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Ruled surfaces in $mathbbP^3$ of high degree
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Suppose $Ssubseteq mathbbP^3(x:y:z:w)$ is a surface of degree $d$, which I assume to be integral as a scheme. So $S$ is the zero locus of a homogeneous polynomial in $k[x,y,z,w]$ of degree $d$ which generates a prime ideal. Here $k$ is the base field and is assumed to be algebraically closed.
If $S$ is ruled (every point of $S$ is contained in a line $Lsubseteq mathbbP^3$ which is contained in $S$) and $d>2$, must $S$ be a cone? If not, is there a classification of such surfaces?
The only smooth ruled surfaces are planes and smooth quadrics. So any surface satisfying the above conditions must be singular, but beyond that I've been having trouble determining further properties they must satisfy.
algebraic-geometry
asked 7 hours ago
catfish
842158
add a comment |Â
up vote
2
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Suppose $Ssubseteq mathbbP^3(x:y:z:w)$ is a surface of degree $d$, which I assume to be integral as a scheme. So $S$ is the zero locus of a homogeneous polynomial in $k[x,y,z,w]$ of degree $d$ which generates a prime ideal. Here $k$ is the base field and is assumed to be algebraically closed.
If $S$ is ruled (every point of $S$ is contained in a line $Lsubseteq mathbbP^3$ which is contained in $S$) and $d>2$, must $S$ be a cone? If not, is there a classification of such surfaces?
The only smooth ruled surfaces are planes and smooth quadrics. So any surface satisfying the above conditions must be singular, but beyond that I've been having trouble determining further properties they must satisfy.
algebraic-geometry
asked 7 hours ago
catfish
842158
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $Ssubseteq mathbbP^3(x:y:z:w)$ is a surface of degree $d$, which I assume to be integral as a scheme. So $S$ is the zero locus of a homogeneous polynomial in $k[x,y,z,w]$ of degree $d$ which generates a prime ideal. Here $k$ is the base field and is assumed to be algebraically closed.
If $S$ is ruled (every point of $S$ is contained in a line $Lsubseteq mathbbP^3$ which is contained in $S$) and $d>2$, must $S$ be a cone? If not, is there a classification of such surfaces?
The only smooth ruled surfaces are planes and smooth quadrics. So any surface satisfying the above conditions must be singular, but beyond that I've been having trouble determining further properties they must satisfy.
algebraic-geometry
asked 7 hours ago
catfish
842158
Suppose $Ssubseteq mathbbP^3(x:y:z:w)$ is a surface of degree $d$, which I assume to be integral as a scheme. So $S$ is the zero locus of a homogeneous polynomial in $k[x,y,z,w]$ of degree $d$ which generates a prime ideal. Here $k$ is the base field and is assumed to be algebraically closed.
If $S$ is ruled (every point of $S$ is contained in a line $Lsubseteq mathbbP^3$ which is contained in $S$) and $d>2$, must $S$ be a cone? If not, is there a classification of such surfaces?
The only smooth ruled surfaces are planes and smooth quadrics. So any surface satisfying the above conditions must be singular, but beyond that I've been having trouble determining further properties they must satisfy.
algebraic-geometry
asked 7 hours ago
catfish
842158
asked 7 hours ago
catfish
842158
asked 7 hours ago
catfish
842158
asked 7 hours ago
catfish
842158
842158
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
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up vote
2
down vote
accepted
Here's one for you to play with: Consider the surface
$$z^2w^2+4x^3z-6xyzw+4y^3w - 3x^2y^2 = 0.$$
It's called a tangent developable, the surface of lines tangent to a smooth curve in $Bbb P^3$. Can you find the curve?
answered 6 hours ago
Ted Shifrin
59.2k44286
Is the second term not supposed to have a w? With that edit, I think I understand. The dual of a smooth curve will be a surface, and is formed as the set of all tangent planes to the curve (planes containing the tangent lines). Each tangent line of the curve determines a line in dual projective space of the planes tangent to it, and thus the dual variety of the curve is a surface ruled by these lines. My computer algebra program can compute the dual of the given surface using elimination, and this gives a twisted cubic curve.
– catfish
1 hour ago
I also noticed that the underlying set of the singular locus of this surface is the twisted cubic defined by $(y^2 - xz, xy - zw, x^2 - yw)$. My guess is that this is the curve, but a proof that the singular loci of these tangent developable surfaces are the curves they came from escapes me at the moment. Either way, thanks for the example!
– catfish
1 hour ago
1
Yes, the $w$ in the second term was a typo. My apologies. Yes, this is the tangent developable of the twisted cubic. You can prove locally that the tangent developable is smooth away from the curve itself (when the curve is not degenerate). It is a local differential geometry exercise (either in Euclidean space or in projective space) that a ruled surface whose Gauss map is degenerate must be locally a cone, a cylinder, a plane, or a tangent developable. I haven't thought about a global characterization of generic ruled surfaces in $Bbb P^3$.
– Ted Shifrin
37 mins ago
By the way, you may find this pretty paper of Griffiths and Harris interesting. J.M. Landsberg has published numerous papers following various ideas in this paper, too.
– Ted Shifrin
32 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here's one for you to play with: Consider the surface
$$z^2w^2+4x^3z-6xyzw+4y^3w - 3x^2y^2 = 0.$$
It's called a tangent developable, the surface of lines tangent to a smooth curve in $Bbb P^3$. Can you find the curve?
answered 6 hours ago
Ted Shifrin
59.2k44286
Is the second term not supposed to have a w? With that edit, I think I understand. The dual of a smooth curve will be a surface, and is formed as the set of all tangent planes to the curve (planes containing the tangent lines). Each tangent line of the curve determines a line in dual projective space of the planes tangent to it, and thus the dual variety of the curve is a surface ruled by these lines. My computer algebra program can compute the dual of the given surface using elimination, and this gives a twisted cubic curve.
– catfish
1 hour ago
I also noticed that the underlying set of the singular locus of this surface is the twisted cubic defined by $(y^2 - xz, xy - zw, x^2 - yw)$. My guess is that this is the curve, but a proof that the singular loci of these tangent developable surfaces are the curves they came from escapes me at the moment. Either way, thanks for the example!
– catfish
1 hour ago
1
Yes, the $w$ in the second term was a typo. My apologies. Yes, this is the tangent developable of the twisted cubic. You can prove locally that the tangent developable is smooth away from the curve itself (when the curve is not degenerate). It is a local differential geometry exercise (either in Euclidean space or in projective space) that a ruled surface whose Gauss map is degenerate must be locally a cone, a cylinder, a plane, or a tangent developable. I haven't thought about a global characterization of generic ruled surfaces in $Bbb P^3$.
– Ted Shifrin
37 mins ago
By the way, you may find this pretty paper of Griffiths and Harris interesting. J.M. Landsberg has published numerous papers following various ideas in this paper, too.
– Ted Shifrin
32 mins ago
add a comment |Â
up vote
2
down vote
accepted
Here's one for you to play with: Consider the surface
$$z^2w^2+4x^3z-6xyzw+4y^3w - 3x^2y^2 = 0.$$
It's called a tangent developable, the surface of lines tangent to a smooth curve in $Bbb P^3$. Can you find the curve?
answered 6 hours ago
Ted Shifrin
59.2k44286
Is the second term not supposed to have a w? With that edit, I think I understand. The dual of a smooth curve will be a surface, and is formed as the set of all tangent planes to the curve (planes containing the tangent lines). Each tangent line of the curve determines a line in dual projective space of the planes tangent to it, and thus the dual variety of the curve is a surface ruled by these lines. My computer algebra program can compute the dual of the given surface using elimination, and this gives a twisted cubic curve.
– catfish
1 hour ago
I also noticed that the underlying set of the singular locus of this surface is the twisted cubic defined by $(y^2 - xz, xy - zw, x^2 - yw)$. My guess is that this is the curve, but a proof that the singular loci of these tangent developable surfaces are the curves they came from escapes me at the moment. Either way, thanks for the example!
– catfish
1 hour ago
1
Yes, the $w$ in the second term was a typo. My apologies. Yes, this is the tangent developable of the twisted cubic. You can prove locally that the tangent developable is smooth away from the curve itself (when the curve is not degenerate). It is a local differential geometry exercise (either in Euclidean space or in projective space) that a ruled surface whose Gauss map is degenerate must be locally a cone, a cylinder, a plane, or a tangent developable. I haven't thought about a global characterization of generic ruled surfaces in $Bbb P^3$.
– Ted Shifrin
37 mins ago
By the way, you may find this pretty paper of Griffiths and Harris interesting. J.M. Landsberg has published numerous papers following various ideas in this paper, too.
– Ted Shifrin
32 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here's one for you to play with: Consider the surface
$$z^2w^2+4x^3z-6xyzw+4y^3w - 3x^2y^2 = 0.$$
It's called a tangent developable, the surface of lines tangent to a smooth curve in $Bbb P^3$. Can you find the curve?
answered 6 hours ago
Ted Shifrin
59.2k44286
Here's one for you to play with: Consider the surface
$$z^2w^2+4x^3z-6xyzw+4y^3w - 3x^2y^2 = 0.$$
It's called a tangent developable, the surface of lines tangent to a smooth curve in $Bbb P^3$. Can you find the curve?
answered 6 hours ago
Ted Shifrin
59.2k44286
edited 39 mins ago
answered 6 hours ago
Ted Shifrin
59.2k44286
answered 6 hours ago
Ted Shifrin
59.2k44286
answered 6 hours ago
Ted Shifrin
59.2k44286
59.2k44286
Is the second term not supposed to have a w? With that edit, I think I understand. The dual of a smooth curve will be a surface, and is formed as the set of all tangent planes to the curve (planes containing the tangent lines). Each tangent line of the curve determines a line in dual projective space of the planes tangent to it, and thus the dual variety of the curve is a surface ruled by these lines. My computer algebra program can compute the dual of the given surface using elimination, and this gives a twisted cubic curve.
– catfish
1 hour ago
I also noticed that the underlying set of the singular locus of this surface is the twisted cubic defined by $(y^2 - xz, xy - zw, x^2 - yw)$. My guess is that this is the curve, but a proof that the singular loci of these tangent developable surfaces are the curves they came from escapes me at the moment. Either way, thanks for the example!
– catfish
1 hour ago
1
Yes, the $w$ in the second term was a typo. My apologies. Yes, this is the tangent developable of the twisted cubic. You can prove locally that the tangent developable is smooth away from the curve itself (when the curve is not degenerate). It is a local differential geometry exercise (either in Euclidean space or in projective space) that a ruled surface whose Gauss map is degenerate must be locally a cone, a cylinder, a plane, or a tangent developable. I haven't thought about a global characterization of generic ruled surfaces in $Bbb P^3$.
– Ted Shifrin
37 mins ago
By the way, you may find this pretty paper of Griffiths and Harris interesting. J.M. Landsberg has published numerous papers following various ideas in this paper, too.
– Ted Shifrin
32 mins ago
add a comment |Â
Is the second term not supposed to have a w? With that edit, I think I understand. The dual of a smooth curve will be a surface, and is formed as the set of all tangent planes to the curve (planes containing the tangent lines). Each tangent line of the curve determines a line in dual projective space of the planes tangent to it, and thus the dual variety of the curve is a surface ruled by these lines. My computer algebra program can compute the dual of the given surface using elimination, and this gives a twisted cubic curve.
– catfish
1 hour ago
I also noticed that the underlying set of the singular locus of this surface is the twisted cubic defined by $(y^2 - xz, xy - zw, x^2 - yw)$. My guess is that this is the curve, but a proof that the singular loci of these tangent developable surfaces are the curves they came from escapes me at the moment. Either way, thanks for the example!
– catfish
1 hour ago
1
Yes, the $w$ in the second term was a typo. My apologies. Yes, this is the tangent developable of the twisted cubic. You can prove locally that the tangent developable is smooth away from the curve itself (when the curve is not degenerate). It is a local differential geometry exercise (either in Euclidean space or in projective space) that a ruled surface whose Gauss map is degenerate must be locally a cone, a cylinder, a plane, or a tangent developable. I haven't thought about a global characterization of generic ruled surfaces in $Bbb P^3$.
– Ted Shifrin
37 mins ago
By the way, you may find this pretty paper of Griffiths and Harris interesting. J.M. Landsberg has published numerous papers following various ideas in this paper, too.
– Ted Shifrin
32 mins ago
Is the second term not supposed to have a w? With that edit, I think I understand. The dual of a smooth curve will be a surface, and is formed as the set of all tangent planes to the curve (planes containing the tangent lines). Each tangent line of the curve determines a line in dual projective space of the planes tangent to it, and thus the dual variety of the curve is a surface ruled by these lines. My computer algebra program can compute the dual of the given surface using elimination, and this gives a twisted cubic curve.
– catfish
1 hour ago
Is the second term not supposed to have a w? With that edit, I think I understand. The dual of a smooth curve will be a surface, and is formed as the set of all tangent planes to the curve (planes containing the tangent lines). Each tangent line of the curve determines a line in dual projective space of the planes tangent to it, and thus the dual variety of the curve is a surface ruled by these lines. My computer algebra program can compute the dual of the given surface using elimination, and this gives a twisted cubic curve.
– catfish
1 hour ago
I also noticed that the underlying set of the singular locus of this surface is the twisted cubic defined by $(y^2 - xz, xy - zw, x^2 - yw)$. My guess is that this is the curve, but a proof that the singular loci of these tangent developable surfaces are the curves they came from escapes me at the moment. Either way, thanks for the example!
– catfish
1 hour ago
I also noticed that the underlying set of the singular locus of this surface is the twisted cubic defined by $(y^2 - xz, xy - zw, x^2 - yw)$. My guess is that this is the curve, but a proof that the singular loci of these tangent developable surfaces are the curves they came from escapes me at the moment. Either way, thanks for the example!
– catfish
1 hour ago
1
1
Yes, the $w$ in the second term was a typo. My apologies. Yes, this is the tangent developable of the twisted cubic. You can prove locally that the tangent developable is smooth away from the curve itself (when the curve is not degenerate). It is a local differential geometry exercise (either in Euclidean space or in projective space) that a ruled surface whose Gauss map is degenerate must be locally a cone, a cylinder, a plane, or a tangent developable. I haven't thought about a global characterization of generic ruled surfaces in $Bbb P^3$.
– Ted Shifrin
37 mins ago
Yes, the $w$ in the second term was a typo. My apologies. Yes, this is the tangent developable of the twisted cubic. You can prove locally that the tangent developable is smooth away from the curve itself (when the curve is not degenerate). It is a local differential geometry exercise (either in Euclidean space or in projective space) that a ruled surface whose Gauss map is degenerate must be locally a cone, a cylinder, a plane, or a tangent developable. I haven't thought about a global characterization of generic ruled surfaces in $Bbb P^3$.
– Ted Shifrin
37 mins ago
By the way, you may find this pretty paper of Griffiths and Harris interesting. J.M. Landsberg has published numerous papers following various ideas in this paper, too.
– Ted Shifrin
32 mins ago
By the way, you may find this pretty paper of Griffiths and Harris interesting. J.M. Landsberg has published numerous papers following various ideas in this paper, too.
– Ted Shifrin
32 mins ago
add a comment |Â
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