Simple groups and fundamental groups
$begingroup$
What does it mean, topologically, to have a simple fundamental group?
For instance, the torus $S^1 times S^1$ has $mathbb Z times mathbb Z$not simple. The case of $S^1$ is $mathbb Z$, not simple either.
But if instead, in $S^1$ we identify antipodal points, or even, if we identify points after rotating 120 degrees, we would get $mathbb Z/2mathbb Z, mathbb Z/3mathbb Z$...
We can obtain the torus from a $2$ dimensional square by two identifications of a segment, whereas we may construct S$^1$ by doing so from one segment.
Is there a relationship between what we identify and being or not simple the fundamental group?
Another construction of $S^1$ is by identifying, in $mathbb R$, integers...May be if we identify elements of some infinite simple group (and still obtain a topological space)...
general-topology topological-groups
asked Dec 5 '18 at 15:35
vanmerivanmeri
658
$endgroup$
add a comment |
$begingroup$
What does it mean, topologically, to have a simple fundamental group?
For instance, the torus $S^1 times S^1$ has $mathbb Z times mathbb Z$not simple. The case of $S^1$ is $mathbb Z$, not simple either.
But if instead, in $S^1$ we identify antipodal points, or even, if we identify points after rotating 120 degrees, we would get $mathbb Z/2mathbb Z, mathbb Z/3mathbb Z$...
We can obtain the torus from a $2$ dimensional square by two identifications of a segment, whereas we may construct S$^1$ by doing so from one segment.
Is there a relationship between what we identify and being or not simple the fundamental group?
Another construction of $S^1$ is by identifying, in $mathbb R$, integers...May be if we identify elements of some infinite simple group (and still obtain a topological space)...
general-topology topological-groups
asked Dec 5 '18 at 15:35
vanmerivanmeri
658
$endgroup$
add a comment |
$begingroup$
What does it mean, topologically, to have a simple fundamental group?
For instance, the torus $S^1 times S^1$ has $mathbb Z times mathbb Z$not simple. The case of $S^1$ is $mathbb Z$, not simple either.
But if instead, in $S^1$ we identify antipodal points, or even, if we identify points after rotating 120 degrees, we would get $mathbb Z/2mathbb Z, mathbb Z/3mathbb Z$...
We can obtain the torus from a $2$ dimensional square by two identifications of a segment, whereas we may construct S$^1$ by doing so from one segment.
Is there a relationship between what we identify and being or not simple the fundamental group?
Another construction of $S^1$ is by identifying, in $mathbb R$, integers...May be if we identify elements of some infinite simple group (and still obtain a topological space)...
general-topology topological-groups
asked Dec 5 '18 at 15:35
vanmerivanmeri
658
$endgroup$
What does it mean, topologically, to have a simple fundamental group?
For instance, the torus $S^1 times S^1$ has $mathbb Z times mathbb Z$not simple. The case of $S^1$ is $mathbb Z$, not simple either.
But if instead, in $S^1$ we identify antipodal points, or even, if we identify points after rotating 120 degrees, we would get $mathbb Z/2mathbb Z, mathbb Z/3mathbb Z$...
We can obtain the torus from a $2$ dimensional square by two identifications of a segment, whereas we may construct S$^1$ by doing so from one segment.
Is there a relationship between what we identify and being or not simple the fundamental group?
Another construction of $S^1$ is by identifying, in $mathbb R$, integers...May be if we identify elements of some infinite simple group (and still obtain a topological space)...
general-topology topological-groups
general-topology topological-groups
asked Dec 5 '18 at 15:35
vanmerivanmeri
658
asked Dec 5 '18 at 15:35
vanmerivanmeri
658
edited Dec 5 '18 at 16:03
vanmeri
asked Dec 5 '18 at 15:35
vanmerivanmeri
658
asked Dec 5 '18 at 15:35
vanmerivanmeri
658
asked Dec 5 '18 at 15:35
vanmerivanmeri
658
658
add a comment |
add a comment |
1 Answer
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I'm not sure where you're going with the identification idea. In general there are many ways to construct a given space by making an identification from some other space. Also, the identifications you suggest on $S^1$ do not produce the fundamental groups $Bbb{Z}/n$; the resulting space is still $S^1$, so the fundamental group is still $Bbb{Z}$. Maybe it would help you to gather more examples. (You might consider the lens spaces.)
However, there is a correspondence between covering spaces and subgroups of the fundamental group. This gives a topological description of spaces with simple fundamental group: they are spaces which have only two covering spaces, the trivial cover and the universal cover.
answered Dec 9 '18 at 5:04
Hew WolffHew Wolff
2,260716
$endgroup$
$begingroup$
You need "niceness" assumptions on spaces. In general, a space does not have a universal cover.
$endgroup$
– Paul Frost
Dec 9 '18 at 17:36
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True, but the examples in the question suggest that it is basically about manifolds.
$endgroup$
– Hew Wolff
Dec 12 '18 at 0:41
add a comment |
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$begingroup$
I'm not sure where you're going with the identification idea. In general there are many ways to construct a given space by making an identification from some other space. Also, the identifications you suggest on $S^1$ do not produce the fundamental groups $Bbb{Z}/n$; the resulting space is still $S^1$, so the fundamental group is still $Bbb{Z}$. Maybe it would help you to gather more examples. (You might consider the lens spaces.)
However, there is a correspondence between covering spaces and subgroups of the fundamental group. This gives a topological description of spaces with simple fundamental group: they are spaces which have only two covering spaces, the trivial cover and the universal cover.
answered Dec 9 '18 at 5:04
Hew WolffHew Wolff
2,260716
$endgroup$
$begingroup$
You need "niceness" assumptions on spaces. In general, a space does not have a universal cover.
$endgroup$
– Paul Frost
Dec 9 '18 at 17:36
$begingroup$
True, but the examples in the question suggest that it is basically about manifolds.
$endgroup$
– Hew Wolff
Dec 12 '18 at 0:41
add a comment |
$begingroup$
I'm not sure where you're going with the identification idea. In general there are many ways to construct a given space by making an identification from some other space. Also, the identifications you suggest on $S^1$ do not produce the fundamental groups $Bbb{Z}/n$; the resulting space is still $S^1$, so the fundamental group is still $Bbb{Z}$. Maybe it would help you to gather more examples. (You might consider the lens spaces.)
However, there is a correspondence between covering spaces and subgroups of the fundamental group. This gives a topological description of spaces with simple fundamental group: they are spaces which have only two covering spaces, the trivial cover and the universal cover.
answered Dec 9 '18 at 5:04
Hew WolffHew Wolff
2,260716
$endgroup$
$begingroup$
You need "niceness" assumptions on spaces. In general, a space does not have a universal cover.
$endgroup$
– Paul Frost
Dec 9 '18 at 17:36
$begingroup$
True, but the examples in the question suggest that it is basically about manifolds.
$endgroup$
– Hew Wolff
Dec 12 '18 at 0:41
add a comment |
$begingroup$
I'm not sure where you're going with the identification idea. In general there are many ways to construct a given space by making an identification from some other space. Also, the identifications you suggest on $S^1$ do not produce the fundamental groups $Bbb{Z}/n$; the resulting space is still $S^1$, so the fundamental group is still $Bbb{Z}$. Maybe it would help you to gather more examples. (You might consider the lens spaces.)
However, there is a correspondence between covering spaces and subgroups of the fundamental group. This gives a topological description of spaces with simple fundamental group: they are spaces which have only two covering spaces, the trivial cover and the universal cover.
answered Dec 9 '18 at 5:04
Hew WolffHew Wolff
2,260716
$endgroup$
I'm not sure where you're going with the identification idea. In general there are many ways to construct a given space by making an identification from some other space. Also, the identifications you suggest on $S^1$ do not produce the fundamental groups $Bbb{Z}/n$; the resulting space is still $S^1$, so the fundamental group is still $Bbb{Z}$. Maybe it would help you to gather more examples. (You might consider the lens spaces.)
However, there is a correspondence between covering spaces and subgroups of the fundamental group. This gives a topological description of spaces with simple fundamental group: they are spaces which have only two covering spaces, the trivial cover and the universal cover.
answered Dec 9 '18 at 5:04
Hew WolffHew Wolff
2,260716
answered Dec 9 '18 at 5:04
Hew WolffHew Wolff
2,260716
answered Dec 9 '18 at 5:04
Hew WolffHew Wolff
2,260716
answered Dec 9 '18 at 5:04
Hew WolffHew Wolff
2,260716
2,260716
$begingroup$
You need "niceness" assumptions on spaces. In general, a space does not have a universal cover.
$endgroup$
– Paul Frost
Dec 9 '18 at 17:36
$begingroup$
True, but the examples in the question suggest that it is basically about manifolds.
$endgroup$
– Hew Wolff
Dec 12 '18 at 0:41
add a comment |
$begingroup$
You need "niceness" assumptions on spaces. In general, a space does not have a universal cover.
$endgroup$
– Paul Frost
Dec 9 '18 at 17:36
$begingroup$
True, but the examples in the question suggest that it is basically about manifolds.
$endgroup$
– Hew Wolff
Dec 12 '18 at 0:41
$begingroup$
You need "niceness" assumptions on spaces. In general, a space does not have a universal cover.
$endgroup$
– Paul Frost
Dec 9 '18 at 17:36
$begingroup$
You need "niceness" assumptions on spaces. In general, a space does not have a universal cover.
$endgroup$
– Paul Frost
Dec 9 '18 at 17:36
$begingroup$
True, but the examples in the question suggest that it is basically about manifolds.
$endgroup$
– Hew Wolff
Dec 12 '18 at 0:41
$begingroup$
True, but the examples in the question suggest that it is basically about manifolds.
$endgroup$
– Hew Wolff
Dec 12 '18 at 0:41
add a comment |
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