Projective/ Finite Geometric Basics!
$begingroup$
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
edited Dec 4 '18 at 23:34
Morgan Rodgers
9,77021440
asked Dec 4 '18 at 22:37
TransMITTransMIT
115
$endgroup$
add a comment |
$begingroup$
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
edited Dec 4 '18 at 23:34
Morgan Rodgers
9,77021440
asked Dec 4 '18 at 22:37
TransMITTransMIT
115
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 4 '18 at 23:34
Morgan Rodgers
9,77021440
asked Dec 4 '18 at 22:37
TransMITTransMIT
115
$endgroup$
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
edited Dec 4 '18 at 23:34
Morgan Rodgers
9,77021440
asked Dec 4 '18 at 22:37
TransMITTransMIT
115
edited Dec 4 '18 at 23:34
Morgan Rodgers
9,77021440
asked Dec 4 '18 at 22:37
TransMITTransMIT
115
edited Dec 4 '18 at 23:34
Morgan Rodgers
9,77021440
edited Dec 4 '18 at 23:34
Morgan Rodgers
9,77021440
edited Dec 4 '18 at 23:34
Morgan Rodgers
9,77021440
9,77021440
asked Dec 4 '18 at 22:37
TransMITTransMIT
115
asked Dec 4 '18 at 22:37
TransMITTransMIT
115
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 4 '18 at 23:30
Morgan RodgersMorgan Rodgers
9,77021440
$endgroup$
$begingroup$
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026277%2fprojective-finite-geometric-basics%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 4 '18 at 23:30
Morgan RodgersMorgan Rodgers
9,77021440
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 23:30
Morgan RodgersMorgan Rodgers
9,77021440
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 23:30
Morgan RodgersMorgan Rodgers
9,77021440
$endgroup$
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.
answered Dec 4 '18 at 23:30
Morgan RodgersMorgan Rodgers
9,77021440
answered Dec 4 '18 at 23:30
Morgan RodgersMorgan Rodgers
9,77021440
answered Dec 4 '18 at 23:30
Morgan RodgersMorgan Rodgers
9,77021440
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
add a comment |
$begingroup$
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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026277%2fprojective-finite-geometric-basics%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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