Mixing
How to characterize the group of homeomorphisms of unit disk in terms of the group of homeomorphisms of...
$begingroup$
Let $G$ be the group of homeomorphisms of unit disk $(D)$ fixing boundary point wise and $P$ be the group of homeomorphisms of plane $mathbb{R}^2$.
Can we characterize $G$ in terms of $P$. Speaking precisely, can we have the result of the form: $gamma$ is in $G$ if and only if $F(gamma)$ is in $P$, where $F$ is some condition or some function (for example it could be in terms of euclidean norm in $mathbb{R}^2$).
We can easily embedd $G$ in $P$ as:
Define a homeomorphism $g:Int(D)rightarrowmathbb{R}^2$ in polar coordinates as $g(r,theta)=(frac{r}{1-r},theta)$. Then the function $f:Grightarrow P$ defined by $f(gamma)=g^{-1}gamma g$ is an embedding. Through $f$ we can study the image of $G$ in $P$ rather than $G$ itself.
real-analysis general-topology group-theory geometric-topology geometric-group-theory
asked Jan 18 at 23:35
ershersh
423113
$endgroup$
add a comment |
$begingroup$
Let $G$ be the group of homeomorphisms of unit disk $(D)$ fixing boundary point wise and $P$ be the group of homeomorphisms of plane $mathbb{R}^2$.
Can we characterize $G$ in terms of $P$. Speaking precisely, can we have the result of the form: $gamma$ is in $G$ if and only if $F(gamma)$ is in $P$, where $F$ is some condition or some function (for example it could be in terms of euclidean norm in $mathbb{R}^2$).
We can easily embedd $G$ in $P$ as:
Define a homeomorphism $g:Int(D)rightarrowmathbb{R}^2$ in polar coordinates as $g(r,theta)=(frac{r}{1-r},theta)$. Then the function $f:Grightarrow P$ defined by $f(gamma)=g^{-1}gamma g$ is an embedding. Through $f$ we can study the image of $G$ in $P$ rather than $G$ itself.
real-analysis general-topology group-theory geometric-topology geometric-group-theory
asked Jan 18 at 23:35
ershersh
423113
$endgroup$
$begingroup$
How can $g$ be a homeo? One is compact and one is not.
$endgroup$
– Randall
Jan 18 at 23:58
$begingroup$
Sorry, $g$ is defined on the interior of $D$. Corrected.
$endgroup$
– ersh
Jan 19 at 0:00
add a comment |
$begingroup$
Let $G$ be the group of homeomorphisms of unit disk $(D)$ fixing boundary point wise and $P$ be the group of homeomorphisms of plane $mathbb{R}^2$.
Can we characterize $G$ in terms of $P$. Speaking precisely, can we have the result of the form: $gamma$ is in $G$ if and only if $F(gamma)$ is in $P$, where $F$ is some condition or some function (for example it could be in terms of euclidean norm in $mathbb{R}^2$).
We can easily embedd $G$ in $P$ as:
Define a homeomorphism $g:Int(D)rightarrowmathbb{R}^2$ in polar coordinates as $g(r,theta)=(frac{r}{1-r},theta)$. Then the function $f:Grightarrow P$ defined by $f(gamma)=g^{-1}gamma g$ is an embedding. Through $f$ we can study the image of $G$ in $P$ rather than $G$ itself.
real-analysis general-topology group-theory geometric-topology geometric-group-theory
asked Jan 18 at 23:35
ershersh
423113
$endgroup$
Let $G$ be the group of homeomorphisms of unit disk $(D)$ fixing boundary point wise and $P$ be the group of homeomorphisms of plane $mathbb{R}^2$.
Can we characterize $G$ in terms of $P$. Speaking precisely, can we have the result of the form: $gamma$ is in $G$ if and only if $F(gamma)$ is in $P$, where $F$ is some condition or some function (for example it could be in terms of euclidean norm in $mathbb{R}^2$).
We can easily embedd $G$ in $P$ as:
Define a homeomorphism $g:Int(D)rightarrowmathbb{R}^2$ in polar coordinates as $g(r,theta)=(frac{r}{1-r},theta)$. Then the function $f:Grightarrow P$ defined by $f(gamma)=g^{-1}gamma g$ is an embedding. Through $f$ we can study the image of $G$ in $P$ rather than $G$ itself.
real-analysis general-topology group-theory geometric-topology geometric-group-theory
real-analysis general-topology group-theory geometric-topology geometric-group-theory
asked Jan 18 at 23:35
ershersh
423113
asked Jan 18 at 23:35
ershersh
423113
edited Jan 21 at 19:00
ersh
asked Jan 18 at 23:35
ershersh
423113
asked Jan 18 at 23:35
ershersh
423113
asked Jan 18 at 23:35
ershersh
423113
423113
$begingroup$
How can $g$ be a homeo? One is compact and one is not.
$endgroup$
– Randall
Jan 18 at 23:58
$begingroup$
Sorry, $g$ is defined on the interior of $D$. Corrected.
$endgroup$
– ersh
Jan 19 at 0:00
add a comment |
$begingroup$
How can $g$ be a homeo? One is compact and one is not.
$endgroup$
– Randall
Jan 18 at 23:58
$begingroup$
Sorry, $g$ is defined on the interior of $D$. Corrected.
$endgroup$
– ersh
Jan 19 at 0:00
$begingroup$
How can $g$ be a homeo? One is compact and one is not.
$endgroup$
– Randall
Jan 18 at 23:58
$begingroup$
How can $g$ be a homeo? One is compact and one is not.
$endgroup$
– Randall
Jan 18 at 23:58
$begingroup$
Sorry, $g$ is defined on the interior of $D$. Corrected.
$endgroup$
– ersh
Jan 19 at 0:00
$begingroup$
Sorry, $g$ is defined on the interior of $D$. Corrected.
$endgroup$
– ersh
Jan 19 at 0:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The only answer is this.
For $h in P$, let $phi(h) : D to phi(D)$ denote the restriction of $h$ (which is a homeomorphism). Then $phi(h) in G$ if and only if $h(x) = x$ for all $x in S^1 = partial D$.
The "only if" part is trivial. Conversely, assume $h(x) = x$ for all $x in S^1$. We know that $S^1$ separates the plane into two connected components $mathring{D}$ = open unit disk (which is contractible) and $D' = mathbb{R}^2 setminus D$ (which is not contractible). $h(mathring{D})$ must be one of these connected components. But $h(mathring{D}) approx mathring{D}$ is contractible, whereas $h(D') approx D'$ is not. This shows that $h(mathring{D}) = mathring{D}$, i.e. $h(D) = D$. Thus $phi(h) in G$.
Edited:
Here are some examples of homeomorphisms of the plane which fix $S^1$ pointwise.
1) Let $f : (0,infty) to mathbb{R}$ be any continuous map such that $f(1) = 0$. Define
$$h(z) =
begin{cases}
e^{if(lvert z rvert)}z & z ne 0 \
0 & z = 0
end{cases}
$$
In the first line we use the complex multiplication on $mathbb{C} = mathbb{R}^2$. Note that $lvert e^{if(lvert z rvert)} rvert = 1$, i.e $h$ maps each circle with radius $r$ around $0$ homeomorphically onto itself.
2) Let $g : [0,infty) to [0,infty)$ be any continuous strictly monotonic increasing function such that $g(0) = 0, g(1) = 1$ and $g(t) to infty$ as $t to infty$. Define
$$h(z) = g(lvert z rvert) z .$$
answered Jan 19 at 0:47
Paul FrostPaul Frost
11.3k3934
$endgroup$
$begingroup$
It seems you are only characterizing those maps of disc which arise as the restrictions of the homeomorphisms $h$ of a plane. That is, such restricted maps will be in $G$ if and only if $h$ fixes circle point wise. Isn't it so? What about those maps of a disc which are not restrictions of the homeomorphisms of plane? Won't they generate their own class in the group $G$, in a similar manner you have provided the characterization for the 'restricted maps?
$endgroup$
– ersh
Jan 19 at 1:37
1
$begingroup$
Each $g in G$ extends (non.uniquley) to an element of $P$. The elements of $P$ obtained in that way are precisely those in my answer.
$endgroup$
– Paul Frost
Jan 19 at 9:25
$begingroup$
What would be an explicite examples of such homeomorphisms $h$ of a plane which fix circle point wise? If I take $h=(x,y)$ on $S^1$ and $(x+2,y+2)$ everywherelse, is this the homeomorphism of plane?
$endgroup$
– ersh
Jan 19 at 15:44
1
$begingroup$
I shall add some examples to my answer.
$endgroup$
– Paul Frost
Jan 19 at 16:27
1
$begingroup$
By the way, the map $t(x,y) = (x+2,y+2)$ is a translation. It does not keep $S^1$ fixed So the map which fixes $S^1$ and is $= t$ everywhere else is neither continuous nor bijective.
$endgroup$
– Paul Frost
Jan 19 at 16:54
|
show 3 more comments
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The only answer is this.
For $h in P$, let $phi(h) : D to phi(D)$ denote the restriction of $h$ (which is a homeomorphism). Then $phi(h) in G$ if and only if $h(x) = x$ for all $x in S^1 = partial D$.
The "only if" part is trivial. Conversely, assume $h(x) = x$ for all $x in S^1$. We know that $S^1$ separates the plane into two connected components $mathring{D}$ = open unit disk (which is contractible) and $D' = mathbb{R}^2 setminus D$ (which is not contractible). $h(mathring{D})$ must be one of these connected components. But $h(mathring{D}) approx mathring{D}$ is contractible, whereas $h(D') approx D'$ is not. This shows that $h(mathring{D}) = mathring{D}$, i.e. $h(D) = D$. Thus $phi(h) in G$.
Edited:
Here are some examples of homeomorphisms of the plane which fix $S^1$ pointwise.
1) Let $f : (0,infty) to mathbb{R}$ be any continuous map such that $f(1) = 0$. Define
$$h(z) =
begin{cases}
e^{if(lvert z rvert)}z & z ne 0 \
0 & z = 0
end{cases}
$$
In the first line we use the complex multiplication on $mathbb{C} = mathbb{R}^2$. Note that $lvert e^{if(lvert z rvert)} rvert = 1$, i.e $h$ maps each circle with radius $r$ around $0$ homeomorphically onto itself.
2) Let $g : [0,infty) to [0,infty)$ be any continuous strictly monotonic increasing function such that $g(0) = 0, g(1) = 1$ and $g(t) to infty$ as $t to infty$. Define
$$h(z) = g(lvert z rvert) z .$$
answered Jan 19 at 0:47
Paul FrostPaul Frost
11.3k3934
$endgroup$
$begingroup$
It seems you are only characterizing those maps of disc which arise as the restrictions of the homeomorphisms $h$ of a plane. That is, such restricted maps will be in $G$ if and only if $h$ fixes circle point wise. Isn't it so? What about those maps of a disc which are not restrictions of the homeomorphisms of plane? Won't they generate their own class in the group $G$, in a similar manner you have provided the characterization for the 'restricted maps?
$endgroup$
– ersh
Jan 19 at 1:37
1
$begingroup$
Each $g in G$ extends (non.uniquley) to an element of $P$. The elements of $P$ obtained in that way are precisely those in my answer.
$endgroup$
– Paul Frost
Jan 19 at 9:25
$begingroup$
What would be an explicite examples of such homeomorphisms $h$ of a plane which fix circle point wise? If I take $h=(x,y)$ on $S^1$ and $(x+2,y+2)$ everywherelse, is this the homeomorphism of plane?
$endgroup$
– ersh
Jan 19 at 15:44
1
$begingroup$
I shall add some examples to my answer.
$endgroup$
– Paul Frost
Jan 19 at 16:27
1
$begingroup$
By the way, the map $t(x,y) = (x+2,y+2)$ is a translation. It does not keep $S^1$ fixed So the map which fixes $S^1$ and is $= t$ everywhere else is neither continuous nor bijective.
$endgroup$
– Paul Frost
Jan 19 at 16:54
|
show 3 more comments
$begingroup$
The only answer is this.
For $h in P$, let $phi(h) : D to phi(D)$ denote the restriction of $h$ (which is a homeomorphism). Then $phi(h) in G$ if and only if $h(x) = x$ for all $x in S^1 = partial D$.
The "only if" part is trivial. Conversely, assume $h(x) = x$ for all $x in S^1$. We know that $S^1$ separates the plane into two connected components $mathring{D}$ = open unit disk (which is contractible) and $D' = mathbb{R}^2 setminus D$ (which is not contractible). $h(mathring{D})$ must be one of these connected components. But $h(mathring{D}) approx mathring{D}$ is contractible, whereas $h(D') approx D'$ is not. This shows that $h(mathring{D}) = mathring{D}$, i.e. $h(D) = D$. Thus $phi(h) in G$.
Edited:
Here are some examples of homeomorphisms of the plane which fix $S^1$ pointwise.
1) Let $f : (0,infty) to mathbb{R}$ be any continuous map such that $f(1) = 0$. Define
$$h(z) =
begin{cases}
e^{if(lvert z rvert)}z & z ne 0 \
0 & z = 0
end{cases}
$$
In the first line we use the complex multiplication on $mathbb{C} = mathbb{R}^2$. Note that $lvert e^{if(lvert z rvert)} rvert = 1$, i.e $h$ maps each circle with radius $r$ around $0$ homeomorphically onto itself.
2) Let $g : [0,infty) to [0,infty)$ be any continuous strictly monotonic increasing function such that $g(0) = 0, g(1) = 1$ and $g(t) to infty$ as $t to infty$. Define
$$h(z) = g(lvert z rvert) z .$$
answered Jan 19 at 0:47
Paul FrostPaul Frost
11.3k3934
$endgroup$
$begingroup$
It seems you are only characterizing those maps of disc which arise as the restrictions of the homeomorphisms $h$ of a plane. That is, such restricted maps will be in $G$ if and only if $h$ fixes circle point wise. Isn't it so? What about those maps of a disc which are not restrictions of the homeomorphisms of plane? Won't they generate their own class in the group $G$, in a similar manner you have provided the characterization for the 'restricted maps?
$endgroup$
– ersh
Jan 19 at 1:37
1
$begingroup$
Each $g in G$ extends (non.uniquley) to an element of $P$. The elements of $P$ obtained in that way are precisely those in my answer.
$endgroup$
– Paul Frost
Jan 19 at 9:25
$begingroup$
What would be an explicite examples of such homeomorphisms $h$ of a plane which fix circle point wise? If I take $h=(x,y)$ on $S^1$ and $(x+2,y+2)$ everywherelse, is this the homeomorphism of plane?
$endgroup$
– ersh
Jan 19 at 15:44
1
$begingroup$
I shall add some examples to my answer.
$endgroup$
– Paul Frost
Jan 19 at 16:27
1
$begingroup$
By the way, the map $t(x,y) = (x+2,y+2)$ is a translation. It does not keep $S^1$ fixed So the map which fixes $S^1$ and is $= t$ everywhere else is neither continuous nor bijective.
$endgroup$
– Paul Frost
Jan 19 at 16:54
|
show 3 more comments
$begingroup$
The only answer is this.
For $h in P$, let $phi(h) : D to phi(D)$ denote the restriction of $h$ (which is a homeomorphism). Then $phi(h) in G$ if and only if $h(x) = x$ for all $x in S^1 = partial D$.
The "only if" part is trivial. Conversely, assume $h(x) = x$ for all $x in S^1$. We know that $S^1$ separates the plane into two connected components $mathring{D}$ = open unit disk (which is contractible) and $D' = mathbb{R}^2 setminus D$ (which is not contractible). $h(mathring{D})$ must be one of these connected components. But $h(mathring{D}) approx mathring{D}$ is contractible, whereas $h(D') approx D'$ is not. This shows that $h(mathring{D}) = mathring{D}$, i.e. $h(D) = D$. Thus $phi(h) in G$.
Edited:
Here are some examples of homeomorphisms of the plane which fix $S^1$ pointwise.
1) Let $f : (0,infty) to mathbb{R}$ be any continuous map such that $f(1) = 0$. Define
$$h(z) =
begin{cases}
e^{if(lvert z rvert)}z & z ne 0 \
0 & z = 0
end{cases}
$$
In the first line we use the complex multiplication on $mathbb{C} = mathbb{R}^2$. Note that $lvert e^{if(lvert z rvert)} rvert = 1$, i.e $h$ maps each circle with radius $r$ around $0$ homeomorphically onto itself.
2) Let $g : [0,infty) to [0,infty)$ be any continuous strictly monotonic increasing function such that $g(0) = 0, g(1) = 1$ and $g(t) to infty$ as $t to infty$. Define
$$h(z) = g(lvert z rvert) z .$$
answered Jan 19 at 0:47
Paul FrostPaul Frost
11.3k3934
$endgroup$
The only answer is this.
For $h in P$, let $phi(h) : D to phi(D)$ denote the restriction of $h$ (which is a homeomorphism). Then $phi(h) in G$ if and only if $h(x) = x$ for all $x in S^1 = partial D$.
The "only if" part is trivial. Conversely, assume $h(x) = x$ for all $x in S^1$. We know that $S^1$ separates the plane into two connected components $mathring{D}$ = open unit disk (which is contractible) and $D' = mathbb{R}^2 setminus D$ (which is not contractible). $h(mathring{D})$ must be one of these connected components. But $h(mathring{D}) approx mathring{D}$ is contractible, whereas $h(D') approx D'$ is not. This shows that $h(mathring{D}) = mathring{D}$, i.e. $h(D) = D$. Thus $phi(h) in G$.
Edited:
Here are some examples of homeomorphisms of the plane which fix $S^1$ pointwise.
1) Let $f : (0,infty) to mathbb{R}$ be any continuous map such that $f(1) = 0$. Define
$$h(z) =
begin{cases}
e^{if(lvert z rvert)}z & z ne 0 \
0 & z = 0
end{cases}
$$
In the first line we use the complex multiplication on $mathbb{C} = mathbb{R}^2$. Note that $lvert e^{if(lvert z rvert)} rvert = 1$, i.e $h$ maps each circle with radius $r$ around $0$ homeomorphically onto itself.
2) Let $g : [0,infty) to [0,infty)$ be any continuous strictly monotonic increasing function such that $g(0) = 0, g(1) = 1$ and $g(t) to infty$ as $t to infty$. Define
$$h(z) = g(lvert z rvert) z .$$
answered Jan 19 at 0:47
Paul FrostPaul Frost
11.3k3934
edited Jan 19 at 16:46
answered Jan 19 at 0:47
Paul FrostPaul Frost
11.3k3934
answered Jan 19 at 0:47
Paul FrostPaul Frost
11.3k3934
answered Jan 19 at 0:47
Paul FrostPaul Frost
11.3k3934
11.3k3934
$begingroup$
It seems you are only characterizing those maps of disc which arise as the restrictions of the homeomorphisms $h$ of a plane. That is, such restricted maps will be in $G$ if and only if $h$ fixes circle point wise. Isn't it so? What about those maps of a disc which are not restrictions of the homeomorphisms of plane? Won't they generate their own class in the group $G$, in a similar manner you have provided the characterization for the 'restricted maps?
$endgroup$
– ersh
Jan 19 at 1:37
1
$begingroup$
Each $g in G$ extends (non.uniquley) to an element of $P$. The elements of $P$ obtained in that way are precisely those in my answer.
$endgroup$
– Paul Frost
Jan 19 at 9:25
$begingroup$
What would be an explicite examples of such homeomorphisms $h$ of a plane which fix circle point wise? If I take $h=(x,y)$ on $S^1$ and $(x+2,y+2)$ everywherelse, is this the homeomorphism of plane?
$endgroup$
– ersh
Jan 19 at 15:44
1
$begingroup$
I shall add some examples to my answer.
$endgroup$
– Paul Frost
Jan 19 at 16:27
1
$begingroup$
By the way, the map $t(x,y) = (x+2,y+2)$ is a translation. It does not keep $S^1$ fixed So the map which fixes $S^1$ and is $= t$ everywhere else is neither continuous nor bijective.
$endgroup$
– Paul Frost
Jan 19 at 16:54
|
show 3 more comments
$begingroup$
It seems you are only characterizing those maps of disc which arise as the restrictions of the homeomorphisms $h$ of a plane. That is, such restricted maps will be in $G$ if and only if $h$ fixes circle point wise. Isn't it so? What about those maps of a disc which are not restrictions of the homeomorphisms of plane? Won't they generate their own class in the group $G$, in a similar manner you have provided the characterization for the 'restricted maps?
$endgroup$
– ersh
Jan 19 at 1:37
1
$begingroup$
Each $g in G$ extends (non.uniquley) to an element of $P$. The elements of $P$ obtained in that way are precisely those in my answer.
$endgroup$
– Paul Frost
Jan 19 at 9:25
$begingroup$
What would be an explicite examples of such homeomorphisms $h$ of a plane which fix circle point wise? If I take $h=(x,y)$ on $S^1$ and $(x+2,y+2)$ everywherelse, is this the homeomorphism of plane?
$endgroup$
– ersh
Jan 19 at 15:44
1
$begingroup$
I shall add some examples to my answer.
$endgroup$
– Paul Frost
Jan 19 at 16:27
1
$begingroup$
By the way, the map $t(x,y) = (x+2,y+2)$ is a translation. It does not keep $S^1$ fixed So the map which fixes $S^1$ and is $= t$ everywhere else is neither continuous nor bijective.
$endgroup$
– Paul Frost
Jan 19 at 16:54
$begingroup$
It seems you are only characterizing those maps of disc which arise as the restrictions of the homeomorphisms $h$ of a plane. That is, such restricted maps will be in $G$ if and only if $h$ fixes circle point wise. Isn't it so? What about those maps of a disc which are not restrictions of the homeomorphisms of plane? Won't they generate their own class in the group $G$, in a similar manner you have provided the characterization for the 'restricted maps?
$endgroup$
– ersh
Jan 19 at 1:37
$begingroup$
It seems you are only characterizing those maps of disc which arise as the restrictions of the homeomorphisms $h$ of a plane. That is, such restricted maps will be in $G$ if and only if $h$ fixes circle point wise. Isn't it so? What about those maps of a disc which are not restrictions of the homeomorphisms of plane? Won't they generate their own class in the group $G$, in a similar manner you have provided the characterization for the 'restricted maps?
$endgroup$
– ersh
Jan 19 at 1:37
1
1
$begingroup$
Each $g in G$ extends (non.uniquley) to an element of $P$. The elements of $P$ obtained in that way are precisely those in my answer.
$endgroup$
– Paul Frost
Jan 19 at 9:25
$begingroup$
Each $g in G$ extends (non.uniquley) to an element of $P$. The elements of $P$ obtained in that way are precisely those in my answer.
$endgroup$
– Paul Frost
Jan 19 at 9:25
$begingroup$
What would be an explicite examples of such homeomorphisms $h$ of a plane which fix circle point wise? If I take $h=(x,y)$ on $S^1$ and $(x+2,y+2)$ everywherelse, is this the homeomorphism of plane?
$endgroup$
– ersh
Jan 19 at 15:44
$begingroup$
What would be an explicite examples of such homeomorphisms $h$ of a plane which fix circle point wise? If I take $h=(x,y)$ on $S^1$ and $(x+2,y+2)$ everywherelse, is this the homeomorphism of plane?
$endgroup$
– ersh
Jan 19 at 15:44
1
1
$begingroup$
I shall add some examples to my answer.
$endgroup$
– Paul Frost
Jan 19 at 16:27
$begingroup$
I shall add some examples to my answer.
$endgroup$
– Paul Frost
Jan 19 at 16:27
1
1
$begingroup$
By the way, the map $t(x,y) = (x+2,y+2)$ is a translation. It does not keep $S^1$ fixed So the map which fixes $S^1$ and is $= t$ everywhere else is neither continuous nor bijective.
$endgroup$
– Paul Frost
Jan 19 at 16:54
$begingroup$
By the way, the map $t(x,y) = (x+2,y+2)$ is a translation. It does not keep $S^1$ fixed So the map which fixes $S^1$ and is $= t$ everywhere else is neither continuous nor bijective.
$endgroup$
– Paul Frost
Jan 19 at 16:54
|
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$begingroup$
How can $g$ be a homeo? One is compact and one is not.
$endgroup$
– Randall
Jan 18 at 23:58
$begingroup$
Sorry, $g$ is defined on the interior of $D$. Corrected.
$endgroup$
– ersh
Jan 19 at 0:00