Is there an example of a manifold with fundamental group $mathbb Z/3 mathbb Z$?
I feel a little confused because I was told that there exist some manifolds with fundamental group $mathbb Z/3 mathbb Z$, but I can’t find an example, On the other hand, since any manifold $M$ has an orientable $2$-cover, which means $pi_1(M)$ has a subgroup of index $2$ which seems to be a contradiction with $pi_1(M)$ can be $mathbb Z/3 mathbb Z$.
algebraic-topology manifolds geometric-topology
asked Nov 30 '18 at 20:01
DannyDanny
1,062412
add a comment |
I feel a little confused because I was told that there exist some manifolds with fundamental group $mathbb Z/3 mathbb Z$, but I can’t find an example, On the other hand, since any manifold $M$ has an orientable $2$-cover, which means $pi_1(M)$ has a subgroup of index $2$ which seems to be a contradiction with $pi_1(M)$ can be $mathbb Z/3 mathbb Z$.
algebraic-topology manifolds geometric-topology
asked Nov 30 '18 at 20:01
DannyDanny
1,062412
5
What your covering argument shows is that the orientation cover of a manifold with fundamental group $mathbb{Z}/3mathbb{Z}$ must not be connected, as indeed happens for every orientable manifold.
– Kevin Carlson
Nov 30 '18 at 20:09
1
I would like to point out that any finitely presented group is the fundamental group of some manifolds (in a similar way that any group is the fundamental group of a 2-complex).
– Paul Plummer
Nov 30 '18 at 20:17
add a comment |
I feel a little confused because I was told that there exist some manifolds with fundamental group $mathbb Z/3 mathbb Z$, but I can’t find an example, On the other hand, since any manifold $M$ has an orientable $2$-cover, which means $pi_1(M)$ has a subgroup of index $2$ which seems to be a contradiction with $pi_1(M)$ can be $mathbb Z/3 mathbb Z$.
algebraic-topology manifolds geometric-topology
asked Nov 30 '18 at 20:01
DannyDanny
1,062412
I feel a little confused because I was told that there exist some manifolds with fundamental group $mathbb Z/3 mathbb Z$, but I can’t find an example, On the other hand, since any manifold $M$ has an orientable $2$-cover, which means $pi_1(M)$ has a subgroup of index $2$ which seems to be a contradiction with $pi_1(M)$ can be $mathbb Z/3 mathbb Z$.
algebraic-topology manifolds geometric-topology
algebraic-topology manifolds geometric-topology
asked Nov 30 '18 at 20:01
DannyDanny
1,062412
asked Nov 30 '18 at 20:01
DannyDanny
1,062412
edited Nov 30 '18 at 20:04
user621469
asked Nov 30 '18 at 20:01
DannyDanny
1,062412
asked Nov 30 '18 at 20:01
DannyDanny
1,062412
asked Nov 30 '18 at 20:01
DannyDanny
1,062412
1,062412
5
What your covering argument shows is that the orientation cover of a manifold with fundamental group $mathbb{Z}/3mathbb{Z}$ must not be connected, as indeed happens for every orientable manifold.
– Kevin Carlson
Nov 30 '18 at 20:09
1
I would like to point out that any finitely presented group is the fundamental group of some manifolds (in a similar way that any group is the fundamental group of a 2-complex).
– Paul Plummer
Nov 30 '18 at 20:17
add a comment |
5
What your covering argument shows is that the orientation cover of a manifold with fundamental group $mathbb{Z}/3mathbb{Z}$ must not be connected, as indeed happens for every orientable manifold.
– Kevin Carlson
Nov 30 '18 at 20:09
1
I would like to point out that any finitely presented group is the fundamental group of some manifolds (in a similar way that any group is the fundamental group of a 2-complex).
– Paul Plummer
Nov 30 '18 at 20:17
5
5
What your covering argument shows is that the orientation cover of a manifold with fundamental group $mathbb{Z}/3mathbb{Z}$ must not be connected, as indeed happens for every orientable manifold.
– Kevin Carlson
Nov 30 '18 at 20:09
What your covering argument shows is that the orientation cover of a manifold with fundamental group $mathbb{Z}/3mathbb{Z}$ must not be connected, as indeed happens for every orientable manifold.
– Kevin Carlson
Nov 30 '18 at 20:09
1
1
I would like to point out that any finitely presented group is the fundamental group of some manifolds (in a similar way that any group is the fundamental group of a 2-complex).
– Paul Plummer
Nov 30 '18 at 20:17
I would like to point out that any finitely presented group is the fundamental group of some manifolds (in a similar way that any group is the fundamental group of a 2-complex).
– Paul Plummer
Nov 30 '18 at 20:17
add a comment |
1 Answer
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There are infinitely many $3$-manifolds whose fundamental groups isomorphic to $mathbb Z / 3mathbb Z$.
Indeed, for $mathrm gcd(3,q)=1$, take any Lens space $L(3,q)$.
6
all $L(3,q)$ are diffeomorphic. there is only one closed 3-manifold with fundamental group $Bbb Z/3$ up to diffeomorphism, and two up to oriented diffeomorphism.
– Mike Miller
Nov 30 '18 at 21:00
add a comment |
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1 Answer
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1 Answer
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active
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There are infinitely many $3$-manifolds whose fundamental groups isomorphic to $mathbb Z / 3mathbb Z$.
Indeed, for $mathrm gcd(3,q)=1$, take any Lens space $L(3,q)$.
6
all $L(3,q)$ are diffeomorphic. there is only one closed 3-manifold with fundamental group $Bbb Z/3$ up to diffeomorphism, and two up to oriented diffeomorphism.
– Mike Miller
Nov 30 '18 at 21:00
add a comment |
There are infinitely many $3$-manifolds whose fundamental groups isomorphic to $mathbb Z / 3mathbb Z$.
Indeed, for $mathrm gcd(3,q)=1$, take any Lens space $L(3,q)$.
6
all $L(3,q)$ are diffeomorphic. there is only one closed 3-manifold with fundamental group $Bbb Z/3$ up to diffeomorphism, and two up to oriented diffeomorphism.
– Mike Miller
Nov 30 '18 at 21:00
add a comment |
There are infinitely many $3$-manifolds whose fundamental groups isomorphic to $mathbb Z / 3mathbb Z$.
Indeed, for $mathrm gcd(3,q)=1$, take any Lens space $L(3,q)$.
There are infinitely many $3$-manifolds whose fundamental groups isomorphic to $mathbb Z / 3mathbb Z$.
Indeed, for $mathrm gcd(3,q)=1$, take any Lens space $L(3,q)$.
edited Nov 30 '18 at 20:18
answered Nov 30 '18 at 20:13
user621469
6
all $L(3,q)$ are diffeomorphic. there is only one closed 3-manifold with fundamental group $Bbb Z/3$ up to diffeomorphism, and two up to oriented diffeomorphism.
– Mike Miller
Nov 30 '18 at 21:00
add a comment |
6
all $L(3,q)$ are diffeomorphic. there is only one closed 3-manifold with fundamental group $Bbb Z/3$ up to diffeomorphism, and two up to oriented diffeomorphism.
– Mike Miller
Nov 30 '18 at 21:00
6
6
all $L(3,q)$ are diffeomorphic. there is only one closed 3-manifold with fundamental group $Bbb Z/3$ up to diffeomorphism, and two up to oriented diffeomorphism.
– Mike Miller
Nov 30 '18 at 21:00
all $L(3,q)$ are diffeomorphic. there is only one closed 3-manifold with fundamental group $Bbb Z/3$ up to diffeomorphism, and two up to oriented diffeomorphism.
– Mike Miller
Nov 30 '18 at 21:00
add a comment |
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5
What your covering argument shows is that the orientation cover of a manifold with fundamental group $mathbb{Z}/3mathbb{Z}$ must not be connected, as indeed happens for every orientable manifold.
– Kevin Carlson
Nov 30 '18 at 20:09
1
I would like to point out that any finitely presented group is the fundamental group of some manifolds (in a similar way that any group is the fundamental group of a 2-complex).
– Paul Plummer
Nov 30 '18 at 20:17