Branch points of a projection to $[X:Y] = Bbb{CP}^1$ using homogeneous coordinates
$begingroup$
I have a question that I've been stuck on for some time now, and although I'm able to understand similar questions I'm stuck on this particular type of projection to the complex projective line.
The problem is, given a curve
$$
Z^2Y^2 = X^4 + Y^4 + Z^4
$$
Consider the holomorphic map
$$
f: [X:Y:Z] → [X:Y]
$$
What are the branch points of f?
If I consider a projection to the Riemann Sphere of the affine and non affine components respectively:
$$g([x,y,1]) → x $$
$$g([x,y,0]) → infty $$
I'm able to compute that the $deg(g) = 4$, since it has four branching points whose preimages each contain a ramification point of index four, which agrees with what I would expect.
On reflection, I don't think that $f$ and $g$ are the same map. But I'm unsure of how to proceed. If we look at the points on the affine curve:
$$
y^2(1 - y^2) - x^4 - 1 = 0
$$
We see that
$$
y^2(1 - y^2) - 1 = x^4
$$
And so, for the four values
$$
{ y_i | y_i^2(1 - y_i^2) = -1}
$$
I had previously thought that their images might be the branching values, e.g.:
$$ f^{-1}([0:y_i]) = {[0: y_i: 1]} $$
but (as I can now see) I'm incorrect since all of the $y_i$ in the image are equal up to scaling by a complex number. Any help specifically with this particular projection would be greatly appreciated as I seem to be going round in circles.
algebraic-geometry algebraic-curves riemann-surfaces
asked Nov 27 '18 at 14:51
b_dobresb_dobres
83
$endgroup$
add a comment |
$begingroup$
I have a question that I've been stuck on for some time now, and although I'm able to understand similar questions I'm stuck on this particular type of projection to the complex projective line.
The problem is, given a curve
$$
Z^2Y^2 = X^4 + Y^4 + Z^4
$$
Consider the holomorphic map
$$
f: [X:Y:Z] → [X:Y]
$$
What are the branch points of f?
If I consider a projection to the Riemann Sphere of the affine and non affine components respectively:
$$g([x,y,1]) → x $$
$$g([x,y,0]) → infty $$
I'm able to compute that the $deg(g) = 4$, since it has four branching points whose preimages each contain a ramification point of index four, which agrees with what I would expect.
On reflection, I don't think that $f$ and $g$ are the same map. But I'm unsure of how to proceed. If we look at the points on the affine curve:
$$
y^2(1 - y^2) - x^4 - 1 = 0
$$
We see that
$$
y^2(1 - y^2) - 1 = x^4
$$
And so, for the four values
$$
{ y_i | y_i^2(1 - y_i^2) = -1}
$$
I had previously thought that their images might be the branching values, e.g.:
$$ f^{-1}([0:y_i]) = {[0: y_i: 1]} $$
but (as I can now see) I'm incorrect since all of the $y_i$ in the image are equal up to scaling by a complex number. Any help specifically with this particular projection would be greatly appreciated as I seem to be going round in circles.
algebraic-geometry algebraic-curves riemann-surfaces
asked Nov 27 '18 at 14:51
b_dobresb_dobres
83
$endgroup$
add a comment |
$begingroup$
I have a question that I've been stuck on for some time now, and although I'm able to understand similar questions I'm stuck on this particular type of projection to the complex projective line.
The problem is, given a curve
$$
Z^2Y^2 = X^4 + Y^4 + Z^4
$$
Consider the holomorphic map
$$
f: [X:Y:Z] → [X:Y]
$$
What are the branch points of f?
If I consider a projection to the Riemann Sphere of the affine and non affine components respectively:
$$g([x,y,1]) → x $$
$$g([x,y,0]) → infty $$
I'm able to compute that the $deg(g) = 4$, since it has four branching points whose preimages each contain a ramification point of index four, which agrees with what I would expect.
On reflection, I don't think that $f$ and $g$ are the same map. But I'm unsure of how to proceed. If we look at the points on the affine curve:
$$
y^2(1 - y^2) - x^4 - 1 = 0
$$
We see that
$$
y^2(1 - y^2) - 1 = x^4
$$
And so, for the four values
$$
{ y_i | y_i^2(1 - y_i^2) = -1}
$$
I had previously thought that their images might be the branching values, e.g.:
$$ f^{-1}([0:y_i]) = {[0: y_i: 1]} $$
but (as I can now see) I'm incorrect since all of the $y_i$ in the image are equal up to scaling by a complex number. Any help specifically with this particular projection would be greatly appreciated as I seem to be going round in circles.
algebraic-geometry algebraic-curves riemann-surfaces
asked Nov 27 '18 at 14:51
b_dobresb_dobres
83
$endgroup$
I have a question that I've been stuck on for some time now, and although I'm able to understand similar questions I'm stuck on this particular type of projection to the complex projective line.
The problem is, given a curve
$$
Z^2Y^2 = X^4 + Y^4 + Z^4
$$
Consider the holomorphic map
$$
f: [X:Y:Z] → [X:Y]
$$
What are the branch points of f?
If I consider a projection to the Riemann Sphere of the affine and non affine components respectively:
$$g([x,y,1]) → x $$
$$g([x,y,0]) → infty $$
I'm able to compute that the $deg(g) = 4$, since it has four branching points whose preimages each contain a ramification point of index four, which agrees with what I would expect.
On reflection, I don't think that $f$ and $g$ are the same map. But I'm unsure of how to proceed. If we look at the points on the affine curve:
$$
y^2(1 - y^2) - x^4 - 1 = 0
$$
We see that
$$
y^2(1 - y^2) - 1 = x^4
$$
And so, for the four values
$$
{ y_i | y_i^2(1 - y_i^2) = -1}
$$
I had previously thought that their images might be the branching values, e.g.:
$$ f^{-1}([0:y_i]) = {[0: y_i: 1]} $$
but (as I can now see) I'm incorrect since all of the $y_i$ in the image are equal up to scaling by a complex number. Any help specifically with this particular projection would be greatly appreciated as I seem to be going round in circles.
algebraic-geometry algebraic-curves riemann-surfaces
algebraic-geometry algebraic-curves riemann-surfaces
asked Nov 27 '18 at 14:51
b_dobresb_dobres
83
asked Nov 27 '18 at 14:51
b_dobresb_dobres
83
edited Nov 27 '18 at 15:36
b_dobres
asked Nov 27 '18 at 14:51
b_dobresb_dobres
83
asked Nov 27 '18 at 14:51
b_dobresb_dobres
83
asked Nov 27 '18 at 14:51
b_dobresb_dobres
83
83
add a comment |
add a comment |
1 Answer
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I don't think you should be taking the affine chart $Zne 0$ in this case. You have a map $Crightarrow mathbb{P}^1$ and if you want to understand this map affine locally, you should take an affine chart on $mathbb{P}^1$ and consider its preimage instead.
Let's call homogeneous coordinates on $mathbb{P}^2$ by $X,Y,Z$ and homogeneous coordinates on $mathbb{P}^1$ to be $U,V$. Then the pre-image of $D(U) ={ [U:V]: Une 0} cong mathbb{A}^1$ is $D(X)cap C = {[X:Y:Z]in C: Xne 0 }$. In particular the corresponding affine curve is $z^2y^2 = 1 + y^4 + z^4$, and in this case your map is given by projection to $y$-coordinate. To determine the branch points then it suffices to determine all $y$ such that there are less than 4 distinct roots for $z$, which now becomes a problem in solving quadratic equations.
Then you can also look at the other chart $D(V)$ and the corresponding affine curve and determine if $infty$ (which corresponds to the point $[0:1]$) is a branch point by the same argument.
answered Nov 27 '18 at 16:47
lochloch
1,9681810
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$begingroup$
Thank you very much! Taking the chart on the image is the key piece of logic I was missing.
$endgroup$
– b_dobres
Nov 27 '18 at 18:26
add a comment |
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$begingroup$
I don't think you should be taking the affine chart $Zne 0$ in this case. You have a map $Crightarrow mathbb{P}^1$ and if you want to understand this map affine locally, you should take an affine chart on $mathbb{P}^1$ and consider its preimage instead.
Let's call homogeneous coordinates on $mathbb{P}^2$ by $X,Y,Z$ and homogeneous coordinates on $mathbb{P}^1$ to be $U,V$. Then the pre-image of $D(U) ={ [U:V]: Une 0} cong mathbb{A}^1$ is $D(X)cap C = {[X:Y:Z]in C: Xne 0 }$. In particular the corresponding affine curve is $z^2y^2 = 1 + y^4 + z^4$, and in this case your map is given by projection to $y$-coordinate. To determine the branch points then it suffices to determine all $y$ such that there are less than 4 distinct roots for $z$, which now becomes a problem in solving quadratic equations.
Then you can also look at the other chart $D(V)$ and the corresponding affine curve and determine if $infty$ (which corresponds to the point $[0:1]$) is a branch point by the same argument.
answered Nov 27 '18 at 16:47
lochloch
1,9681810
$endgroup$
$begingroup$
Thank you very much! Taking the chart on the image is the key piece of logic I was missing.
$endgroup$
– b_dobres
Nov 27 '18 at 18:26
add a comment |
$begingroup$
I don't think you should be taking the affine chart $Zne 0$ in this case. You have a map $Crightarrow mathbb{P}^1$ and if you want to understand this map affine locally, you should take an affine chart on $mathbb{P}^1$ and consider its preimage instead.
Let's call homogeneous coordinates on $mathbb{P}^2$ by $X,Y,Z$ and homogeneous coordinates on $mathbb{P}^1$ to be $U,V$. Then the pre-image of $D(U) ={ [U:V]: Une 0} cong mathbb{A}^1$ is $D(X)cap C = {[X:Y:Z]in C: Xne 0 }$. In particular the corresponding affine curve is $z^2y^2 = 1 + y^4 + z^4$, and in this case your map is given by projection to $y$-coordinate. To determine the branch points then it suffices to determine all $y$ such that there are less than 4 distinct roots for $z$, which now becomes a problem in solving quadratic equations.
Then you can also look at the other chart $D(V)$ and the corresponding affine curve and determine if $infty$ (which corresponds to the point $[0:1]$) is a branch point by the same argument.
answered Nov 27 '18 at 16:47
lochloch
1,9681810
$endgroup$
$begingroup$
Thank you very much! Taking the chart on the image is the key piece of logic I was missing.
$endgroup$
– b_dobres
Nov 27 '18 at 18:26
add a comment |
$begingroup$
I don't think you should be taking the affine chart $Zne 0$ in this case. You have a map $Crightarrow mathbb{P}^1$ and if you want to understand this map affine locally, you should take an affine chart on $mathbb{P}^1$ and consider its preimage instead.
Let's call homogeneous coordinates on $mathbb{P}^2$ by $X,Y,Z$ and homogeneous coordinates on $mathbb{P}^1$ to be $U,V$. Then the pre-image of $D(U) ={ [U:V]: Une 0} cong mathbb{A}^1$ is $D(X)cap C = {[X:Y:Z]in C: Xne 0 }$. In particular the corresponding affine curve is $z^2y^2 = 1 + y^4 + z^4$, and in this case your map is given by projection to $y$-coordinate. To determine the branch points then it suffices to determine all $y$ such that there are less than 4 distinct roots for $z$, which now becomes a problem in solving quadratic equations.
Then you can also look at the other chart $D(V)$ and the corresponding affine curve and determine if $infty$ (which corresponds to the point $[0:1]$) is a branch point by the same argument.
answered Nov 27 '18 at 16:47
lochloch
1,9681810
$endgroup$
I don't think you should be taking the affine chart $Zne 0$ in this case. You have a map $Crightarrow mathbb{P}^1$ and if you want to understand this map affine locally, you should take an affine chart on $mathbb{P}^1$ and consider its preimage instead.
Let's call homogeneous coordinates on $mathbb{P}^2$ by $X,Y,Z$ and homogeneous coordinates on $mathbb{P}^1$ to be $U,V$. Then the pre-image of $D(U) ={ [U:V]: Une 0} cong mathbb{A}^1$ is $D(X)cap C = {[X:Y:Z]in C: Xne 0 }$. In particular the corresponding affine curve is $z^2y^2 = 1 + y^4 + z^4$, and in this case your map is given by projection to $y$-coordinate. To determine the branch points then it suffices to determine all $y$ such that there are less than 4 distinct roots for $z$, which now becomes a problem in solving quadratic equations.
Then you can also look at the other chart $D(V)$ and the corresponding affine curve and determine if $infty$ (which corresponds to the point $[0:1]$) is a branch point by the same argument.
answered Nov 27 '18 at 16:47
lochloch
1,9681810
answered Nov 27 '18 at 16:47
lochloch
1,9681810
answered Nov 27 '18 at 16:47
lochloch
1,9681810
answered Nov 27 '18 at 16:47
lochloch
1,9681810
1,9681810
$begingroup$
Thank you very much! Taking the chart on the image is the key piece of logic I was missing.
$endgroup$
– b_dobres
Nov 27 '18 at 18:26
add a comment |
$begingroup$
Thank you very much! Taking the chart on the image is the key piece of logic I was missing.
$endgroup$
– b_dobres
Nov 27 '18 at 18:26
$begingroup$
Thank you very much! Taking the chart on the image is the key piece of logic I was missing.
$endgroup$
– b_dobres
Nov 27 '18 at 18:26
$begingroup$
Thank you very much! Taking the chart on the image is the key piece of logic I was missing.
$endgroup$
– b_dobres
Nov 27 '18 at 18:26
add a comment |
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