Connected Sum Surgery
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Is there any relationship between the connected sum operation and surgery theory? Is it possible to use surgery theory to "sew" two manifolds together and if so how is doing it by that approach different from the connected sum?
algebraic-topology manifolds surgery-theory
edited Dec 3 '18 at 23:56
Michael Albanese
63.5k1599305
asked Dec 3 '18 at 20:53
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add a comment |
$begingroup$
Is there any relationship between the connected sum operation and surgery theory? Is it possible to use surgery theory to "sew" two manifolds together and if so how is doing it by that approach different from the connected sum?
algebraic-topology manifolds surgery-theory
edited Dec 3 '18 at 23:56
Michael Albanese
63.5k1599305
asked Dec 3 '18 at 20:53
Canonical MomentaCanonical Momenta
283
$endgroup$
add a comment |
$begingroup$
Is there any relationship between the connected sum operation and surgery theory? Is it possible to use surgery theory to "sew" two manifolds together and if so how is doing it by that approach different from the connected sum?
algebraic-topology manifolds surgery-theory
edited Dec 3 '18 at 23:56
Michael Albanese
63.5k1599305
asked Dec 3 '18 at 20:53
Canonical MomentaCanonical Momenta
283
$endgroup$
Is there any relationship between the connected sum operation and surgery theory? Is it possible to use surgery theory to "sew" two manifolds together and if so how is doing it by that approach different from the connected sum?
algebraic-topology manifolds surgery-theory
algebraic-topology manifolds surgery-theory
edited Dec 3 '18 at 23:56
Michael Albanese
63.5k1599305
asked Dec 3 '18 at 20:53
Canonical MomentaCanonical Momenta
283
edited Dec 3 '18 at 23:56
Michael Albanese
63.5k1599305
asked Dec 3 '18 at 20:53
Canonical MomentaCanonical Momenta
283
edited Dec 3 '18 at 23:56
Michael Albanese
63.5k1599305
edited Dec 3 '18 at 23:56
Michael Albanese
63.5k1599305
edited Dec 3 '18 at 23:56
Michael Albanese
63.5k1599305
63.5k1599305
asked Dec 3 '18 at 20:53
Canonical MomentaCanonical Momenta
283
asked Dec 3 '18 at 20:53
Canonical MomentaCanonical Momenta
283
asked Dec 3 '18 at 20:53
Canonical MomentaCanonical Momenta
283
283
add a comment |
add a comment |
1 Answer
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If the manifolds involved are smooth, one has to be careful when doing connected sums and surgeries - I will not be so careful. These details are carried out in full in Kosinski's Differential Manifolds.
Let $M$ be an $n$-dimensional manifold and suppose there is an embedded $S^ktimes D^{n-k}$. Note that $Msetminus (S^ktimes D^{n-k})$ is an $n$-dimensional manifold with boundary $S^ktimes S^{n-k-1}$, as is $D^{k+1}times S^{n-k-1}$. We can glue these two together along their boundary to form a new $n$-manifold
$$M' := (Msetminus(S^ktimes D^{n-k-1}))cup_{S^ktimes S^{n-k-1}}(D^{k+1}times S^{n-k-1}).$$
We say that $M'$ is obtained from $M$ by a $k$-surgery.
The connected sum operation is just a special type of $0$-surgery. Let $X$ and $Y$ be connected $n$-dimensional manifolds, and consider the manifold $M := X sqcup Y$. We can find an embedded $S^0times D^n = D^nsqcup D^n$ in $M$ by choosing an open disc in $X$ and an open disc in $Y$. Doing a $0$-surgery on this embedded $S^0times D^n$ means removing those discs and gluing in $D^1times S^{n-1} = [0, 1]times S^{n-1}$ (i.e. a cylinder) to join the resulting boundaries; this is exactly the connected sum of $X$ and $Y$.
Note, if we had chosen an embedding of $S^0times D^n = D^nsqcup D^n$ such that both discs are in either $X$ or $Y$, the result of the $0$-surgery would not be a connected sum. A $0$-surgery where both discs lie in $X$ would correspond to adding a one-handle to $X$, so the resulting manifold would be $X#(S^1times S^{n-1}) sqcup Y$.
answered Dec 3 '18 at 23:52
Michael AlbaneseMichael Albanese
63.5k1599305
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$begingroup$
If the manifolds involved are smooth, one has to be careful when doing connected sums and surgeries - I will not be so careful. These details are carried out in full in Kosinski's Differential Manifolds.
Let $M$ be an $n$-dimensional manifold and suppose there is an embedded $S^ktimes D^{n-k}$. Note that $Msetminus (S^ktimes D^{n-k})$ is an $n$-dimensional manifold with boundary $S^ktimes S^{n-k-1}$, as is $D^{k+1}times S^{n-k-1}$. We can glue these two together along their boundary to form a new $n$-manifold
$$M' := (Msetminus(S^ktimes D^{n-k-1}))cup_{S^ktimes S^{n-k-1}}(D^{k+1}times S^{n-k-1}).$$
We say that $M'$ is obtained from $M$ by a $k$-surgery.
The connected sum operation is just a special type of $0$-surgery. Let $X$ and $Y$ be connected $n$-dimensional manifolds, and consider the manifold $M := X sqcup Y$. We can find an embedded $S^0times D^n = D^nsqcup D^n$ in $M$ by choosing an open disc in $X$ and an open disc in $Y$. Doing a $0$-surgery on this embedded $S^0times D^n$ means removing those discs and gluing in $D^1times S^{n-1} = [0, 1]times S^{n-1}$ (i.e. a cylinder) to join the resulting boundaries; this is exactly the connected sum of $X$ and $Y$.
Note, if we had chosen an embedding of $S^0times D^n = D^nsqcup D^n$ such that both discs are in either $X$ or $Y$, the result of the $0$-surgery would not be a connected sum. A $0$-surgery where both discs lie in $X$ would correspond to adding a one-handle to $X$, so the resulting manifold would be $X#(S^1times S^{n-1}) sqcup Y$.
answered Dec 3 '18 at 23:52
Michael AlbaneseMichael Albanese
63.5k1599305
$endgroup$
add a comment |
$begingroup$
If the manifolds involved are smooth, one has to be careful when doing connected sums and surgeries - I will not be so careful. These details are carried out in full in Kosinski's Differential Manifolds.
Let $M$ be an $n$-dimensional manifold and suppose there is an embedded $S^ktimes D^{n-k}$. Note that $Msetminus (S^ktimes D^{n-k})$ is an $n$-dimensional manifold with boundary $S^ktimes S^{n-k-1}$, as is $D^{k+1}times S^{n-k-1}$. We can glue these two together along their boundary to form a new $n$-manifold
$$M' := (Msetminus(S^ktimes D^{n-k-1}))cup_{S^ktimes S^{n-k-1}}(D^{k+1}times S^{n-k-1}).$$
We say that $M'$ is obtained from $M$ by a $k$-surgery.
The connected sum operation is just a special type of $0$-surgery. Let $X$ and $Y$ be connected $n$-dimensional manifolds, and consider the manifold $M := X sqcup Y$. We can find an embedded $S^0times D^n = D^nsqcup D^n$ in $M$ by choosing an open disc in $X$ and an open disc in $Y$. Doing a $0$-surgery on this embedded $S^0times D^n$ means removing those discs and gluing in $D^1times S^{n-1} = [0, 1]times S^{n-1}$ (i.e. a cylinder) to join the resulting boundaries; this is exactly the connected sum of $X$ and $Y$.
Note, if we had chosen an embedding of $S^0times D^n = D^nsqcup D^n$ such that both discs are in either $X$ or $Y$, the result of the $0$-surgery would not be a connected sum. A $0$-surgery where both discs lie in $X$ would correspond to adding a one-handle to $X$, so the resulting manifold would be $X#(S^1times S^{n-1}) sqcup Y$.
answered Dec 3 '18 at 23:52
Michael AlbaneseMichael Albanese
63.5k1599305
$endgroup$
add a comment |
$begingroup$
If the manifolds involved are smooth, one has to be careful when doing connected sums and surgeries - I will not be so careful. These details are carried out in full in Kosinski's Differential Manifolds.
Let $M$ be an $n$-dimensional manifold and suppose there is an embedded $S^ktimes D^{n-k}$. Note that $Msetminus (S^ktimes D^{n-k})$ is an $n$-dimensional manifold with boundary $S^ktimes S^{n-k-1}$, as is $D^{k+1}times S^{n-k-1}$. We can glue these two together along their boundary to form a new $n$-manifold
$$M' := (Msetminus(S^ktimes D^{n-k-1}))cup_{S^ktimes S^{n-k-1}}(D^{k+1}times S^{n-k-1}).$$
We say that $M'$ is obtained from $M$ by a $k$-surgery.
The connected sum operation is just a special type of $0$-surgery. Let $X$ and $Y$ be connected $n$-dimensional manifolds, and consider the manifold $M := X sqcup Y$. We can find an embedded $S^0times D^n = D^nsqcup D^n$ in $M$ by choosing an open disc in $X$ and an open disc in $Y$. Doing a $0$-surgery on this embedded $S^0times D^n$ means removing those discs and gluing in $D^1times S^{n-1} = [0, 1]times S^{n-1}$ (i.e. a cylinder) to join the resulting boundaries; this is exactly the connected sum of $X$ and $Y$.
Note, if we had chosen an embedding of $S^0times D^n = D^nsqcup D^n$ such that both discs are in either $X$ or $Y$, the result of the $0$-surgery would not be a connected sum. A $0$-surgery where both discs lie in $X$ would correspond to adding a one-handle to $X$, so the resulting manifold would be $X#(S^1times S^{n-1}) sqcup Y$.
answered Dec 3 '18 at 23:52
Michael AlbaneseMichael Albanese
63.5k1599305
$endgroup$
If the manifolds involved are smooth, one has to be careful when doing connected sums and surgeries - I will not be so careful. These details are carried out in full in Kosinski's Differential Manifolds.
Let $M$ be an $n$-dimensional manifold and suppose there is an embedded $S^ktimes D^{n-k}$. Note that $Msetminus (S^ktimes D^{n-k})$ is an $n$-dimensional manifold with boundary $S^ktimes S^{n-k-1}$, as is $D^{k+1}times S^{n-k-1}$. We can glue these two together along their boundary to form a new $n$-manifold
$$M' := (Msetminus(S^ktimes D^{n-k-1}))cup_{S^ktimes S^{n-k-1}}(D^{k+1}times S^{n-k-1}).$$
We say that $M'$ is obtained from $M$ by a $k$-surgery.
The connected sum operation is just a special type of $0$-surgery. Let $X$ and $Y$ be connected $n$-dimensional manifolds, and consider the manifold $M := X sqcup Y$. We can find an embedded $S^0times D^n = D^nsqcup D^n$ in $M$ by choosing an open disc in $X$ and an open disc in $Y$. Doing a $0$-surgery on this embedded $S^0times D^n$ means removing those discs and gluing in $D^1times S^{n-1} = [0, 1]times S^{n-1}$ (i.e. a cylinder) to join the resulting boundaries; this is exactly the connected sum of $X$ and $Y$.
Note, if we had chosen an embedding of $S^0times D^n = D^nsqcup D^n$ such that both discs are in either $X$ or $Y$, the result of the $0$-surgery would not be a connected sum. A $0$-surgery where both discs lie in $X$ would correspond to adding a one-handle to $X$, so the resulting manifold would be $X#(S^1times S^{n-1}) sqcup Y$.
answered Dec 3 '18 at 23:52
Michael AlbaneseMichael Albanese
63.5k1599305
answered Dec 3 '18 at 23:52
Michael AlbaneseMichael Albanese
63.5k1599305
answered Dec 3 '18 at 23:52
Michael AlbaneseMichael Albanese
63.5k1599305
answered Dec 3 '18 at 23:52
Michael AlbaneseMichael Albanese
63.5k1599305
63.5k1599305
add a comment |
add a comment |
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