Number of branch points for a projection to $Bbb{CP}^1$
Based on a helpful response to my previous post which offered advice on how to split the image of a map into affine and non-affine components, I've come up with a solution to the following problem. However the solutions provided by the TA indicate that there are in fact 9 branching points (my solution gives exactly 4).
I would be very grateful if someone could point out where I've made my mistake?
The problem
Given a projective curve C defined by
$$Z^2Y^2 = X^4 + Y^4 + Z^4$$
Consider the holomorphic map$$
f: [X:Y:Z] mapsto [X:Y] = mathbb{CP}^1
$$
What are the branching points of f?
My solution
For $[X:Y] in mathbb{CP}^1$, let $[U:V]:=[X:Y] $. First we consider the preimage of the affine component ${[U:V] | U = 1 }$. We have that:
$$
f^{-1} ( [1:V] ) = C cap [1:Y:Z]
$$
Which gives us the corresponding affine curve:
$$
z^2y^2 - 1 - y^4 - z^4 = 0
$$
Rewriting, we have:
$$
(-1)(z^2)^2 + (y^2)(z^2) + (- y^4 - 1) = 0
$$
Which gives us coefficients $ a = -1, b=y^2, c = -y^4 - 1$ for the quadratic formula:
$${z^2=frac{-y^2pmsqrt{-3y^4 -4}}{-2}}$$
Writing $sqrt[4]{-1} = omega_i $ for a 4th root of negative unity, we then have four branching points
$$y in { omega_1, omega_2, omega_3, omega_4 }$$ each with brancing index 4.
Finally, consider the preimage of the non-affine component:
$$f^{-1} ( [0:V] ) = C cap [0:Y:Z] $$
which gives the corresponding curve
$$z^2y^2 - y^4 - z^4 = 0.$$
Now, since $[0:0:0] notin mathbb{CP}^2$, it follows that $y=z=0$ is not a solution, and we can therefore scale the co-ordinates $[0:Y:Z]$ to $[0:1:Z]$ by multiplication by some nonzero factor. This then gives us:
$$ z^4 -1 -z^2 = 0$$
Which clearly has four distinct roots $z_1, z_2, z_3, z_4$ , with:
$$ f(z_i) = [0:V] = infty $$
It follows that the preimage of $infty$ contains four points, each with ramification index 1 and therefore it is not a branching point. Since there are no more possible candidates for branching points, we are done.
algebraic-geometry algebraic-curves riemann-surfaces
asked Nov 29 at 14:32
b_dobres
133
add a comment |
Based on a helpful response to my previous post which offered advice on how to split the image of a map into affine and non-affine components, I've come up with a solution to the following problem. However the solutions provided by the TA indicate that there are in fact 9 branching points (my solution gives exactly 4).
I would be very grateful if someone could point out where I've made my mistake?
The problem
Given a projective curve C defined by
$$Z^2Y^2 = X^4 + Y^4 + Z^4$$
Consider the holomorphic map$$
f: [X:Y:Z] mapsto [X:Y] = mathbb{CP}^1
$$
What are the branching points of f?
My solution
For $[X:Y] in mathbb{CP}^1$, let $[U:V]:=[X:Y] $. First we consider the preimage of the affine component ${[U:V] | U = 1 }$. We have that:
$$
f^{-1} ( [1:V] ) = C cap [1:Y:Z]
$$
Which gives us the corresponding affine curve:
$$
z^2y^2 - 1 - y^4 - z^4 = 0
$$
Rewriting, we have:
$$
(-1)(z^2)^2 + (y^2)(z^2) + (- y^4 - 1) = 0
$$
Which gives us coefficients $ a = -1, b=y^2, c = -y^4 - 1$ for the quadratic formula:
$${z^2=frac{-y^2pmsqrt{-3y^4 -4}}{-2}}$$
Writing $sqrt[4]{-1} = omega_i $ for a 4th root of negative unity, we then have four branching points
$$y in { omega_1, omega_2, omega_3, omega_4 }$$ each with brancing index 4.
Finally, consider the preimage of the non-affine component:
$$f^{-1} ( [0:V] ) = C cap [0:Y:Z] $$
which gives the corresponding curve
$$z^2y^2 - y^4 - z^4 = 0.$$
Now, since $[0:0:0] notin mathbb{CP}^2$, it follows that $y=z=0$ is not a solution, and we can therefore scale the co-ordinates $[0:Y:Z]$ to $[0:1:Z]$ by multiplication by some nonzero factor. This then gives us:
$$ z^4 -1 -z^2 = 0$$
Which clearly has four distinct roots $z_1, z_2, z_3, z_4$ , with:
$$ f(z_i) = [0:V] = infty $$
It follows that the preimage of $infty$ contains four points, each with ramification index 1 and therefore it is not a branching point. Since there are no more possible candidates for branching points, we are done.
algebraic-geometry algebraic-curves riemann-surfaces
asked Nov 29 at 14:32
b_dobres
133
Hmm are you sure the answer is not 8? Anyway when you had $z^2 = frac{y^2 pm sqrt{-3y^4 -4}}{2}$, you want to consider $y$ such that $-3y^4-4 = 0$ (so you get 4 $y$'s), and also $y$ such that $y^2 pm sqrt{-3y^4-4} = 0$ (which also gives you 4 $y$'s). This gives 8 branch points, but I think this is what you should expect from Riemann-Hurwitz..
– loch
Nov 29 at 17:29
This looks useful: math.stackexchange.com/questions/1511153/…
– André 3000
Nov 29 at 19:15
@loch yes thankyou, I see where my mistake is now. It seems like the solution given to us by the TA was also incorrect, which has contributed to my confusion!
– b_dobres
Nov 29 at 21:55
add a comment |
Based on a helpful response to my previous post which offered advice on how to split the image of a map into affine and non-affine components, I've come up with a solution to the following problem. However the solutions provided by the TA indicate that there are in fact 9 branching points (my solution gives exactly 4).
I would be very grateful if someone could point out where I've made my mistake?
The problem
Given a projective curve C defined by
$$Z^2Y^2 = X^4 + Y^4 + Z^4$$
Consider the holomorphic map$$
f: [X:Y:Z] mapsto [X:Y] = mathbb{CP}^1
$$
What are the branching points of f?
My solution
For $[X:Y] in mathbb{CP}^1$, let $[U:V]:=[X:Y] $. First we consider the preimage of the affine component ${[U:V] | U = 1 }$. We have that:
$$
f^{-1} ( [1:V] ) = C cap [1:Y:Z]
$$
Which gives us the corresponding affine curve:
$$
z^2y^2 - 1 - y^4 - z^4 = 0
$$
Rewriting, we have:
$$
(-1)(z^2)^2 + (y^2)(z^2) + (- y^4 - 1) = 0
$$
Which gives us coefficients $ a = -1, b=y^2, c = -y^4 - 1$ for the quadratic formula:
$${z^2=frac{-y^2pmsqrt{-3y^4 -4}}{-2}}$$
Writing $sqrt[4]{-1} = omega_i $ for a 4th root of negative unity, we then have four branching points
$$y in { omega_1, omega_2, omega_3, omega_4 }$$ each with brancing index 4.
Finally, consider the preimage of the non-affine component:
$$f^{-1} ( [0:V] ) = C cap [0:Y:Z] $$
which gives the corresponding curve
$$z^2y^2 - y^4 - z^4 = 0.$$
Now, since $[0:0:0] notin mathbb{CP}^2$, it follows that $y=z=0$ is not a solution, and we can therefore scale the co-ordinates $[0:Y:Z]$ to $[0:1:Z]$ by multiplication by some nonzero factor. This then gives us:
$$ z^4 -1 -z^2 = 0$$
Which clearly has four distinct roots $z_1, z_2, z_3, z_4$ , with:
$$ f(z_i) = [0:V] = infty $$
It follows that the preimage of $infty$ contains four points, each with ramification index 1 and therefore it is not a branching point. Since there are no more possible candidates for branching points, we are done.
algebraic-geometry algebraic-curves riemann-surfaces
asked Nov 29 at 14:32
b_dobres
133
Based on a helpful response to my previous post which offered advice on how to split the image of a map into affine and non-affine components, I've come up with a solution to the following problem. However the solutions provided by the TA indicate that there are in fact 9 branching points (my solution gives exactly 4).
I would be very grateful if someone could point out where I've made my mistake?
The problem
Given a projective curve C defined by
$$Z^2Y^2 = X^4 + Y^4 + Z^4$$
Consider the holomorphic map$$
f: [X:Y:Z] mapsto [X:Y] = mathbb{CP}^1
$$
What are the branching points of f?
My solution
For $[X:Y] in mathbb{CP}^1$, let $[U:V]:=[X:Y] $. First we consider the preimage of the affine component ${[U:V] | U = 1 }$. We have that:
$$
f^{-1} ( [1:V] ) = C cap [1:Y:Z]
$$
Which gives us the corresponding affine curve:
$$
z^2y^2 - 1 - y^4 - z^4 = 0
$$
Rewriting, we have:
$$
(-1)(z^2)^2 + (y^2)(z^2) + (- y^4 - 1) = 0
$$
Which gives us coefficients $ a = -1, b=y^2, c = -y^4 - 1$ for the quadratic formula:
$${z^2=frac{-y^2pmsqrt{-3y^4 -4}}{-2}}$$
Writing $sqrt[4]{-1} = omega_i $ for a 4th root of negative unity, we then have four branching points
$$y in { omega_1, omega_2, omega_3, omega_4 }$$ each with brancing index 4.
Finally, consider the preimage of the non-affine component:
$$f^{-1} ( [0:V] ) = C cap [0:Y:Z] $$
which gives the corresponding curve
$$z^2y^2 - y^4 - z^4 = 0.$$
Now, since $[0:0:0] notin mathbb{CP}^2$, it follows that $y=z=0$ is not a solution, and we can therefore scale the co-ordinates $[0:Y:Z]$ to $[0:1:Z]$ by multiplication by some nonzero factor. This then gives us:
$$ z^4 -1 -z^2 = 0$$
Which clearly has four distinct roots $z_1, z_2, z_3, z_4$ , with:
$$ f(z_i) = [0:V] = infty $$
It follows that the preimage of $infty$ contains four points, each with ramification index 1 and therefore it is not a branching point. Since there are no more possible candidates for branching points, we are done.
algebraic-geometry algebraic-curves riemann-surfaces
algebraic-geometry algebraic-curves riemann-surfaces
asked Nov 29 at 14:32
b_dobres
133
asked Nov 29 at 14:32
b_dobres
133
edited Nov 29 at 15:22
asked Nov 29 at 14:32
b_dobres
133
asked Nov 29 at 14:32
b_dobres
133
asked Nov 29 at 14:32
b_dobres
133
133
Hmm are you sure the answer is not 8? Anyway when you had $z^2 = frac{y^2 pm sqrt{-3y^4 -4}}{2}$, you want to consider $y$ such that $-3y^4-4 = 0$ (so you get 4 $y$'s), and also $y$ such that $y^2 pm sqrt{-3y^4-4} = 0$ (which also gives you 4 $y$'s). This gives 8 branch points, but I think this is what you should expect from Riemann-Hurwitz..
– loch
Nov 29 at 17:29
This looks useful: math.stackexchange.com/questions/1511153/…
– André 3000
Nov 29 at 19:15
@loch yes thankyou, I see where my mistake is now. It seems like the solution given to us by the TA was also incorrect, which has contributed to my confusion!
– b_dobres
Nov 29 at 21:55
add a comment |
Hmm are you sure the answer is not 8? Anyway when you had $z^2 = frac{y^2 pm sqrt{-3y^4 -4}}{2}$, you want to consider $y$ such that $-3y^4-4 = 0$ (so you get 4 $y$'s), and also $y$ such that $y^2 pm sqrt{-3y^4-4} = 0$ (which also gives you 4 $y$'s). This gives 8 branch points, but I think this is what you should expect from Riemann-Hurwitz..
– loch
Nov 29 at 17:29
This looks useful: math.stackexchange.com/questions/1511153/…
– André 3000
Nov 29 at 19:15
@loch yes thankyou, I see where my mistake is now. It seems like the solution given to us by the TA was also incorrect, which has contributed to my confusion!
– b_dobres
Nov 29 at 21:55
Hmm are you sure the answer is not 8? Anyway when you had $z^2 = frac{y^2 pm sqrt{-3y^4 -4}}{2}$, you want to consider $y$ such that $-3y^4-4 = 0$ (so you get 4 $y$'s), and also $y$ such that $y^2 pm sqrt{-3y^4-4} = 0$ (which also gives you 4 $y$'s). This gives 8 branch points, but I think this is what you should expect from Riemann-Hurwitz..
– loch
Nov 29 at 17:29
Hmm are you sure the answer is not 8? Anyway when you had $z^2 = frac{y^2 pm sqrt{-3y^4 -4}}{2}$, you want to consider $y$ such that $-3y^4-4 = 0$ (so you get 4 $y$'s), and also $y$ such that $y^2 pm sqrt{-3y^4-4} = 0$ (which also gives you 4 $y$'s). This gives 8 branch points, but I think this is what you should expect from Riemann-Hurwitz..
– loch
Nov 29 at 17:29
This looks useful: math.stackexchange.com/questions/1511153/…
– André 3000
Nov 29 at 19:15
This looks useful: math.stackexchange.com/questions/1511153/…
– André 3000
Nov 29 at 19:15
@loch yes thankyou, I see where my mistake is now. It seems like the solution given to us by the TA was also incorrect, which has contributed to my confusion!
– b_dobres
Nov 29 at 21:55
@loch yes thankyou, I see where my mistake is now. It seems like the solution given to us by the TA was also incorrect, which has contributed to my confusion!
– b_dobres
Nov 29 at 21:55
add a comment |
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Hmm are you sure the answer is not 8? Anyway when you had $z^2 = frac{y^2 pm sqrt{-3y^4 -4}}{2}$, you want to consider $y$ such that $-3y^4-4 = 0$ (so you get 4 $y$'s), and also $y$ such that $y^2 pm sqrt{-3y^4-4} = 0$ (which also gives you 4 $y$'s). This gives 8 branch points, but I think this is what you should expect from Riemann-Hurwitz..
– loch
Nov 29 at 17:29
This looks useful: math.stackexchange.com/questions/1511153/…
– André 3000
Nov 29 at 19:15
@loch yes thankyou, I see where my mistake is now. It seems like the solution given to us by the TA was also incorrect, which has contributed to my confusion!
– b_dobres
Nov 29 at 21:55