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Diff$(S^1)$ is deformation retracts to $O(2)$ [duplicate]
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This question already has an answer here:
Space of homeomorphisms Homeo$(S^1)$ of $S^1$ deformation retracts onto $O(2)$
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I have proved the following Diff$^+(S^1$) is path connected. Now I want to prove it is deformation retracts to $O(2)$.
What I tried is the following:
I define an onto homomorphism $$ f:textDiff^+(mathbbD^2)to textDiff^+(S^1) $$ by $$ phimapsto phi|_partial mathbbD^2=S^1$$
This is onto because any diffeomorphism of $S^1$ can be extended to a diffeomorphism of $mathbbD^2$. The kernel of this homomorphism is $textDiff^+(mathbbD^2textrel partial)$. After that I don't know what to do.
analysis algebraic-topology
asked Jul 28 at 16:23
Sachchidanand Prasad
1,642619
marked as duplicate by Andres Mejia, José Carlos Santos, amWhy, Adrian Keister, Alexander Gruber♦ Jul 29 at 5:50
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
2
down vote
favorite
This question already has an answer here:
Space of homeomorphisms Homeo$(S^1)$ of $S^1$ deformation retracts onto $O(2)$
1 answer
I have proved the following Diff$^+(S^1$) is path connected. Now I want to prove it is deformation retracts to $O(2)$.
What I tried is the following:
I define an onto homomorphism $$ f:textDiff^+(mathbbD^2)to textDiff^+(S^1) $$ by $$ phimapsto phi|_partial mathbbD^2=S^1$$
This is onto because any diffeomorphism of $S^1$ can be extended to a diffeomorphism of $mathbbD^2$. The kernel of this homomorphism is $textDiff^+(mathbbD^2textrel partial)$. After that I don't know what to do.
analysis algebraic-topology
asked Jul 28 at 16:23
Sachchidanand Prasad
1,642619
marked as duplicate by Andres Mejia, José Carlos Santos, amWhy, Adrian Keister, Alexander Gruber♦ Jul 29 at 5:50
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
If you are using $Diff^+(S^1$, then I think you want $SO(2)$, since $O(2)$ is not path connected and does not act in an orientation-preserving manner.
– Tyrone
Jul 28 at 16:48
Yeah, that's fine but I want to prove Diff$(S^1)$ is deformation retracts to $O(2)$.
– Sachchidanand Prasad
Jul 28 at 17:01
Think of this as fibration. If you assume continuous category is same as differential category in low dimension (2&3)...then try to prove that each of the fiber Diff+(Drelpartial) are contractible by using Alexander trick. And thus complete the proof.
– Anubhav Mukherjee
Jul 28 at 17:01
@SachchidanandPrasad, you may have to rethink, then, since $Diff^+(S^1)$ does not deformation retract onto $O(2)$, but rather onto $SO(2)$.
– Tyrone
Jul 28 at 17:33
@Tyrone I know that Diff$^+(S^1)$ is deformation reatract to $SO(2)$. But I want to prove the result for Diff$(S^1)$
– Sachchidanand Prasad
Jul 28 at 19:48
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question already has an answer here:
Space of homeomorphisms Homeo$(S^1)$ of $S^1$ deformation retracts onto $O(2)$
1 answer
I have proved the following Diff$^+(S^1$) is path connected. Now I want to prove it is deformation retracts to $O(2)$.
What I tried is the following:
I define an onto homomorphism $$ f:textDiff^+(mathbbD^2)to textDiff^+(S^1) $$ by $$ phimapsto phi|_partial mathbbD^2=S^1$$
This is onto because any diffeomorphism of $S^1$ can be extended to a diffeomorphism of $mathbbD^2$. The kernel of this homomorphism is $textDiff^+(mathbbD^2textrel partial)$. After that I don't know what to do.
analysis algebraic-topology
asked Jul 28 at 16:23
Sachchidanand Prasad
1,642619
This question already has an answer here:
Space of homeomorphisms Homeo$(S^1)$ of $S^1$ deformation retracts onto $O(2)$
1 answer
I have proved the following Diff$^+(S^1$) is path connected. Now I want to prove it is deformation retracts to $O(2)$.
What I tried is the following:
I define an onto homomorphism $$ f:textDiff^+(mathbbD^2)to textDiff^+(S^1) $$ by $$ phimapsto phi|_partial mathbbD^2=S^1$$
This is onto because any diffeomorphism of $S^1$ can be extended to a diffeomorphism of $mathbbD^2$. The kernel of this homomorphism is $textDiff^+(mathbbD^2textrel partial)$. After that I don't know what to do.
This question already has an answer here:
Space of homeomorphisms Homeo$(S^1)$ of $S^1$ deformation retracts onto $O(2)$
1 answer
analysis algebraic-topology
asked Jul 28 at 16:23
Sachchidanand Prasad
1,642619
asked Jul 28 at 16:23
Sachchidanand Prasad
1,642619
asked Jul 28 at 16:23
Sachchidanand Prasad
1,642619
asked Jul 28 at 16:23
Sachchidanand Prasad
1,642619
1,642619
marked as duplicate by Andres Mejia, José Carlos Santos, amWhy, Adrian Keister, Alexander Gruber♦ Jul 29 at 5:50
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Andres Mejia, José Carlos Santos, amWhy, Adrian Keister, Alexander Gruber♦ Jul 29 at 5:50
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
If you are using $Diff^+(S^1$, then I think you want $SO(2)$, since $O(2)$ is not path connected and does not act in an orientation-preserving manner.
– Tyrone
Jul 28 at 16:48
Yeah, that's fine but I want to prove Diff$(S^1)$ is deformation retracts to $O(2)$.
– Sachchidanand Prasad
Jul 28 at 17:01
Think of this as fibration. If you assume continuous category is same as differential category in low dimension (2&3)...then try to prove that each of the fiber Diff+(Drelpartial) are contractible by using Alexander trick. And thus complete the proof.
– Anubhav Mukherjee
Jul 28 at 17:01
@SachchidanandPrasad, you may have to rethink, then, since $Diff^+(S^1)$ does not deformation retract onto $O(2)$, but rather onto $SO(2)$.
– Tyrone
Jul 28 at 17:33
@Tyrone I know that Diff$^+(S^1)$ is deformation reatract to $SO(2)$. But I want to prove the result for Diff$(S^1)$
– Sachchidanand Prasad
Jul 28 at 19:48
 |Â
show 1 more comment
If you are using $Diff^+(S^1$, then I think you want $SO(2)$, since $O(2)$ is not path connected and does not act in an orientation-preserving manner.
– Tyrone
Jul 28 at 16:48
Yeah, that's fine but I want to prove Diff$(S^1)$ is deformation retracts to $O(2)$.
– Sachchidanand Prasad
Jul 28 at 17:01
Think of this as fibration. If you assume continuous category is same as differential category in low dimension (2&3)...then try to prove that each of the fiber Diff+(Drelpartial) are contractible by using Alexander trick. And thus complete the proof.
– Anubhav Mukherjee
Jul 28 at 17:01
@SachchidanandPrasad, you may have to rethink, then, since $Diff^+(S^1)$ does not deformation retract onto $O(2)$, but rather onto $SO(2)$.
– Tyrone
Jul 28 at 17:33
@Tyrone I know that Diff$^+(S^1)$ is deformation reatract to $SO(2)$. But I want to prove the result for Diff$(S^1)$
– Sachchidanand Prasad
Jul 28 at 19:48
If you are using $Diff^+(S^1$, then I think you want $SO(2)$, since $O(2)$ is not path connected and does not act in an orientation-preserving manner.
– Tyrone
Jul 28 at 16:48
If you are using $Diff^+(S^1$, then I think you want $SO(2)$, since $O(2)$ is not path connected and does not act in an orientation-preserving manner.
– Tyrone
Jul 28 at 16:48
Yeah, that's fine but I want to prove Diff$(S^1)$ is deformation retracts to $O(2)$.
– Sachchidanand Prasad
Jul 28 at 17:01
Yeah, that's fine but I want to prove Diff$(S^1)$ is deformation retracts to $O(2)$.
– Sachchidanand Prasad
Jul 28 at 17:01
Think of this as fibration. If you assume continuous category is same as differential category in low dimension (2&3)...then try to prove that each of the fiber Diff+(Drelpartial) are contractible by using Alexander trick. And thus complete the proof.
– Anubhav Mukherjee
Jul 28 at 17:01
Think of this as fibration. If you assume continuous category is same as differential category in low dimension (2&3)...then try to prove that each of the fiber Diff+(Drelpartial) are contractible by using Alexander trick. And thus complete the proof.
– Anubhav Mukherjee
Jul 28 at 17:01
@SachchidanandPrasad, you may have to rethink, then, since $Diff^+(S^1)$ does not deformation retract onto $O(2)$, but rather onto $SO(2)$.
– Tyrone
Jul 28 at 17:33
@SachchidanandPrasad, you may have to rethink, then, since $Diff^+(S^1)$ does not deformation retract onto $O(2)$, but rather onto $SO(2)$.
– Tyrone
Jul 28 at 17:33
@Tyrone I know that Diff$^+(S^1)$ is deformation reatract to $SO(2)$. But I want to prove the result for Diff$(S^1)$
– Sachchidanand Prasad
Jul 28 at 19:48
@Tyrone I know that Diff$^+(S^1)$ is deformation reatract to $SO(2)$. But I want to prove the result for Diff$(S^1)$
– Sachchidanand Prasad
Jul 28 at 19:48
 |Â
show 1 more comment
1 Answer
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We will show the result for homotopy equivalences of $S^1$, denoted $Self(S^1)$, and then show that we can restrict down to either the homeomorphism or diffeomorphism group. This is all discussed in Lurie's notes, but I'll reproduce the argument for completeness.
both $Diff(S^1)$ and $Self(S^1)$ can be decomposed into two pieces, orientation preserving and orientation reversing, and likewise so can $O(2)$.
Hence, it is enough to show that $SO(2)$ is a deformation retract of $Diff^+(S^1)$, as noted in the comments.
First, note that there are decompositions $Diff^+(S^1)=Diff_0^+(S^1)SO(2)$ and $Self^+=Self_0^+(S^1)SO(2)$, where the subscript $0$ indicates that they fix a base point of $S^1$. In other words, every diffeomorphism can be obtained from one that fixes a base point, and then rotates appropriately.
First, we can regard $S^1= mathbb R/mathbf Z$, so a base-point perserving map $f:S^1 to S^1$ can be regarded as a map $tildef:mathbb R to mathbb R$ where $f(x+1)=f(x) pm d$ and $tildef(0)=0$ (take the perserved fixed point to be the image of $0$.) Note that $f$ is a homotopy equivalence if and only if $d= pm 1$, since $d$ is the degree of the map, and since it is orientation perserving, $d=1$.
Hence, we have an identification $Self_0^+(S^1)=tildef:mathbb R to mathbb R mid tildef(0)=0, f(x+1)= x+1$.
However, this lets us form the straight line homotopy $F_t:=(1-t)tildef(x)+tx$, which shows the space to be contractible (retractible) (as all functions are homotopic to identity)
$Diff_0^+(S^1) subset Self_0^+(S^1)$ in the above identification by taking the subset of smooth maps without vanishing derivative. We can follow the same homotopy.
In particular, this shows a retract via the straight line homotopy from $Diff^+(S^1)$ to $SO(2)$
answered Jul 28 at 20:15
Andres Mejia
14.3k11242
Will you please tell me why $textDiff^+(S^1)=textDiff_0^+(S^1)SO(2)$ what is the meaning of $textDiff_0^+(S^1)SO(2)$?
– Sachchidanand Prasad
Jul 29 at 8:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We will show the result for homotopy equivalences of $S^1$, denoted $Self(S^1)$, and then show that we can restrict down to either the homeomorphism or diffeomorphism group. This is all discussed in Lurie's notes, but I'll reproduce the argument for completeness.
both $Diff(S^1)$ and $Self(S^1)$ can be decomposed into two pieces, orientation preserving and orientation reversing, and likewise so can $O(2)$.
Hence, it is enough to show that $SO(2)$ is a deformation retract of $Diff^+(S^1)$, as noted in the comments.
First, note that there are decompositions $Diff^+(S^1)=Diff_0^+(S^1)SO(2)$ and $Self^+=Self_0^+(S^1)SO(2)$, where the subscript $0$ indicates that they fix a base point of $S^1$. In other words, every diffeomorphism can be obtained from one that fixes a base point, and then rotates appropriately.
First, we can regard $S^1= mathbb R/mathbf Z$, so a base-point perserving map $f:S^1 to S^1$ can be regarded as a map $tildef:mathbb R to mathbb R$ where $f(x+1)=f(x) pm d$ and $tildef(0)=0$ (take the perserved fixed point to be the image of $0$.) Note that $f$ is a homotopy equivalence if and only if $d= pm 1$, since $d$ is the degree of the map, and since it is orientation perserving, $d=1$.
Hence, we have an identification $Self_0^+(S^1)=tildef:mathbb R to mathbb R mid tildef(0)=0, f(x+1)= x+1$.
However, this lets us form the straight line homotopy $F_t:=(1-t)tildef(x)+tx$, which shows the space to be contractible (retractible) (as all functions are homotopic to identity)
$Diff_0^+(S^1) subset Self_0^+(S^1)$ in the above identification by taking the subset of smooth maps without vanishing derivative. We can follow the same homotopy.
In particular, this shows a retract via the straight line homotopy from $Diff^+(S^1)$ to $SO(2)$
answered Jul 28 at 20:15
Andres Mejia
14.3k11242
Will you please tell me why $textDiff^+(S^1)=textDiff_0^+(S^1)SO(2)$ what is the meaning of $textDiff_0^+(S^1)SO(2)$?
– Sachchidanand Prasad
Jul 29 at 8:04
add a comment |Â
up vote
0
down vote
We will show the result for homotopy equivalences of $S^1$, denoted $Self(S^1)$, and then show that we can restrict down to either the homeomorphism or diffeomorphism group. This is all discussed in Lurie's notes, but I'll reproduce the argument for completeness.
both $Diff(S^1)$ and $Self(S^1)$ can be decomposed into two pieces, orientation preserving and orientation reversing, and likewise so can $O(2)$.
Hence, it is enough to show that $SO(2)$ is a deformation retract of $Diff^+(S^1)$, as noted in the comments.
First, note that there are decompositions $Diff^+(S^1)=Diff_0^+(S^1)SO(2)$ and $Self^+=Self_0^+(S^1)SO(2)$, where the subscript $0$ indicates that they fix a base point of $S^1$. In other words, every diffeomorphism can be obtained from one that fixes a base point, and then rotates appropriately.
First, we can regard $S^1= mathbb R/mathbf Z$, so a base-point perserving map $f:S^1 to S^1$ can be regarded as a map $tildef:mathbb R to mathbb R$ where $f(x+1)=f(x) pm d$ and $tildef(0)=0$ (take the perserved fixed point to be the image of $0$.) Note that $f$ is a homotopy equivalence if and only if $d= pm 1$, since $d$ is the degree of the map, and since it is orientation perserving, $d=1$.
Hence, we have an identification $Self_0^+(S^1)=tildef:mathbb R to mathbb R mid tildef(0)=0, f(x+1)= x+1$.
However, this lets us form the straight line homotopy $F_t:=(1-t)tildef(x)+tx$, which shows the space to be contractible (retractible) (as all functions are homotopic to identity)
$Diff_0^+(S^1) subset Self_0^+(S^1)$ in the above identification by taking the subset of smooth maps without vanishing derivative. We can follow the same homotopy.
In particular, this shows a retract via the straight line homotopy from $Diff^+(S^1)$ to $SO(2)$
answered Jul 28 at 20:15
Andres Mejia
14.3k11242
Will you please tell me why $textDiff^+(S^1)=textDiff_0^+(S^1)SO(2)$ what is the meaning of $textDiff_0^+(S^1)SO(2)$?
– Sachchidanand Prasad
Jul 29 at 8:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We will show the result for homotopy equivalences of $S^1$, denoted $Self(S^1)$, and then show that we can restrict down to either the homeomorphism or diffeomorphism group. This is all discussed in Lurie's notes, but I'll reproduce the argument for completeness.
both $Diff(S^1)$ and $Self(S^1)$ can be decomposed into two pieces, orientation preserving and orientation reversing, and likewise so can $O(2)$.
Hence, it is enough to show that $SO(2)$ is a deformation retract of $Diff^+(S^1)$, as noted in the comments.
First, note that there are decompositions $Diff^+(S^1)=Diff_0^+(S^1)SO(2)$ and $Self^+=Self_0^+(S^1)SO(2)$, where the subscript $0$ indicates that they fix a base point of $S^1$. In other words, every diffeomorphism can be obtained from one that fixes a base point, and then rotates appropriately.
First, we can regard $S^1= mathbb R/mathbf Z$, so a base-point perserving map $f:S^1 to S^1$ can be regarded as a map $tildef:mathbb R to mathbb R$ where $f(x+1)=f(x) pm d$ and $tildef(0)=0$ (take the perserved fixed point to be the image of $0$.) Note that $f$ is a homotopy equivalence if and only if $d= pm 1$, since $d$ is the degree of the map, and since it is orientation perserving, $d=1$.
Hence, we have an identification $Self_0^+(S^1)=tildef:mathbb R to mathbb R mid tildef(0)=0, f(x+1)= x+1$.
However, this lets us form the straight line homotopy $F_t:=(1-t)tildef(x)+tx$, which shows the space to be contractible (retractible) (as all functions are homotopic to identity)
$Diff_0^+(S^1) subset Self_0^+(S^1)$ in the above identification by taking the subset of smooth maps without vanishing derivative. We can follow the same homotopy.
In particular, this shows a retract via the straight line homotopy from $Diff^+(S^1)$ to $SO(2)$
answered Jul 28 at 20:15
Andres Mejia
14.3k11242
We will show the result for homotopy equivalences of $S^1$, denoted $Self(S^1)$, and then show that we can restrict down to either the homeomorphism or diffeomorphism group. This is all discussed in Lurie's notes, but I'll reproduce the argument for completeness.
both $Diff(S^1)$ and $Self(S^1)$ can be decomposed into two pieces, orientation preserving and orientation reversing, and likewise so can $O(2)$.
Hence, it is enough to show that $SO(2)$ is a deformation retract of $Diff^+(S^1)$, as noted in the comments.
First, note that there are decompositions $Diff^+(S^1)=Diff_0^+(S^1)SO(2)$ and $Self^+=Self_0^+(S^1)SO(2)$, where the subscript $0$ indicates that they fix a base point of $S^1$. In other words, every diffeomorphism can be obtained from one that fixes a base point, and then rotates appropriately.
First, we can regard $S^1= mathbb R/mathbf Z$, so a base-point perserving map $f:S^1 to S^1$ can be regarded as a map $tildef:mathbb R to mathbb R$ where $f(x+1)=f(x) pm d$ and $tildef(0)=0$ (take the perserved fixed point to be the image of $0$.) Note that $f$ is a homotopy equivalence if and only if $d= pm 1$, since $d$ is the degree of the map, and since it is orientation perserving, $d=1$.
Hence, we have an identification $Self_0^+(S^1)=tildef:mathbb R to mathbb R mid tildef(0)=0, f(x+1)= x+1$.
However, this lets us form the straight line homotopy $F_t:=(1-t)tildef(x)+tx$, which shows the space to be contractible (retractible) (as all functions are homotopic to identity)
$Diff_0^+(S^1) subset Self_0^+(S^1)$ in the above identification by taking the subset of smooth maps without vanishing derivative. We can follow the same homotopy.
In particular, this shows a retract via the straight line homotopy from $Diff^+(S^1)$ to $SO(2)$
answered Jul 28 at 20:15
Andres Mejia
14.3k11242
answered Jul 28 at 20:15
Andres Mejia
14.3k11242
answered Jul 28 at 20:15
Andres Mejia
14.3k11242
answered Jul 28 at 20:15
Andres Mejia
14.3k11242
14.3k11242
Will you please tell me why $textDiff^+(S^1)=textDiff_0^+(S^1)SO(2)$ what is the meaning of $textDiff_0^+(S^1)SO(2)$?
– Sachchidanand Prasad
Jul 29 at 8:04
add a comment |Â
Will you please tell me why $textDiff^+(S^1)=textDiff_0^+(S^1)SO(2)$ what is the meaning of $textDiff_0^+(S^1)SO(2)$?
– Sachchidanand Prasad
Jul 29 at 8:04
Will you please tell me why $textDiff^+(S^1)=textDiff_0^+(S^1)SO(2)$ what is the meaning of $textDiff_0^+(S^1)SO(2)$?
– Sachchidanand Prasad
Jul 29 at 8:04
Will you please tell me why $textDiff^+(S^1)=textDiff_0^+(S^1)SO(2)$ what is the meaning of $textDiff_0^+(S^1)SO(2)$?
– Sachchidanand Prasad
Jul 29 at 8:04
add a comment |Â
If you are using $Diff^+(S^1$, then I think you want $SO(2)$, since $O(2)$ is not path connected and does not act in an orientation-preserving manner.
– Tyrone
Jul 28 at 16:48
Yeah, that's fine but I want to prove Diff$(S^1)$ is deformation retracts to $O(2)$.
– Sachchidanand Prasad
Jul 28 at 17:01
Think of this as fibration. If you assume continuous category is same as differential category in low dimension (2&3)...then try to prove that each of the fiber Diff+(Drelpartial) are contractible by using Alexander trick. And thus complete the proof.
– Anubhav Mukherjee
Jul 28 at 17:01
@SachchidanandPrasad, you may have to rethink, then, since $Diff^+(S^1)$ does not deformation retract onto $O(2)$, but rather onto $SO(2)$.
– Tyrone
Jul 28 at 17:33
@Tyrone I know that Diff$^+(S^1)$ is deformation reatract to $SO(2)$. But I want to prove the result for Diff$(S^1)$
– Sachchidanand Prasad
Jul 28 at 19:48