Visualizing covers of $S^1vee S^1$
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In looking back at my university's past topology qualifying exams, it seems that in most years there is a question of the form
Let $X=S^1vee S^1$. Draw a picture of the cover of $X$ corresponding to the subgroup $H=langletextsome generatorsrangle$ of $pi_1(X)=langle a,brangle$.
When I see a solution to this type of problem, I can see why it satisfies the right conditions, typically having to with the fact that traversing a series of loops corresponding to a generator of $H$ brings you back to the basepoint, etc. However, I'm not good at coming up with these pictures myself. Is there a general method or "trick"?
Maybe I can give some explicit examples of $H$ and people could share their "thought process" for the solution they come to. For example,
(a) take $H=langle ab,barangle$
(b) take $H=langle(a^2b)^3,(ab)^3,bab,a^2ba^-1rangle$
(c) take $H=langle aba^-1b^-1,ab^-1a^-1b,a^-1bab^-1,a^-1b^-1abrangle$
algebraic-topology covering-spaces
asked Aug 24 at 4:48
Alex Mathers
10.2k21343
add a comment |Â
up vote
3
down vote
favorite
In looking back at my university's past topology qualifying exams, it seems that in most years there is a question of the form
Let $X=S^1vee S^1$. Draw a picture of the cover of $X$ corresponding to the subgroup $H=langletextsome generatorsrangle$ of $pi_1(X)=langle a,brangle$.
When I see a solution to this type of problem, I can see why it satisfies the right conditions, typically having to with the fact that traversing a series of loops corresponding to a generator of $H$ brings you back to the basepoint, etc. However, I'm not good at coming up with these pictures myself. Is there a general method or "trick"?
Maybe I can give some explicit examples of $H$ and people could share their "thought process" for the solution they come to. For example,
(a) take $H=langle ab,barangle$
(b) take $H=langle(a^2b)^3,(ab)^3,bab,a^2ba^-1rangle$
(c) take $H=langle aba^-1b^-1,ab^-1a^-1b,a^-1bab^-1,a^-1b^-1abrangle$
algebraic-topology covering-spaces
asked Aug 24 at 4:48
Alex Mathers
10.2k21343
1
It's gonna be hard to draw a picture for (a), since it is infinite-sheeted....
– Steve D
Aug 24 at 4:58
In general though, I would do this by first drawing the vertices [there will be $[G:H]$ of them]. Then, to be a covering, you just need each vertex to have two incoming edges (labeled $a$ and $b$), and two outgoing edges (also labeled $a$ and $b$). You do this in a way that the only loops at a particular vertex are the words in $H$.
– Steve D
Aug 24 at 5:07
Hatcher does a nice writeup on this in the opening parts of his chapter on covering spaces if I recall.
– Tyrone
Aug 25 at 5:44
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
In looking back at my university's past topology qualifying exams, it seems that in most years there is a question of the form
Let $X=S^1vee S^1$. Draw a picture of the cover of $X$ corresponding to the subgroup $H=langletextsome generatorsrangle$ of $pi_1(X)=langle a,brangle$.
When I see a solution to this type of problem, I can see why it satisfies the right conditions, typically having to with the fact that traversing a series of loops corresponding to a generator of $H$ brings you back to the basepoint, etc. However, I'm not good at coming up with these pictures myself. Is there a general method or "trick"?
Maybe I can give some explicit examples of $H$ and people could share their "thought process" for the solution they come to. For example,
(a) take $H=langle ab,barangle$
(b) take $H=langle(a^2b)^3,(ab)^3,bab,a^2ba^-1rangle$
(c) take $H=langle aba^-1b^-1,ab^-1a^-1b,a^-1bab^-1,a^-1b^-1abrangle$
algebraic-topology covering-spaces
asked Aug 24 at 4:48
Alex Mathers
10.2k21343
In looking back at my university's past topology qualifying exams, it seems that in most years there is a question of the form
Let $X=S^1vee S^1$. Draw a picture of the cover of $X$ corresponding to the subgroup $H=langletextsome generatorsrangle$ of $pi_1(X)=langle a,brangle$.
When I see a solution to this type of problem, I can see why it satisfies the right conditions, typically having to with the fact that traversing a series of loops corresponding to a generator of $H$ brings you back to the basepoint, etc. However, I'm not good at coming up with these pictures myself. Is there a general method or "trick"?
Maybe I can give some explicit examples of $H$ and people could share their "thought process" for the solution they come to. For example,
(a) take $H=langle ab,barangle$
(b) take $H=langle(a^2b)^3,(ab)^3,bab,a^2ba^-1rangle$
(c) take $H=langle aba^-1b^-1,ab^-1a^-1b,a^-1bab^-1,a^-1b^-1abrangle$
algebraic-topology covering-spaces
asked Aug 24 at 4:48
Alex Mathers
10.2k21343
asked Aug 24 at 4:48
Alex Mathers
10.2k21343
asked Aug 24 at 4:48
Alex Mathers
10.2k21343
asked Aug 24 at 4:48
Alex Mathers
10.2k21343
10.2k21343
1
It's gonna be hard to draw a picture for (a), since it is infinite-sheeted....
– Steve D
Aug 24 at 4:58
In general though, I would do this by first drawing the vertices [there will be $[G:H]$ of them]. Then, to be a covering, you just need each vertex to have two incoming edges (labeled $a$ and $b$), and two outgoing edges (also labeled $a$ and $b$). You do this in a way that the only loops at a particular vertex are the words in $H$.
– Steve D
Aug 24 at 5:07
Hatcher does a nice writeup on this in the opening parts of his chapter on covering spaces if I recall.
– Tyrone
Aug 25 at 5:44
add a comment |Â
1
It's gonna be hard to draw a picture for (a), since it is infinite-sheeted....
– Steve D
Aug 24 at 4:58
In general though, I would do this by first drawing the vertices [there will be $[G:H]$ of them]. Then, to be a covering, you just need each vertex to have two incoming edges (labeled $a$ and $b$), and two outgoing edges (also labeled $a$ and $b$). You do this in a way that the only loops at a particular vertex are the words in $H$.
– Steve D
Aug 24 at 5:07
Hatcher does a nice writeup on this in the opening parts of his chapter on covering spaces if I recall.
– Tyrone
Aug 25 at 5:44
1
1
It's gonna be hard to draw a picture for (a), since it is infinite-sheeted....
– Steve D
Aug 24 at 4:58
It's gonna be hard to draw a picture for (a), since it is infinite-sheeted....
– Steve D
Aug 24 at 4:58
In general though, I would do this by first drawing the vertices [there will be $[G:H]$ of them]. Then, to be a covering, you just need each vertex to have two incoming edges (labeled $a$ and $b$), and two outgoing edges (also labeled $a$ and $b$). You do this in a way that the only loops at a particular vertex are the words in $H$.
– Steve D
Aug 24 at 5:07
In general though, I would do this by first drawing the vertices [there will be $[G:H]$ of them]. Then, to be a covering, you just need each vertex to have two incoming edges (labeled $a$ and $b$), and two outgoing edges (also labeled $a$ and $b$). You do this in a way that the only loops at a particular vertex are the words in $H$.
– Steve D
Aug 24 at 5:07
Hatcher does a nice writeup on this in the opening parts of his chapter on covering spaces if I recall.
– Tyrone
Aug 25 at 5:44
Hatcher does a nice writeup on this in the opening parts of his chapter on covering spaces if I recall.
– Tyrone
Aug 25 at 5:44
add a comment |Â
1 Answer
1
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2
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Here's a general method.
To draw the covering corresponding to $H$, you let your vertices be indexed by the cosets of $H$ in $G$; your basepoint will correspond to the coset $H$.Then you draw an $a$ edge from $C$ to $Ca$ for every coset $C$; similarly for $b$.
I can't imagine there's an easier way to handle the general case. Something helpful here is that $H$ is a free group, so it has some number $n$ of generators. Thus your covering should be homotopic to the wedge of $n$ circles. For example, that helps me realize that the covering for (a) should look like
where each tree extends infinitely (just like the Cayley graph / universal cover for $G$).
answered Aug 24 at 5:59
Steve D
2,5471620
Awesome answer, thanks
– Alex Mathers
Aug 24 at 22:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Here's a general method.
To draw the covering corresponding to $H$, you let your vertices be indexed by the cosets of $H$ in $G$; your basepoint will correspond to the coset $H$.Then you draw an $a$ edge from $C$ to $Ca$ for every coset $C$; similarly for $b$.
I can't imagine there's an easier way to handle the general case. Something helpful here is that $H$ is a free group, so it has some number $n$ of generators. Thus your covering should be homotopic to the wedge of $n$ circles. For example, that helps me realize that the covering for (a) should look like
where each tree extends infinitely (just like the Cayley graph / universal cover for $G$).
answered Aug 24 at 5:59
Steve D
2,5471620
Awesome answer, thanks
– Alex Mathers
Aug 24 at 22:39
add a comment |Â
up vote
2
down vote
Here's a general method.
To draw the covering corresponding to $H$, you let your vertices be indexed by the cosets of $H$ in $G$; your basepoint will correspond to the coset $H$.Then you draw an $a$ edge from $C$ to $Ca$ for every coset $C$; similarly for $b$.
I can't imagine there's an easier way to handle the general case. Something helpful here is that $H$ is a free group, so it has some number $n$ of generators. Thus your covering should be homotopic to the wedge of $n$ circles. For example, that helps me realize that the covering for (a) should look like
where each tree extends infinitely (just like the Cayley graph / universal cover for $G$).
answered Aug 24 at 5:59
Steve D
2,5471620
Awesome answer, thanks
– Alex Mathers
Aug 24 at 22:39
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here's a general method.
To draw the covering corresponding to $H$, you let your vertices be indexed by the cosets of $H$ in $G$; your basepoint will correspond to the coset $H$.Then you draw an $a$ edge from $C$ to $Ca$ for every coset $C$; similarly for $b$.
I can't imagine there's an easier way to handle the general case. Something helpful here is that $H$ is a free group, so it has some number $n$ of generators. Thus your covering should be homotopic to the wedge of $n$ circles. For example, that helps me realize that the covering for (a) should look like
where each tree extends infinitely (just like the Cayley graph / universal cover for $G$).
answered Aug 24 at 5:59
Steve D
2,5471620
Here's a general method.
To draw the covering corresponding to $H$, you let your vertices be indexed by the cosets of $H$ in $G$; your basepoint will correspond to the coset $H$.Then you draw an $a$ edge from $C$ to $Ca$ for every coset $C$; similarly for $b$.
I can't imagine there's an easier way to handle the general case. Something helpful here is that $H$ is a free group, so it has some number $n$ of generators. Thus your covering should be homotopic to the wedge of $n$ circles. For example, that helps me realize that the covering for (a) should look like
where each tree extends infinitely (just like the Cayley graph / universal cover for $G$).
answered Aug 24 at 5:59
Steve D
2,5471620
answered Aug 24 at 5:59
Steve D
2,5471620
answered Aug 24 at 5:59
Steve D
2,5471620
answered Aug 24 at 5:59
Steve D
2,5471620
2,5471620
Awesome answer, thanks
– Alex Mathers
Aug 24 at 22:39
add a comment |Â
Awesome answer, thanks
– Alex Mathers
Aug 24 at 22:39
Awesome answer, thanks
– Alex Mathers
Aug 24 at 22:39
Awesome answer, thanks
– Alex Mathers
Aug 24 at 22:39
add a comment |Â
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1
It's gonna be hard to draw a picture for (a), since it is infinite-sheeted....
– Steve D
Aug 24 at 4:58
In general though, I would do this by first drawing the vertices [there will be $[G:H]$ of them]. Then, to be a covering, you just need each vertex to have two incoming edges (labeled $a$ and $b$), and two outgoing edges (also labeled $a$ and $b$). You do this in a way that the only loops at a particular vertex are the words in $H$.
– Steve D
Aug 24 at 5:07
Hatcher does a nice writeup on this in the opening parts of his chapter on covering spaces if I recall.
– Tyrone
Aug 25 at 5:44