“line at infinity” in projective plane
$begingroup$
("Algebraic Geometry: A Problem Solving Approach" by Thomas Garrity)
I am struggling with Exercise 1.4.12.1 in the above, which I quote with some context:
Here is my intuitive thinking:
(a) lines in $mathbb{C}^3$ with some elevation from the x-y plane (ie where $z neq 0$) are reduced to a single point, where that point is the intersection of the line with the plane $z =1$. This point is unique (which is why $phi$ in 1.4.9 is a bijection), and it is labelled $(x:y:1)$
(b) the lines in the x-y plane in 1.4.11/1.4.12 do not hit $z=1$. Those lines are of the form $( x,frac{-ax-c}{b},0 )$ , or alternatively $( bx,-ax-c,0 )$
But then :
I do not understand why under $phi$ these lines are assigned the point stated in 1.4.12.1
even so, why are such points necessarily distinct from those assigned to lines for the prior case, where $z neq 0$ ?
EDIT #1: I add the rest of the exercise (ie 1.14.12.3) to show the conclusion reached by the author. Personally, I feel able to reach that conclusion directly from my (b) above:
algebraic-geometry projective-geometry projective-space
asked Dec 9 '18 at 4:27
user3203476user3203476
727612
$endgroup$
add a comment |
$begingroup$
("Algebraic Geometry: A Problem Solving Approach" by Thomas Garrity)
I am struggling with Exercise 1.4.12.1 in the above, which I quote with some context:
Here is my intuitive thinking:
(a) lines in $mathbb{C}^3$ with some elevation from the x-y plane (ie where $z neq 0$) are reduced to a single point, where that point is the intersection of the line with the plane $z =1$. This point is unique (which is why $phi$ in 1.4.9 is a bijection), and it is labelled $(x:y:1)$
(b) the lines in the x-y plane in 1.4.11/1.4.12 do not hit $z=1$. Those lines are of the form $( x,frac{-ax-c}{b},0 )$ , or alternatively $( bx,-ax-c,0 )$
But then :
I do not understand why under $phi$ these lines are assigned the point stated in 1.4.12.1
even so, why are such points necessarily distinct from those assigned to lines for the prior case, where $z neq 0$ ?
EDIT #1: I add the rest of the exercise (ie 1.14.12.3) to show the conclusion reached by the author. Personally, I feel able to reach that conclusion directly from my (b) above:
algebraic-geometry projective-geometry projective-space
asked Dec 9 '18 at 4:27
user3203476user3203476
727612
$endgroup$
$begingroup$
Note that $(x:y:z)=(lambda x:lambda y:lambda z)$ if $lambdaneq 0$ (in your problem, take advantage of $bneq 0$)
$endgroup$
– Hamed
Dec 9 '18 at 4:48
$begingroup$
Personally, i think that the point assigned to (bx,−ax−c,0) should just be the one assigned to (1,(−ax−c)/(bx),0) , ie (1:(−ax−c)/(bx):0) , with (1:−a/b:0) the point assigned to the case x = infinity. Then i would understand why such points are distinct from those for lines with elevation, since the last component is (::0) rather than (::1)
$endgroup$
– user3203476
Dec 9 '18 at 5:03
$begingroup$
I'm not sure I follow.... The point is not $(bx:-ax-c: {color{red} 0})$, it is $(bx:-ax-c: {color{red} b})$, which is the same thing as $(x:-(ax+c)/b: {color{red} 1})$.
$endgroup$
– Hamed
Dec 9 '18 at 5:06
$begingroup$
I have added the conclusion reached by the author in an edit. Personally I feel able to skip 1.4.12.1/2 and reach 1.4.12.3 directly given my (b), although i am probably wrong in doing so :)
$endgroup$
– user3203476
Dec 9 '18 at 5:14
add a comment |
$begingroup$
("Algebraic Geometry: A Problem Solving Approach" by Thomas Garrity)
I am struggling with Exercise 1.4.12.1 in the above, which I quote with some context:
Here is my intuitive thinking:
(a) lines in $mathbb{C}^3$ with some elevation from the x-y plane (ie where $z neq 0$) are reduced to a single point, where that point is the intersection of the line with the plane $z =1$. This point is unique (which is why $phi$ in 1.4.9 is a bijection), and it is labelled $(x:y:1)$
(b) the lines in the x-y plane in 1.4.11/1.4.12 do not hit $z=1$. Those lines are of the form $( x,frac{-ax-c}{b},0 )$ , or alternatively $( bx,-ax-c,0 )$
But then :
I do not understand why under $phi$ these lines are assigned the point stated in 1.4.12.1
even so, why are such points necessarily distinct from those assigned to lines for the prior case, where $z neq 0$ ?
EDIT #1: I add the rest of the exercise (ie 1.14.12.3) to show the conclusion reached by the author. Personally, I feel able to reach that conclusion directly from my (b) above:
algebraic-geometry projective-geometry projective-space
asked Dec 9 '18 at 4:27
user3203476user3203476
727612
$endgroup$
("Algebraic Geometry: A Problem Solving Approach" by Thomas Garrity)
I am struggling with Exercise 1.4.12.1 in the above, which I quote with some context:
Here is my intuitive thinking:
(a) lines in $mathbb{C}^3$ with some elevation from the x-y plane (ie where $z neq 0$) are reduced to a single point, where that point is the intersection of the line with the plane $z =1$. This point is unique (which is why $phi$ in 1.4.9 is a bijection), and it is labelled $(x:y:1)$
(b) the lines in the x-y plane in 1.4.11/1.4.12 do not hit $z=1$. Those lines are of the form $( x,frac{-ax-c}{b},0 )$ , or alternatively $( bx,-ax-c,0 )$
But then :
I do not understand why under $phi$ these lines are assigned the point stated in 1.4.12.1
even so, why are such points necessarily distinct from those assigned to lines for the prior case, where $z neq 0$ ?
EDIT #1: I add the rest of the exercise (ie 1.14.12.3) to show the conclusion reached by the author. Personally, I feel able to reach that conclusion directly from my (b) above:
algebraic-geometry projective-geometry projective-space
algebraic-geometry projective-geometry projective-space
asked Dec 9 '18 at 4:27
user3203476user3203476
727612
asked Dec 9 '18 at 4:27
user3203476user3203476
727612
edited Dec 9 '18 at 5:13
user3203476
asked Dec 9 '18 at 4:27
user3203476user3203476
727612
asked Dec 9 '18 at 4:27
user3203476user3203476
727612
asked Dec 9 '18 at 4:27
user3203476user3203476
727612
727612
$begingroup$
Note that $(x:y:z)=(lambda x:lambda y:lambda z)$ if $lambdaneq 0$ (in your problem, take advantage of $bneq 0$)
$endgroup$
– Hamed
Dec 9 '18 at 4:48
$begingroup$
Personally, i think that the point assigned to (bx,−ax−c,0) should just be the one assigned to (1,(−ax−c)/(bx),0) , ie (1:(−ax−c)/(bx):0) , with (1:−a/b:0) the point assigned to the case x = infinity. Then i would understand why such points are distinct from those for lines with elevation, since the last component is (::0) rather than (::1)
$endgroup$
– user3203476
Dec 9 '18 at 5:03
$begingroup$
I'm not sure I follow.... The point is not $(bx:-ax-c: {color{red} 0})$, it is $(bx:-ax-c: {color{red} b})$, which is the same thing as $(x:-(ax+c)/b: {color{red} 1})$.
$endgroup$
– Hamed
Dec 9 '18 at 5:06
$begingroup$
I have added the conclusion reached by the author in an edit. Personally I feel able to skip 1.4.12.1/2 and reach 1.4.12.3 directly given my (b), although i am probably wrong in doing so :)
$endgroup$
– user3203476
Dec 9 '18 at 5:14
add a comment |
$begingroup$
Note that $(x:y:z)=(lambda x:lambda y:lambda z)$ if $lambdaneq 0$ (in your problem, take advantage of $bneq 0$)
$endgroup$
– Hamed
Dec 9 '18 at 4:48
$begingroup$
Personally, i think that the point assigned to (bx,−ax−c,0) should just be the one assigned to (1,(−ax−c)/(bx),0) , ie (1:(−ax−c)/(bx):0) , with (1:−a/b:0) the point assigned to the case x = infinity. Then i would understand why such points are distinct from those for lines with elevation, since the last component is (::0) rather than (::1)
$endgroup$
– user3203476
Dec 9 '18 at 5:03
$begingroup$
I'm not sure I follow.... The point is not $(bx:-ax-c: {color{red} 0})$, it is $(bx:-ax-c: {color{red} b})$, which is the same thing as $(x:-(ax+c)/b: {color{red} 1})$.
$endgroup$
– Hamed
Dec 9 '18 at 5:06
$begingroup$
I have added the conclusion reached by the author in an edit. Personally I feel able to skip 1.4.12.1/2 and reach 1.4.12.3 directly given my (b), although i am probably wrong in doing so :)
$endgroup$
– user3203476
Dec 9 '18 at 5:14
$begingroup$
Note that $(x:y:z)=(lambda x:lambda y:lambda z)$ if $lambdaneq 0$ (in your problem, take advantage of $bneq 0$)
$endgroup$
– Hamed
Dec 9 '18 at 4:48
$begingroup$
Note that $(x:y:z)=(lambda x:lambda y:lambda z)$ if $lambdaneq 0$ (in your problem, take advantage of $bneq 0$)
$endgroup$
– Hamed
Dec 9 '18 at 4:48
$begingroup$
Personally, i think that the point assigned to (bx,−ax−c,0) should just be the one assigned to (1,(−ax−c)/(bx),0) , ie (1:(−ax−c)/(bx):0) , with (1:−a/b:0) the point assigned to the case x = infinity. Then i would understand why such points are distinct from those for lines with elevation, since the last component is (::0) rather than (::1)
$endgroup$
– user3203476
Dec 9 '18 at 5:03
$begingroup$
Personally, i think that the point assigned to (bx,−ax−c,0) should just be the one assigned to (1,(−ax−c)/(bx),0) , ie (1:(−ax−c)/(bx):0) , with (1:−a/b:0) the point assigned to the case x = infinity. Then i would understand why such points are distinct from those for lines with elevation, since the last component is (::0) rather than (::1)
$endgroup$
– user3203476
Dec 9 '18 at 5:03
$begingroup$
I'm not sure I follow.... The point is not $(bx:-ax-c: {color{red} 0})$, it is $(bx:-ax-c: {color{red} b})$, which is the same thing as $(x:-(ax+c)/b: {color{red} 1})$.
$endgroup$
– Hamed
Dec 9 '18 at 5:06
$begingroup$
I'm not sure I follow.... The point is not $(bx:-ax-c: {color{red} 0})$, it is $(bx:-ax-c: {color{red} b})$, which is the same thing as $(x:-(ax+c)/b: {color{red} 1})$.
$endgroup$
– Hamed
Dec 9 '18 at 5:06
$begingroup$
I have added the conclusion reached by the author in an edit. Personally I feel able to skip 1.4.12.1/2 and reach 1.4.12.3 directly given my (b), although i am probably wrong in doing so :)
$endgroup$
– user3203476
Dec 9 '18 at 5:14
$begingroup$
I have added the conclusion reached by the author in an edit. Personally I feel able to skip 1.4.12.1/2 and reach 1.4.12.3 directly given my (b), although i am probably wrong in doing so :)
$endgroup$
– user3203476
Dec 9 '18 at 5:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The map $phi: mathbb{C}^2 rightarrow mathbb{P}^2 setminus {z = 0}$ sends $(x,y)$ to $[x:y:1]$.
Your line $ell$ is defined as $y = (-ax-c)/b$, and therefore
$$ phi((x, (-ax-c)/b)) = [x: (-ax-c)/b: 1] = [bx : -ax-c: b].$$
In your intuitive thinking, you should think on the affine plane as having $z$ coordinate equal to 1. In other words, in $mathbb{C}^3$, you are in the plane $z=1$!
An affine line in $mathbb{C}^2$ therefore corresponds to a plane in $mathbb{C}^3$: each point $(x_0, y_0)$ of the line in $mathbb{C}^2$ corresponds in $mathbb{C}^3$ to the line passing through the origin and the point $(x_0, y_0, 1)$, and putting together all the points in $mathbb{C}^2$ (resp. lines in $mathbb{C}^3$) you get a line (resp. a plane). So the affine line in $mathbb{C}^2$ not only intersects the plane $z=1$, but is cointaned in it!
answered Dec 10 '18 at 0:50
Pedro A. CastillejoPedro A. Castillejo
836415
$endgroup$
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%2f3032007%2fline-at-infinity-in-projective-plane%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
$begingroup$
The map $phi: mathbb{C}^2 rightarrow mathbb{P}^2 setminus {z = 0}$ sends $(x,y)$ to $[x:y:1]$.
Your line $ell$ is defined as $y = (-ax-c)/b$, and therefore
$$ phi((x, (-ax-c)/b)) = [x: (-ax-c)/b: 1] = [bx : -ax-c: b].$$
In your intuitive thinking, you should think on the affine plane as having $z$ coordinate equal to 1. In other words, in $mathbb{C}^3$, you are in the plane $z=1$!
An affine line in $mathbb{C}^2$ therefore corresponds to a plane in $mathbb{C}^3$: each point $(x_0, y_0)$ of the line in $mathbb{C}^2$ corresponds in $mathbb{C}^3$ to the line passing through the origin and the point $(x_0, y_0, 1)$, and putting together all the points in $mathbb{C}^2$ (resp. lines in $mathbb{C}^3$) you get a line (resp. a plane). So the affine line in $mathbb{C}^2$ not only intersects the plane $z=1$, but is cointaned in it!
answered Dec 10 '18 at 0:50
Pedro A. CastillejoPedro A. Castillejo
836415
$endgroup$
add a comment |
$begingroup$
The map $phi: mathbb{C}^2 rightarrow mathbb{P}^2 setminus {z = 0}$ sends $(x,y)$ to $[x:y:1]$.
Your line $ell$ is defined as $y = (-ax-c)/b$, and therefore
$$ phi((x, (-ax-c)/b)) = [x: (-ax-c)/b: 1] = [bx : -ax-c: b].$$
In your intuitive thinking, you should think on the affine plane as having $z$ coordinate equal to 1. In other words, in $mathbb{C}^3$, you are in the plane $z=1$!
An affine line in $mathbb{C}^2$ therefore corresponds to a plane in $mathbb{C}^3$: each point $(x_0, y_0)$ of the line in $mathbb{C}^2$ corresponds in $mathbb{C}^3$ to the line passing through the origin and the point $(x_0, y_0, 1)$, and putting together all the points in $mathbb{C}^2$ (resp. lines in $mathbb{C}^3$) you get a line (resp. a plane). So the affine line in $mathbb{C}^2$ not only intersects the plane $z=1$, but is cointaned in it!
answered Dec 10 '18 at 0:50
Pedro A. CastillejoPedro A. Castillejo
836415
$endgroup$
add a comment |
$begingroup$
The map $phi: mathbb{C}^2 rightarrow mathbb{P}^2 setminus {z = 0}$ sends $(x,y)$ to $[x:y:1]$.
Your line $ell$ is defined as $y = (-ax-c)/b$, and therefore
$$ phi((x, (-ax-c)/b)) = [x: (-ax-c)/b: 1] = [bx : -ax-c: b].$$
In your intuitive thinking, you should think on the affine plane as having $z$ coordinate equal to 1. In other words, in $mathbb{C}^3$, you are in the plane $z=1$!
An affine line in $mathbb{C}^2$ therefore corresponds to a plane in $mathbb{C}^3$: each point $(x_0, y_0)$ of the line in $mathbb{C}^2$ corresponds in $mathbb{C}^3$ to the line passing through the origin and the point $(x_0, y_0, 1)$, and putting together all the points in $mathbb{C}^2$ (resp. lines in $mathbb{C}^3$) you get a line (resp. a plane). So the affine line in $mathbb{C}^2$ not only intersects the plane $z=1$, but is cointaned in it!
answered Dec 10 '18 at 0:50
Pedro A. CastillejoPedro A. Castillejo
836415
$endgroup$
The map $phi: mathbb{C}^2 rightarrow mathbb{P}^2 setminus {z = 0}$ sends $(x,y)$ to $[x:y:1]$.
Your line $ell$ is defined as $y = (-ax-c)/b$, and therefore
$$ phi((x, (-ax-c)/b)) = [x: (-ax-c)/b: 1] = [bx : -ax-c: b].$$
In your intuitive thinking, you should think on the affine plane as having $z$ coordinate equal to 1. In other words, in $mathbb{C}^3$, you are in the plane $z=1$!
An affine line in $mathbb{C}^2$ therefore corresponds to a plane in $mathbb{C}^3$: each point $(x_0, y_0)$ of the line in $mathbb{C}^2$ corresponds in $mathbb{C}^3$ to the line passing through the origin and the point $(x_0, y_0, 1)$, and putting together all the points in $mathbb{C}^2$ (resp. lines in $mathbb{C}^3$) you get a line (resp. a plane). So the affine line in $mathbb{C}^2$ not only intersects the plane $z=1$, but is cointaned in it!
answered Dec 10 '18 at 0:50
Pedro A. CastillejoPedro A. Castillejo
836415
answered Dec 10 '18 at 0:50
Pedro A. CastillejoPedro A. Castillejo
836415
answered Dec 10 '18 at 0:50
Pedro A. CastillejoPedro A. Castillejo
836415
answered Dec 10 '18 at 0:50
Pedro A. CastillejoPedro A. Castillejo
836415
836415
add a comment |
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%2f3032007%2fline-at-infinity-in-projective-plane%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
$begingroup$
Note that $(x:y:z)=(lambda x:lambda y:lambda z)$ if $lambdaneq 0$ (in your problem, take advantage of $bneq 0$)
$endgroup$
– Hamed
Dec 9 '18 at 4:48
$begingroup$
Personally, i think that the point assigned to (bx,−ax−c,0) should just be the one assigned to (1,(−ax−c)/(bx),0) , ie (1:(−ax−c)/(bx):0) , with (1:−a/b:0) the point assigned to the case x = infinity. Then i would understand why such points are distinct from those for lines with elevation, since the last component is (::0) rather than (::1)
$endgroup$
– user3203476
Dec 9 '18 at 5:03
$begingroup$
I'm not sure I follow.... The point is not $(bx:-ax-c: {color{red} 0})$, it is $(bx:-ax-c: {color{red} b})$, which is the same thing as $(x:-(ax+c)/b: {color{red} 1})$.
$endgroup$
– Hamed
Dec 9 '18 at 5:06
$begingroup$
I have added the conclusion reached by the author in an edit. Personally I feel able to skip 1.4.12.1/2 and reach 1.4.12.3 directly given my (b), although i am probably wrong in doing so :)
$endgroup$
– user3203476
Dec 9 '18 at 5:14