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What is the role of the genus in the double cover of a rational normal curve by a hyperelliptic curve?
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My question concerns the following problem from Rick Miranda's Algebraic Curves and Riemann Surfaces (p. 167):
[S]how that if $v^2 = h(u)$ defines a hyperelliptic curve of genus $g$, then $phi = [1 colon u colon u^2 colon cdots colon u^g-1]$ defines a degree 2 map onto a rational normal curve of degree $g-1$ in $mathbbP^g-1$, and that the hyperplane divisors of $phi$ have degree $2g-2$.
Constructing $phi$ is easy: just compose the projection of the hyperelliptic curve onto $u$ with the standard map from $mathbbP^1$ onto the rational normal curve of degree $g-1$.
What I don't understand is why the genus is of importance. Can't we simply obtain a degree 2 map from any hyperelliptic curve onto any rational normal curve in exactly the same way, even if the genus and degree don't match up?
algebraic-geometry algebraic-curves riemann-surfaces
asked Jul 23 at 21:50
merle
536
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up vote
3
down vote
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My question concerns the following problem from Rick Miranda's Algebraic Curves and Riemann Surfaces (p. 167):
[S]how that if $v^2 = h(u)$ defines a hyperelliptic curve of genus $g$, then $phi = [1 colon u colon u^2 colon cdots colon u^g-1]$ defines a degree 2 map onto a rational normal curve of degree $g-1$ in $mathbbP^g-1$, and that the hyperplane divisors of $phi$ have degree $2g-2$.
Constructing $phi$ is easy: just compose the projection of the hyperelliptic curve onto $u$ with the standard map from $mathbbP^1$ onto the rational normal curve of degree $g-1$.
What I don't understand is why the genus is of importance. Can't we simply obtain a degree 2 map from any hyperelliptic curve onto any rational normal curve in exactly the same way, even if the genus and degree don't match up?
algebraic-geometry algebraic-curves riemann-surfaces
asked Jul 23 at 21:50
merle
536
1
You are right the degree and the genus are independant here. However I think it's probably used to show that the canonical map is not an embedding for an hyperelliptic curve.
– Nicolas Hemelsoet
Jul 23 at 23:08
Thanks, that must be it. I have found the theorem you are referring to - it is stated two chapters later.
– merle
Jul 23 at 23:28
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
My question concerns the following problem from Rick Miranda's Algebraic Curves and Riemann Surfaces (p. 167):
[S]how that if $v^2 = h(u)$ defines a hyperelliptic curve of genus $g$, then $phi = [1 colon u colon u^2 colon cdots colon u^g-1]$ defines a degree 2 map onto a rational normal curve of degree $g-1$ in $mathbbP^g-1$, and that the hyperplane divisors of $phi$ have degree $2g-2$.
Constructing $phi$ is easy: just compose the projection of the hyperelliptic curve onto $u$ with the standard map from $mathbbP^1$ onto the rational normal curve of degree $g-1$.
What I don't understand is why the genus is of importance. Can't we simply obtain a degree 2 map from any hyperelliptic curve onto any rational normal curve in exactly the same way, even if the genus and degree don't match up?
algebraic-geometry algebraic-curves riemann-surfaces
asked Jul 23 at 21:50
merle
536
My question concerns the following problem from Rick Miranda's Algebraic Curves and Riemann Surfaces (p. 167):
[S]how that if $v^2 = h(u)$ defines a hyperelliptic curve of genus $g$, then $phi = [1 colon u colon u^2 colon cdots colon u^g-1]$ defines a degree 2 map onto a rational normal curve of degree $g-1$ in $mathbbP^g-1$, and that the hyperplane divisors of $phi$ have degree $2g-2$.
Constructing $phi$ is easy: just compose the projection of the hyperelliptic curve onto $u$ with the standard map from $mathbbP^1$ onto the rational normal curve of degree $g-1$.
What I don't understand is why the genus is of importance. Can't we simply obtain a degree 2 map from any hyperelliptic curve onto any rational normal curve in exactly the same way, even if the genus and degree don't match up?
algebraic-geometry algebraic-curves riemann-surfaces
asked Jul 23 at 21:50
merle
536
asked Jul 23 at 21:50
merle
536
asked Jul 23 at 21:50
merle
536
asked Jul 23 at 21:50
merle
536
536
1
You are right the degree and the genus are independant here. However I think it's probably used to show that the canonical map is not an embedding for an hyperelliptic curve.
– Nicolas Hemelsoet
Jul 23 at 23:08
Thanks, that must be it. I have found the theorem you are referring to - it is stated two chapters later.
– merle
Jul 23 at 23:28
add a comment |Â
1
You are right the degree and the genus are independant here. However I think it's probably used to show that the canonical map is not an embedding for an hyperelliptic curve.
– Nicolas Hemelsoet
Jul 23 at 23:08
Thanks, that must be it. I have found the theorem you are referring to - it is stated two chapters later.
– merle
Jul 23 at 23:28
1
1
You are right the degree and the genus are independant here. However I think it's probably used to show that the canonical map is not an embedding for an hyperelliptic curve.
– Nicolas Hemelsoet
Jul 23 at 23:08
You are right the degree and the genus are independant here. However I think it's probably used to show that the canonical map is not an embedding for an hyperelliptic curve.
– Nicolas Hemelsoet
Jul 23 at 23:08
Thanks, that must be it. I have found the theorem you are referring to - it is stated two chapters later.
– merle
Jul 23 at 23:28
Thanks, that must be it. I have found the theorem you are referring to - it is stated two chapters later.
– merle
Jul 23 at 23:28
add a comment |Â
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1
You are right the degree and the genus are independant here. However I think it's probably used to show that the canonical map is not an embedding for an hyperelliptic curve.
– Nicolas Hemelsoet
Jul 23 at 23:08
Thanks, that must be it. I have found the theorem you are referring to - it is stated two chapters later.
– merle
Jul 23 at 23:28