Projective/ Finite Geometric Basics!












1












$begingroup$


I'm taking intro to coding theory and am having some trouble understanding the basics of Projective Geometry, since our text does not give it much discussion. Namely, if PG(r-1,q) is the set of all subspaces of V(r,q), then how to I denote the lines in PG(3,q)?



I understand that PG(2,q) has lines that can be "cut-out" by single linear equations such as x1=0 for homogeneous coordinates (x1:x2:x3) (this is how prof described it in class). In fact, the coefficients of such equations can be represented as the points in PG(2,q).



BUT for PG(3,q) one equation leaves us with a plane. So how do I go about making a line in PG(3,q)? Using two linear equations? This makes two planes, but how I know that they intersect and aren't parallel? How can I generate all of them? Ex: How to draw all lines in PG(3,2)?



Unfortunately, I can't find a simple online resource on this anywhere so a direct explanation would be very helpful. Especially if it used the above terms.



Thanks!
Mike






geometry projective-geometry projective-space finite-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    There's no such thing as "parallel" in projective space. A line should be a one-dimensional subspace, so scalar multiples of a given element.
    $endgroup$
    – Gerry Myerson
    Dec 4 '18 at 23:20
















1












$begingroup$


I'm taking intro to coding theory and am having some trouble understanding the basics of Projective Geometry, since our text does not give it much discussion. Namely, if PG(r-1,q) is the set of all subspaces of V(r,q), then how to I denote the lines in PG(3,q)?



I understand that PG(2,q) has lines that can be "cut-out" by single linear equations such as x1=0 for homogeneous coordinates (x1:x2:x3) (this is how prof described it in class). In fact, the coefficients of such equations can be represented as the points in PG(2,q).



BUT for PG(3,q) one equation leaves us with a plane. So how do I go about making a line in PG(3,q)? Using two linear equations? This makes two planes, but how I know that they intersect and aren't parallel? How can I generate all of them? Ex: How to draw all lines in PG(3,2)?



Unfortunately, I can't find a simple online resource on this anywhere so a direct explanation would be very helpful. Especially if it used the above terms.



Thanks!
Mike






geometry projective-geometry projective-space finite-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    There's no such thing as "parallel" in projective space. A line should be a one-dimensional subspace, so scalar multiples of a given element.
    $endgroup$
    – Gerry Myerson
    Dec 4 '18 at 23:20














1












1








1





$begingroup$


I'm taking intro to coding theory and am having some trouble understanding the basics of Projective Geometry, since our text does not give it much discussion. Namely, if PG(r-1,q) is the set of all subspaces of V(r,q), then how to I denote the lines in PG(3,q)?



I understand that PG(2,q) has lines that can be "cut-out" by single linear equations such as x1=0 for homogeneous coordinates (x1:x2:x3) (this is how prof described it in class). In fact, the coefficients of such equations can be represented as the points in PG(2,q).



BUT for PG(3,q) one equation leaves us with a plane. So how do I go about making a line in PG(3,q)? Using two linear equations? This makes two planes, but how I know that they intersect and aren't parallel? How can I generate all of them? Ex: How to draw all lines in PG(3,2)?



Unfortunately, I can't find a simple online resource on this anywhere so a direct explanation would be very helpful. Especially if it used the above terms.



Thanks!
Mike






geometry projective-geometry projective-space finite-geometry







share|cite|improve this question











$endgroup$




I'm taking intro to coding theory and am having some trouble understanding the basics of Projective Geometry, since our text does not give it much discussion. Namely, if PG(r-1,q) is the set of all subspaces of V(r,q), then how to I denote the lines in PG(3,q)?



I understand that PG(2,q) has lines that can be "cut-out" by single linear equations such as x1=0 for homogeneous coordinates (x1:x2:x3) (this is how prof described it in class). In fact, the coefficients of such equations can be represented as the points in PG(2,q).



BUT for PG(3,q) one equation leaves us with a plane. So how do I go about making a line in PG(3,q)? Using two linear equations? This makes two planes, but how I know that they intersect and aren't parallel? How can I generate all of them? Ex: How to draw all lines in PG(3,2)?



Unfortunately, I can't find a simple online resource on this anywhere so a direct explanation would be very helpful. Especially if it used the above terms.



Thanks!
Mike






geometry projective-geometry projective-space finite-geometry




geometry projective-geometry projective-space finite-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 23:34









Morgan Rodgers

9,77021440




9,77021440










asked Dec 4 '18 at 22:37









TransMITTransMIT

115




115












  • $begingroup$
    There's no such thing as "parallel" in projective space. A line should be a one-dimensional subspace, so scalar multiples of a given element.
    $endgroup$
    – Gerry Myerson
    Dec 4 '18 at 23:20


















  • $begingroup$
    There's no such thing as "parallel" in projective space. A line should be a one-dimensional subspace, so scalar multiples of a given element.
    $endgroup$
    – Gerry Myerson
    Dec 4 '18 at 23:20
















$begingroup$
There's no such thing as "parallel" in projective space. A line should be a one-dimensional subspace, so scalar multiples of a given element.
$endgroup$
– Gerry Myerson
Dec 4 '18 at 23:20




$begingroup$
There's no such thing as "parallel" in projective space. A line should be a one-dimensional subspace, so scalar multiples of a given element.
$endgroup$
– Gerry Myerson
Dec 4 '18 at 23:20










1 Answer
1






active

oldest

votes


















0












$begingroup$

A line of $mathrm{PG}(3,q)$ can be represented as a span of two points, or as the nullspace of a rank 2 $2times 4$ matrix. This is the same as using two linear equations; each gives a plane, and the planes of $mathrm{PG}(3,q)$ either coincide or intersect in a line. The planes will coincide precisely when the two equations are multiples of each other.



In $mathrm{PG}(r-1,q)$, you can still represent a line as a span of two points. If you want to use the nullspace of a matrix, you will need to use a full rank $(r-2)times r$ matrix (this represents a line as an intersection of $r-2$ hyperplanes).



If you want all the lines of $mathrm{PG}(3,q)$ (each represented exactly once), you need to consider a set of $2 times 4$ matrices having rank 2, no pair being row-equivalent. To do this, you can consider that matrices are row-equivalent if and only if they have the same reduced echelon form.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So would PG(3,2) have (2^4-1)*(2^4-2) / 6 = 30 lines? One for each permutation of two rows, divided by the groups with row-equivalent matrices?
    $endgroup$
    – TransMIT
    Dec 5 '18 at 1:20












  • $begingroup$
    You can't count the groups with row-equivalent matrices so easily. If you want the easiest way to count the number of lines of $mathrm{PG}(3,q)$, you should use the fact that there are $q^{3}+q^{2}+q+1$ points, each line has $q+1$ points, and each pair of points is on a unique line. This gives $(15cdot 14)/(3cdot 2) = 35$ lines for $mathrm{PG}(3,2)$.
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:39












  • $begingroup$
    (counting the number of reduced echelon matrices of a given rank over $mathbb{F}_{q}$ involves looking at Ferrers diagrams)
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:44













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1 Answer
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$begingroup$

A line of $mathrm{PG}(3,q)$ can be represented as a span of two points, or as the nullspace of a rank 2 $2times 4$ matrix. This is the same as using two linear equations; each gives a plane, and the planes of $mathrm{PG}(3,q)$ either coincide or intersect in a line. The planes will coincide precisely when the two equations are multiples of each other.



In $mathrm{PG}(r-1,q)$, you can still represent a line as a span of two points. If you want to use the nullspace of a matrix, you will need to use a full rank $(r-2)times r$ matrix (this represents a line as an intersection of $r-2$ hyperplanes).



If you want all the lines of $mathrm{PG}(3,q)$ (each represented exactly once), you need to consider a set of $2 times 4$ matrices having rank 2, no pair being row-equivalent. To do this, you can consider that matrices are row-equivalent if and only if they have the same reduced echelon form.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So would PG(3,2) have (2^4-1)*(2^4-2) / 6 = 30 lines? One for each permutation of two rows, divided by the groups with row-equivalent matrices?
    $endgroup$
    – TransMIT
    Dec 5 '18 at 1:20












  • $begingroup$
    You can't count the groups with row-equivalent matrices so easily. If you want the easiest way to count the number of lines of $mathrm{PG}(3,q)$, you should use the fact that there are $q^{3}+q^{2}+q+1$ points, each line has $q+1$ points, and each pair of points is on a unique line. This gives $(15cdot 14)/(3cdot 2) = 35$ lines for $mathrm{PG}(3,2)$.
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:39












  • $begingroup$
    (counting the number of reduced echelon matrices of a given rank over $mathbb{F}_{q}$ involves looking at Ferrers diagrams)
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:44


















0












$begingroup$

A line of $mathrm{PG}(3,q)$ can be represented as a span of two points, or as the nullspace of a rank 2 $2times 4$ matrix. This is the same as using two linear equations; each gives a plane, and the planes of $mathrm{PG}(3,q)$ either coincide or intersect in a line. The planes will coincide precisely when the two equations are multiples of each other.



In $mathrm{PG}(r-1,q)$, you can still represent a line as a span of two points. If you want to use the nullspace of a matrix, you will need to use a full rank $(r-2)times r$ matrix (this represents a line as an intersection of $r-2$ hyperplanes).



If you want all the lines of $mathrm{PG}(3,q)$ (each represented exactly once), you need to consider a set of $2 times 4$ matrices having rank 2, no pair being row-equivalent. To do this, you can consider that matrices are row-equivalent if and only if they have the same reduced echelon form.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So would PG(3,2) have (2^4-1)*(2^4-2) / 6 = 30 lines? One for each permutation of two rows, divided by the groups with row-equivalent matrices?
    $endgroup$
    – TransMIT
    Dec 5 '18 at 1:20












  • $begingroup$
    You can't count the groups with row-equivalent matrices so easily. If you want the easiest way to count the number of lines of $mathrm{PG}(3,q)$, you should use the fact that there are $q^{3}+q^{2}+q+1$ points, each line has $q+1$ points, and each pair of points is on a unique line. This gives $(15cdot 14)/(3cdot 2) = 35$ lines for $mathrm{PG}(3,2)$.
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:39












  • $begingroup$
    (counting the number of reduced echelon matrices of a given rank over $mathbb{F}_{q}$ involves looking at Ferrers diagrams)
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:44
















0












0








0





$begingroup$

A line of $mathrm{PG}(3,q)$ can be represented as a span of two points, or as the nullspace of a rank 2 $2times 4$ matrix. This is the same as using two linear equations; each gives a plane, and the planes of $mathrm{PG}(3,q)$ either coincide or intersect in a line. The planes will coincide precisely when the two equations are multiples of each other.



In $mathrm{PG}(r-1,q)$, you can still represent a line as a span of two points. If you want to use the nullspace of a matrix, you will need to use a full rank $(r-2)times r$ matrix (this represents a line as an intersection of $r-2$ hyperplanes).



If you want all the lines of $mathrm{PG}(3,q)$ (each represented exactly once), you need to consider a set of $2 times 4$ matrices having rank 2, no pair being row-equivalent. To do this, you can consider that matrices are row-equivalent if and only if they have the same reduced echelon form.






share|cite|improve this answer









$endgroup$



A line of $mathrm{PG}(3,q)$ can be represented as a span of two points, or as the nullspace of a rank 2 $2times 4$ matrix. This is the same as using two linear equations; each gives a plane, and the planes of $mathrm{PG}(3,q)$ either coincide or intersect in a line. The planes will coincide precisely when the two equations are multiples of each other.



In $mathrm{PG}(r-1,q)$, you can still represent a line as a span of two points. If you want to use the nullspace of a matrix, you will need to use a full rank $(r-2)times r$ matrix (this represents a line as an intersection of $r-2$ hyperplanes).



If you want all the lines of $mathrm{PG}(3,q)$ (each represented exactly once), you need to consider a set of $2 times 4$ matrices having rank 2, no pair being row-equivalent. To do this, you can consider that matrices are row-equivalent if and only if they have the same reduced echelon form.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 23:30









Morgan RodgersMorgan Rodgers

9,77021440




9,77021440












  • $begingroup$
    So would PG(3,2) have (2^4-1)*(2^4-2) / 6 = 30 lines? One for each permutation of two rows, divided by the groups with row-equivalent matrices?
    $endgroup$
    – TransMIT
    Dec 5 '18 at 1:20












  • $begingroup$
    You can't count the groups with row-equivalent matrices so easily. If you want the easiest way to count the number of lines of $mathrm{PG}(3,q)$, you should use the fact that there are $q^{3}+q^{2}+q+1$ points, each line has $q+1$ points, and each pair of points is on a unique line. This gives $(15cdot 14)/(3cdot 2) = 35$ lines for $mathrm{PG}(3,2)$.
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:39












  • $begingroup$
    (counting the number of reduced echelon matrices of a given rank over $mathbb{F}_{q}$ involves looking at Ferrers diagrams)
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:44




















  • $begingroup$
    So would PG(3,2) have (2^4-1)*(2^4-2) / 6 = 30 lines? One for each permutation of two rows, divided by the groups with row-equivalent matrices?
    $endgroup$
    – TransMIT
    Dec 5 '18 at 1:20












  • $begingroup$
    You can't count the groups with row-equivalent matrices so easily. If you want the easiest way to count the number of lines of $mathrm{PG}(3,q)$, you should use the fact that there are $q^{3}+q^{2}+q+1$ points, each line has $q+1$ points, and each pair of points is on a unique line. This gives $(15cdot 14)/(3cdot 2) = 35$ lines for $mathrm{PG}(3,2)$.
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:39












  • $begingroup$
    (counting the number of reduced echelon matrices of a given rank over $mathbb{F}_{q}$ involves looking at Ferrers diagrams)
    $endgroup$
    – Morgan Rodgers
    Dec 5 '18 at 6:44


















$begingroup$
So would PG(3,2) have (2^4-1)*(2^4-2) / 6 = 30 lines? One for each permutation of two rows, divided by the groups with row-equivalent matrices?
$endgroup$
– TransMIT
Dec 5 '18 at 1:20






$begingroup$
So would PG(3,2) have (2^4-1)*(2^4-2) / 6 = 30 lines? One for each permutation of two rows, divided by the groups with row-equivalent matrices?
$endgroup$
– TransMIT
Dec 5 '18 at 1:20














$begingroup$
You can't count the groups with row-equivalent matrices so easily. If you want the easiest way to count the number of lines of $mathrm{PG}(3,q)$, you should use the fact that there are $q^{3}+q^{2}+q+1$ points, each line has $q+1$ points, and each pair of points is on a unique line. This gives $(15cdot 14)/(3cdot 2) = 35$ lines for $mathrm{PG}(3,2)$.
$endgroup$
– Morgan Rodgers
Dec 5 '18 at 6:39






$begingroup$
You can't count the groups with row-equivalent matrices so easily. If you want the easiest way to count the number of lines of $mathrm{PG}(3,q)$, you should use the fact that there are $q^{3}+q^{2}+q+1$ points, each line has $q+1$ points, and each pair of points is on a unique line. This gives $(15cdot 14)/(3cdot 2) = 35$ lines for $mathrm{PG}(3,2)$.
$endgroup$
– Morgan Rodgers
Dec 5 '18 at 6:39














$begingroup$
(counting the number of reduced echelon matrices of a given rank over $mathbb{F}_{q}$ involves looking at Ferrers diagrams)
$endgroup$
– Morgan Rodgers
Dec 5 '18 at 6:44






$begingroup$
(counting the number of reduced echelon matrices of a given rank over $mathbb{F}_{q}$ involves looking at Ferrers diagrams)
$endgroup$
– Morgan Rodgers
Dec 5 '18 at 6:44




















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