Fundamental group of complement of $n$ lines through the origin in $mathbb{R}^3$
up vote
9
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Just a quick question to verify whether I'm right.
Claim: The fundamental group of the complement of $n$ lines through the origin in $mathbb{R}^3$ is $F_n$, the free group on $n$ generators.
Proof: remove a line from $mathbb{R}^3$. We may deformation retract the remaining space onto a cylinder radius $epsilon$ about the line, and thence to a circle $S^1$. There is no trouble repeating this process with a second distinct line, except that then we will be a wedge union $S^1 vee S^1$. Continue inductively, and recall that the wedge union of $n$ circles has the stated fundamental group.
I'm only just starting to really get my head around this stuff, so any feedback would be really useful!
Thanks!
algebraic-topology homotopy-theory
asked Apr 8 '12 at 19:30
Edward Hughes
1,9701550
|
show 2 more comments
up vote
9
down vote
favorite
Just a quick question to verify whether I'm right.
Claim: The fundamental group of the complement of $n$ lines through the origin in $mathbb{R}^3$ is $F_n$, the free group on $n$ generators.
Proof: remove a line from $mathbb{R}^3$. We may deformation retract the remaining space onto a cylinder radius $epsilon$ about the line, and thence to a circle $S^1$. There is no trouble repeating this process with a second distinct line, except that then we will be a wedge union $S^1 vee S^1$. Continue inductively, and recall that the wedge union of $n$ circles has the stated fundamental group.
I'm only just starting to really get my head around this stuff, so any feedback would be really useful!
Thanks!
algebraic-topology homotopy-theory
asked Apr 8 '12 at 19:30
Edward Hughes
1,9701550
1
Sounds right : )
– Rudy the Reindeer
Apr 8 '12 at 19:33
1
This is basically right, though writing down all the details could be messy.
– Cheerful Parsnip
Apr 8 '12 at 19:39
Thanks! I agree the details could be messy, but now at least I know I have the right idea.
– Edward Hughes
Apr 8 '12 at 19:41
1
Do you mean "minus $n$ lines" like in the title, or "minus $n$ lines through the origin"? There is a significant difference: if the $n$ lines are disjoint then the fundamental group is $F_{n}$ (seen by deformation retracting onto ($mathbb{R}^2$ minus $n$ points)), but if $ngeq 2$ and they all intersect at the same point then the fundamental group is $F_{2n-1}$ (as shown by user8268)
– you
Apr 8 '12 at 22:48
3
You say «There is no trouble repeating this process with a second distinct line». What process? WHat you explain in the case of one line cannot be done when there are two of them!
– Mariano Suárez-Álvarez
Apr 8 '12 at 23:19
|
show 2 more comments
up vote
9
down vote
favorite
up vote
9
down vote
favorite
Just a quick question to verify whether I'm right.
Claim: The fundamental group of the complement of $n$ lines through the origin in $mathbb{R}^3$ is $F_n$, the free group on $n$ generators.
Proof: remove a line from $mathbb{R}^3$. We may deformation retract the remaining space onto a cylinder radius $epsilon$ about the line, and thence to a circle $S^1$. There is no trouble repeating this process with a second distinct line, except that then we will be a wedge union $S^1 vee S^1$. Continue inductively, and recall that the wedge union of $n$ circles has the stated fundamental group.
I'm only just starting to really get my head around this stuff, so any feedback would be really useful!
Thanks!
algebraic-topology homotopy-theory
asked Apr 8 '12 at 19:30
Edward Hughes
1,9701550
Just a quick question to verify whether I'm right.
Claim: The fundamental group of the complement of $n$ lines through the origin in $mathbb{R}^3$ is $F_n$, the free group on $n$ generators.
Proof: remove a line from $mathbb{R}^3$. We may deformation retract the remaining space onto a cylinder radius $epsilon$ about the line, and thence to a circle $S^1$. There is no trouble repeating this process with a second distinct line, except that then we will be a wedge union $S^1 vee S^1$. Continue inductively, and recall that the wedge union of $n$ circles has the stated fundamental group.
I'm only just starting to really get my head around this stuff, so any feedback would be really useful!
Thanks!
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
asked Apr 8 '12 at 19:30
Edward Hughes
1,9701550
asked Apr 8 '12 at 19:30
Edward Hughes
1,9701550
edited Jun 2 '12 at 16:03
asked Apr 8 '12 at 19:30
Edward Hughes
1,9701550
asked Apr 8 '12 at 19:30
Edward Hughes
1,9701550
asked Apr 8 '12 at 19:30
Edward Hughes
1,9701550
1,9701550
1
Sounds right : )
– Rudy the Reindeer
Apr 8 '12 at 19:33
1
This is basically right, though writing down all the details could be messy.
– Cheerful Parsnip
Apr 8 '12 at 19:39
Thanks! I agree the details could be messy, but now at least I know I have the right idea.
– Edward Hughes
Apr 8 '12 at 19:41
1
Do you mean "minus $n$ lines" like in the title, or "minus $n$ lines through the origin"? There is a significant difference: if the $n$ lines are disjoint then the fundamental group is $F_{n}$ (seen by deformation retracting onto ($mathbb{R}^2$ minus $n$ points)), but if $ngeq 2$ and they all intersect at the same point then the fundamental group is $F_{2n-1}$ (as shown by user8268)
– you
Apr 8 '12 at 22:48
3
You say «There is no trouble repeating this process with a second distinct line». What process? WHat you explain in the case of one line cannot be done when there are two of them!
– Mariano Suárez-Álvarez
Apr 8 '12 at 23:19
|
show 2 more comments
1
Sounds right : )
– Rudy the Reindeer
Apr 8 '12 at 19:33
1
This is basically right, though writing down all the details could be messy.
– Cheerful Parsnip
Apr 8 '12 at 19:39
Thanks! I agree the details could be messy, but now at least I know I have the right idea.
– Edward Hughes
Apr 8 '12 at 19:41
1
Do you mean "minus $n$ lines" like in the title, or "minus $n$ lines through the origin"? There is a significant difference: if the $n$ lines are disjoint then the fundamental group is $F_{n}$ (seen by deformation retracting onto ($mathbb{R}^2$ minus $n$ points)), but if $ngeq 2$ and they all intersect at the same point then the fundamental group is $F_{2n-1}$ (as shown by user8268)
– you
Apr 8 '12 at 22:48
3
You say «There is no trouble repeating this process with a second distinct line». What process? WHat you explain in the case of one line cannot be done when there are two of them!
– Mariano Suárez-Álvarez
Apr 8 '12 at 23:19
1
1
Sounds right : )
– Rudy the Reindeer
Apr 8 '12 at 19:33
Sounds right : )
– Rudy the Reindeer
Apr 8 '12 at 19:33
1
1
This is basically right, though writing down all the details could be messy.
– Cheerful Parsnip
Apr 8 '12 at 19:39
This is basically right, though writing down all the details could be messy.
– Cheerful Parsnip
Apr 8 '12 at 19:39
Thanks! I agree the details could be messy, but now at least I know I have the right idea.
– Edward Hughes
Apr 8 '12 at 19:41
Thanks! I agree the details could be messy, but now at least I know I have the right idea.
– Edward Hughes
Apr 8 '12 at 19:41
1
1
Do you mean "minus $n$ lines" like in the title, or "minus $n$ lines through the origin"? There is a significant difference: if the $n$ lines are disjoint then the fundamental group is $F_{n}$ (seen by deformation retracting onto ($mathbb{R}^2$ minus $n$ points)), but if $ngeq 2$ and they all intersect at the same point then the fundamental group is $F_{2n-1}$ (as shown by user8268)
– you
Apr 8 '12 at 22:48
Do you mean "minus $n$ lines" like in the title, or "minus $n$ lines through the origin"? There is a significant difference: if the $n$ lines are disjoint then the fundamental group is $F_{n}$ (seen by deformation retracting onto ($mathbb{R}^2$ minus $n$ points)), but if $ngeq 2$ and they all intersect at the same point then the fundamental group is $F_{2n-1}$ (as shown by user8268)
– you
Apr 8 '12 at 22:48
3
3
You say «There is no trouble repeating this process with a second distinct line». What process? WHat you explain in the case of one line cannot be done when there are two of them!
– Mariano Suárez-Álvarez
Apr 8 '12 at 23:19
You say «There is no trouble repeating this process with a second distinct line». What process? WHat you explain in the case of one line cannot be done when there are two of them!
– Mariano Suárez-Álvarez
Apr 8 '12 at 23:19
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
13
down vote
accepted
There is a deformation retraction of ($mathbb{R}^3$ minus $n$ lines through the origin) to (the unit sphere with $2n$ points removed). The $2n$ points are the intersections of the lines with the sphere, the deformation retraction is along the rays from the origin.
As a result, the fundamental group is actually $F_{2n-1}$, not $F_n$.
answered Apr 8 '12 at 22:34
user8268
16.7k12746
1
How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks!
– Edward Hughes
Apr 10 '12 at 23:45
1
Have you heard of stereographic projection? It is a homeomorphism from the complement of a point on $S^n$ to $mathbb{R}^n$. Relevant link: en.wikipedia.org/wiki/Stereographic_projection
– John Stalfos
Apr 12 '12 at 2:14
4
I was asking this question myself too ; I guess the right way to do it is to use one of the $2n$ points as a "north pole" for the stereographic projection, which leaves us with $mathbb R^2$ with $2n-1$ points removed. Using van Kampen's theorem, we get $F_{2n-1}$. :)
– Patrick Da Silva
Sep 1 '13 at 12:11
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
There is a deformation retraction of ($mathbb{R}^3$ minus $n$ lines through the origin) to (the unit sphere with $2n$ points removed). The $2n$ points are the intersections of the lines with the sphere, the deformation retraction is along the rays from the origin.
As a result, the fundamental group is actually $F_{2n-1}$, not $F_n$.
answered Apr 8 '12 at 22:34
user8268
16.7k12746
1
How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks!
– Edward Hughes
Apr 10 '12 at 23:45
1
Have you heard of stereographic projection? It is a homeomorphism from the complement of a point on $S^n$ to $mathbb{R}^n$. Relevant link: en.wikipedia.org/wiki/Stereographic_projection
– John Stalfos
Apr 12 '12 at 2:14
4
I was asking this question myself too ; I guess the right way to do it is to use one of the $2n$ points as a "north pole" for the stereographic projection, which leaves us with $mathbb R^2$ with $2n-1$ points removed. Using van Kampen's theorem, we get $F_{2n-1}$. :)
– Patrick Da Silva
Sep 1 '13 at 12:11
add a comment |
up vote
13
down vote
accepted
There is a deformation retraction of ($mathbb{R}^3$ minus $n$ lines through the origin) to (the unit sphere with $2n$ points removed). The $2n$ points are the intersections of the lines with the sphere, the deformation retraction is along the rays from the origin.
As a result, the fundamental group is actually $F_{2n-1}$, not $F_n$.
answered Apr 8 '12 at 22:34
user8268
16.7k12746
1
How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks!
– Edward Hughes
Apr 10 '12 at 23:45
1
Have you heard of stereographic projection? It is a homeomorphism from the complement of a point on $S^n$ to $mathbb{R}^n$. Relevant link: en.wikipedia.org/wiki/Stereographic_projection
– John Stalfos
Apr 12 '12 at 2:14
4
I was asking this question myself too ; I guess the right way to do it is to use one of the $2n$ points as a "north pole" for the stereographic projection, which leaves us with $mathbb R^2$ with $2n-1$ points removed. Using van Kampen's theorem, we get $F_{2n-1}$. :)
– Patrick Da Silva
Sep 1 '13 at 12:11
add a comment |
up vote
13
down vote
accepted
up vote
13
down vote
accepted
There is a deformation retraction of ($mathbb{R}^3$ minus $n$ lines through the origin) to (the unit sphere with $2n$ points removed). The $2n$ points are the intersections of the lines with the sphere, the deformation retraction is along the rays from the origin.
As a result, the fundamental group is actually $F_{2n-1}$, not $F_n$.
answered Apr 8 '12 at 22:34
user8268
16.7k12746
There is a deformation retraction of ($mathbb{R}^3$ minus $n$ lines through the origin) to (the unit sphere with $2n$ points removed). The $2n$ points are the intersections of the lines with the sphere, the deformation retraction is along the rays from the origin.
As a result, the fundamental group is actually $F_{2n-1}$, not $F_n$.
answered Apr 8 '12 at 22:34
user8268
16.7k12746
answered Apr 8 '12 at 22:34
user8268
16.7k12746
answered Apr 8 '12 at 22:34
user8268
16.7k12746
answered Apr 8 '12 at 22:34
user8268
16.7k12746
16.7k12746
1
How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks!
– Edward Hughes
Apr 10 '12 at 23:45
1
Have you heard of stereographic projection? It is a homeomorphism from the complement of a point on $S^n$ to $mathbb{R}^n$. Relevant link: en.wikipedia.org/wiki/Stereographic_projection
– John Stalfos
Apr 12 '12 at 2:14
4
I was asking this question myself too ; I guess the right way to do it is to use one of the $2n$ points as a "north pole" for the stereographic projection, which leaves us with $mathbb R^2$ with $2n-1$ points removed. Using van Kampen's theorem, we get $F_{2n-1}$. :)
– Patrick Da Silva
Sep 1 '13 at 12:11
add a comment |
1
How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks!
– Edward Hughes
Apr 10 '12 at 23:45
1
Have you heard of stereographic projection? It is a homeomorphism from the complement of a point on $S^n$ to $mathbb{R}^n$. Relevant link: en.wikipedia.org/wiki/Stereographic_projection
– John Stalfos
Apr 12 '12 at 2:14
4
I was asking this question myself too ; I guess the right way to do it is to use one of the $2n$ points as a "north pole" for the stereographic projection, which leaves us with $mathbb R^2$ with $2n-1$ points removed. Using van Kampen's theorem, we get $F_{2n-1}$. :)
– Patrick Da Silva
Sep 1 '13 at 12:11
1
1
How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks!
– Edward Hughes
Apr 10 '12 at 23:45
How do you show that the unit sphere with 2$n$ points removed has fundamental group $F_{2n-1}$? Thanks!
– Edward Hughes
Apr 10 '12 at 23:45
1
1
Have you heard of stereographic projection? It is a homeomorphism from the complement of a point on $S^n$ to $mathbb{R}^n$. Relevant link: en.wikipedia.org/wiki/Stereographic_projection
– John Stalfos
Apr 12 '12 at 2:14
Have you heard of stereographic projection? It is a homeomorphism from the complement of a point on $S^n$ to $mathbb{R}^n$. Relevant link: en.wikipedia.org/wiki/Stereographic_projection
– John Stalfos
Apr 12 '12 at 2:14
4
4
I was asking this question myself too ; I guess the right way to do it is to use one of the $2n$ points as a "north pole" for the stereographic projection, which leaves us with $mathbb R^2$ with $2n-1$ points removed. Using van Kampen's theorem, we get $F_{2n-1}$. :)
– Patrick Da Silva
Sep 1 '13 at 12:11
I was asking this question myself too ; I guess the right way to do it is to use one of the $2n$ points as a "north pole" for the stereographic projection, which leaves us with $mathbb R^2$ with $2n-1$ points removed. Using van Kampen's theorem, we get $F_{2n-1}$. :)
– Patrick Da Silva
Sep 1 '13 at 12:11
add a comment |
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1
Sounds right : )
– Rudy the Reindeer
Apr 8 '12 at 19:33
1
This is basically right, though writing down all the details could be messy.
– Cheerful Parsnip
Apr 8 '12 at 19:39
Thanks! I agree the details could be messy, but now at least I know I have the right idea.
– Edward Hughes
Apr 8 '12 at 19:41
1
Do you mean "minus $n$ lines" like in the title, or "minus $n$ lines through the origin"? There is a significant difference: if the $n$ lines are disjoint then the fundamental group is $F_{n}$ (seen by deformation retracting onto ($mathbb{R}^2$ minus $n$ points)), but if $ngeq 2$ and they all intersect at the same point then the fundamental group is $F_{2n-1}$ (as shown by user8268)
– you
Apr 8 '12 at 22:48
3
You say «There is no trouble repeating this process with a second distinct line». What process? WHat you explain in the case of one line cannot be done when there are two of them!
– Mariano Suárez-Álvarez
Apr 8 '12 at 23:19