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Invariant framing of $S^1$ and “the other oneâ€
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It is known that the left invariant framing of $S^1$ generates $Omega_1^fr$ (the stably framed bordism group of 1-dimensional manifolds). I do not understand how "the other" framing looks like which gives the zero element in $Omega_1^fr$. Can someone explain this to me?
(First I thought it would be the framing obtained by reversing orientation, but this should give the same element in $Omega_1^fr$.)
algebraic-topology differential-topology homotopy-theory
asked Aug 6 at 13:54
Wilhelm L.
31215
add a comment |Â
up vote
3
down vote
favorite
It is known that the left invariant framing of $S^1$ generates $Omega_1^fr$ (the stably framed bordism group of 1-dimensional manifolds). I do not understand how "the other" framing looks like which gives the zero element in $Omega_1^fr$. Can someone explain this to me?
(First I thought it would be the framing obtained by reversing orientation, but this should give the same element in $Omega_1^fr$.)
algebraic-topology differential-topology homotopy-theory
asked Aug 6 at 13:54
Wilhelm L.
31215
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
It is known that the left invariant framing of $S^1$ generates $Omega_1^fr$ (the stably framed bordism group of 1-dimensional manifolds). I do not understand how "the other" framing looks like which gives the zero element in $Omega_1^fr$. Can someone explain this to me?
(First I thought it would be the framing obtained by reversing orientation, but this should give the same element in $Omega_1^fr$.)
algebraic-topology differential-topology homotopy-theory
asked Aug 6 at 13:54
Wilhelm L.
31215
It is known that the left invariant framing of $S^1$ generates $Omega_1^fr$ (the stably framed bordism group of 1-dimensional manifolds). I do not understand how "the other" framing looks like which gives the zero element in $Omega_1^fr$. Can someone explain this to me?
(First I thought it would be the framing obtained by reversing orientation, but this should give the same element in $Omega_1^fr$.)
algebraic-topology differential-topology homotopy-theory
asked Aug 6 at 13:54
Wilhelm L.
31215
asked Aug 6 at 13:54
Wilhelm L.
31215
asked Aug 6 at 13:54
Wilhelm L.
31215
asked Aug 6 at 13:54
Wilhelm L.
31215
31215
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
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accepted
Non-stably the zero element is easy to see i think. just imagine a circle in $mathbbR^2$ with an outward (or inward) pointing normal. You can see by hand that this is framed cobordant to the empty set. But now you can just suspend: Take the same circle with this framing and see it sitting inside $mathbbR^2timesmathbbR^2$ with a constant basis of the normal space in the other two directions. Thus the framing of the standard circle $(x,y,0,0)in mathbbR^4,x^2+y^2=1$ at the point $(x,y,0,0)$ is $e_1=(x,y,0,0),e_2=(0,0,1,0),e_3=(0,0,0,1)$.
The non-zero element in framed cobordism can be seen as follows: take a non-zero contractible loop $S^1rightarrow SO(3)$ and let it act on the basis above pointwise.
EDIT: Consider the circle $S^1subset mathbb R^2$ with the outward pointing normal as a framing. Consider the $W=,x^2+y^2+2z^2=1.$ with framing $(x,y,2z)$. This is a framed nullbordism. I encourage you to make a drawing. You can even go one dimension lower: Take $S^0subset mathbb R$ (two points) with outward normal. This extends over an arc $gammasubset mathbb Rtimes[0,1]$ with boundary on $S^0times0$.
answered Aug 7 at 13:02
Thomas Rot
6,5101642
I dont see that the standard normal framing of $S^1 subset mathbb R^2$ is framed cobordant to the empty set. This would mean, that this framing can be extended to the unit disk, which is clearly not possibly, since this is the identity map on $S^1$.
– Wilhelm L.
2 days ago
The framing needs to be extended to a disc $D^2subset mathbb R^3$. Does that help? Or you can consider the map $S^2rightarrow S^1$ by $(x,y,z)mapsto (z,sqrt(1-z^2))$. The framed submanifold that one obtains from a regular value is the one I describe in my answer. But the map is contractible, hence the framed submanifold is framed nullbordant.
– Thomas Rot
2 days ago
Sorry, I still have problems to understand. This framing is the identity map on $S^1$. So there is no extension to any disk, right?
– Wilhelm L.
2 days ago
@WilhelmL.: Does this edit help?
– Thomas Rot
2 days ago
The map is the identify map as seen as a map $S^1rightarrow S^1$. But the framing is a map $S^1$ to the normal bundle of $S^1$. This is trivial.
– Thomas Rot
2 days ago
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Non-stably the zero element is easy to see i think. just imagine a circle in $mathbbR^2$ with an outward (or inward) pointing normal. You can see by hand that this is framed cobordant to the empty set. But now you can just suspend: Take the same circle with this framing and see it sitting inside $mathbbR^2timesmathbbR^2$ with a constant basis of the normal space in the other two directions. Thus the framing of the standard circle $(x,y,0,0)in mathbbR^4,x^2+y^2=1$ at the point $(x,y,0,0)$ is $e_1=(x,y,0,0),e_2=(0,0,1,0),e_3=(0,0,0,1)$.
The non-zero element in framed cobordism can be seen as follows: take a non-zero contractible loop $S^1rightarrow SO(3)$ and let it act on the basis above pointwise.
EDIT: Consider the circle $S^1subset mathbb R^2$ with the outward pointing normal as a framing. Consider the $W=,x^2+y^2+2z^2=1.$ with framing $(x,y,2z)$. This is a framed nullbordism. I encourage you to make a drawing. You can even go one dimension lower: Take $S^0subset mathbb R$ (two points) with outward normal. This extends over an arc $gammasubset mathbb Rtimes[0,1]$ with boundary on $S^0times0$.
answered Aug 7 at 13:02
Thomas Rot
6,5101642
I dont see that the standard normal framing of $S^1 subset mathbb R^2$ is framed cobordant to the empty set. This would mean, that this framing can be extended to the unit disk, which is clearly not possibly, since this is the identity map on $S^1$.
– Wilhelm L.
2 days ago
The framing needs to be extended to a disc $D^2subset mathbb R^3$. Does that help? Or you can consider the map $S^2rightarrow S^1$ by $(x,y,z)mapsto (z,sqrt(1-z^2))$. The framed submanifold that one obtains from a regular value is the one I describe in my answer. But the map is contractible, hence the framed submanifold is framed nullbordant.
– Thomas Rot
2 days ago
Sorry, I still have problems to understand. This framing is the identity map on $S^1$. So there is no extension to any disk, right?
– Wilhelm L.
2 days ago
@WilhelmL.: Does this edit help?
– Thomas Rot
2 days ago
The map is the identify map as seen as a map $S^1rightarrow S^1$. But the framing is a map $S^1$ to the normal bundle of $S^1$. This is trivial.
– Thomas Rot
2 days ago
 |Â
show 2 more comments
up vote
3
down vote
accepted
Non-stably the zero element is easy to see i think. just imagine a circle in $mathbbR^2$ with an outward (or inward) pointing normal. You can see by hand that this is framed cobordant to the empty set. But now you can just suspend: Take the same circle with this framing and see it sitting inside $mathbbR^2timesmathbbR^2$ with a constant basis of the normal space in the other two directions. Thus the framing of the standard circle $(x,y,0,0)in mathbbR^4,x^2+y^2=1$ at the point $(x,y,0,0)$ is $e_1=(x,y,0,0),e_2=(0,0,1,0),e_3=(0,0,0,1)$.
The non-zero element in framed cobordism can be seen as follows: take a non-zero contractible loop $S^1rightarrow SO(3)$ and let it act on the basis above pointwise.
EDIT: Consider the circle $S^1subset mathbb R^2$ with the outward pointing normal as a framing. Consider the $W=,x^2+y^2+2z^2=1.$ with framing $(x,y,2z)$. This is a framed nullbordism. I encourage you to make a drawing. You can even go one dimension lower: Take $S^0subset mathbb R$ (two points) with outward normal. This extends over an arc $gammasubset mathbb Rtimes[0,1]$ with boundary on $S^0times0$.
answered Aug 7 at 13:02
Thomas Rot
6,5101642
I dont see that the standard normal framing of $S^1 subset mathbb R^2$ is framed cobordant to the empty set. This would mean, that this framing can be extended to the unit disk, which is clearly not possibly, since this is the identity map on $S^1$.
– Wilhelm L.
2 days ago
The framing needs to be extended to a disc $D^2subset mathbb R^3$. Does that help? Or you can consider the map $S^2rightarrow S^1$ by $(x,y,z)mapsto (z,sqrt(1-z^2))$. The framed submanifold that one obtains from a regular value is the one I describe in my answer. But the map is contractible, hence the framed submanifold is framed nullbordant.
– Thomas Rot
2 days ago
Sorry, I still have problems to understand. This framing is the identity map on $S^1$. So there is no extension to any disk, right?
– Wilhelm L.
2 days ago
@WilhelmL.: Does this edit help?
– Thomas Rot
2 days ago
The map is the identify map as seen as a map $S^1rightarrow S^1$. But the framing is a map $S^1$ to the normal bundle of $S^1$. This is trivial.
– Thomas Rot
2 days ago
 |Â
show 2 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Non-stably the zero element is easy to see i think. just imagine a circle in $mathbbR^2$ with an outward (or inward) pointing normal. You can see by hand that this is framed cobordant to the empty set. But now you can just suspend: Take the same circle with this framing and see it sitting inside $mathbbR^2timesmathbbR^2$ with a constant basis of the normal space in the other two directions. Thus the framing of the standard circle $(x,y,0,0)in mathbbR^4,x^2+y^2=1$ at the point $(x,y,0,0)$ is $e_1=(x,y,0,0),e_2=(0,0,1,0),e_3=(0,0,0,1)$.
The non-zero element in framed cobordism can be seen as follows: take a non-zero contractible loop $S^1rightarrow SO(3)$ and let it act on the basis above pointwise.
EDIT: Consider the circle $S^1subset mathbb R^2$ with the outward pointing normal as a framing. Consider the $W=,x^2+y^2+2z^2=1.$ with framing $(x,y,2z)$. This is a framed nullbordism. I encourage you to make a drawing. You can even go one dimension lower: Take $S^0subset mathbb R$ (two points) with outward normal. This extends over an arc $gammasubset mathbb Rtimes[0,1]$ with boundary on $S^0times0$.
answered Aug 7 at 13:02
Thomas Rot
6,5101642
Non-stably the zero element is easy to see i think. just imagine a circle in $mathbbR^2$ with an outward (or inward) pointing normal. You can see by hand that this is framed cobordant to the empty set. But now you can just suspend: Take the same circle with this framing and see it sitting inside $mathbbR^2timesmathbbR^2$ with a constant basis of the normal space in the other two directions. Thus the framing of the standard circle $(x,y,0,0)in mathbbR^4,x^2+y^2=1$ at the point $(x,y,0,0)$ is $e_1=(x,y,0,0),e_2=(0,0,1,0),e_3=(0,0,0,1)$.
The non-zero element in framed cobordism can be seen as follows: take a non-zero contractible loop $S^1rightarrow SO(3)$ and let it act on the basis above pointwise.
EDIT: Consider the circle $S^1subset mathbb R^2$ with the outward pointing normal as a framing. Consider the $W=,x^2+y^2+2z^2=1.$ with framing $(x,y,2z)$. This is a framed nullbordism. I encourage you to make a drawing. You can even go one dimension lower: Take $S^0subset mathbb R$ (two points) with outward normal. This extends over an arc $gammasubset mathbb Rtimes[0,1]$ with boundary on $S^0times0$.
answered Aug 7 at 13:02
Thomas Rot
6,5101642
edited 2 days ago
answered Aug 7 at 13:02
Thomas Rot
6,5101642
answered Aug 7 at 13:02
Thomas Rot
6,5101642
answered Aug 7 at 13:02
Thomas Rot
6,5101642
6,5101642
I dont see that the standard normal framing of $S^1 subset mathbb R^2$ is framed cobordant to the empty set. This would mean, that this framing can be extended to the unit disk, which is clearly not possibly, since this is the identity map on $S^1$.
– Wilhelm L.
2 days ago
The framing needs to be extended to a disc $D^2subset mathbb R^3$. Does that help? Or you can consider the map $S^2rightarrow S^1$ by $(x,y,z)mapsto (z,sqrt(1-z^2))$. The framed submanifold that one obtains from a regular value is the one I describe in my answer. But the map is contractible, hence the framed submanifold is framed nullbordant.
– Thomas Rot
2 days ago
Sorry, I still have problems to understand. This framing is the identity map on $S^1$. So there is no extension to any disk, right?
– Wilhelm L.
2 days ago
@WilhelmL.: Does this edit help?
– Thomas Rot
2 days ago
The map is the identify map as seen as a map $S^1rightarrow S^1$. But the framing is a map $S^1$ to the normal bundle of $S^1$. This is trivial.
– Thomas Rot
2 days ago
 |Â
show 2 more comments
I dont see that the standard normal framing of $S^1 subset mathbb R^2$ is framed cobordant to the empty set. This would mean, that this framing can be extended to the unit disk, which is clearly not possibly, since this is the identity map on $S^1$.
– Wilhelm L.
2 days ago
The framing needs to be extended to a disc $D^2subset mathbb R^3$. Does that help? Or you can consider the map $S^2rightarrow S^1$ by $(x,y,z)mapsto (z,sqrt(1-z^2))$. The framed submanifold that one obtains from a regular value is the one I describe in my answer. But the map is contractible, hence the framed submanifold is framed nullbordant.
– Thomas Rot
2 days ago
Sorry, I still have problems to understand. This framing is the identity map on $S^1$. So there is no extension to any disk, right?
– Wilhelm L.
2 days ago
@WilhelmL.: Does this edit help?
– Thomas Rot
2 days ago
The map is the identify map as seen as a map $S^1rightarrow S^1$. But the framing is a map $S^1$ to the normal bundle of $S^1$. This is trivial.
– Thomas Rot
2 days ago
I dont see that the standard normal framing of $S^1 subset mathbb R^2$ is framed cobordant to the empty set. This would mean, that this framing can be extended to the unit disk, which is clearly not possibly, since this is the identity map on $S^1$.
– Wilhelm L.
2 days ago
I dont see that the standard normal framing of $S^1 subset mathbb R^2$ is framed cobordant to the empty set. This would mean, that this framing can be extended to the unit disk, which is clearly not possibly, since this is the identity map on $S^1$.
– Wilhelm L.
2 days ago
The framing needs to be extended to a disc $D^2subset mathbb R^3$. Does that help? Or you can consider the map $S^2rightarrow S^1$ by $(x,y,z)mapsto (z,sqrt(1-z^2))$. The framed submanifold that one obtains from a regular value is the one I describe in my answer. But the map is contractible, hence the framed submanifold is framed nullbordant.
– Thomas Rot
2 days ago
The framing needs to be extended to a disc $D^2subset mathbb R^3$. Does that help? Or you can consider the map $S^2rightarrow S^1$ by $(x,y,z)mapsto (z,sqrt(1-z^2))$. The framed submanifold that one obtains from a regular value is the one I describe in my answer. But the map is contractible, hence the framed submanifold is framed nullbordant.
– Thomas Rot
2 days ago
Sorry, I still have problems to understand. This framing is the identity map on $S^1$. So there is no extension to any disk, right?
– Wilhelm L.
2 days ago
Sorry, I still have problems to understand. This framing is the identity map on $S^1$. So there is no extension to any disk, right?
– Wilhelm L.
2 days ago
@WilhelmL.: Does this edit help?
– Thomas Rot
2 days ago
@WilhelmL.: Does this edit help?
– Thomas Rot
2 days ago
The map is the identify map as seen as a map $S^1rightarrow S^1$. But the framing is a map $S^1$ to the normal bundle of $S^1$. This is trivial.
– Thomas Rot
2 days ago
The map is the identify map as seen as a map $S^1rightarrow S^1$. But the framing is a map $S^1$ to the normal bundle of $S^1$. This is trivial.
– Thomas Rot
2 days ago
 |Â
show 2 more comments
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