$begingroup$
I know what is Euler's characteristic but I learned differential forms from Rick Miranda's book which defines them simply as collections of expressions of the form $f_alpha :mathrmdz_alpha$ for each chart.
$endgroup$
– Gabriel
Apr 8 at 6:55
$begingroup$
@Thomas See my answer there
$endgroup$
– reuns
Apr 8 at 6:58
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If you know what is the Euler characteristic, this is easy. A holomorphic form is a section of the cotangent bundle, whose Euler characteristic is $-(2-2g)$ (this bundle is the dual of the tangent bundle), and all zeroes of this section have to be counted with a $+1$, as we choose an holomorphic form.
answered Apr 8 at 6:46
ThomasThomas
4,182510
$endgroup$
$begingroup$
I know what is Euler's characteristic but I learned differential forms from Rick Miranda's book which defines them simply as collections of expressions of the form $f_alpha :mathrmdz_alpha$ for each chart.
$endgroup$
– Gabriel
Apr 8 at 6:55
$begingroup$
@Thomas See my answer there
$endgroup$
– reuns
Apr 8 at 6:58
add a comment |
$begingroup$
If you know what is the Euler characteristic, this is easy. A holomorphic form is a section of the cotangent bundle, whose Euler characteristic is $-(2-2g)$ (this bundle is the dual of the tangent bundle), and all zeroes of this section have to be counted with a $+1$, as we choose an holomorphic form.
answered Apr 8 at 6:46
ThomasThomas
4,182510
$endgroup$
$begingroup$
I know what is Euler's characteristic but I learned differential forms from Rick Miranda's book which defines them simply as collections of expressions of the form $f_alpha :mathrmdz_alpha$ for each chart.
$endgroup$
– Gabriel
Apr 8 at 6:55
$begingroup$
@Thomas See my answer there
$endgroup$
– reuns
Apr 8 at 6:58
add a comment |
$begingroup$
If you know what is the Euler characteristic, this is easy. A holomorphic form is a section of the cotangent bundle, whose Euler characteristic is $-(2-2g)$ (this bundle is the dual of the tangent bundle), and all zeroes of this section have to be counted with a $+1$, as we choose an holomorphic form.
answered Apr 8 at 6:46
ThomasThomas
4,182510
$endgroup$
If you know what is the Euler characteristic, this is easy. A holomorphic form is a section of the cotangent bundle, whose Euler characteristic is $-(2-2g)$ (this bundle is the dual of the tangent bundle), and all zeroes of this section have to be counted with a $+1$, as we choose an holomorphic form.
answered Apr 8 at 6:46
ThomasThomas
4,182510
answered Apr 8 at 6:46
ThomasThomas
4,182510
answered Apr 8 at 6:46
ThomasThomas
4,182510
answered Apr 8 at 6:46
ThomasThomas
4,182510
4,182510
$begingroup$
I know what is Euler's characteristic but I learned differential forms from Rick Miranda's book which defines them simply as collections of expressions of the form $f_alpha :mathrmdz_alpha$ for each chart.
$endgroup$
– Gabriel
Apr 8 at 6:55
$begingroup$
@Thomas See my answer there
$endgroup$
– reuns
Apr 8 at 6:58
add a comment |
$begingroup$
I know what is Euler's characteristic but I learned differential forms from Rick Miranda's book which defines them simply as collections of expressions of the form $f_alpha :mathrmdz_alpha$ for each chart.
$endgroup$
– Gabriel
Apr 8 at 6:55
$begingroup$
@Thomas See my answer there
$endgroup$
– reuns
Apr 8 at 6:58
$begingroup$
I know what is Euler's characteristic but I learned differential forms from Rick Miranda's book which defines them simply as collections of expressions of the form $f_alpha :mathrmdz_alpha$ for each chart.
$endgroup$
– Gabriel
Apr 8 at 6:55
$begingroup$
I know what is Euler's characteristic but I learned differential forms from Rick Miranda's book which defines them simply as collections of expressions of the form $f_alpha :mathrmdz_alpha$ for each chart.
$endgroup$
– Gabriel
Apr 8 at 6:55
$begingroup$
@Thomas See my answer there
$endgroup$
– reuns
Apr 8 at 6:58
$begingroup$
@Thomas See my answer there
$endgroup$
– reuns
Apr 8 at 6:58
add a comment |
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$begingroup$
Is it a joke, I answered to that question two days ago for nothing ? This is not heavy machinery but the strict minimum to define the different objects and obtain the result. math.stackexchange.com/questions/3176319/…
$endgroup$
– reuns
Apr 8 at 6:57
$begingroup$
Before Riemann-Hurwitz you need to show a meromorphic function $u$ has the same number of zeros/poles, showing $int_gamma fracduu=0$ for $gamma$ enclosing the whole of $X$ and applying the argument principle (look first at $X = BbbCP^1$ and $X=BbbC/(Z+iZ)$)
$endgroup$
– reuns
Apr 8 at 7:06
$begingroup$
@reuns they are not the same question. It first I thought this result would be valid for holomorphic functions but your answer helped clarify that.
$endgroup$
– Gabriel
Apr 8 at 7:15
$begingroup$
I'll do some thinking here and try to understand what you mean. Thanks a lot for your help
$endgroup$
– Gabriel
Apr 8 at 7:16
1
$begingroup$
For $X=mathbbC/(mathbbZ+imathbbZ)$ then $D = (0,1)+i(0,1)$ is a fundamental domain that is an open set such that every point of $X$ is contained in $overlineD$ and $D$ doesn't contain any element twice. For $u$ having no pole/zero in $partial D$ then look at $gamma = partial D$ and show the edges of $gamma$ cancel so $int_gamma fracduu=0$. For more complicated $X$ you'll have $gamma= bigcup_j=1^J Gamma_jbigcup_j=1^J Gamma_sigma(j)^-$ for some curves $ Gamma_j$ and some permutation $sigma$
$endgroup$
– reuns
Apr 8 at 7:48