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The holomorphic map from a compact smooth curve $C$ to $mathbbCP^1$ when $H^0(C,mathcalO(p))=2$

Differential of a smooth map on holomorphic tangent spaceDivisor of meromorphic section of point bundle over a Riemann surfaceIsomorphisms (and non-isomorphisms) of holomorphic degree $1$ line bundles on $mathbbCP^1$ and elliptic curvesFinding holomorphic map on Riemann surface from a map between two Riemann surfacesShow $f$ is an immersion, where $f$ holomorphic map between compact Riemann SurfacesGlobal sections of holomorphic line bundles $mathcalO(n)$ over $mathbbP^1(mathbbC)$Low dimensional classifying map determining $mathcalO(k)$ on $mathbbP^n$Proving a nonconstant holomorphic map from $mathbbC_infty to Y$ gives a homeomorphismHow to properly deduce the Holomorphic Implicit Function Theorem from the Smooth Real Implicit Function Theorem?If $f$ is holomorphic in a compact Riemann surface, it is constant. Why doesn't this work for compact subsets of $mathbbC$?

1

$begingroup$

Let $C$ be a compact smooth complex curve with $H^0(C,mathcalO(p))=2$.

I feel confused with the following words:

Denote by $a$ and $b$ two non-collinear sections in $H^0(C,mathcalO(p))$. Then one can consider the ratio $f=a/b$ as a holomorphic map from $C$ to $mathbbCP^1$.

$bullet$ If both sections vanish at the same point $q$ then $f$ does not take the value $0$ so is constant. This contradicts the non-collinearity of $a$ and $b$.

$bullet$ If $a$ but not $b$ vanishes at $q$ then looking at the fiber of $0in mathbbCP^1$ the degree of the map $f$ is $1$.

I don't understand "If both sections vanish at the same point $q$ then $f$ does not take the value $0$ so is constant." And does non-collinear means $a,b$ are not linear?

I also wonder how to deduce the degree is $1$ from $a(q)=0,b(q)not=0$?

At last, how is that related to $p$?

complex-geometry riemann-surfaces complex-manifolds

share|cite|improve this question

asked Mar 30 at 3:27

DannyDanny

1,159412

$endgroup$

• 
  
  
  

  
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  $begingroup$
  $H^0(C,O(p))$ is the set of meromorphic functions $f$ such that for every $U, fin O(p)(U)$ thus it is the meromorphic functions with only one pole at $p$. Since $C$ is compact this means $f$ has only one zero. If $dim H^0(C,O(p))=2$ then $H^0(C,O(p)) = BbbC f_1+BbbCf_2$ where $Div(f_1) = q_1-p,Div(f_2) = q_2-p$. If $q_1=q_2$ then $Div(f_1/f_2) =emptyset$ so $f_1/f_2$ is constant, contradiction. Thus $zin Cmapsto[f_1(z):f_2(z)]$ is a surjective holomorphic map $CtoBbbCP^1$ of degree $1$, it is an isomorphism and $CcongBbbCP^1$.
  $endgroup$
  – reuns
  Mar 30 at 5:01
  
  

  
  
  
  


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  $begingroup$
  It is of degree $1$ because $Div(f_1/f_2) = q_1-q_2$ so for any $s in BbbC, Div(f_1/f_2-s) = r-q_2$ and hence $f=f_1/f_2$ takes the value $s$ only once. This proves the function field of $C$ is $BbbC(f)$. Also $z mapsto [f_1:f_2]$ is locally biholomorphic, to prove it is globally biholomorphic you can look at the Cauchy integral formula for $f^-1$. Why don't you search for elementary texts ?
  $endgroup$
  – reuns
  Mar 30 at 5:11
  
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  With $C = BbbCP^1=BbbCcup infty$ and $p in BbbC$ then $H^0(C,O(p)) = (z mapsto a+fracbz-p), a,b in BbbC= BbbC f_1+BbbCf_2= BbbC (f_1+f_2)+BbbCf_2$ where $f_1(z)=1, f_2(z)= frac1z-p$ and $Div(f_2) = infty-p$. For any $qin C - p$ then $Div( f_2-f_2(q)) = q-p$ which is what you need to say for any $g$ meromorphic with $Div(g) = sum_j n_j q_j$ and the $q_j ne p$ then $g =cprod_j (f_2-f_2(q_j))^n_j$
  $endgroup$
  – reuns
  Mar 30 at 6:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $BbbC$ is a non-compact Riemann surface. How many zeros and poles does $fracz^2z-1$ and $fracsin zz-1$ have. For $C$ compact $f$ has the same number of zeros and poles by the argument principle : let $gamma$ a closed curve enclosing all of them, then $fracdff$ is a meromorphic one-form, you can modify the curve to $gamma=bigcup_l=1^m gamma_lbigcup_l=1^mgamma_sigma(l)^-$ for some permutation $sigma$, thus $int_gamma fracdff=sum_l=1^m int_gamma_lfracdff-sum_l=1^m int_gamma_sigma(l)fracdff=0$
  $endgroup$
  – reuns
  Mar 30 at 6:33
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Of course you are supposed to look first at the examples given by $BbbCP^1$ and $BbbC/(BbbZ+iZ)$ en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions
  $endgroup$
  – reuns
  Mar 30 at 6:37
  
  

  
  
  
  

 | 
show 3 more comments

1

$begingroup$

Let $C$ be a compact smooth complex curve with $H^0(C,mathcalO(p))=2$.

I feel confused with the following words:

Denote by $a$ and $b$ two non-collinear sections in $H^0(C,mathcalO(p))$. Then one can consider the ratio $f=a/b$ as a holomorphic map from $C$ to $mathbbCP^1$.

$bullet$ If both sections vanish at the same point $q$ then $f$ does not take the value $0$ so is constant. This contradicts the non-collinearity of $a$ and $b$.

$bullet$ If $a$ but not $b$ vanishes at $q$ then looking at the fiber of $0in mathbbCP^1$ the degree of the map $f$ is $1$.

I don't understand "If both sections vanish at the same point $q$ then $f$ does not take the value $0$ so is constant." And does non-collinear means $a,b$ are not linear?

I also wonder how to deduce the degree is $1$ from $a(q)=0,b(q)not=0$?

At last, how is that related to $p$?

complex-geometry riemann-surfaces complex-manifolds

share|cite|improve this question

asked Mar 30 at 3:27

DannyDanny

1,159412

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $H^0(C,O(p))$ is the set of meromorphic functions $f$ such that for every $U, fin O(p)(U)$ thus it is the meromorphic functions with only one pole at $p$. Since $C$ is compact this means $f$ has only one zero. If $dim H^0(C,O(p))=2$ then $H^0(C,O(p)) = BbbC f_1+BbbCf_2$ where $Div(f_1) = q_1-p,Div(f_2) = q_2-p$. If $q_1=q_2$ then $Div(f_1/f_2) =emptyset$ so $f_1/f_2$ is constant, contradiction. Thus $zin Cmapsto[f_1(z):f_2(z)]$ is a surjective holomorphic map $CtoBbbCP^1$ of degree $1$, it is an isomorphism and $CcongBbbCP^1$.
  $endgroup$
  – reuns
  Mar 30 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is of degree $1$ because $Div(f_1/f_2) = q_1-q_2$ so for any $s in BbbC, Div(f_1/f_2-s) = r-q_2$ and hence $f=f_1/f_2$ takes the value $s$ only once. This proves the function field of $C$ is $BbbC(f)$. Also $z mapsto [f_1:f_2]$ is locally biholomorphic, to prove it is globally biholomorphic you can look at the Cauchy integral formula for $f^-1$. Why don't you search for elementary texts ?
  $endgroup$
  – reuns
  Mar 30 at 5:11
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  With $C = BbbCP^1=BbbCcup infty$ and $p in BbbC$ then $H^0(C,O(p)) = (z mapsto a+fracbz-p), a,b in BbbC= BbbC f_1+BbbCf_2= BbbC (f_1+f_2)+BbbCf_2$ where $f_1(z)=1, f_2(z)= frac1z-p$ and $Div(f_2) = infty-p$. For any $qin C - p$ then $Div( f_2-f_2(q)) = q-p$ which is what you need to say for any $g$ meromorphic with $Div(g) = sum_j n_j q_j$ and the $q_j ne p$ then $g =cprod_j (f_2-f_2(q_j))^n_j$
  $endgroup$
  – reuns
  Mar 30 at 6:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $BbbC$ is a non-compact Riemann surface. How many zeros and poles does $fracz^2z-1$ and $fracsin zz-1$ have. For $C$ compact $f$ has the same number of zeros and poles by the argument principle : let $gamma$ a closed curve enclosing all of them, then $fracdff$ is a meromorphic one-form, you can modify the curve to $gamma=bigcup_l=1^m gamma_lbigcup_l=1^mgamma_sigma(l)^-$ for some permutation $sigma$, thus $int_gamma fracdff=sum_l=1^m int_gamma_lfracdff-sum_l=1^m int_gamma_sigma(l)fracdff=0$
  $endgroup$
  – reuns
  Mar 30 at 6:33
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Of course you are supposed to look first at the examples given by $BbbCP^1$ and $BbbC/(BbbZ+iZ)$ en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions
  $endgroup$
  – reuns
  Mar 30 at 6:37
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Let $C$ be a compact smooth complex curve with $H^0(C,mathcalO(p))=2$.

I feel confused with the following words:

Denote by $a$ and $b$ two non-collinear sections in $H^0(C,mathcalO(p))$. Then one can consider the ratio $f=a/b$ as a holomorphic map from $C$ to $mathbbCP^1$.

$bullet$ If both sections vanish at the same point $q$ then $f$ does not take the value $0$ so is constant. This contradicts the non-collinearity of $a$ and $b$.

$bullet$ If $a$ but not $b$ vanishes at $q$ then looking at the fiber of $0in mathbbCP^1$ the degree of the map $f$ is $1$.

I don't understand "If both sections vanish at the same point $q$ then $f$ does not take the value $0$ so is constant." And does non-collinear means $a,b$ are not linear?

I also wonder how to deduce the degree is $1$ from $a(q)=0,b(q)not=0$?

At last, how is that related to $p$?

complex-geometry riemann-surfaces complex-manifolds

share|cite|improve this question

asked Mar 30 at 3:27

DannyDanny

1,159412

$endgroup$

share|cite|improve this question

asked Mar 30 at 3:27

DannyDanny

1,159412

share|cite|improve this question

asked Mar 30 at 3:27

DannyDanny

1,159412

asked Mar 30 at 3:27

DannyDanny

1,159412

asked Mar 30 at 3:27

DannyDanny

1,159412