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edited Feb 13 2011 at 23:35 |
The intersection of the two quadrics in $\mathbb{P}^3$ is a complete intersection and defines an elliptic curve, so the genus should be is 1. A way to see it this is to pick a point $p$ on $C$ and project from $p$ onto a general hyperplane. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have genus 1, we are done.
Another way to see this that $g(C)=1$ is by computing cohomology of the sequence
$$
0 \to O_{P^3}(-4) \to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0
$$(This is the resolution of $O_C$ as an $O_{P^3}$ module, which is easy to write down for complete intersections). Using this and the standard formulae for cohomology on $P^n$, we get $g=h^1(O_C)=1$.
Yet another way to see it is by looking at the curve as a divisor of type $(2,2)$ on the quartic surface $X_0X_3-2X_1X_2$. In general, by the adjunction formula, divisors of type $(a,b)$ have arithmetic genus $(a-1)(b-1)$, so again we get g=1.
The twised cubic $C$ is the (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using a projection, and $C'$ would be a plane curve of degree 2.
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edited Feb 13 2011 at 18:01 |
The intersection of the two quadrics in $\mathbb{P}^3$ is a complete intersection and defines an elliptic curve, so the genus should be 1. One A way to see it is to pick a point $p$ on $C$ and project from $p$ onto a general hyperplane. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have genus 1, we are done.
Another way to see this is by computing cohomology of the sequence
$$
0 \to O_{P^3}(-4) \to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0
$$(This is the resolution of $O_C$ as an $O_{P^3}$ module, which is easy to write down for complete intersections). Using this and the standard formulae for cohomology on $P^n$, we get $g=h^1(O_C)=1$.
Another
Yet another way to see it is by looking at the curve as a divisor of type $(2,2)$ on the quartic surface $X_0X_3-2X_1X_2$. In general, by the adjunction formula, divisors of type $(a,b)$ have arithmetic genus $(a-1)(b-1)$, so again we get g=1.
EDIT: Yet another way to see it is to pick a point $p$ on $C$ and project from $p$ onto a general hyperplane. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have arithmetic genus 1, we are done.
The twised cubic $C$ is the (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using a projection, and $C'$ would be a plane curve of degree 2.
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edited Feb 13 2011 at 17:51 |
The intersection of the two quadrics in $\mathbb{P}^3$ is a complete intersection and defines an elliptic curve, so the genus should be 1. One way to see this is by computing cohomology of the sequence
$$
0 \to O_{P^3}(-4) \to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0
$$(This is the resolution of $O_C$ as an $O_{P^3}$ module, which is easy to write down for complete intersections). Using this and the standard formulae for cohomology on $P^n$, we get $g=h^1(O_C)=1$.
Another way to see it is by looking at the curve as a divisor of type $(2,2)$ on the quartic surface $X_0X_3-2X_1X_2$. In general, by the adjunction formula, divisors of type $(a,b)$ have arithmetic genus $(a-1)(b-1)$, so again we get g=1.
EDIT: Yet another way to see it is to pick a general point $p$ on $C$ and project from $p$ onto the a general hyperplaneat infinity. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have arithmetic genus 1, we are done.
The twised cubic $C$ is the (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using a projection, and $C'$ would be a plane curve of degree 2.
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edited Feb 13 2011 at 17:32 |
The intersection of the two quadrics in $\mathbb{P}^3$ is a complete intersection and defines an elliptic curve, so the genus should be 1. One way to see this is by computing cohomology of the sequence
$$
0 \to O_{P^3}(-4) \to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0
$$(This is the resolution of $O_C$ as an $O_{P^3}$ module, which is easy to write down for complete intersections). Using this and the standard formulae for cohomology on $P^n$, we get $g=h^1(O_C)=1$.
Another way to see it is by looking at the curve as a divisor of type $(2,2)$ on the quartic surface $X_0X_3-2X_1X_2$. In general, by the adjunction formula, divisors of type $(a,b)$ have arithmetic genus $(a-1)(b-1)$, so again we get g=1.
EDIT: Yet another way to see it is to pick a general point $p$ on $C$ and project from $p$ onto the hyperplane at infinity. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have arithmetic genus 1, we are done.
The twised cubic $C$ is the (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using a resolution as aboveprojection, but in this case and $C$ is not C'$ would be a complete intersectionplane curve of degree 2.
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edited Feb 13 2011 at 13:50 |
The (transverse) intersection of the two smooth quadrics in $\mathbb{P}^3$ is a complete intersection and defines an elliptic curve, so the genus should be 1. One way to see this is by computing cohomology of the sequence
$$
0 \to O_{P^3}(-4) \to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0
$$This $(This is the resolution of $O_X$ O_C$ as an $O_{P^3}$ module, which is easy to write down for complete intersections). Using this and the standard formulae for cohomology on $P^n$, we get $g=h^1(O_C)=1$.
Another way to see it is since by looking at the curve defines as a divisor of type $(2,2)$ on the quartic surface $X_0X3-2X_1X_2$. X_0X_3-2X_1X_2$. In general, by the adjunction formula, divisors of type $(a,b)$ have arithmetic genus $(a-1)(b-1)$, so again we get g=1.
The twised cubic $C$ is the (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using the a resolution as above, but in this case $C$ is not a complete intersection.
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edited Feb 13 2011 at 13:44 |
The (transverse) intersection of two smooth quadrics in $\mathbb{P}^3$ is an elliptic curve, so the genus should be 1. One way to see this is by computing cohomology of the sequence
$$
0 \to O_{P^3}(-4)\to {O}{P^3}(-2) O_{P^3}(-4) \oplus {O}{P^3}(2)\to {O}{P^3}\to {O}{X}(-4)\to to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0
.
$$This gives is the resolution of $h^1(O_X)=1$ using O_X$ as an $O_{P^3}$ module. Using this and the standard formulae for cohomology on $P^n$.P^n$, we get $g=h^1(O_C)=1$.
Another way to see it is since the curve defines a divisor of type $(2,2)$ on the quartic surface $X_0X3-2X_1X_2$. In general, it is known that by the adjunction formula, divisors of type $(a,b)$ have genus $(a-1)(b-1)$, so again we get g=1. This computation can be seen from the adjunction formula.
The twised cubic $C$ is the isomorphic (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using the resolution as above, but in this case $C$ is not a complete intersection.
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answered Feb 13 2011 at 13:37 |
The (transverse) intersection of two smooth quadrics in $\mathbb{P}^3$ is an elliptic curve, so the genus should be 1. One way to see this is by computing cohomology of the sequence
$$
0 \to O_{P^3}(-4)\to {O}{P^3}(-2) \oplus {O}{P^3}(2)\to {O}{P^3}\to {O}{X}(-4)\to 0.
$$This gives $h^1(O_X)=1$ using the standard formulae for cohomology on $P^n$.
Another way to see it is since the curve defines a divisor of type $(2,2)$ on the quartic surface $X_0X3-2X_1X_2$. In general, it is known that divisors of type $(a,b)$ have genus $(a-1)(b-1)$, so again we get g=1. This computation can be seen from the adjunction formula.
The twised cubic $C$ is the isomorphic image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0.
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