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Classification of maps $f:mathbbS^1timesmathbbS^2tomathbbS^2$ up to homotopy and geometrical meaning. [closed]
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I have two questions.
First, how can one classify maps $f:mathbbS^1timesmathbbS^2tomathbbS^2$ up to homotopy?
Second, how to interpret geometrically what distinguishes them? I mean, if two maps are non homotopic (let's say they appear to be non homotopic because some invariants distinguish them), I would be really interested in the geometrical interpretation of these invariants. Thank you in advance.
algebraic-topology geometric-topology
edited Jul 15 at 8:55
Arnaud Mortier
19.2k22159
asked Jul 14 at 16:15
galois1989
307
closed as off-topic by amWhy, Chris Custer, John Ma, José Carlos Santos, Xander Henderson Jul 15 at 0:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Chris Custer, John Ma, Jos̩ Carlos Santos, Xander Henderson
add a comment |Â
up vote
2
down vote
favorite
I have two questions.
First, how can one classify maps $f:mathbbS^1timesmathbbS^2tomathbbS^2$ up to homotopy?
Second, how to interpret geometrically what distinguishes them? I mean, if two maps are non homotopic (let's say they appear to be non homotopic because some invariants distinguish them), I would be really interested in the geometrical interpretation of these invariants. Thank you in advance.
algebraic-topology geometric-topology
edited Jul 15 at 8:55
Arnaud Mortier
19.2k22159
asked Jul 14 at 16:15
galois1989
307
closed as off-topic by amWhy, Chris Custer, John Ma, José Carlos Santos, Xander Henderson Jul 15 at 0:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Chris Custer, John Ma, Jos̩ Carlos Santos, Xander Henderson
If $M$ is an oriented 3-manifold, then $TM$ is trivial, and so picking a trivialization, a vector field of unit norm is the same as a map to $S^2$. A vector field of unit norm has an orthogonal complement, which is oriented; and changing the map to $S^2$ by a homotopy changes this plane field by a homotopy. So $[M, S^2]$ is oriented plane fields on $M$ up to homotopy.
– Mike Miller
Jul 14 at 16:38
I misread this as $S^1 times S^1$. Apologies.
– Balarka Sen
Jul 14 at 16:41
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have two questions.
First, how can one classify maps $f:mathbbS^1timesmathbbS^2tomathbbS^2$ up to homotopy?
Second, how to interpret geometrically what distinguishes them? I mean, if two maps are non homotopic (let's say they appear to be non homotopic because some invariants distinguish them), I would be really interested in the geometrical interpretation of these invariants. Thank you in advance.
algebraic-topology geometric-topology
edited Jul 15 at 8:55
Arnaud Mortier
19.2k22159
asked Jul 14 at 16:15
galois1989
307
I have two questions.
First, how can one classify maps $f:mathbbS^1timesmathbbS^2tomathbbS^2$ up to homotopy?
Second, how to interpret geometrically what distinguishes them? I mean, if two maps are non homotopic (let's say they appear to be non homotopic because some invariants distinguish them), I would be really interested in the geometrical interpretation of these invariants. Thank you in advance.
algebraic-topology geometric-topology
edited Jul 15 at 8:55
Arnaud Mortier
19.2k22159
asked Jul 14 at 16:15
galois1989
307
edited Jul 15 at 8:55
Arnaud Mortier
19.2k22159
edited Jul 15 at 8:55
Arnaud Mortier
19.2k22159
edited Jul 15 at 8:55
Arnaud Mortier
19.2k22159
19.2k22159
asked Jul 14 at 16:15
galois1989
307
asked Jul 14 at 16:15
galois1989
307
asked Jul 14 at 16:15
galois1989
307
307
closed as off-topic by amWhy, Chris Custer, John Ma, José Carlos Santos, Xander Henderson Jul 15 at 0:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Chris Custer, John Ma, Jos̩ Carlos Santos, Xander Henderson
closed as off-topic by amWhy, Chris Custer, John Ma, José Carlos Santos, Xander Henderson Jul 15 at 0:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Chris Custer, John Ma, Jos̩ Carlos Santos, Xander Henderson
If $M$ is an oriented 3-manifold, then $TM$ is trivial, and so picking a trivialization, a vector field of unit norm is the same as a map to $S^2$. A vector field of unit norm has an orthogonal complement, which is oriented; and changing the map to $S^2$ by a homotopy changes this plane field by a homotopy. So $[M, S^2]$ is oriented plane fields on $M$ up to homotopy.
– Mike Miller
Jul 14 at 16:38
I misread this as $S^1 times S^1$. Apologies.
– Balarka Sen
Jul 14 at 16:41
add a comment |Â
If $M$ is an oriented 3-manifold, then $TM$ is trivial, and so picking a trivialization, a vector field of unit norm is the same as a map to $S^2$. A vector field of unit norm has an orthogonal complement, which is oriented; and changing the map to $S^2$ by a homotopy changes this plane field by a homotopy. So $[M, S^2]$ is oriented plane fields on $M$ up to homotopy.
– Mike Miller
Jul 14 at 16:38
I misread this as $S^1 times S^1$. Apologies.
– Balarka Sen
Jul 14 at 16:41
If $M$ is an oriented 3-manifold, then $TM$ is trivial, and so picking a trivialization, a vector field of unit norm is the same as a map to $S^2$. A vector field of unit norm has an orthogonal complement, which is oriented; and changing the map to $S^2$ by a homotopy changes this plane field by a homotopy. So $[M, S^2]$ is oriented plane fields on $M$ up to homotopy.
– Mike Miller
Jul 14 at 16:38
If $M$ is an oriented 3-manifold, then $TM$ is trivial, and so picking a trivialization, a vector field of unit norm is the same as a map to $S^2$. A vector field of unit norm has an orthogonal complement, which is oriented; and changing the map to $S^2$ by a homotopy changes this plane field by a homotopy. So $[M, S^2]$ is oriented plane fields on $M$ up to homotopy.
– Mike Miller
Jul 14 at 16:38
I misread this as $S^1 times S^1$. Apologies.
– Balarka Sen
Jul 14 at 16:41
I misread this as $S^1 times S^1$. Apologies.
– Balarka Sen
Jul 14 at 16:41
add a comment |Â
1 Answer
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oldest
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up vote
3
down vote
accepted
A map $Bbb S^1times Bbb S^2toBbb S^2$ is a free loop in the space of maps $Bbb S^2toBbb S^2$.
This means that you have to look at
- the set of connected components of maps $Bbb S^2toBbb S^2$: they are indexed by $pi_2(Bbb S^2)=Bbb Z$. I write "=" rather than "$simeq$" because the identity map is your favourite generator.
- for each connected component $C_i, iin Bbb Z$, consider $pi_1(C_i)$.
To see in which connected component $C_i$ a map $Bbb S^1times Bbb S^2toBbb S^2$ is, you just need the degree $i$ of the induced map $cdottimes Bbb S^2toBbb S^2$ where $cdot$ is an arbitrary point of the circle.
Now to compute $pi_1(C_i)$, I can't see just now how to do that easily for all $i$, but at least it is clear that $pi_1(C_0)=operatorname1$, and I believe that one should have $pi_1(C_1)simeqpi_1(SO(3))simeq Bbb Z/2Bbb Z$.
answered Jul 14 at 16:41
Arnaud Mortier
19.2k22159
1
Good idea. In fact the space of maps are essentially computed here; for the component of degree $k$ one has $pi_1 textMap_k = Bbb Z/2|k|$.
– Mike Miller
Jul 14 at 17:06
@MikeMiller Good reference, thanks!
– Arnaud Mortier
Jul 14 at 17:48
Thank you all a lot! Could you please explain me why is it obviuos that $pi_1(C_0)$ is the trivial group? And, also, the intuition between $pi_1(C_1)=pi_1(SO(3))?$
– galois1989
Jul 15 at 8:58
Ok, I think that the first question is obvious because being a degree 0 map gives you the homotopy to the constant loop just by the nullhomotopy. But, I still do not see the intuition behind the equality $pi_1(C_1)=pi_1(SO(3))$
– galois1989
Jul 15 at 9:17
1
@galois1989 Elements of $SO(3)$ are particular maps $Bbb S^2to Bbb S^2$ of degree $1$, namely isometries. My guts told me that the space of maps of degree $1$ should have the same homotopy type as the subspace of isometries.
– Arnaud Mortier
Jul 15 at 10:06
 |Â
show 7 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
A map $Bbb S^1times Bbb S^2toBbb S^2$ is a free loop in the space of maps $Bbb S^2toBbb S^2$.
This means that you have to look at
- the set of connected components of maps $Bbb S^2toBbb S^2$: they are indexed by $pi_2(Bbb S^2)=Bbb Z$. I write "=" rather than "$simeq$" because the identity map is your favourite generator.
- for each connected component $C_i, iin Bbb Z$, consider $pi_1(C_i)$.
To see in which connected component $C_i$ a map $Bbb S^1times Bbb S^2toBbb S^2$ is, you just need the degree $i$ of the induced map $cdottimes Bbb S^2toBbb S^2$ where $cdot$ is an arbitrary point of the circle.
Now to compute $pi_1(C_i)$, I can't see just now how to do that easily for all $i$, but at least it is clear that $pi_1(C_0)=operatorname1$, and I believe that one should have $pi_1(C_1)simeqpi_1(SO(3))simeq Bbb Z/2Bbb Z$.
answered Jul 14 at 16:41
Arnaud Mortier
19.2k22159
1
Good idea. In fact the space of maps are essentially computed here; for the component of degree $k$ one has $pi_1 textMap_k = Bbb Z/2|k|$.
– Mike Miller
Jul 14 at 17:06
@MikeMiller Good reference, thanks!
– Arnaud Mortier
Jul 14 at 17:48
Thank you all a lot! Could you please explain me why is it obviuos that $pi_1(C_0)$ is the trivial group? And, also, the intuition between $pi_1(C_1)=pi_1(SO(3))?$
– galois1989
Jul 15 at 8:58
Ok, I think that the first question is obvious because being a degree 0 map gives you the homotopy to the constant loop just by the nullhomotopy. But, I still do not see the intuition behind the equality $pi_1(C_1)=pi_1(SO(3))$
– galois1989
Jul 15 at 9:17
1
@galois1989 Elements of $SO(3)$ are particular maps $Bbb S^2to Bbb S^2$ of degree $1$, namely isometries. My guts told me that the space of maps of degree $1$ should have the same homotopy type as the subspace of isometries.
– Arnaud Mortier
Jul 15 at 10:06
 |Â
show 7 more comments
up vote
3
down vote
accepted
A map $Bbb S^1times Bbb S^2toBbb S^2$ is a free loop in the space of maps $Bbb S^2toBbb S^2$.
This means that you have to look at
- the set of connected components of maps $Bbb S^2toBbb S^2$: they are indexed by $pi_2(Bbb S^2)=Bbb Z$. I write "=" rather than "$simeq$" because the identity map is your favourite generator.
- for each connected component $C_i, iin Bbb Z$, consider $pi_1(C_i)$.
To see in which connected component $C_i$ a map $Bbb S^1times Bbb S^2toBbb S^2$ is, you just need the degree $i$ of the induced map $cdottimes Bbb S^2toBbb S^2$ where $cdot$ is an arbitrary point of the circle.
Now to compute $pi_1(C_i)$, I can't see just now how to do that easily for all $i$, but at least it is clear that $pi_1(C_0)=operatorname1$, and I believe that one should have $pi_1(C_1)simeqpi_1(SO(3))simeq Bbb Z/2Bbb Z$.
answered Jul 14 at 16:41
Arnaud Mortier
19.2k22159
1
Good idea. In fact the space of maps are essentially computed here; for the component of degree $k$ one has $pi_1 textMap_k = Bbb Z/2|k|$.
– Mike Miller
Jul 14 at 17:06
@MikeMiller Good reference, thanks!
– Arnaud Mortier
Jul 14 at 17:48
Thank you all a lot! Could you please explain me why is it obviuos that $pi_1(C_0)$ is the trivial group? And, also, the intuition between $pi_1(C_1)=pi_1(SO(3))?$
– galois1989
Jul 15 at 8:58
Ok, I think that the first question is obvious because being a degree 0 map gives you the homotopy to the constant loop just by the nullhomotopy. But, I still do not see the intuition behind the equality $pi_1(C_1)=pi_1(SO(3))$
– galois1989
Jul 15 at 9:17
1
@galois1989 Elements of $SO(3)$ are particular maps $Bbb S^2to Bbb S^2$ of degree $1$, namely isometries. My guts told me that the space of maps of degree $1$ should have the same homotopy type as the subspace of isometries.
– Arnaud Mortier
Jul 15 at 10:06
 |Â
show 7 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
A map $Bbb S^1times Bbb S^2toBbb S^2$ is a free loop in the space of maps $Bbb S^2toBbb S^2$.
This means that you have to look at
- the set of connected components of maps $Bbb S^2toBbb S^2$: they are indexed by $pi_2(Bbb S^2)=Bbb Z$. I write "=" rather than "$simeq$" because the identity map is your favourite generator.
- for each connected component $C_i, iin Bbb Z$, consider $pi_1(C_i)$.
To see in which connected component $C_i$ a map $Bbb S^1times Bbb S^2toBbb S^2$ is, you just need the degree $i$ of the induced map $cdottimes Bbb S^2toBbb S^2$ where $cdot$ is an arbitrary point of the circle.
Now to compute $pi_1(C_i)$, I can't see just now how to do that easily for all $i$, but at least it is clear that $pi_1(C_0)=operatorname1$, and I believe that one should have $pi_1(C_1)simeqpi_1(SO(3))simeq Bbb Z/2Bbb Z$.
answered Jul 14 at 16:41
Arnaud Mortier
19.2k22159
A map $Bbb S^1times Bbb S^2toBbb S^2$ is a free loop in the space of maps $Bbb S^2toBbb S^2$.
This means that you have to look at
- the set of connected components of maps $Bbb S^2toBbb S^2$: they are indexed by $pi_2(Bbb S^2)=Bbb Z$. I write "=" rather than "$simeq$" because the identity map is your favourite generator.
- for each connected component $C_i, iin Bbb Z$, consider $pi_1(C_i)$.
To see in which connected component $C_i$ a map $Bbb S^1times Bbb S^2toBbb S^2$ is, you just need the degree $i$ of the induced map $cdottimes Bbb S^2toBbb S^2$ where $cdot$ is an arbitrary point of the circle.
Now to compute $pi_1(C_i)$, I can't see just now how to do that easily for all $i$, but at least it is clear that $pi_1(C_0)=operatorname1$, and I believe that one should have $pi_1(C_1)simeqpi_1(SO(3))simeq Bbb Z/2Bbb Z$.
answered Jul 14 at 16:41
Arnaud Mortier
19.2k22159
answered Jul 14 at 16:41
Arnaud Mortier
19.2k22159
answered Jul 14 at 16:41
Arnaud Mortier
19.2k22159
answered Jul 14 at 16:41
Arnaud Mortier
19.2k22159
19.2k22159
1
Good idea. In fact the space of maps are essentially computed here; for the component of degree $k$ one has $pi_1 textMap_k = Bbb Z/2|k|$.
– Mike Miller
Jul 14 at 17:06
@MikeMiller Good reference, thanks!
– Arnaud Mortier
Jul 14 at 17:48
Thank you all a lot! Could you please explain me why is it obviuos that $pi_1(C_0)$ is the trivial group? And, also, the intuition between $pi_1(C_1)=pi_1(SO(3))?$
– galois1989
Jul 15 at 8:58
Ok, I think that the first question is obvious because being a degree 0 map gives you the homotopy to the constant loop just by the nullhomotopy. But, I still do not see the intuition behind the equality $pi_1(C_1)=pi_1(SO(3))$
– galois1989
Jul 15 at 9:17
1
@galois1989 Elements of $SO(3)$ are particular maps $Bbb S^2to Bbb S^2$ of degree $1$, namely isometries. My guts told me that the space of maps of degree $1$ should have the same homotopy type as the subspace of isometries.
– Arnaud Mortier
Jul 15 at 10:06
 |Â
show 7 more comments
1
Good idea. In fact the space of maps are essentially computed here; for the component of degree $k$ one has $pi_1 textMap_k = Bbb Z/2|k|$.
– Mike Miller
Jul 14 at 17:06
@MikeMiller Good reference, thanks!
– Arnaud Mortier
Jul 14 at 17:48
Thank you all a lot! Could you please explain me why is it obviuos that $pi_1(C_0)$ is the trivial group? And, also, the intuition between $pi_1(C_1)=pi_1(SO(3))?$
– galois1989
Jul 15 at 8:58
Ok, I think that the first question is obvious because being a degree 0 map gives you the homotopy to the constant loop just by the nullhomotopy. But, I still do not see the intuition behind the equality $pi_1(C_1)=pi_1(SO(3))$
– galois1989
Jul 15 at 9:17
1
@galois1989 Elements of $SO(3)$ are particular maps $Bbb S^2to Bbb S^2$ of degree $1$, namely isometries. My guts told me that the space of maps of degree $1$ should have the same homotopy type as the subspace of isometries.
– Arnaud Mortier
Jul 15 at 10:06
1
1
Good idea. In fact the space of maps are essentially computed here; for the component of degree $k$ one has $pi_1 textMap_k = Bbb Z/2|k|$.
– Mike Miller
Jul 14 at 17:06
Good idea. In fact the space of maps are essentially computed here; for the component of degree $k$ one has $pi_1 textMap_k = Bbb Z/2|k|$.
– Mike Miller
Jul 14 at 17:06
@MikeMiller Good reference, thanks!
– Arnaud Mortier
Jul 14 at 17:48
@MikeMiller Good reference, thanks!
– Arnaud Mortier
Jul 14 at 17:48
Thank you all a lot! Could you please explain me why is it obviuos that $pi_1(C_0)$ is the trivial group? And, also, the intuition between $pi_1(C_1)=pi_1(SO(3))?$
– galois1989
Jul 15 at 8:58
Thank you all a lot! Could you please explain me why is it obviuos that $pi_1(C_0)$ is the trivial group? And, also, the intuition between $pi_1(C_1)=pi_1(SO(3))?$
– galois1989
Jul 15 at 8:58
Ok, I think that the first question is obvious because being a degree 0 map gives you the homotopy to the constant loop just by the nullhomotopy. But, I still do not see the intuition behind the equality $pi_1(C_1)=pi_1(SO(3))$
– galois1989
Jul 15 at 9:17
Ok, I think that the first question is obvious because being a degree 0 map gives you the homotopy to the constant loop just by the nullhomotopy. But, I still do not see the intuition behind the equality $pi_1(C_1)=pi_1(SO(3))$
– galois1989
Jul 15 at 9:17
1
1
@galois1989 Elements of $SO(3)$ are particular maps $Bbb S^2to Bbb S^2$ of degree $1$, namely isometries. My guts told me that the space of maps of degree $1$ should have the same homotopy type as the subspace of isometries.
– Arnaud Mortier
Jul 15 at 10:06
@galois1989 Elements of $SO(3)$ are particular maps $Bbb S^2to Bbb S^2$ of degree $1$, namely isometries. My guts told me that the space of maps of degree $1$ should have the same homotopy type as the subspace of isometries.
– Arnaud Mortier
Jul 15 at 10:06
 |Â
show 7 more comments
If $M$ is an oriented 3-manifold, then $TM$ is trivial, and so picking a trivialization, a vector field of unit norm is the same as a map to $S^2$. A vector field of unit norm has an orthogonal complement, which is oriented; and changing the map to $S^2$ by a homotopy changes this plane field by a homotopy. So $[M, S^2]$ is oriented plane fields on $M$ up to homotopy.
– Mike Miller
Jul 14 at 16:38
I misread this as $S^1 times S^1$. Apologies.
– Balarka Sen
Jul 14 at 16:41