Lusternik-Schnirelmann category of $mathbb{RP}^2 = 3$ (or $2$ depending on the definition) using only...
$begingroup$
I want to prove that $LS(mathbb{RP}^2)=3$, where $$LS(X) = min{lbrace k in mathbb{N} hspace{3pt} | hspace{3pt} text{there exists an open cover }U_{1},...,U_{k} text{ s.t. } i_{l}:U_{l} hookrightarrow X text{ is null-homotopic} rbrace}$$
It was easy enough to prove that $LS(mathbb{RP^2})leq 3$, by finding three orthogonal great circles on the sphere $S^2$, taking small belts around them $B_{1}$, $B_{2}$ and $B_{3}$, and setting $U_{i} = p(S^2 setminus B_{i})$, where $p:S^2 to mathbb{RP}^2$ is the standard covering map.
However, I can't find a proof for $LS(mathbb{RP^2}) geq 3$.
I start by supposing that there exist open $A$ and $B$ whose inclusions are null-homotopic. Then $(mathbb{RP}^2; A, B)$ is an excisive triad (https://en.wikipedia.org/wiki/Excisive_triad), so by excision we have that $j:(A, A cap B) hookrightarrow (mathbb{RP^2}, B)$ induces an isomorphism in the homology, so $H_{n}(A, A cap B) cong H_{n}(mathbb{RP^2}, B)$. Also, from the long exact sequence of $(mathbb{RP}^2, B)$ and null-homotopic-ness of $i_{l}$ we have that the following sequences are short and exact (for all $n in mathbb{N}$):
$$0 to H_{n}(mathbb{RP^2}) to H_{n}(mathbb{RP^2}, B) to H_{n-1}(B) to 0.$$
However, I can't connect these two facts, because I have no idea how to find $H_{n}(A, A cap B)$, $H_{n}(A)$, $H_{n}(B)$ or $H_{n}(B, A cap B)$.
Is it possible to derive a contradiction from here?
algebraic-topology homology-cohomology
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,019517
$endgroup$
add a comment |
$begingroup$
I want to prove that $LS(mathbb{RP}^2)=3$, where $$LS(X) = min{lbrace k in mathbb{N} hspace{3pt} | hspace{3pt} text{there exists an open cover }U_{1},...,U_{k} text{ s.t. } i_{l}:U_{l} hookrightarrow X text{ is null-homotopic} rbrace}$$
It was easy enough to prove that $LS(mathbb{RP^2})leq 3$, by finding three orthogonal great circles on the sphere $S^2$, taking small belts around them $B_{1}$, $B_{2}$ and $B_{3}$, and setting $U_{i} = p(S^2 setminus B_{i})$, where $p:S^2 to mathbb{RP}^2$ is the standard covering map.
However, I can't find a proof for $LS(mathbb{RP^2}) geq 3$.
I start by supposing that there exist open $A$ and $B$ whose inclusions are null-homotopic. Then $(mathbb{RP}^2; A, B)$ is an excisive triad (https://en.wikipedia.org/wiki/Excisive_triad), so by excision we have that $j:(A, A cap B) hookrightarrow (mathbb{RP^2}, B)$ induces an isomorphism in the homology, so $H_{n}(A, A cap B) cong H_{n}(mathbb{RP^2}, B)$. Also, from the long exact sequence of $(mathbb{RP}^2, B)$ and null-homotopic-ness of $i_{l}$ we have that the following sequences are short and exact (for all $n in mathbb{N}$):
$$0 to H_{n}(mathbb{RP^2}) to H_{n}(mathbb{RP^2}, B) to H_{n-1}(B) to 0.$$
However, I can't connect these two facts, because I have no idea how to find $H_{n}(A, A cap B)$, $H_{n}(A)$, $H_{n}(B)$ or $H_{n}(B, A cap B)$.
Is it possible to derive a contradiction from here?
algebraic-topology homology-cohomology
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,019517
$endgroup$
2
$begingroup$
You could use cohomology rather than homology, and the criteria that the cup-length is a lower bound for the LS category. If you know the mod-2 cohomology ring of $mathbb{R}P^3$ then you get the bound you are looking for.
$endgroup$
– Tyrone
Jan 8 at 18:34
$begingroup$
I saw a cohomological solution, but I'm taking an intro to algebraic topology course which only covers homology theory. I haven't learned anything about cohomology yet, so I'm looking for a homological solution.
$endgroup$
– Matija Sreckovic
Jan 8 at 19:18
add a comment |
$begingroup$
I want to prove that $LS(mathbb{RP}^2)=3$, where $$LS(X) = min{lbrace k in mathbb{N} hspace{3pt} | hspace{3pt} text{there exists an open cover }U_{1},...,U_{k} text{ s.t. } i_{l}:U_{l} hookrightarrow X text{ is null-homotopic} rbrace}$$
It was easy enough to prove that $LS(mathbb{RP^2})leq 3$, by finding three orthogonal great circles on the sphere $S^2$, taking small belts around them $B_{1}$, $B_{2}$ and $B_{3}$, and setting $U_{i} = p(S^2 setminus B_{i})$, where $p:S^2 to mathbb{RP}^2$ is the standard covering map.
However, I can't find a proof for $LS(mathbb{RP^2}) geq 3$.
I start by supposing that there exist open $A$ and $B$ whose inclusions are null-homotopic. Then $(mathbb{RP}^2; A, B)$ is an excisive triad (https://en.wikipedia.org/wiki/Excisive_triad), so by excision we have that $j:(A, A cap B) hookrightarrow (mathbb{RP^2}, B)$ induces an isomorphism in the homology, so $H_{n}(A, A cap B) cong H_{n}(mathbb{RP^2}, B)$. Also, from the long exact sequence of $(mathbb{RP}^2, B)$ and null-homotopic-ness of $i_{l}$ we have that the following sequences are short and exact (for all $n in mathbb{N}$):
$$0 to H_{n}(mathbb{RP^2}) to H_{n}(mathbb{RP^2}, B) to H_{n-1}(B) to 0.$$
However, I can't connect these two facts, because I have no idea how to find $H_{n}(A, A cap B)$, $H_{n}(A)$, $H_{n}(B)$ or $H_{n}(B, A cap B)$.
Is it possible to derive a contradiction from here?
algebraic-topology homology-cohomology
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,019517
$endgroup$
I want to prove that $LS(mathbb{RP}^2)=3$, where $$LS(X) = min{lbrace k in mathbb{N} hspace{3pt} | hspace{3pt} text{there exists an open cover }U_{1},...,U_{k} text{ s.t. } i_{l}:U_{l} hookrightarrow X text{ is null-homotopic} rbrace}$$
It was easy enough to prove that $LS(mathbb{RP^2})leq 3$, by finding three orthogonal great circles on the sphere $S^2$, taking small belts around them $B_{1}$, $B_{2}$ and $B_{3}$, and setting $U_{i} = p(S^2 setminus B_{i})$, where $p:S^2 to mathbb{RP}^2$ is the standard covering map.
However, I can't find a proof for $LS(mathbb{RP^2}) geq 3$.
I start by supposing that there exist open $A$ and $B$ whose inclusions are null-homotopic. Then $(mathbb{RP}^2; A, B)$ is an excisive triad (https://en.wikipedia.org/wiki/Excisive_triad), so by excision we have that $j:(A, A cap B) hookrightarrow (mathbb{RP^2}, B)$ induces an isomorphism in the homology, so $H_{n}(A, A cap B) cong H_{n}(mathbb{RP^2}, B)$. Also, from the long exact sequence of $(mathbb{RP}^2, B)$ and null-homotopic-ness of $i_{l}$ we have that the following sequences are short and exact (for all $n in mathbb{N}$):
$$0 to H_{n}(mathbb{RP^2}) to H_{n}(mathbb{RP^2}, B) to H_{n-1}(B) to 0.$$
However, I can't connect these two facts, because I have no idea how to find $H_{n}(A, A cap B)$, $H_{n}(A)$, $H_{n}(B)$ or $H_{n}(B, A cap B)$.
Is it possible to derive a contradiction from here?
algebraic-topology homology-cohomology
algebraic-topology homology-cohomology
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,019517
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,019517
edited Jan 8 at 19:21
Matija Sreckovic
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,019517
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,019517
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,019517
1,019517
2
$begingroup$
You could use cohomology rather than homology, and the criteria that the cup-length is a lower bound for the LS category. If you know the mod-2 cohomology ring of $mathbb{R}P^3$ then you get the bound you are looking for.
$endgroup$
– Tyrone
Jan 8 at 18:34
$begingroup$
I saw a cohomological solution, but I'm taking an intro to algebraic topology course which only covers homology theory. I haven't learned anything about cohomology yet, so I'm looking for a homological solution.
$endgroup$
– Matija Sreckovic
Jan 8 at 19:18
add a comment |
2
$begingroup$
You could use cohomology rather than homology, and the criteria that the cup-length is a lower bound for the LS category. If you know the mod-2 cohomology ring of $mathbb{R}P^3$ then you get the bound you are looking for.
$endgroup$
– Tyrone
Jan 8 at 18:34
$begingroup$
I saw a cohomological solution, but I'm taking an intro to algebraic topology course which only covers homology theory. I haven't learned anything about cohomology yet, so I'm looking for a homological solution.
$endgroup$
– Matija Sreckovic
Jan 8 at 19:18
2
2
$begingroup$
You could use cohomology rather than homology, and the criteria that the cup-length is a lower bound for the LS category. If you know the mod-2 cohomology ring of $mathbb{R}P^3$ then you get the bound you are looking for.
$endgroup$
– Tyrone
Jan 8 at 18:34
$begingroup$
You could use cohomology rather than homology, and the criteria that the cup-length is a lower bound for the LS category. If you know the mod-2 cohomology ring of $mathbb{R}P^3$ then you get the bound you are looking for.
$endgroup$
– Tyrone
Jan 8 at 18:34
$begingroup$
I saw a cohomological solution, but I'm taking an intro to algebraic topology course which only covers homology theory. I haven't learned anything about cohomology yet, so I'm looking for a homological solution.
$endgroup$
– Matija Sreckovic
Jan 8 at 19:18
$begingroup$
I saw a cohomological solution, but I'm taking an intro to algebraic topology course which only covers homology theory. I haven't learned anything about cohomology yet, so I'm looking for a homological solution.
$endgroup$
– Matija Sreckovic
Jan 8 at 19:18
add a comment |
0
active
oldest
votes
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%2f3066417%2flusternik-schnirelmann-category-of-mathbbrp2-3-or-2-depending-on-the%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3066417%2flusternik-schnirelmann-category-of-mathbbrp2-3-or-2-depending-on-the%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
2
$begingroup$
You could use cohomology rather than homology, and the criteria that the cup-length is a lower bound for the LS category. If you know the mod-2 cohomology ring of $mathbb{R}P^3$ then you get the bound you are looking for.
$endgroup$
– Tyrone
Jan 8 at 18:34
$begingroup$
I saw a cohomological solution, but I'm taking an intro to algebraic topology course which only covers homology theory. I haven't learned anything about cohomology yet, so I'm looking for a homological solution.
$endgroup$
– Matija Sreckovic
Jan 8 at 19:18