Why isn't $P^2$ the real projective plane homeomorphic to $mathbb{R}^2$
up vote
1
down vote
favorite
The real projective plane $P^2$ is the set of linear subspaces in $mathbb{R}^3$, with the quotient topology. So this means a set of linear subspaces $U in P^2$ is open if and only if the points on the union of points in $U$ (regarded as points in $mathbb{R}^3$)
So if I define the following map:
$f: P^2 rightarrow mathbb{R}^2 times 1$ (Where $mathbb{R}^2 times 1$ is the subset of $mathbb{R}^3$ with z coordinate = $1$.
by $f([x] = x$ where $x$ is the point in the linear subspace $[x]$ with $z$ coordinate as $1$.
This map is well defined, and is continuous because for any open set $V$ in the subspace topology of $mathbb{R}^2$, $f^{-1}(V)$ is the set of linear subspace that defined by a point in $V$ and the origin. and $f^{-1}(V)$ is open if the union of those these points in the union of the linear subspace is open in $R^3$. Which shouldn't be too hard to show by showing this on the basis for the subspace topology of $mathbb{R}^2$.
Here I already have the fraudulent result that says there's a surjective continuous function from $P^2$ to $mathbb{R}^2$. And I have no idea what I am doing wrong here.
Can someone point out why the reasoning above is false?
general-topology manifolds
asked Nov 23 at 6:23
Ecotistician
30818
add a comment |
up vote
1
down vote
favorite
The real projective plane $P^2$ is the set of linear subspaces in $mathbb{R}^3$, with the quotient topology. So this means a set of linear subspaces $U in P^2$ is open if and only if the points on the union of points in $U$ (regarded as points in $mathbb{R}^3$)
So if I define the following map:
$f: P^2 rightarrow mathbb{R}^2 times 1$ (Where $mathbb{R}^2 times 1$ is the subset of $mathbb{R}^3$ with z coordinate = $1$.
by $f([x] = x$ where $x$ is the point in the linear subspace $[x]$ with $z$ coordinate as $1$.
This map is well defined, and is continuous because for any open set $V$ in the subspace topology of $mathbb{R}^2$, $f^{-1}(V)$ is the set of linear subspace that defined by a point in $V$ and the origin. and $f^{-1}(V)$ is open if the union of those these points in the union of the linear subspace is open in $R^3$. Which shouldn't be too hard to show by showing this on the basis for the subspace topology of $mathbb{R}^2$.
Here I already have the fraudulent result that says there's a surjective continuous function from $P^2$ to $mathbb{R}^2$. And I have no idea what I am doing wrong here.
Can someone point out why the reasoning above is false?
general-topology manifolds
asked Nov 23 at 6:23
Ecotistician
30818
4
A linear subspace may have no point with $z$-coordinate 1.
– Gerry Myerson
Nov 23 at 6:26
oh shoot...you are right. Then what can one do to show that P^n is a n manifold in general. I had a "proof" that P^n is locally euclidean using this map. So I guess my approach is completely wrong then...
– Ecotistician
Nov 23 at 6:29
1
Some interesting topological ways in which the projective plane differs from the other plane: it is nonorientable; you can embed $K_5$ and $K_{3,3}$ in it without edge-crossings.
– Gerry Myerson
Nov 23 at 6:29
2
So to show that it is an n-manifold, can we still use a similar "projecting map" by showing that for every linear subspace in $[x] in P^2$, there exists a point $x in [x]$ s.t. some coordinate of $x$ is equals to 1, and it should be easy to show that this is true for some "neighborhood" of $[x]$, and then define a similar map above only on such a neighborhood of $[x]$?
– Ecotistician
Nov 23 at 6:36
Probably a better idea to post a new question about showing it's a manifold (first checking to see whether such a question has already been asked and answered here).
– Gerry Myerson
Nov 23 at 6:39
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The real projective plane $P^2$ is the set of linear subspaces in $mathbb{R}^3$, with the quotient topology. So this means a set of linear subspaces $U in P^2$ is open if and only if the points on the union of points in $U$ (regarded as points in $mathbb{R}^3$)
So if I define the following map:
$f: P^2 rightarrow mathbb{R}^2 times 1$ (Where $mathbb{R}^2 times 1$ is the subset of $mathbb{R}^3$ with z coordinate = $1$.
by $f([x] = x$ where $x$ is the point in the linear subspace $[x]$ with $z$ coordinate as $1$.
This map is well defined, and is continuous because for any open set $V$ in the subspace topology of $mathbb{R}^2$, $f^{-1}(V)$ is the set of linear subspace that defined by a point in $V$ and the origin. and $f^{-1}(V)$ is open if the union of those these points in the union of the linear subspace is open in $R^3$. Which shouldn't be too hard to show by showing this on the basis for the subspace topology of $mathbb{R}^2$.
Here I already have the fraudulent result that says there's a surjective continuous function from $P^2$ to $mathbb{R}^2$. And I have no idea what I am doing wrong here.
Can someone point out why the reasoning above is false?
general-topology manifolds
asked Nov 23 at 6:23
Ecotistician
30818
The real projective plane $P^2$ is the set of linear subspaces in $mathbb{R}^3$, with the quotient topology. So this means a set of linear subspaces $U in P^2$ is open if and only if the points on the union of points in $U$ (regarded as points in $mathbb{R}^3$)
So if I define the following map:
$f: P^2 rightarrow mathbb{R}^2 times 1$ (Where $mathbb{R}^2 times 1$ is the subset of $mathbb{R}^3$ with z coordinate = $1$.
by $f([x] = x$ where $x$ is the point in the linear subspace $[x]$ with $z$ coordinate as $1$.
This map is well defined, and is continuous because for any open set $V$ in the subspace topology of $mathbb{R}^2$, $f^{-1}(V)$ is the set of linear subspace that defined by a point in $V$ and the origin. and $f^{-1}(V)$ is open if the union of those these points in the union of the linear subspace is open in $R^3$. Which shouldn't be too hard to show by showing this on the basis for the subspace topology of $mathbb{R}^2$.
Here I already have the fraudulent result that says there's a surjective continuous function from $P^2$ to $mathbb{R}^2$. And I have no idea what I am doing wrong here.
Can someone point out why the reasoning above is false?
general-topology manifolds
general-topology manifolds
asked Nov 23 at 6:23
Ecotistician
30818
asked Nov 23 at 6:23
Ecotistician
30818
asked Nov 23 at 6:23
Ecotistician
30818
asked Nov 23 at 6:23
Ecotistician
30818
asked Nov 23 at 6:23
Ecotistician
30818
30818
4
A linear subspace may have no point with $z$-coordinate 1.
– Gerry Myerson
Nov 23 at 6:26
oh shoot...you are right. Then what can one do to show that P^n is a n manifold in general. I had a "proof" that P^n is locally euclidean using this map. So I guess my approach is completely wrong then...
– Ecotistician
Nov 23 at 6:29
1
Some interesting topological ways in which the projective plane differs from the other plane: it is nonorientable; you can embed $K_5$ and $K_{3,3}$ in it without edge-crossings.
– Gerry Myerson
Nov 23 at 6:29
2
So to show that it is an n-manifold, can we still use a similar "projecting map" by showing that for every linear subspace in $[x] in P^2$, there exists a point $x in [x]$ s.t. some coordinate of $x$ is equals to 1, and it should be easy to show that this is true for some "neighborhood" of $[x]$, and then define a similar map above only on such a neighborhood of $[x]$?
– Ecotistician
Nov 23 at 6:36
Probably a better idea to post a new question about showing it's a manifold (first checking to see whether such a question has already been asked and answered here).
– Gerry Myerson
Nov 23 at 6:39
add a comment |
4
A linear subspace may have no point with $z$-coordinate 1.
– Gerry Myerson
Nov 23 at 6:26
oh shoot...you are right. Then what can one do to show that P^n is a n manifold in general. I had a "proof" that P^n is locally euclidean using this map. So I guess my approach is completely wrong then...
– Ecotistician
Nov 23 at 6:29
1
Some interesting topological ways in which the projective plane differs from the other plane: it is nonorientable; you can embed $K_5$ and $K_{3,3}$ in it without edge-crossings.
– Gerry Myerson
Nov 23 at 6:29
2
So to show that it is an n-manifold, can we still use a similar "projecting map" by showing that for every linear subspace in $[x] in P^2$, there exists a point $x in [x]$ s.t. some coordinate of $x$ is equals to 1, and it should be easy to show that this is true for some "neighborhood" of $[x]$, and then define a similar map above only on such a neighborhood of $[x]$?
– Ecotistician
Nov 23 at 6:36
Probably a better idea to post a new question about showing it's a manifold (first checking to see whether such a question has already been asked and answered here).
– Gerry Myerson
Nov 23 at 6:39
4
4
A linear subspace may have no point with $z$-coordinate 1.
– Gerry Myerson
Nov 23 at 6:26
A linear subspace may have no point with $z$-coordinate 1.
– Gerry Myerson
Nov 23 at 6:26
oh shoot...you are right. Then what can one do to show that P^n is a n manifold in general. I had a "proof" that P^n is locally euclidean using this map. So I guess my approach is completely wrong then...
– Ecotistician
Nov 23 at 6:29
oh shoot...you are right. Then what can one do to show that P^n is a n manifold in general. I had a "proof" that P^n is locally euclidean using this map. So I guess my approach is completely wrong then...
– Ecotistician
Nov 23 at 6:29
1
1
Some interesting topological ways in which the projective plane differs from the other plane: it is nonorientable; you can embed $K_5$ and $K_{3,3}$ in it without edge-crossings.
– Gerry Myerson
Nov 23 at 6:29
Some interesting topological ways in which the projective plane differs from the other plane: it is nonorientable; you can embed $K_5$ and $K_{3,3}$ in it without edge-crossings.
– Gerry Myerson
Nov 23 at 6:29
2
2
So to show that it is an n-manifold, can we still use a similar "projecting map" by showing that for every linear subspace in $[x] in P^2$, there exists a point $x in [x]$ s.t. some coordinate of $x$ is equals to 1, and it should be easy to show that this is true for some "neighborhood" of $[x]$, and then define a similar map above only on such a neighborhood of $[x]$?
– Ecotistician
Nov 23 at 6:36
So to show that it is an n-manifold, can we still use a similar "projecting map" by showing that for every linear subspace in $[x] in P^2$, there exists a point $x in [x]$ s.t. some coordinate of $x$ is equals to 1, and it should be easy to show that this is true for some "neighborhood" of $[x]$, and then define a similar map above only on such a neighborhood of $[x]$?
– Ecotistician
Nov 23 at 6:36
Probably a better idea to post a new question about showing it's a manifold (first checking to see whether such a question has already been asked and answered here).
– Gerry Myerson
Nov 23 at 6:39
Probably a better idea to post a new question about showing it's a manifold (first checking to see whether such a question has already been asked and answered here).
– Gerry Myerson
Nov 23 at 6:39
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Your $f$ function is a projection onto $mathbb{R}^2$. However (as noted in comments) the map is not defined for points without $z=1$ coordinate. In the real projetive plane these are points with the third coordinate $0$, i.e. $[x,y,0]$. And there is no way to fix that in order to obtain a homeomorphism as I will explain soon.
What you've actually describing is very similar to the sterographic projection. Similarly the sterographic projection maps homeomorphically a sphere without a point to $mathbb{R}^n$. But a sphere itself is not homeomoprhic to $mathbb{R}^n$.
This also is true for $P^n$ simply because $P^n$ can be defined as a quotient:
$$q:S^nto P^n$$
under the antipodal points identification. In particular this implies that $P^n$ is compact (as an image of a sphere) and so it cannot be homeomorphic to an Euclidean space. And indeed, by the same argument $mathbb{R}^m$ can't even be an image of $P^n$.
answered Nov 23 at 15:01
freakish
11k1527
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',
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%2f3010035%2fwhy-isnt-p2-the-real-projective-plane-homeomorphic-to-mathbbr2%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
up vote
1
down vote
Your $f$ function is a projection onto $mathbb{R}^2$. However (as noted in comments) the map is not defined for points without $z=1$ coordinate. In the real projetive plane these are points with the third coordinate $0$, i.e. $[x,y,0]$. And there is no way to fix that in order to obtain a homeomorphism as I will explain soon.
What you've actually describing is very similar to the sterographic projection. Similarly the sterographic projection maps homeomorphically a sphere without a point to $mathbb{R}^n$. But a sphere itself is not homeomoprhic to $mathbb{R}^n$.
This also is true for $P^n$ simply because $P^n$ can be defined as a quotient:
$$q:S^nto P^n$$
under the antipodal points identification. In particular this implies that $P^n$ is compact (as an image of a sphere) and so it cannot be homeomorphic to an Euclidean space. And indeed, by the same argument $mathbb{R}^m$ can't even be an image of $P^n$.
answered Nov 23 at 15:01
freakish
11k1527
add a comment |
up vote
1
down vote
Your $f$ function is a projection onto $mathbb{R}^2$. However (as noted in comments) the map is not defined for points without $z=1$ coordinate. In the real projetive plane these are points with the third coordinate $0$, i.e. $[x,y,0]$. And there is no way to fix that in order to obtain a homeomorphism as I will explain soon.
What you've actually describing is very similar to the sterographic projection. Similarly the sterographic projection maps homeomorphically a sphere without a point to $mathbb{R}^n$. But a sphere itself is not homeomoprhic to $mathbb{R}^n$.
This also is true for $P^n$ simply because $P^n$ can be defined as a quotient:
$$q:S^nto P^n$$
under the antipodal points identification. In particular this implies that $P^n$ is compact (as an image of a sphere) and so it cannot be homeomorphic to an Euclidean space. And indeed, by the same argument $mathbb{R}^m$ can't even be an image of $P^n$.
answered Nov 23 at 15:01
freakish
11k1527
add a comment |
up vote
1
down vote
up vote
1
down vote
Your $f$ function is a projection onto $mathbb{R}^2$. However (as noted in comments) the map is not defined for points without $z=1$ coordinate. In the real projetive plane these are points with the third coordinate $0$, i.e. $[x,y,0]$. And there is no way to fix that in order to obtain a homeomorphism as I will explain soon.
What you've actually describing is very similar to the sterographic projection. Similarly the sterographic projection maps homeomorphically a sphere without a point to $mathbb{R}^n$. But a sphere itself is not homeomoprhic to $mathbb{R}^n$.
This also is true for $P^n$ simply because $P^n$ can be defined as a quotient:
$$q:S^nto P^n$$
under the antipodal points identification. In particular this implies that $P^n$ is compact (as an image of a sphere) and so it cannot be homeomorphic to an Euclidean space. And indeed, by the same argument $mathbb{R}^m$ can't even be an image of $P^n$.
answered Nov 23 at 15:01
freakish
11k1527
Your $f$ function is a projection onto $mathbb{R}^2$. However (as noted in comments) the map is not defined for points without $z=1$ coordinate. In the real projetive plane these are points with the third coordinate $0$, i.e. $[x,y,0]$. And there is no way to fix that in order to obtain a homeomorphism as I will explain soon.
What you've actually describing is very similar to the sterographic projection. Similarly the sterographic projection maps homeomorphically a sphere without a point to $mathbb{R}^n$. But a sphere itself is not homeomoprhic to $mathbb{R}^n$.
This also is true for $P^n$ simply because $P^n$ can be defined as a quotient:
$$q:S^nto P^n$$
under the antipodal points identification. In particular this implies that $P^n$ is compact (as an image of a sphere) and so it cannot be homeomorphic to an Euclidean space. And indeed, by the same argument $mathbb{R}^m$ can't even be an image of $P^n$.
answered Nov 23 at 15:01
freakish
11k1527
edited Nov 23 at 15:06
answered Nov 23 at 15:01
freakish
11k1527
answered Nov 23 at 15:01
freakish
11k1527
answered Nov 23 at 15:01
freakish
11k1527
11k1527
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3010035%2fwhy-isnt-p2-the-real-projective-plane-homeomorphic-to-mathbbr2%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
4
A linear subspace may have no point with $z$-coordinate 1.
– Gerry Myerson
Nov 23 at 6:26
oh shoot...you are right. Then what can one do to show that P^n is a n manifold in general. I had a "proof" that P^n is locally euclidean using this map. So I guess my approach is completely wrong then...
– Ecotistician
Nov 23 at 6:29
1
Some interesting topological ways in which the projective plane differs from the other plane: it is nonorientable; you can embed $K_5$ and $K_{3,3}$ in it without edge-crossings.
– Gerry Myerson
Nov 23 at 6:29
2
So to show that it is an n-manifold, can we still use a similar "projecting map" by showing that for every linear subspace in $[x] in P^2$, there exists a point $x in [x]$ s.t. some coordinate of $x$ is equals to 1, and it should be easy to show that this is true for some "neighborhood" of $[x]$, and then define a similar map above only on such a neighborhood of $[x]$?
– Ecotistician
Nov 23 at 6:36
Probably a better idea to post a new question about showing it's a manifold (first checking to see whether such a question has already been asked and answered here).
– Gerry Myerson
Nov 23 at 6:39