Mixing
Explicit equation for the dual hypersurface
0
$begingroup$
Let $X$ be a hypersurface in $mathbb P^n$. Assume $X$ is defined by some homogenous polynomial $F$. Then, is there a way to write the equation for $X^vee$ explicitly?
(This is related to my previous question)
algebraic-geometry reference-request complex-geometry
asked Jan 13 at 11:31
User XUser X
32911
$endgroup$
add a comment |
0
$begingroup$
Let $X$ be a hypersurface in $mathbb P^n$. Assume $X$ is defined by some homogenous polynomial $F$. Then, is there a way to write the equation for $X^vee$ explicitly?
(This is related to my previous question)
algebraic-geometry reference-request complex-geometry
asked Jan 13 at 11:31
User XUser X
32911
$endgroup$
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Dual_curve#Equations and try to mimic the technique there.
$endgroup$
– DKS
Jan 13 at 15:23
$begingroup$
@DKS Thanks! But other from the conic, for hypersurfaces in higher degree, it is not easy to eliminate the parameters. For example, for $X$ defined by $x^3+y^3+z^3=0$, in the way you suggest, we can get the equation for the dual curve $X^{3/2}+Y^{3/2}+Z^{3/2}=0$ . On the other hand, it should be a curve of degree $6$, so I expect an explicit polynomial of degree $6$, which is not clear to me. Is there some way to do this?
$endgroup$
– User X
Jan 13 at 17:15
$begingroup$
Use $(X^{3/2}+Y^{3/2}+Z^{3/2})(X^{3/2}-Y^{3/2}+Z^{3/2})(X^{3/2}+Y^{3/2}-Z^{3/2})(X^{3/2}-Y^{3/2}-Z^{3/2})=0$ ?
$endgroup$
– random
Jan 14 at 12:30
$begingroup$
@random Yes, I think it is in the special case. But in general, it is not easy to get the polynomial from the parametric equations.
$endgroup$
– User X
Jan 17 at 17:38
add a comment |
0
0
0
$begingroup$
Let $X$ be a hypersurface in $mathbb P^n$. Assume $X$ is defined by some homogenous polynomial $F$. Then, is there a way to write the equation for $X^vee$ explicitly?
(This is related to my previous question)
algebraic-geometry reference-request complex-geometry
asked Jan 13 at 11:31
User XUser X
32911
$endgroup$
Let $X$ be a hypersurface in $mathbb P^n$. Assume $X$ is defined by some homogenous polynomial $F$. Then, is there a way to write the equation for $X^vee$ explicitly?
(This is related to my previous question)
algebraic-geometry reference-request complex-geometry
algebraic-geometry reference-request complex-geometry
asked Jan 13 at 11:31
User XUser X
32911
asked Jan 13 at 11:31
User XUser X
32911
asked Jan 13 at 11:31
User XUser X
32911
asked Jan 13 at 11:31
User XUser X
32911
asked Jan 13 at 11:31
User XUser X
32911
32911
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Dual_curve#Equations and try to mimic the technique there.
$endgroup$
– DKS
Jan 13 at 15:23
$begingroup$
@DKS Thanks! But other from the conic, for hypersurfaces in higher degree, it is not easy to eliminate the parameters. For example, for $X$ defined by $x^3+y^3+z^3=0$, in the way you suggest, we can get the equation for the dual curve $X^{3/2}+Y^{3/2}+Z^{3/2}=0$ . On the other hand, it should be a curve of degree $6$, so I expect an explicit polynomial of degree $6$, which is not clear to me. Is there some way to do this?
$endgroup$
– User X
Jan 13 at 17:15
$begingroup$
Use $(X^{3/2}+Y^{3/2}+Z^{3/2})(X^{3/2}-Y^{3/2}+Z^{3/2})(X^{3/2}+Y^{3/2}-Z^{3/2})(X^{3/2}-Y^{3/2}-Z^{3/2})=0$ ?
$endgroup$
– random
Jan 14 at 12:30
$begingroup$
@random Yes, I think it is in the special case. But in general, it is not easy to get the polynomial from the parametric equations.
$endgroup$
– User X
Jan 17 at 17:38
add a comment |
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Dual_curve#Equations and try to mimic the technique there.
$endgroup$
– DKS
Jan 13 at 15:23
$begingroup$
@DKS Thanks! But other from the conic, for hypersurfaces in higher degree, it is not easy to eliminate the parameters. For example, for $X$ defined by $x^3+y^3+z^3=0$, in the way you suggest, we can get the equation for the dual curve $X^{3/2}+Y^{3/2}+Z^{3/2}=0$ . On the other hand, it should be a curve of degree $6$, so I expect an explicit polynomial of degree $6$, which is not clear to me. Is there some way to do this?
$endgroup$
– User X
Jan 13 at 17:15
$begingroup$
Use $(X^{3/2}+Y^{3/2}+Z^{3/2})(X^{3/2}-Y^{3/2}+Z^{3/2})(X^{3/2}+Y^{3/2}-Z^{3/2})(X^{3/2}-Y^{3/2}-Z^{3/2})=0$ ?
$endgroup$
– random
Jan 14 at 12:30
$begingroup$
@random Yes, I think it is in the special case. But in general, it is not easy to get the polynomial from the parametric equations.
$endgroup$
– User X
Jan 17 at 17:38
2
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Dual_curve#Equations and try to mimic the technique there.
$endgroup$
– DKS
Jan 13 at 15:23
$begingroup$
Take a look at en.wikipedia.org/wiki/Dual_curve#Equations and try to mimic the technique there.
$endgroup$
– DKS
Jan 13 at 15:23
$begingroup$
@DKS Thanks! But other from the conic, for hypersurfaces in higher degree, it is not easy to eliminate the parameters. For example, for $X$ defined by $x^3+y^3+z^3=0$, in the way you suggest, we can get the equation for the dual curve $X^{3/2}+Y^{3/2}+Z^{3/2}=0$ . On the other hand, it should be a curve of degree $6$, so I expect an explicit polynomial of degree $6$, which is not clear to me. Is there some way to do this?
$endgroup$
– User X
Jan 13 at 17:15
$begingroup$
@DKS Thanks! But other from the conic, for hypersurfaces in higher degree, it is not easy to eliminate the parameters. For example, for $X$ defined by $x^3+y^3+z^3=0$, in the way you suggest, we can get the equation for the dual curve $X^{3/2}+Y^{3/2}+Z^{3/2}=0$ . On the other hand, it should be a curve of degree $6$, so I expect an explicit polynomial of degree $6$, which is not clear to me. Is there some way to do this?
$endgroup$
– User X
Jan 13 at 17:15
$begingroup$
Use $(X^{3/2}+Y^{3/2}+Z^{3/2})(X^{3/2}-Y^{3/2}+Z^{3/2})(X^{3/2}+Y^{3/2}-Z^{3/2})(X^{3/2}-Y^{3/2}-Z^{3/2})=0$ ?
$endgroup$
– random
Jan 14 at 12:30
$begingroup$
Use $(X^{3/2}+Y^{3/2}+Z^{3/2})(X^{3/2}-Y^{3/2}+Z^{3/2})(X^{3/2}+Y^{3/2}-Z^{3/2})(X^{3/2}-Y^{3/2}-Z^{3/2})=0$ ?
$endgroup$
– random
Jan 14 at 12:30
$begingroup$
@random Yes, I think it is in the special case. But in general, it is not easy to get the polynomial from the parametric equations.
$endgroup$
– User X
Jan 17 at 17:38
$begingroup$
@random Yes, I think it is in the special case. But in general, it is not easy to get the polynomial from the parametric equations.
$endgroup$
– User X
Jan 17 at 17:38
add a comment |
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2
$begingroup$
Take a look at en.wikipedia.org/wiki/Dual_curve#Equations and try to mimic the technique there.
$endgroup$
– DKS
Jan 13 at 15:23
$begingroup$
@DKS Thanks! But other from the conic, for hypersurfaces in higher degree, it is not easy to eliminate the parameters. For example, for $X$ defined by $x^3+y^3+z^3=0$, in the way you suggest, we can get the equation for the dual curve $X^{3/2}+Y^{3/2}+Z^{3/2}=0$ . On the other hand, it should be a curve of degree $6$, so I expect an explicit polynomial of degree $6$, which is not clear to me. Is there some way to do this?
$endgroup$
– User X
Jan 13 at 17:15
$begingroup$
Use $(X^{3/2}+Y^{3/2}+Z^{3/2})(X^{3/2}-Y^{3/2}+Z^{3/2})(X^{3/2}+Y^{3/2}-Z^{3/2})(X^{3/2}-Y^{3/2}-Z^{3/2})=0$ ?
$endgroup$
– random
Jan 14 at 12:30
$begingroup$
@random Yes, I think it is in the special case. But in general, it is not easy to get the polynomial from the parametric equations.
$endgroup$
– User X
Jan 17 at 17:38