Mixing
Cell complex structure of real projective plane
$begingroup$
A Cell Complex structure of $RP^2$, the real protective plane, is $e^0∪e^1∪e^2$. But I am unable to find this cell complex structure for $RP^2$ from iterative method: What should be the set $X^0$ and then how the set $X^1$ will look like ? Anybody please.
algebraic-topology
edited Jan 11 at 15:41
Paul Frost
10.4k3933
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
615210
$endgroup$
add a comment |
$begingroup$
A Cell Complex structure of $RP^2$, the real protective plane, is $e^0∪e^1∪e^2$. But I am unable to find this cell complex structure for $RP^2$ from iterative method: What should be the set $X^0$ and then how the set $X^1$ will look like ? Anybody please.
algebraic-topology
edited Jan 11 at 15:41
Paul Frost
10.4k3933
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
615210
$endgroup$
add a comment |
$begingroup$
A Cell Complex structure of $RP^2$, the real protective plane, is $e^0∪e^1∪e^2$. But I am unable to find this cell complex structure for $RP^2$ from iterative method: What should be the set $X^0$ and then how the set $X^1$ will look like ? Anybody please.
algebraic-topology
edited Jan 11 at 15:41
Paul Frost
10.4k3933
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
615210
$endgroup$
A Cell Complex structure of $RP^2$, the real protective plane, is $e^0∪e^1∪e^2$. But I am unable to find this cell complex structure for $RP^2$ from iterative method: What should be the set $X^0$ and then how the set $X^1$ will look like ? Anybody please.
algebraic-topology
algebraic-topology
edited Jan 11 at 15:41
Paul Frost
10.4k3933
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
615210
edited Jan 11 at 15:41
Paul Frost
10.4k3933
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
615210
edited Jan 11 at 15:41
Paul Frost
10.4k3933
edited Jan 11 at 15:41
Paul Frost
10.4k3933
edited Jan 11 at 15:41
Paul Frost
10.4k3933
10.4k3933
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
615210
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
615210
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
615210
615210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
10.4k3933
$endgroup$
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%2f3069800%2fcell-complex-structure-of-real-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
$begingroup$
$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
10.4k3933
$endgroup$
add a comment |
$begingroup$
$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
10.4k3933
$endgroup$
add a comment |
$begingroup$
$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
10.4k3933
$endgroup$
$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
10.4k3933
edited Jan 11 at 17:47
answered Jan 11 at 15:39
Paul FrostPaul Frost
10.4k3933
answered Jan 11 at 15:39
Paul FrostPaul Frost
10.4k3933
answered Jan 11 at 15:39
Paul FrostPaul Frost
10.4k3933
10.4k3933
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%2f3069800%2fcell-complex-structure-of-real-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
