Fundamental Group of Connected Sum of two $3$-manifolds
$begingroup$
I've been asked to find the fundamental group of $(S^1times S^2)#(S^1times S^2)$. However, connected sum, as briefly defined in my class (in a vague way) was as follows: You remove a $2$-disk from each surface and concatenate them. My questions are:
- In this case, do we remove a $3$-disk?
- I have been told to use the Van Kampen theorem. How do I make the explicit partitions?
- I would imagine that I would at least need the fundamental group of $(S^1times S^2)$, which is $mathbb{Z}$. How do I proceed from here?
algebraic-topology low-dimensional-topology
edited Dec 2 '18 at 15:58
Lee Mosher
48.4k33681
asked Dec 2 '18 at 14:50
BoshuBoshu
705315
$endgroup$
add a comment |
$begingroup$
I've been asked to find the fundamental group of $(S^1times S^2)#(S^1times S^2)$. However, connected sum, as briefly defined in my class (in a vague way) was as follows: You remove a $2$-disk from each surface and concatenate them. My questions are:
- In this case, do we remove a $3$-disk?
- I have been told to use the Van Kampen theorem. How do I make the explicit partitions?
- I would imagine that I would at least need the fundamental group of $(S^1times S^2)$, which is $mathbb{Z}$. How do I proceed from here?
algebraic-topology low-dimensional-topology
edited Dec 2 '18 at 15:58
Lee Mosher
48.4k33681
asked Dec 2 '18 at 14:50
BoshuBoshu
705315
$endgroup$
$begingroup$
What you do is to remove a $3$-ball from each $3$-manifold and then glue the two manifolds along the boundaries of the 3-balls. As a sphere is simply connected Van Kampen theorem implies that the fundamental group is the coproduct of the fundamental groups of the manifolds.
$endgroup$
– Camilo Arosemena-Serrato
Dec 2 '18 at 15:01
add a comment |
$begingroup$
I've been asked to find the fundamental group of $(S^1times S^2)#(S^1times S^2)$. However, connected sum, as briefly defined in my class (in a vague way) was as follows: You remove a $2$-disk from each surface and concatenate them. My questions are:
- In this case, do we remove a $3$-disk?
- I have been told to use the Van Kampen theorem. How do I make the explicit partitions?
- I would imagine that I would at least need the fundamental group of $(S^1times S^2)$, which is $mathbb{Z}$. How do I proceed from here?
algebraic-topology low-dimensional-topology
edited Dec 2 '18 at 15:58
Lee Mosher
48.4k33681
asked Dec 2 '18 at 14:50
BoshuBoshu
705315
$endgroup$
I've been asked to find the fundamental group of $(S^1times S^2)#(S^1times S^2)$. However, connected sum, as briefly defined in my class (in a vague way) was as follows: You remove a $2$-disk from each surface and concatenate them. My questions are:
- In this case, do we remove a $3$-disk?
- I have been told to use the Van Kampen theorem. How do I make the explicit partitions?
- I would imagine that I would at least need the fundamental group of $(S^1times S^2)$, which is $mathbb{Z}$. How do I proceed from here?
algebraic-topology low-dimensional-topology
algebraic-topology low-dimensional-topology
edited Dec 2 '18 at 15:58
Lee Mosher
48.4k33681
asked Dec 2 '18 at 14:50
BoshuBoshu
705315
edited Dec 2 '18 at 15:58
Lee Mosher
48.4k33681
asked Dec 2 '18 at 14:50
BoshuBoshu
705315
edited Dec 2 '18 at 15:58
Lee Mosher
48.4k33681
edited Dec 2 '18 at 15:58
Lee Mosher
48.4k33681
edited Dec 2 '18 at 15:58
Lee Mosher
48.4k33681
48.4k33681
asked Dec 2 '18 at 14:50
BoshuBoshu
705315
asked Dec 2 '18 at 14:50
BoshuBoshu
705315
asked Dec 2 '18 at 14:50
BoshuBoshu
705315
705315
$begingroup$
What you do is to remove a $3$-ball from each $3$-manifold and then glue the two manifolds along the boundaries of the 3-balls. As a sphere is simply connected Van Kampen theorem implies that the fundamental group is the coproduct of the fundamental groups of the manifolds.
$endgroup$
– Camilo Arosemena-Serrato
Dec 2 '18 at 15:01
add a comment |
$begingroup$
What you do is to remove a $3$-ball from each $3$-manifold and then glue the two manifolds along the boundaries of the 3-balls. As a sphere is simply connected Van Kampen theorem implies that the fundamental group is the coproduct of the fundamental groups of the manifolds.
$endgroup$
– Camilo Arosemena-Serrato
Dec 2 '18 at 15:01
$begingroup$
What you do is to remove a $3$-ball from each $3$-manifold and then glue the two manifolds along the boundaries of the 3-balls. As a sphere is simply connected Van Kampen theorem implies that the fundamental group is the coproduct of the fundamental groups of the manifolds.
$endgroup$
– Camilo Arosemena-Serrato
Dec 2 '18 at 15:01
$begingroup$
What you do is to remove a $3$-ball from each $3$-manifold and then glue the two manifolds along the boundaries of the 3-balls. As a sphere is simply connected Van Kampen theorem implies that the fundamental group is the coproduct of the fundamental groups of the manifolds.
$endgroup$
– Camilo Arosemena-Serrato
Dec 2 '18 at 15:01
add a comment |
1 Answer
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$begingroup$
Yes, you remove a 3-disk from each manifold.
More specifically, and to address question 2, let me set up some notation. Denote $M_1$, $M_2$ to be the two $S^1 times S^2$ summands. Let $D_i subset M_i$ be a subset homeomorphic to the open 3-disk $mathbb B = {x in mathbb R^3 mid |x| < 1}$, and let $f_i : D_i to mathbb B$ be a homeomorphism. Denote $mathbb B^{1/2} = {x in mathbb R^3 mid |x| < 1/2}$, and denote $B_i^{1/2} = f_i^{-1}(mathbb B^{1/2})$. Let $S^i = partial (M_i - B_i^{1/2})$ which also equals $f_i^{-1}$ of the sphere of radius $1/2$.
As a topological space, the connected sum $M_1 # M_2$ is the quotient space of the disjoint union $M_1 - B^{1/2}_i$ and $M_2 - B^{1/2}_i$ where $S_1$ is glued to $S_2$ using the homeomorphism $f_2^{-1} circ f_1 : S_1 to S_2$.
The open sets needed for application of Van Kampen's theorem are
$$U_1 = (M_1 - B^{1/2}_1) cup f_2^{-1}(mathbb B - mathbb B^{1/2})
$$
and
$$U_2 = (M_2 - B^{1/2}_2) cup f_1^{-1}(mathbb B - mathbb B^{1/2})
$$
To address question 3, what you need for Van Kampen's Theorem is $pi_1 U_1$ and $pi_1 U_2$ and $pi_1 (U_1 cap U_2)$. It should be straightforward to make the following deductions:
$pi_1 U_i approx pi_1 M_i$ (this is, itself, a Van Kampen's Theorem application)
$U_1 cap U_2$ is homeomorphic to $S^2 times (-1,+1)$ and hence is simply connected.
answered Dec 2 '18 at 16:10
Lee MosherLee Mosher
48.4k33681
$endgroup$
$begingroup$
That was very helpful, thanks a lot!
$endgroup$
– Boshu
Dec 2 '18 at 16:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, you remove a 3-disk from each manifold.
More specifically, and to address question 2, let me set up some notation. Denote $M_1$, $M_2$ to be the two $S^1 times S^2$ summands. Let $D_i subset M_i$ be a subset homeomorphic to the open 3-disk $mathbb B = {x in mathbb R^3 mid |x| < 1}$, and let $f_i : D_i to mathbb B$ be a homeomorphism. Denote $mathbb B^{1/2} = {x in mathbb R^3 mid |x| < 1/2}$, and denote $B_i^{1/2} = f_i^{-1}(mathbb B^{1/2})$. Let $S^i = partial (M_i - B_i^{1/2})$ which also equals $f_i^{-1}$ of the sphere of radius $1/2$.
As a topological space, the connected sum $M_1 # M_2$ is the quotient space of the disjoint union $M_1 - B^{1/2}_i$ and $M_2 - B^{1/2}_i$ where $S_1$ is glued to $S_2$ using the homeomorphism $f_2^{-1} circ f_1 : S_1 to S_2$.
The open sets needed for application of Van Kampen's theorem are
$$U_1 = (M_1 - B^{1/2}_1) cup f_2^{-1}(mathbb B - mathbb B^{1/2})
$$
and
$$U_2 = (M_2 - B^{1/2}_2) cup f_1^{-1}(mathbb B - mathbb B^{1/2})
$$
To address question 3, what you need for Van Kampen's Theorem is $pi_1 U_1$ and $pi_1 U_2$ and $pi_1 (U_1 cap U_2)$. It should be straightforward to make the following deductions:
$pi_1 U_i approx pi_1 M_i$ (this is, itself, a Van Kampen's Theorem application)
$U_1 cap U_2$ is homeomorphic to $S^2 times (-1,+1)$ and hence is simply connected.
answered Dec 2 '18 at 16:10
Lee MosherLee Mosher
48.4k33681
$endgroup$
$begingroup$
That was very helpful, thanks a lot!
$endgroup$
– Boshu
Dec 2 '18 at 16:49
add a comment |
$begingroup$
Yes, you remove a 3-disk from each manifold.
More specifically, and to address question 2, let me set up some notation. Denote $M_1$, $M_2$ to be the two $S^1 times S^2$ summands. Let $D_i subset M_i$ be a subset homeomorphic to the open 3-disk $mathbb B = {x in mathbb R^3 mid |x| < 1}$, and let $f_i : D_i to mathbb B$ be a homeomorphism. Denote $mathbb B^{1/2} = {x in mathbb R^3 mid |x| < 1/2}$, and denote $B_i^{1/2} = f_i^{-1}(mathbb B^{1/2})$. Let $S^i = partial (M_i - B_i^{1/2})$ which also equals $f_i^{-1}$ of the sphere of radius $1/2$.
As a topological space, the connected sum $M_1 # M_2$ is the quotient space of the disjoint union $M_1 - B^{1/2}_i$ and $M_2 - B^{1/2}_i$ where $S_1$ is glued to $S_2$ using the homeomorphism $f_2^{-1} circ f_1 : S_1 to S_2$.
The open sets needed for application of Van Kampen's theorem are
$$U_1 = (M_1 - B^{1/2}_1) cup f_2^{-1}(mathbb B - mathbb B^{1/2})
$$
and
$$U_2 = (M_2 - B^{1/2}_2) cup f_1^{-1}(mathbb B - mathbb B^{1/2})
$$
To address question 3, what you need for Van Kampen's Theorem is $pi_1 U_1$ and $pi_1 U_2$ and $pi_1 (U_1 cap U_2)$. It should be straightforward to make the following deductions:
$pi_1 U_i approx pi_1 M_i$ (this is, itself, a Van Kampen's Theorem application)
$U_1 cap U_2$ is homeomorphic to $S^2 times (-1,+1)$ and hence is simply connected.
answered Dec 2 '18 at 16:10
Lee MosherLee Mosher
48.4k33681
$endgroup$
$begingroup$
That was very helpful, thanks a lot!
$endgroup$
– Boshu
Dec 2 '18 at 16:49
add a comment |
$begingroup$
Yes, you remove a 3-disk from each manifold.
More specifically, and to address question 2, let me set up some notation. Denote $M_1$, $M_2$ to be the two $S^1 times S^2$ summands. Let $D_i subset M_i$ be a subset homeomorphic to the open 3-disk $mathbb B = {x in mathbb R^3 mid |x| < 1}$, and let $f_i : D_i to mathbb B$ be a homeomorphism. Denote $mathbb B^{1/2} = {x in mathbb R^3 mid |x| < 1/2}$, and denote $B_i^{1/2} = f_i^{-1}(mathbb B^{1/2})$. Let $S^i = partial (M_i - B_i^{1/2})$ which also equals $f_i^{-1}$ of the sphere of radius $1/2$.
As a topological space, the connected sum $M_1 # M_2$ is the quotient space of the disjoint union $M_1 - B^{1/2}_i$ and $M_2 - B^{1/2}_i$ where $S_1$ is glued to $S_2$ using the homeomorphism $f_2^{-1} circ f_1 : S_1 to S_2$.
The open sets needed for application of Van Kampen's theorem are
$$U_1 = (M_1 - B^{1/2}_1) cup f_2^{-1}(mathbb B - mathbb B^{1/2})
$$
and
$$U_2 = (M_2 - B^{1/2}_2) cup f_1^{-1}(mathbb B - mathbb B^{1/2})
$$
To address question 3, what you need for Van Kampen's Theorem is $pi_1 U_1$ and $pi_1 U_2$ and $pi_1 (U_1 cap U_2)$. It should be straightforward to make the following deductions:
$pi_1 U_i approx pi_1 M_i$ (this is, itself, a Van Kampen's Theorem application)
$U_1 cap U_2$ is homeomorphic to $S^2 times (-1,+1)$ and hence is simply connected.
answered Dec 2 '18 at 16:10
Lee MosherLee Mosher
48.4k33681
$endgroup$
Yes, you remove a 3-disk from each manifold.
More specifically, and to address question 2, let me set up some notation. Denote $M_1$, $M_2$ to be the two $S^1 times S^2$ summands. Let $D_i subset M_i$ be a subset homeomorphic to the open 3-disk $mathbb B = {x in mathbb R^3 mid |x| < 1}$, and let $f_i : D_i to mathbb B$ be a homeomorphism. Denote $mathbb B^{1/2} = {x in mathbb R^3 mid |x| < 1/2}$, and denote $B_i^{1/2} = f_i^{-1}(mathbb B^{1/2})$. Let $S^i = partial (M_i - B_i^{1/2})$ which also equals $f_i^{-1}$ of the sphere of radius $1/2$.
As a topological space, the connected sum $M_1 # M_2$ is the quotient space of the disjoint union $M_1 - B^{1/2}_i$ and $M_2 - B^{1/2}_i$ where $S_1$ is glued to $S_2$ using the homeomorphism $f_2^{-1} circ f_1 : S_1 to S_2$.
The open sets needed for application of Van Kampen's theorem are
$$U_1 = (M_1 - B^{1/2}_1) cup f_2^{-1}(mathbb B - mathbb B^{1/2})
$$
and
$$U_2 = (M_2 - B^{1/2}_2) cup f_1^{-1}(mathbb B - mathbb B^{1/2})
$$
To address question 3, what you need for Van Kampen's Theorem is $pi_1 U_1$ and $pi_1 U_2$ and $pi_1 (U_1 cap U_2)$. It should be straightforward to make the following deductions:
$pi_1 U_i approx pi_1 M_i$ (this is, itself, a Van Kampen's Theorem application)
$U_1 cap U_2$ is homeomorphic to $S^2 times (-1,+1)$ and hence is simply connected.
answered Dec 2 '18 at 16:10
Lee MosherLee Mosher
48.4k33681
answered Dec 2 '18 at 16:10
Lee MosherLee Mosher
48.4k33681
answered Dec 2 '18 at 16:10
Lee MosherLee Mosher
48.4k33681
answered Dec 2 '18 at 16:10
Lee MosherLee Mosher
48.4k33681
48.4k33681
$begingroup$
That was very helpful, thanks a lot!
$endgroup$
– Boshu
Dec 2 '18 at 16:49
add a comment |
$begingroup$
That was very helpful, thanks a lot!
$endgroup$
– Boshu
Dec 2 '18 at 16:49
$begingroup$
That was very helpful, thanks a lot!
$endgroup$
– Boshu
Dec 2 '18 at 16:49
$begingroup$
That was very helpful, thanks a lot!
$endgroup$
– Boshu
Dec 2 '18 at 16:49
add a comment |
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$begingroup$
What you do is to remove a $3$-ball from each $3$-manifold and then glue the two manifolds along the boundaries of the 3-balls. As a sphere is simply connected Van Kampen theorem implies that the fundamental group is the coproduct of the fundamental groups of the manifolds.
$endgroup$
– Camilo Arosemena-Serrato
Dec 2 '18 at 15:01