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Topic review - homogeneous cordinates |
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Re: homogeneous cordinates |
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thx, but i cant do the change of variables x=X/Z and y=Y/Z, that is, i need tha t. One vector v= x²+y² use subst(v,x,x/z) and subst(v,y,y/z) and the return is v= (x/z)² +(y/z)² gorzel wrote: It would be really helpful if you could give an concrete example. State what you have, how the basering looks like, and what you want to get.
Is m the degree of p and q, or should this also be an additional argument to a possible proc?
In general, dehomogenization i.e. passing from homogenous coordinates X,Y,Z to affine coordinates, where Z = 1, is obtained by substituting Z=1.
This is done by subst(q,Z,1).
The homogenization is done by homog(q,z). If q has degree d, then homog(q,z) is homogenous of degree d.
// --- Or could you point to an article/textbook where this situation is considered?
C. Gorzel
thx, but i cant do the change of variables x=X/Z and y=Y/Z, that is,
i need tha t. One vector v= x²+y² use subst(v,x,x/z) and subst(v,y,y/z)
and the return is v= (x/z)² +(y/z)²
[quote="gorzel"]It would be really helpful if you could give an concrete example. State what you have, how the basering looks like, and what you want to get.
Is [b]m[/b] the degree of p and q, or should this also be an additional argument to a possible proc?
In general, dehomogenization i.e. passing from homogenous coordinates X,Y,Z to affine coordinates, where Z = 1, is obtained by substituting Z=1.
This is done by [b]subst(q,Z,1)[/b].
The homogenization is done by [b]homog(q,z)[/b]. If [b]q[/b] has degree [b]d[/b], then [b]homog(q,z)[/b] is homogenous of degree [b]d[/b].
// --- Or could you point to an article/textbook where this situation is considered?
C. Gorzel[/quote]
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Posted: Fri Apr 16, 2010 3:04 pm |
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Re: homogeneous cordinates |
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It would be really helpful if you could give an concrete example. State what you have, how the basering looks like, and what you want to get.
Is m the degree of p and q, or should this also be an additional argument to a possible proc?
In general, dehomogenization i.e. passing from homogenous coordinates X,Y,Z to affine coordinates, where Z = 1, is obtained by substituting Z=1.
This is done by subst(q,Z,1).
The homogenization is done by homog(q,z). If q has degree d, then homog(q,z) is homogenous of degree d.
// --- Or could you point to an article/textbook where this situation is considered?
C. Gorzel
It would be really helpful if you could give an concrete example. State what you have, how the basering looks like, and what you want to get.
Is [b]m[/b] the degree of p and q, or should this also be an additional argument to a possible proc?
In general, dehomogenization i.e. passing from homogenous coordinates X,Y,Z to affine coordinates, where Z = 1, is obtained by substituting Z=1.
This is done by [b]subst(q,Z,1)[/b].
The homogenization is done by [b]homog(q,z)[/b]. If [b]q[/b] has degree [b]d[/b], then [b]homog(q,z)[/b] is homogenous of degree [b]d[/b].
// --- Or could you point to an article/textbook where this situation is considered?
C. Gorzel
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Posted: Mon Apr 12, 2010 9:32 pm |
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Re: homogeneous cordinates |
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m is the max degree of monomials p, q i need, in true,create a function for desomoneization and homegeneization of polynomials. gorzel wrote: If you define at the beginning already a ring in three variables (x,y,z), where you have two polynomial p and q in the variables x,y, then you can just set Code: q*z^m + p*z^m Otherwise, you have to define a new ring with an additional variable, fetch/imap your input to this ring and proceed then. See: extendring from ring.lib and also the command homog If not enough, give an explicit example. Input: p = ... q = Output: And where comes m = ... from ?
m is the max degree of monomials p, q
i need, in true,create a function for desomoneization and homegeneization of polynomials.
[quote="gorzel"]If you define at the beginning already a ring in three variables (x,y,z), where you have two polynomial p and q in the variables x,y, then you can just set
[code] q*z^m + p*z^m [/code]
Otherwise, you have to define a new ring with an additional variable, fetch/imap your input to this ring and proceed then.
See: [b]extendring[/b] from ring.lib and also the command [b]homog[/b]
If not enough, give an explicit example. Input: p = ... q = Output:
And where comes m = ... from ?[/quote]
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Posted: Mon Apr 12, 2010 4:41 am |
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Re: homogeneous cordinates |
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If you define at the beginning already a ring in three variables (x,y,z), where you have two polynomial p and q in the variables x,y, then you can just set Code: q*z^m + p*z^m Otherwise, you have to define a new ring with an additional variable, fetch/imap your input to this ring and proceed then. See: extendring from ring.lib and also the command homog If not enough, give an explicit example. Input: p = ... q = Output: And where comes m = ... from ?
If you define at the beginning already a ring in three variables (x,y,z), where you have two polynomial p and q in the variables x,y, then you can just set
[code] q*z^m + p*z^m [/code]
Otherwise, you have to define a new ring with an additional variable, fetch/imap your input to this ring and proceed then.
See: [b]extendring[/b] from ring.lib and also the command [b]homog[/b]
If not enough, give an explicit example. Input: p = ... q = Output:
And where comes m = ... from ?
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Posted: Fri Apr 09, 2010 3:39 pm |
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Re: homogeneous cordinates |
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i have one field vector w= -q(x,y)dx +p(x,y)dy
that vector is in P²(polinomial ring with x,y)
then i transform P² in C³\{0} by one equivalence relation
then i have a space (x,y,z), but i need transform the ring such that the result coordinates are in the form (x/z:y/z:1), this is in homogeneous coordinates, because i consider the class of points that pass through (x,y,z) z=1 is the plane that transform C³ in C²,
i need one transformation of this type:
w'= z^(m+2) * w=q(x,y)*z^m +p(x,y)*z^m
i have one field vector w= -q(x,y)dx +p(x,y)dy
that vector is in P²(polinomial ring with x,y)
then i transform P² in C³\{0} by one equivalence relation
then i have a space (x,y,z), but i need transform the ring such that the result coordinates are in the form (x/z:y/z:1), this is in homogeneous coordinates, because i consider the class of points that pass through (x,y,z) z=1 is the plane that transform C³ in C²,
i need one transformation of this type:
w'= z^(m+2) * w=q(x,y)*z^m +p(x,y)*z^m
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Posted: Fri Apr 09, 2010 1:58 pm |
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Re: homogeneous cordinates |
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vpachecu2 wrote: I have a vector field in Cartesian coordinates.
I know also that the ring C ³ in homogeneous coordinates is P ² ie (X, Y, Z) can actually be (x / z, y / z, 1)
I want to get the vector field in these new coordinates then
Not clear what this should mean. Give an explicit mathematical example.
[quote="vpachecu2"] I have a vector field in Cartesian coordinates.
I know also that the ring C ³ in homogeneous coordinates is P ² ie (X, Y, Z) can actually be (x / z, y / z, 1)
I want to get the vector field in these new coordinates then [/quote]
Not clear what this should mean. Give an explicit mathematical example.
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Posted: Thu Apr 08, 2010 6:14 pm |
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homogeneous cordinates |
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Hello again friends I have a problem in algebraic geometry I have a vector field in Cartesian coordinates.
I know also that the ring C ³ in homogeneous coordinates is P ² ie (X, Y, Z) can actually be (x / z, y / z, 1)
I want to get the vector field in these new coordinates then
is there any lib who can help me?
proc transf (matrix a) ( int n = nrows (a); dual matrix [n] [n]; dual [1,1] = a [2,1] / / transform the field in the dual form dual [2,1] = -1 * a [1,1];
return (dual); )
transf (dual (a));
now need to do this function so
Hello again friends I have a problem in algebraic geometry I have a vector field in Cartesian coordinates.
I know also that the ring C ³ in homogeneous coordinates is P ² ie (X, Y, Z) can actually be (x / z, y / z, 1)
I want to get the vector field in these new coordinates then
is there any lib who can help me?
proc transf (matrix a) ( int n = nrows (a); dual matrix [n] [n]; dual [1,1] = a [2,1] / / transform the field in the dual form dual [2,1] = -1 * a [1,1];
return (dual); )
transf (dual (a));
now need to do this function so
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Posted: Mon Apr 05, 2010 4:44 pm |
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