Boundary Map in Mayer Vietoris and Homology of Knot Complement
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Let $K$ be a knot in $S^3$, and N(K) be its tubular neighborhood. I want to compute the homology of $S^3-N(K)$ using Mayer-Vietoris. Let $A$ be $N(K)$ minus a small neighborhood, and $B$ be the complement of N(K) with a small neighborhood added. Note that $Acup B$ is $S^3$, and $Acap B$ deformations retracts onto a torus. $H_1(S^3-N(K))$ and $H_n(S^3-N(K))$, $n>2$, pop out of Mayer Vietoris applied to A and B. However I am struggling with $H_2$. According to Mayer-Vietoris we have the following exact sequence $$H_3(S^3) to H_2(T) to H_2(A)bigoplus H_2(B) to 0$$
The part I don't understand is the map $H_3(S^3) to H_2(T)$. Let $alpha$ be the generator of $H_3(S^3)$. Under the boundary map of Mayer-Vietoris, $alpha$ is subdivided into a part residing in $A$ and a part residing in $B$. Then we take the boundary of the part residing in $A$. Thinking of the boundary map like this makes me think that it is an isomorphism. However, I am not sure of this.
Are there other ways to see what this map does to the generator of $H_3(S^3)$? More generally, how does one work with the boundary map in the MV sequence in higher dimensions, where things cannot be simply visualized?
algebraic-topology knot-theory
asked Dec 6 '18 at 5:00
user2948554user2948554
132
$endgroup$
add a comment |
$begingroup$
Let $K$ be a knot in $S^3$, and N(K) be its tubular neighborhood. I want to compute the homology of $S^3-N(K)$ using Mayer-Vietoris. Let $A$ be $N(K)$ minus a small neighborhood, and $B$ be the complement of N(K) with a small neighborhood added. Note that $Acup B$ is $S^3$, and $Acap B$ deformations retracts onto a torus. $H_1(S^3-N(K))$ and $H_n(S^3-N(K))$, $n>2$, pop out of Mayer Vietoris applied to A and B. However I am struggling with $H_2$. According to Mayer-Vietoris we have the following exact sequence $$H_3(S^3) to H_2(T) to H_2(A)bigoplus H_2(B) to 0$$
The part I don't understand is the map $H_3(S^3) to H_2(T)$. Let $alpha$ be the generator of $H_3(S^3)$. Under the boundary map of Mayer-Vietoris, $alpha$ is subdivided into a part residing in $A$ and a part residing in $B$. Then we take the boundary of the part residing in $A$. Thinking of the boundary map like this makes me think that it is an isomorphism. However, I am not sure of this.
Are there other ways to see what this map does to the generator of $H_3(S^3)$? More generally, how does one work with the boundary map in the MV sequence in higher dimensions, where things cannot be simply visualized?
algebraic-topology knot-theory
asked Dec 6 '18 at 5:00
user2948554user2948554
132
$endgroup$
1
$begingroup$
This is a great question here, and I hope you find Nick L's answer useful :)
$endgroup$
– Mike Miller
Dec 6 '18 at 13:27
add a comment |
$begingroup$
Let $K$ be a knot in $S^3$, and N(K) be its tubular neighborhood. I want to compute the homology of $S^3-N(K)$ using Mayer-Vietoris. Let $A$ be $N(K)$ minus a small neighborhood, and $B$ be the complement of N(K) with a small neighborhood added. Note that $Acup B$ is $S^3$, and $Acap B$ deformations retracts onto a torus. $H_1(S^3-N(K))$ and $H_n(S^3-N(K))$, $n>2$, pop out of Mayer Vietoris applied to A and B. However I am struggling with $H_2$. According to Mayer-Vietoris we have the following exact sequence $$H_3(S^3) to H_2(T) to H_2(A)bigoplus H_2(B) to 0$$
The part I don't understand is the map $H_3(S^3) to H_2(T)$. Let $alpha$ be the generator of $H_3(S^3)$. Under the boundary map of Mayer-Vietoris, $alpha$ is subdivided into a part residing in $A$ and a part residing in $B$. Then we take the boundary of the part residing in $A$. Thinking of the boundary map like this makes me think that it is an isomorphism. However, I am not sure of this.
Are there other ways to see what this map does to the generator of $H_3(S^3)$? More generally, how does one work with the boundary map in the MV sequence in higher dimensions, where things cannot be simply visualized?
algebraic-topology knot-theory
asked Dec 6 '18 at 5:00
user2948554user2948554
132
$endgroup$
Let $K$ be a knot in $S^3$, and N(K) be its tubular neighborhood. I want to compute the homology of $S^3-N(K)$ using Mayer-Vietoris. Let $A$ be $N(K)$ minus a small neighborhood, and $B$ be the complement of N(K) with a small neighborhood added. Note that $Acup B$ is $S^3$, and $Acap B$ deformations retracts onto a torus. $H_1(S^3-N(K))$ and $H_n(S^3-N(K))$, $n>2$, pop out of Mayer Vietoris applied to A and B. However I am struggling with $H_2$. According to Mayer-Vietoris we have the following exact sequence $$H_3(S^3) to H_2(T) to H_2(A)bigoplus H_2(B) to 0$$
The part I don't understand is the map $H_3(S^3) to H_2(T)$. Let $alpha$ be the generator of $H_3(S^3)$. Under the boundary map of Mayer-Vietoris, $alpha$ is subdivided into a part residing in $A$ and a part residing in $B$. Then we take the boundary of the part residing in $A$. Thinking of the boundary map like this makes me think that it is an isomorphism. However, I am not sure of this.
Are there other ways to see what this map does to the generator of $H_3(S^3)$? More generally, how does one work with the boundary map in the MV sequence in higher dimensions, where things cannot be simply visualized?
algebraic-topology knot-theory
algebraic-topology knot-theory
asked Dec 6 '18 at 5:00
user2948554user2948554
132
asked Dec 6 '18 at 5:00
user2948554user2948554
132
asked Dec 6 '18 at 5:00
user2948554user2948554
132
asked Dec 6 '18 at 5:00
user2948554user2948554
132
asked Dec 6 '18 at 5:00
user2948554user2948554
132
132
1
$begingroup$
This is a great question here, and I hope you find Nick L's answer useful :)
$endgroup$
– Mike Miller
Dec 6 '18 at 13:27
add a comment |
1
$begingroup$
This is a great question here, and I hope you find Nick L's answer useful :)
$endgroup$
– Mike Miller
Dec 6 '18 at 13:27
1
1
$begingroup$
This is a great question here, and I hope you find Nick L's answer useful :)
$endgroup$
– Mike Miller
Dec 6 '18 at 13:27
$begingroup$
This is a great question here, and I hope you find Nick L's answer useful :)
$endgroup$
– Mike Miller
Dec 6 '18 at 13:27
add a comment |
1 Answer
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You are correct that it is an isomorphism, by the argument you referred to. The Wikipedia page on MV-sequence has a pretty nice description of the boundary map, explaining why it is well-defined etc. I guess the most involved part is seeing why you can sub-divide cycles so that they lie entirely in $A$ or $B$, after this it is simple algebra to define the boundary map.
For the sake of visualising the map you can assume you have a nice triangulation of $S^{3}$ by tetrahedra such that the boundary torus (say $T = partial N(K)$) is contained in the $2$-skeleton. Then we can partition our set of tetrahedra into two subsets, depending on which connected component of $S^{3} setminus T$ the interior of each tetrahedron is contained in. This gives us our two cycles $U,V$, with
$U+V = [S^{3}]$.
$partial U = - partial V = [T]$.
Then, by definition, the boundary map maps $U+V in H_{3}(S^{3},mathbb{Z})$ to $partial U in H_{2}(T,mathbb{Z})$, hence it maps fundamental class to fundamental class and so is an isomorphism.
answered Dec 6 '18 at 12:59
Nick LNick L
1,269110
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$begingroup$
You are correct that it is an isomorphism, by the argument you referred to. The Wikipedia page on MV-sequence has a pretty nice description of the boundary map, explaining why it is well-defined etc. I guess the most involved part is seeing why you can sub-divide cycles so that they lie entirely in $A$ or $B$, after this it is simple algebra to define the boundary map.
For the sake of visualising the map you can assume you have a nice triangulation of $S^{3}$ by tetrahedra such that the boundary torus (say $T = partial N(K)$) is contained in the $2$-skeleton. Then we can partition our set of tetrahedra into two subsets, depending on which connected component of $S^{3} setminus T$ the interior of each tetrahedron is contained in. This gives us our two cycles $U,V$, with
$U+V = [S^{3}]$.
$partial U = - partial V = [T]$.
Then, by definition, the boundary map maps $U+V in H_{3}(S^{3},mathbb{Z})$ to $partial U in H_{2}(T,mathbb{Z})$, hence it maps fundamental class to fundamental class and so is an isomorphism.
answered Dec 6 '18 at 12:59
Nick LNick L
1,269110
$endgroup$
add a comment |
$begingroup$
You are correct that it is an isomorphism, by the argument you referred to. The Wikipedia page on MV-sequence has a pretty nice description of the boundary map, explaining why it is well-defined etc. I guess the most involved part is seeing why you can sub-divide cycles so that they lie entirely in $A$ or $B$, after this it is simple algebra to define the boundary map.
For the sake of visualising the map you can assume you have a nice triangulation of $S^{3}$ by tetrahedra such that the boundary torus (say $T = partial N(K)$) is contained in the $2$-skeleton. Then we can partition our set of tetrahedra into two subsets, depending on which connected component of $S^{3} setminus T$ the interior of each tetrahedron is contained in. This gives us our two cycles $U,V$, with
$U+V = [S^{3}]$.
$partial U = - partial V = [T]$.
Then, by definition, the boundary map maps $U+V in H_{3}(S^{3},mathbb{Z})$ to $partial U in H_{2}(T,mathbb{Z})$, hence it maps fundamental class to fundamental class and so is an isomorphism.
answered Dec 6 '18 at 12:59
Nick LNick L
1,269110
$endgroup$
add a comment |
$begingroup$
You are correct that it is an isomorphism, by the argument you referred to. The Wikipedia page on MV-sequence has a pretty nice description of the boundary map, explaining why it is well-defined etc. I guess the most involved part is seeing why you can sub-divide cycles so that they lie entirely in $A$ or $B$, after this it is simple algebra to define the boundary map.
For the sake of visualising the map you can assume you have a nice triangulation of $S^{3}$ by tetrahedra such that the boundary torus (say $T = partial N(K)$) is contained in the $2$-skeleton. Then we can partition our set of tetrahedra into two subsets, depending on which connected component of $S^{3} setminus T$ the interior of each tetrahedron is contained in. This gives us our two cycles $U,V$, with
$U+V = [S^{3}]$.
$partial U = - partial V = [T]$.
Then, by definition, the boundary map maps $U+V in H_{3}(S^{3},mathbb{Z})$ to $partial U in H_{2}(T,mathbb{Z})$, hence it maps fundamental class to fundamental class and so is an isomorphism.
answered Dec 6 '18 at 12:59
Nick LNick L
1,269110
$endgroup$
You are correct that it is an isomorphism, by the argument you referred to. The Wikipedia page on MV-sequence has a pretty nice description of the boundary map, explaining why it is well-defined etc. I guess the most involved part is seeing why you can sub-divide cycles so that they lie entirely in $A$ or $B$, after this it is simple algebra to define the boundary map.
For the sake of visualising the map you can assume you have a nice triangulation of $S^{3}$ by tetrahedra such that the boundary torus (say $T = partial N(K)$) is contained in the $2$-skeleton. Then we can partition our set of tetrahedra into two subsets, depending on which connected component of $S^{3} setminus T$ the interior of each tetrahedron is contained in. This gives us our two cycles $U,V$, with
$U+V = [S^{3}]$.
$partial U = - partial V = [T]$.
Then, by definition, the boundary map maps $U+V in H_{3}(S^{3},mathbb{Z})$ to $partial U in H_{2}(T,mathbb{Z})$, hence it maps fundamental class to fundamental class and so is an isomorphism.
answered Dec 6 '18 at 12:59
Nick LNick L
1,269110
edited Dec 6 '18 at 13:39
answered Dec 6 '18 at 12:59
Nick LNick L
1,269110
answered Dec 6 '18 at 12:59
Nick LNick L
1,269110
answered Dec 6 '18 at 12:59
Nick LNick L
1,269110
1,269110
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$begingroup$
This is a great question here, and I hope you find Nick L's answer useful :)
$endgroup$
– Mike Miller
Dec 6 '18 at 13:27