Is it there a nice way of visualising $pi_3(mathbbS^2)$?
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I was wandering if there is a nice geometric way of visualising $pi_3(mathbbS^2)$. Thanks in advance.
algebraic-topology geometric-topology
asked Sep 10 at 9:16
galois1989
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I was wandering if there is a nice geometric way of visualising $pi_3(mathbbS^2)$. Thanks in advance.
algebraic-topology geometric-topology
asked Sep 10 at 9:16
galois1989
808
6
Can you be a little clearer on what you want? Since $pi_3(S^2) = Bbb Z$, one way to visualize it is as a bunch of dots spaced one unit apart on the number line, but I'm pretty sure that's not what you're after.
– John Hughes
Sep 10 at 9:36
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was wandering if there is a nice geometric way of visualising $pi_3(mathbbS^2)$. Thanks in advance.
algebraic-topology geometric-topology
asked Sep 10 at 9:16
galois1989
808
I was wandering if there is a nice geometric way of visualising $pi_3(mathbbS^2)$. Thanks in advance.
algebraic-topology geometric-topology
algebraic-topology geometric-topology
asked Sep 10 at 9:16
galois1989
808
asked Sep 10 at 9:16
galois1989
808
asked Sep 10 at 9:16
galois1989
808
asked Sep 10 at 9:16
galois1989
808
asked Sep 10 at 9:16
galois1989
808
808
6
Can you be a little clearer on what you want? Since $pi_3(S^2) = Bbb Z$, one way to visualize it is as a bunch of dots spaced one unit apart on the number line, but I'm pretty sure that's not what you're after.
– John Hughes
Sep 10 at 9:36
add a comment |Â
6
Can you be a little clearer on what you want? Since $pi_3(S^2) = Bbb Z$, one way to visualize it is as a bunch of dots spaced one unit apart on the number line, but I'm pretty sure that's not what you're after.
– John Hughes
Sep 10 at 9:36
6
6
Can you be a little clearer on what you want? Since $pi_3(S^2) = Bbb Z$, one way to visualize it is as a bunch of dots spaced one unit apart on the number line, but I'm pretty sure that's not what you're after.
– John Hughes
Sep 10 at 9:36
Can you be a little clearer on what you want? Since $pi_3(S^2) = Bbb Z$, one way to visualize it is as a bunch of dots spaced one unit apart on the number line, but I'm pretty sure that's not what you're after.
– John Hughes
Sep 10 at 9:36
add a comment |Â
2 Answers
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4
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Let me make a stab at amplifying @NicolasHemelsoet's answer, using a bunch of unsubstantiated but plausible claims:
Suppose $f : S^3 to S^2$ is a continuous map. Replace $f$ by a new map (which I'll still call $f$) that's actually smooth, but is homotopic to the original.
Most points of $S^2$ will be "nice" (by Sard's Theorem), so that for any two such nice points $p$ and $q$ in the 2-sphere, their preimages are both 1-manifolds. To keep things simple, let's assume that they're actually connected 1-manifolds, hence are closed curves in the 3-sphere.
The preimage $gamma$ of $p$ can be oriented as follows: suppose $P$ is a point with $f(P) = p$. Then $df(P)$ maps the tangent $u$ to $gamma$ at $P$ to the zero vector, but if you pick two vectors $v$ and $w$ that are perpendicular to $u$ and look at $df(P)[v]$ and $df(P)[w]$, you get a basis for the tangent space of $S^2$ at $p$. If it's an oriented basis, good. If not, swap $v$ and $w$. OK, so now you have two vectors, $v, w$ at $P in S^3$, and their images, under $df(P)$, form an oriented basis for $S^2$. Extend these to $v, w, u$; this will either be an oriented basis for the tangent space to $S^3$ at $P$ or anti-oriented. If it's anti-oriented, reverse $u$. Now the vector $u$ gives you an orientation for the curve $gamma$.
Whew!
So you've got two oriented curves in the 3-sphere, one for $P$, one for $Q$. You compute their linking number $s$. Amazingly enough, the value of $s$ is independent of the choice of $P$ and $Q$, and indeed, of $p$ and $q$ (as long as you stick to "nice" points). That linking number $s$ is an integer which tells you which element of $pi_3(S^2)$ the function $f$ belongs to.
For the hopf map, the preimages of any two points are two linking circles, so its degree is $pm 1$.
answered Sep 10 at 10:03
John Hughes
60k23987
add a comment |Â
up vote
4
down vote
A generator is given by the Hopf fibration $p :S^3 to S^2$.
answered Sep 10 at 9:42
Nicolas Hemelsoet
5,225417
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let me make a stab at amplifying @NicolasHemelsoet's answer, using a bunch of unsubstantiated but plausible claims:
Suppose $f : S^3 to S^2$ is a continuous map. Replace $f$ by a new map (which I'll still call $f$) that's actually smooth, but is homotopic to the original.
Most points of $S^2$ will be "nice" (by Sard's Theorem), so that for any two such nice points $p$ and $q$ in the 2-sphere, their preimages are both 1-manifolds. To keep things simple, let's assume that they're actually connected 1-manifolds, hence are closed curves in the 3-sphere.
The preimage $gamma$ of $p$ can be oriented as follows: suppose $P$ is a point with $f(P) = p$. Then $df(P)$ maps the tangent $u$ to $gamma$ at $P$ to the zero vector, but if you pick two vectors $v$ and $w$ that are perpendicular to $u$ and look at $df(P)[v]$ and $df(P)[w]$, you get a basis for the tangent space of $S^2$ at $p$. If it's an oriented basis, good. If not, swap $v$ and $w$. OK, so now you have two vectors, $v, w$ at $P in S^3$, and their images, under $df(P)$, form an oriented basis for $S^2$. Extend these to $v, w, u$; this will either be an oriented basis for the tangent space to $S^3$ at $P$ or anti-oriented. If it's anti-oriented, reverse $u$. Now the vector $u$ gives you an orientation for the curve $gamma$.
Whew!
So you've got two oriented curves in the 3-sphere, one for $P$, one for $Q$. You compute their linking number $s$. Amazingly enough, the value of $s$ is independent of the choice of $P$ and $Q$, and indeed, of $p$ and $q$ (as long as you stick to "nice" points). That linking number $s$ is an integer which tells you which element of $pi_3(S^2)$ the function $f$ belongs to.
For the hopf map, the preimages of any two points are two linking circles, so its degree is $pm 1$.
answered Sep 10 at 10:03
John Hughes
60k23987
add a comment |Â
up vote
4
down vote
accepted
Let me make a stab at amplifying @NicolasHemelsoet's answer, using a bunch of unsubstantiated but plausible claims:
Suppose $f : S^3 to S^2$ is a continuous map. Replace $f$ by a new map (which I'll still call $f$) that's actually smooth, but is homotopic to the original.
Most points of $S^2$ will be "nice" (by Sard's Theorem), so that for any two such nice points $p$ and $q$ in the 2-sphere, their preimages are both 1-manifolds. To keep things simple, let's assume that they're actually connected 1-manifolds, hence are closed curves in the 3-sphere.
The preimage $gamma$ of $p$ can be oriented as follows: suppose $P$ is a point with $f(P) = p$. Then $df(P)$ maps the tangent $u$ to $gamma$ at $P$ to the zero vector, but if you pick two vectors $v$ and $w$ that are perpendicular to $u$ and look at $df(P)[v]$ and $df(P)[w]$, you get a basis for the tangent space of $S^2$ at $p$. If it's an oriented basis, good. If not, swap $v$ and $w$. OK, so now you have two vectors, $v, w$ at $P in S^3$, and their images, under $df(P)$, form an oriented basis for $S^2$. Extend these to $v, w, u$; this will either be an oriented basis for the tangent space to $S^3$ at $P$ or anti-oriented. If it's anti-oriented, reverse $u$. Now the vector $u$ gives you an orientation for the curve $gamma$.
Whew!
So you've got two oriented curves in the 3-sphere, one for $P$, one for $Q$. You compute their linking number $s$. Amazingly enough, the value of $s$ is independent of the choice of $P$ and $Q$, and indeed, of $p$ and $q$ (as long as you stick to "nice" points). That linking number $s$ is an integer which tells you which element of $pi_3(S^2)$ the function $f$ belongs to.
For the hopf map, the preimages of any two points are two linking circles, so its degree is $pm 1$.
answered Sep 10 at 10:03
John Hughes
60k23987
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let me make a stab at amplifying @NicolasHemelsoet's answer, using a bunch of unsubstantiated but plausible claims:
Suppose $f : S^3 to S^2$ is a continuous map. Replace $f$ by a new map (which I'll still call $f$) that's actually smooth, but is homotopic to the original.
Most points of $S^2$ will be "nice" (by Sard's Theorem), so that for any two such nice points $p$ and $q$ in the 2-sphere, their preimages are both 1-manifolds. To keep things simple, let's assume that they're actually connected 1-manifolds, hence are closed curves in the 3-sphere.
The preimage $gamma$ of $p$ can be oriented as follows: suppose $P$ is a point with $f(P) = p$. Then $df(P)$ maps the tangent $u$ to $gamma$ at $P$ to the zero vector, but if you pick two vectors $v$ and $w$ that are perpendicular to $u$ and look at $df(P)[v]$ and $df(P)[w]$, you get a basis for the tangent space of $S^2$ at $p$. If it's an oriented basis, good. If not, swap $v$ and $w$. OK, so now you have two vectors, $v, w$ at $P in S^3$, and their images, under $df(P)$, form an oriented basis for $S^2$. Extend these to $v, w, u$; this will either be an oriented basis for the tangent space to $S^3$ at $P$ or anti-oriented. If it's anti-oriented, reverse $u$. Now the vector $u$ gives you an orientation for the curve $gamma$.
Whew!
So you've got two oriented curves in the 3-sphere, one for $P$, one for $Q$. You compute their linking number $s$. Amazingly enough, the value of $s$ is independent of the choice of $P$ and $Q$, and indeed, of $p$ and $q$ (as long as you stick to "nice" points). That linking number $s$ is an integer which tells you which element of $pi_3(S^2)$ the function $f$ belongs to.
For the hopf map, the preimages of any two points are two linking circles, so its degree is $pm 1$.
answered Sep 10 at 10:03
John Hughes
60k23987
Let me make a stab at amplifying @NicolasHemelsoet's answer, using a bunch of unsubstantiated but plausible claims:
Suppose $f : S^3 to S^2$ is a continuous map. Replace $f$ by a new map (which I'll still call $f$) that's actually smooth, but is homotopic to the original.
Most points of $S^2$ will be "nice" (by Sard's Theorem), so that for any two such nice points $p$ and $q$ in the 2-sphere, their preimages are both 1-manifolds. To keep things simple, let's assume that they're actually connected 1-manifolds, hence are closed curves in the 3-sphere.
The preimage $gamma$ of $p$ can be oriented as follows: suppose $P$ is a point with $f(P) = p$. Then $df(P)$ maps the tangent $u$ to $gamma$ at $P$ to the zero vector, but if you pick two vectors $v$ and $w$ that are perpendicular to $u$ and look at $df(P)[v]$ and $df(P)[w]$, you get a basis for the tangent space of $S^2$ at $p$. If it's an oriented basis, good. If not, swap $v$ and $w$. OK, so now you have two vectors, $v, w$ at $P in S^3$, and their images, under $df(P)$, form an oriented basis for $S^2$. Extend these to $v, w, u$; this will either be an oriented basis for the tangent space to $S^3$ at $P$ or anti-oriented. If it's anti-oriented, reverse $u$. Now the vector $u$ gives you an orientation for the curve $gamma$.
Whew!
So you've got two oriented curves in the 3-sphere, one for $P$, one for $Q$. You compute their linking number $s$. Amazingly enough, the value of $s$ is independent of the choice of $P$ and $Q$, and indeed, of $p$ and $q$ (as long as you stick to "nice" points). That linking number $s$ is an integer which tells you which element of $pi_3(S^2)$ the function $f$ belongs to.
For the hopf map, the preimages of any two points are two linking circles, so its degree is $pm 1$.
answered Sep 10 at 10:03
John Hughes
60k23987
answered Sep 10 at 10:03
John Hughes
60k23987
answered Sep 10 at 10:03
John Hughes
60k23987
answered Sep 10 at 10:03
John Hughes
60k23987
60k23987
add a comment |Â
add a comment |Â
up vote
4
down vote
A generator is given by the Hopf fibration $p :S^3 to S^2$.
answered Sep 10 at 9:42
Nicolas Hemelsoet
5,225417
add a comment |Â
up vote
4
down vote
A generator is given by the Hopf fibration $p :S^3 to S^2$.
answered Sep 10 at 9:42
Nicolas Hemelsoet
5,225417
add a comment |Â
up vote
4
down vote
up vote
4
down vote
A generator is given by the Hopf fibration $p :S^3 to S^2$.
answered Sep 10 at 9:42
Nicolas Hemelsoet
5,225417
A generator is given by the Hopf fibration $p :S^3 to S^2$.
answered Sep 10 at 9:42
Nicolas Hemelsoet
5,225417
answered Sep 10 at 9:42
Nicolas Hemelsoet
5,225417
answered Sep 10 at 9:42
Nicolas Hemelsoet
5,225417
answered Sep 10 at 9:42
Nicolas Hemelsoet
5,225417
5,225417
add a comment |Â
add a comment |Â
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6
Can you be a little clearer on what you want? Since $pi_3(S^2) = Bbb Z$, one way to visualize it is as a bunch of dots spaced one unit apart on the number line, but I'm pretty sure that's not what you're after.
– John Hughes
Sep 10 at 9:36