Lusternik-Schnirelmann category of $mathbb{RP}^2 = 3$ (or $2$ depending on the definition) using only...
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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
1,002517
$endgroup$
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
1,002517
$endgroup$
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
1,002517
$endgroup$
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
1,002517
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,002517
edited Jan 8 at 19:21
Matija Sreckovic
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,002517
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,002517
asked Jan 8 at 16:47
Matija SreckovicMatija Sreckovic
1,002517
1,002517
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 |
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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