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Are the spaces $mathbbH^n$ and $overlinemathbbR^n_+$ homeomorphic?
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Let $mathbbH^n = (x^1, dots , x^n) in mathbbR^n : x^n ge 0$, and $overlinemathbbR^n_+ = (x^1, dots , x^n) in mathbbR^n : x^1 ge 0, dots, x^n ge 0$. Endow each set with the subspace topology it inherits from $mathbbR^n$.
Exercise 16.18 on page 415 of Lee's Introduction to Smooth Manifolds (2nd edition) asks us to show that $mathbbH^n$ and $overlinemathbbR^n_+$ are homeomorphic.
However, it seems to me the spaces are not homeomorphic, as shown by the following argument.
Suppose $f: overlinemathbbR^n_+ to mathbbH^n$ is a homeomorphism. Set $f(0, dots , 0) = (a^1, dots, a^n)$. Then
$$A = f(overlinemathbbR^n_+ setminus (0, dots, 0)) = f((infty,0) times cdots times (infty,0)) = \
((infty, a^1) cup (a^1, - infty)) times cdots times ((infty, a^n) cup (a^n, 0]),$$
where, if $a^n= 0$, then in the last factor we discard $(a^n, 0]$.
The set $A$ is connected as it is the continuous image of the connected set $(infty, 0) times (infty,0)$. But at the same time we see that $A$ is disconnected as it is the union of the disjoint relatively open sets
$(infty, a^1) times cdots times (infty, a^n)$ and $ (a^1, - infty) times cdots times (a^n, 0]$ (or $(infty, a^1) times cdots times (infty, 0)$ and $(a^1, - infty) times cdots times (infty, 0)$ in case $a^n = 0$).
Is there a mistake in my argument or something I am missing? Any comments are greatly appreciated.
general-topology smooth-manifolds connectedness
asked Jul 18 at 1:54
JZShapiro
1,94311023
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up vote
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Let $mathbbH^n = (x^1, dots , x^n) in mathbbR^n : x^n ge 0$, and $overlinemathbbR^n_+ = (x^1, dots , x^n) in mathbbR^n : x^1 ge 0, dots, x^n ge 0$. Endow each set with the subspace topology it inherits from $mathbbR^n$.
Exercise 16.18 on page 415 of Lee's Introduction to Smooth Manifolds (2nd edition) asks us to show that $mathbbH^n$ and $overlinemathbbR^n_+$ are homeomorphic.
However, it seems to me the spaces are not homeomorphic, as shown by the following argument.
Suppose $f: overlinemathbbR^n_+ to mathbbH^n$ is a homeomorphism. Set $f(0, dots , 0) = (a^1, dots, a^n)$. Then
$$A = f(overlinemathbbR^n_+ setminus (0, dots, 0)) = f((infty,0) times cdots times (infty,0)) = \
((infty, a^1) cup (a^1, - infty)) times cdots times ((infty, a^n) cup (a^n, 0]),$$
where, if $a^n= 0$, then in the last factor we discard $(a^n, 0]$.
The set $A$ is connected as it is the continuous image of the connected set $(infty, 0) times (infty,0)$. But at the same time we see that $A$ is disconnected as it is the union of the disjoint relatively open sets
$(infty, a^1) times cdots times (infty, a^n)$ and $ (a^1, - infty) times cdots times (a^n, 0]$ (or $(infty, a^1) times cdots times (infty, 0)$ and $(a^1, - infty) times cdots times (infty, 0)$ in case $a^n = 0$).
Is there a mistake in my argument or something I am missing? Any comments are greatly appreciated.
general-topology smooth-manifolds connectedness
asked Jul 18 at 1:54
JZShapiro
1,94311023
2
I think the problem is the line $$f(overlinemathbbR^n_+ setminus (0,dots,0)) =f( (0,infty) times dotstimes (0,infty)) $$ as the two sets whose f-images you are taking are NOT equal. For example, $(1,0,dots,0) in overlinemathbbR^n_+ setminus (0,dots,0)$ but $(1,0,dots,0) notin (0,infty) times dotstimes (0,infty)$
– itinerantleopard
Jul 18 at 3:12
2
To see that they are indeed homeomorphic, it might help to think about some small-dimensional cases first. When $n=1$, the space $ overlinemathbbR_+^1$ is actually equal to $mathbbH^1$. When $n=2$, the space $overlinemathbbR^2_+ $ is the closed upper right quadrant of the plane and $mathbbH^2$ is the closed upper half plane. The former can be homeomorphically deformed into the latter by "unfolding" the boundary until it lies flat against the $x$-axis. The geometric picture in higher dimensions is similar. Hope that helps!
– itinerantleopard
Jul 18 at 3:18
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $mathbbH^n = (x^1, dots , x^n) in mathbbR^n : x^n ge 0$, and $overlinemathbbR^n_+ = (x^1, dots , x^n) in mathbbR^n : x^1 ge 0, dots, x^n ge 0$. Endow each set with the subspace topology it inherits from $mathbbR^n$.
Exercise 16.18 on page 415 of Lee's Introduction to Smooth Manifolds (2nd edition) asks us to show that $mathbbH^n$ and $overlinemathbbR^n_+$ are homeomorphic.
However, it seems to me the spaces are not homeomorphic, as shown by the following argument.
Suppose $f: overlinemathbbR^n_+ to mathbbH^n$ is a homeomorphism. Set $f(0, dots , 0) = (a^1, dots, a^n)$. Then
$$A = f(overlinemathbbR^n_+ setminus (0, dots, 0)) = f((infty,0) times cdots times (infty,0)) = \
((infty, a^1) cup (a^1, - infty)) times cdots times ((infty, a^n) cup (a^n, 0]),$$
where, if $a^n= 0$, then in the last factor we discard $(a^n, 0]$.
The set $A$ is connected as it is the continuous image of the connected set $(infty, 0) times (infty,0)$. But at the same time we see that $A$ is disconnected as it is the union of the disjoint relatively open sets
$(infty, a^1) times cdots times (infty, a^n)$ and $ (a^1, - infty) times cdots times (a^n, 0]$ (or $(infty, a^1) times cdots times (infty, 0)$ and $(a^1, - infty) times cdots times (infty, 0)$ in case $a^n = 0$).
Is there a mistake in my argument or something I am missing? Any comments are greatly appreciated.
general-topology smooth-manifolds connectedness
asked Jul 18 at 1:54
JZShapiro
1,94311023
Let $mathbbH^n = (x^1, dots , x^n) in mathbbR^n : x^n ge 0$, and $overlinemathbbR^n_+ = (x^1, dots , x^n) in mathbbR^n : x^1 ge 0, dots, x^n ge 0$. Endow each set with the subspace topology it inherits from $mathbbR^n$.
Exercise 16.18 on page 415 of Lee's Introduction to Smooth Manifolds (2nd edition) asks us to show that $mathbbH^n$ and $overlinemathbbR^n_+$ are homeomorphic.
However, it seems to me the spaces are not homeomorphic, as shown by the following argument.
Suppose $f: overlinemathbbR^n_+ to mathbbH^n$ is a homeomorphism. Set $f(0, dots , 0) = (a^1, dots, a^n)$. Then
$$A = f(overlinemathbbR^n_+ setminus (0, dots, 0)) = f((infty,0) times cdots times (infty,0)) = \
((infty, a^1) cup (a^1, - infty)) times cdots times ((infty, a^n) cup (a^n, 0]),$$
where, if $a^n= 0$, then in the last factor we discard $(a^n, 0]$.
The set $A$ is connected as it is the continuous image of the connected set $(infty, 0) times (infty,0)$. But at the same time we see that $A$ is disconnected as it is the union of the disjoint relatively open sets
$(infty, a^1) times cdots times (infty, a^n)$ and $ (a^1, - infty) times cdots times (a^n, 0]$ (or $(infty, a^1) times cdots times (infty, 0)$ and $(a^1, - infty) times cdots times (infty, 0)$ in case $a^n = 0$).
Is there a mistake in my argument or something I am missing? Any comments are greatly appreciated.
general-topology smooth-manifolds connectedness
asked Jul 18 at 1:54
JZShapiro
1,94311023
edited Jul 18 at 2:05
asked Jul 18 at 1:54
JZShapiro
1,94311023
asked Jul 18 at 1:54
JZShapiro
1,94311023
asked Jul 18 at 1:54
JZShapiro
1,94311023
1,94311023
2
I think the problem is the line $$f(overlinemathbbR^n_+ setminus (0,dots,0)) =f( (0,infty) times dotstimes (0,infty)) $$ as the two sets whose f-images you are taking are NOT equal. For example, $(1,0,dots,0) in overlinemathbbR^n_+ setminus (0,dots,0)$ but $(1,0,dots,0) notin (0,infty) times dotstimes (0,infty)$
– itinerantleopard
Jul 18 at 3:12
2
To see that they are indeed homeomorphic, it might help to think about some small-dimensional cases first. When $n=1$, the space $ overlinemathbbR_+^1$ is actually equal to $mathbbH^1$. When $n=2$, the space $overlinemathbbR^2_+ $ is the closed upper right quadrant of the plane and $mathbbH^2$ is the closed upper half plane. The former can be homeomorphically deformed into the latter by "unfolding" the boundary until it lies flat against the $x$-axis. The geometric picture in higher dimensions is similar. Hope that helps!
– itinerantleopard
Jul 18 at 3:18
add a comment |Â
2
I think the problem is the line $$f(overlinemathbbR^n_+ setminus (0,dots,0)) =f( (0,infty) times dotstimes (0,infty)) $$ as the two sets whose f-images you are taking are NOT equal. For example, $(1,0,dots,0) in overlinemathbbR^n_+ setminus (0,dots,0)$ but $(1,0,dots,0) notin (0,infty) times dotstimes (0,infty)$
– itinerantleopard
Jul 18 at 3:12
2
To see that they are indeed homeomorphic, it might help to think about some small-dimensional cases first. When $n=1$, the space $ overlinemathbbR_+^1$ is actually equal to $mathbbH^1$. When $n=2$, the space $overlinemathbbR^2_+ $ is the closed upper right quadrant of the plane and $mathbbH^2$ is the closed upper half plane. The former can be homeomorphically deformed into the latter by "unfolding" the boundary until it lies flat against the $x$-axis. The geometric picture in higher dimensions is similar. Hope that helps!
– itinerantleopard
Jul 18 at 3:18
2
2
I think the problem is the line $$f(overlinemathbbR^n_+ setminus (0,dots,0)) =f( (0,infty) times dotstimes (0,infty)) $$ as the two sets whose f-images you are taking are NOT equal. For example, $(1,0,dots,0) in overlinemathbbR^n_+ setminus (0,dots,0)$ but $(1,0,dots,0) notin (0,infty) times dotstimes (0,infty)$
– itinerantleopard
Jul 18 at 3:12
I think the problem is the line $$f(overlinemathbbR^n_+ setminus (0,dots,0)) =f( (0,infty) times dotstimes (0,infty)) $$ as the two sets whose f-images you are taking are NOT equal. For example, $(1,0,dots,0) in overlinemathbbR^n_+ setminus (0,dots,0)$ but $(1,0,dots,0) notin (0,infty) times dotstimes (0,infty)$
– itinerantleopard
Jul 18 at 3:12
2
2
To see that they are indeed homeomorphic, it might help to think about some small-dimensional cases first. When $n=1$, the space $ overlinemathbbR_+^1$ is actually equal to $mathbbH^1$. When $n=2$, the space $overlinemathbbR^2_+ $ is the closed upper right quadrant of the plane and $mathbbH^2$ is the closed upper half plane. The former can be homeomorphically deformed into the latter by "unfolding" the boundary until it lies flat against the $x$-axis. The geometric picture in higher dimensions is similar. Hope that helps!
– itinerantleopard
Jul 18 at 3:18
To see that they are indeed homeomorphic, it might help to think about some small-dimensional cases first. When $n=1$, the space $ overlinemathbbR_+^1$ is actually equal to $mathbbH^1$. When $n=2$, the space $overlinemathbbR^2_+ $ is the closed upper right quadrant of the plane and $mathbbH^2$ is the closed upper half plane. The former can be homeomorphically deformed into the latter by "unfolding" the boundary until it lies flat against the $x$-axis. The geometric picture in higher dimensions is similar. Hope that helps!
– itinerantleopard
Jul 18 at 3:18
add a comment |Â
2 Answers
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We have Let $mathbbH^n = mathbbR^n-1 times [0,infty)$ and $overlinemathbbR^n_+ = [0,infty)^n$. Let us first construct a homeomorphism $h_2 : [0,infty)^2 to mathbbR times [0,infty)$. Define
$h_2(x_1,x_2) = (x_1,x_2)$ for $x_2 le x_1$,
$h_2(x_1,x_2) = (-x_2 + 2x_1,x_1)$ for $x_1 le x_2$.
$h_2$ maps the line segment connecting $(0,x_2)$ and $(x_2,x_2)$ bijectively onto the line segment connecting $(-x_2,0)$ and $(x_2,x_2)$. It is easy to verify that $h_2$ is continuous. Next define $g: mathbbR times [0,infty) to [0,infty)^2$ by
$g(y_1,y_2) = (y_1,y_2)$ for $y_2 le y_1$,
$g(y_1,y_2) = (y_2,2y_2 - y_1)$ for $y_1 le y_2$.
$g$ is continuous and $g circ h_2 = id$, $h_2 circ g = id$. Therefore $h_2$ is a homeomorphism.
For $n ge 2$ we therefore obtain a homeomorphism
$$h_n = h_2 times id_[0,infty)^n-2 : [0,infty)^n to mathbbR times [0,infty)^n-1 .$$
Then
$$h = (id_mathbbR^n-2 times h_2) circ ... circ (id_mathbbR times h_n-1) circ h_n$$
is the desired homeomorphism.
answered Jul 18 at 11:00
Paul Frost
3,703420
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up vote
2
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I tried to prove using induction as follows:
For case $n=1$, $[0,infty) = BbbH^1$, and for $n=2$ the map $f(z)=z^2$ will do the job, regard $BbbR^2$ as the complex plane $BbbC$. Now assume that $mathbbH^napprox barmathbbR^n_+$ for all dimension less than or equal to $n$. By definition of upper half-space we have
beginalign
barmathbbR^n+1_+ = barmathbbR^n_+ times barmathbbR_+ &approx mathbbH^n times barmathbbR_+ = mathbbR^n-1 times barmathbbR_+ times barmathbbR_+ \ &approx mathbbR^n-1 times mathbbH^2 = mathbbR^n-1 times mathbbR times barmathbbR_+ = mathbbR^n times barmathbbR_+ = mathbbH^n+1.
endalign
answered Jul 18 at 11:20
Sou
2,7062820
The proof of $barmathbbR_+ times barmathbbR_+ approx mathbbH^2$ ($n = 2$) is missing. It cannot be reduced to $n = 1$.
– Paul Frost
Jul 19 at 12:15
@PaulFrost You're right. I didn't see that before. Thank you. I'll amend my answer.
– Sou
Jul 19 at 12:24
I like your proof for $n = 2$. (+1)
– Paul Frost
Jul 19 at 17:38
@PaulFrost : Thanks. I like your proof too.
– Sou
Jul 19 at 17:40
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We have Let $mathbbH^n = mathbbR^n-1 times [0,infty)$ and $overlinemathbbR^n_+ = [0,infty)^n$. Let us first construct a homeomorphism $h_2 : [0,infty)^2 to mathbbR times [0,infty)$. Define
$h_2(x_1,x_2) = (x_1,x_2)$ for $x_2 le x_1$,
$h_2(x_1,x_2) = (-x_2 + 2x_1,x_1)$ for $x_1 le x_2$.
$h_2$ maps the line segment connecting $(0,x_2)$ and $(x_2,x_2)$ bijectively onto the line segment connecting $(-x_2,0)$ and $(x_2,x_2)$. It is easy to verify that $h_2$ is continuous. Next define $g: mathbbR times [0,infty) to [0,infty)^2$ by
$g(y_1,y_2) = (y_1,y_2)$ for $y_2 le y_1$,
$g(y_1,y_2) = (y_2,2y_2 - y_1)$ for $y_1 le y_2$.
$g$ is continuous and $g circ h_2 = id$, $h_2 circ g = id$. Therefore $h_2$ is a homeomorphism.
For $n ge 2$ we therefore obtain a homeomorphism
$$h_n = h_2 times id_[0,infty)^n-2 : [0,infty)^n to mathbbR times [0,infty)^n-1 .$$
Then
$$h = (id_mathbbR^n-2 times h_2) circ ... circ (id_mathbbR times h_n-1) circ h_n$$
is the desired homeomorphism.
answered Jul 18 at 11:00
Paul Frost
3,703420
add a comment |Â
up vote
3
down vote
accepted
We have Let $mathbbH^n = mathbbR^n-1 times [0,infty)$ and $overlinemathbbR^n_+ = [0,infty)^n$. Let us first construct a homeomorphism $h_2 : [0,infty)^2 to mathbbR times [0,infty)$. Define
$h_2(x_1,x_2) = (x_1,x_2)$ for $x_2 le x_1$,
$h_2(x_1,x_2) = (-x_2 + 2x_1,x_1)$ for $x_1 le x_2$.
$h_2$ maps the line segment connecting $(0,x_2)$ and $(x_2,x_2)$ bijectively onto the line segment connecting $(-x_2,0)$ and $(x_2,x_2)$. It is easy to verify that $h_2$ is continuous. Next define $g: mathbbR times [0,infty) to [0,infty)^2$ by
$g(y_1,y_2) = (y_1,y_2)$ for $y_2 le y_1$,
$g(y_1,y_2) = (y_2,2y_2 - y_1)$ for $y_1 le y_2$.
$g$ is continuous and $g circ h_2 = id$, $h_2 circ g = id$. Therefore $h_2$ is a homeomorphism.
For $n ge 2$ we therefore obtain a homeomorphism
$$h_n = h_2 times id_[0,infty)^n-2 : [0,infty)^n to mathbbR times [0,infty)^n-1 .$$
Then
$$h = (id_mathbbR^n-2 times h_2) circ ... circ (id_mathbbR times h_n-1) circ h_n$$
is the desired homeomorphism.
answered Jul 18 at 11:00
Paul Frost
3,703420
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We have Let $mathbbH^n = mathbbR^n-1 times [0,infty)$ and $overlinemathbbR^n_+ = [0,infty)^n$. Let us first construct a homeomorphism $h_2 : [0,infty)^2 to mathbbR times [0,infty)$. Define
$h_2(x_1,x_2) = (x_1,x_2)$ for $x_2 le x_1$,
$h_2(x_1,x_2) = (-x_2 + 2x_1,x_1)$ for $x_1 le x_2$.
$h_2$ maps the line segment connecting $(0,x_2)$ and $(x_2,x_2)$ bijectively onto the line segment connecting $(-x_2,0)$ and $(x_2,x_2)$. It is easy to verify that $h_2$ is continuous. Next define $g: mathbbR times [0,infty) to [0,infty)^2$ by
$g(y_1,y_2) = (y_1,y_2)$ for $y_2 le y_1$,
$g(y_1,y_2) = (y_2,2y_2 - y_1)$ for $y_1 le y_2$.
$g$ is continuous and $g circ h_2 = id$, $h_2 circ g = id$. Therefore $h_2$ is a homeomorphism.
For $n ge 2$ we therefore obtain a homeomorphism
$$h_n = h_2 times id_[0,infty)^n-2 : [0,infty)^n to mathbbR times [0,infty)^n-1 .$$
Then
$$h = (id_mathbbR^n-2 times h_2) circ ... circ (id_mathbbR times h_n-1) circ h_n$$
is the desired homeomorphism.
answered Jul 18 at 11:00
Paul Frost
3,703420
We have Let $mathbbH^n = mathbbR^n-1 times [0,infty)$ and $overlinemathbbR^n_+ = [0,infty)^n$. Let us first construct a homeomorphism $h_2 : [0,infty)^2 to mathbbR times [0,infty)$. Define
$h_2(x_1,x_2) = (x_1,x_2)$ for $x_2 le x_1$,
$h_2(x_1,x_2) = (-x_2 + 2x_1,x_1)$ for $x_1 le x_2$.
$h_2$ maps the line segment connecting $(0,x_2)$ and $(x_2,x_2)$ bijectively onto the line segment connecting $(-x_2,0)$ and $(x_2,x_2)$. It is easy to verify that $h_2$ is continuous. Next define $g: mathbbR times [0,infty) to [0,infty)^2$ by
$g(y_1,y_2) = (y_1,y_2)$ for $y_2 le y_1$,
$g(y_1,y_2) = (y_2,2y_2 - y_1)$ for $y_1 le y_2$.
$g$ is continuous and $g circ h_2 = id$, $h_2 circ g = id$. Therefore $h_2$ is a homeomorphism.
For $n ge 2$ we therefore obtain a homeomorphism
$$h_n = h_2 times id_[0,infty)^n-2 : [0,infty)^n to mathbbR times [0,infty)^n-1 .$$
Then
$$h = (id_mathbbR^n-2 times h_2) circ ... circ (id_mathbbR times h_n-1) circ h_n$$
is the desired homeomorphism.
answered Jul 18 at 11:00
Paul Frost
3,703420
answered Jul 18 at 11:00
Paul Frost
3,703420
answered Jul 18 at 11:00
Paul Frost
3,703420
answered Jul 18 at 11:00
Paul Frost
3,703420
3,703420
add a comment |Â
add a comment |Â
up vote
2
down vote
I tried to prove using induction as follows:
For case $n=1$, $[0,infty) = BbbH^1$, and for $n=2$ the map $f(z)=z^2$ will do the job, regard $BbbR^2$ as the complex plane $BbbC$. Now assume that $mathbbH^napprox barmathbbR^n_+$ for all dimension less than or equal to $n$. By definition of upper half-space we have
beginalign
barmathbbR^n+1_+ = barmathbbR^n_+ times barmathbbR_+ &approx mathbbH^n times barmathbbR_+ = mathbbR^n-1 times barmathbbR_+ times barmathbbR_+ \ &approx mathbbR^n-1 times mathbbH^2 = mathbbR^n-1 times mathbbR times barmathbbR_+ = mathbbR^n times barmathbbR_+ = mathbbH^n+1.
endalign
answered Jul 18 at 11:20
Sou
2,7062820
The proof of $barmathbbR_+ times barmathbbR_+ approx mathbbH^2$ ($n = 2$) is missing. It cannot be reduced to $n = 1$.
– Paul Frost
Jul 19 at 12:15
@PaulFrost You're right. I didn't see that before. Thank you. I'll amend my answer.
– Sou
Jul 19 at 12:24
I like your proof for $n = 2$. (+1)
– Paul Frost
Jul 19 at 17:38
@PaulFrost : Thanks. I like your proof too.
– Sou
Jul 19 at 17:40
add a comment |Â
up vote
2
down vote
I tried to prove using induction as follows:
For case $n=1$, $[0,infty) = BbbH^1$, and for $n=2$ the map $f(z)=z^2$ will do the job, regard $BbbR^2$ as the complex plane $BbbC$. Now assume that $mathbbH^napprox barmathbbR^n_+$ for all dimension less than or equal to $n$. By definition of upper half-space we have
beginalign
barmathbbR^n+1_+ = barmathbbR^n_+ times barmathbbR_+ &approx mathbbH^n times barmathbbR_+ = mathbbR^n-1 times barmathbbR_+ times barmathbbR_+ \ &approx mathbbR^n-1 times mathbbH^2 = mathbbR^n-1 times mathbbR times barmathbbR_+ = mathbbR^n times barmathbbR_+ = mathbbH^n+1.
endalign
answered Jul 18 at 11:20
Sou
2,7062820
The proof of $barmathbbR_+ times barmathbbR_+ approx mathbbH^2$ ($n = 2$) is missing. It cannot be reduced to $n = 1$.
– Paul Frost
Jul 19 at 12:15
@PaulFrost You're right. I didn't see that before. Thank you. I'll amend my answer.
– Sou
Jul 19 at 12:24
I like your proof for $n = 2$. (+1)
– Paul Frost
Jul 19 at 17:38
@PaulFrost : Thanks. I like your proof too.
– Sou
Jul 19 at 17:40
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I tried to prove using induction as follows:
For case $n=1$, $[0,infty) = BbbH^1$, and for $n=2$ the map $f(z)=z^2$ will do the job, regard $BbbR^2$ as the complex plane $BbbC$. Now assume that $mathbbH^napprox barmathbbR^n_+$ for all dimension less than or equal to $n$. By definition of upper half-space we have
beginalign
barmathbbR^n+1_+ = barmathbbR^n_+ times barmathbbR_+ &approx mathbbH^n times barmathbbR_+ = mathbbR^n-1 times barmathbbR_+ times barmathbbR_+ \ &approx mathbbR^n-1 times mathbbH^2 = mathbbR^n-1 times mathbbR times barmathbbR_+ = mathbbR^n times barmathbbR_+ = mathbbH^n+1.
endalign
answered Jul 18 at 11:20
Sou
2,7062820
I tried to prove using induction as follows:
For case $n=1$, $[0,infty) = BbbH^1$, and for $n=2$ the map $f(z)=z^2$ will do the job, regard $BbbR^2$ as the complex plane $BbbC$. Now assume that $mathbbH^napprox barmathbbR^n_+$ for all dimension less than or equal to $n$. By definition of upper half-space we have
beginalign
barmathbbR^n+1_+ = barmathbbR^n_+ times barmathbbR_+ &approx mathbbH^n times barmathbbR_+ = mathbbR^n-1 times barmathbbR_+ times barmathbbR_+ \ &approx mathbbR^n-1 times mathbbH^2 = mathbbR^n-1 times mathbbR times barmathbbR_+ = mathbbR^n times barmathbbR_+ = mathbbH^n+1.
endalign
answered Jul 18 at 11:20
Sou
2,7062820
edited Jul 19 at 12:31
answered Jul 18 at 11:20
Sou
2,7062820
answered Jul 18 at 11:20
Sou
2,7062820
answered Jul 18 at 11:20
Sou
2,7062820
2,7062820
The proof of $barmathbbR_+ times barmathbbR_+ approx mathbbH^2$ ($n = 2$) is missing. It cannot be reduced to $n = 1$.
– Paul Frost
Jul 19 at 12:15
@PaulFrost You're right. I didn't see that before. Thank you. I'll amend my answer.
– Sou
Jul 19 at 12:24
I like your proof for $n = 2$. (+1)
– Paul Frost
Jul 19 at 17:38
@PaulFrost : Thanks. I like your proof too.
– Sou
Jul 19 at 17:40
add a comment |Â
The proof of $barmathbbR_+ times barmathbbR_+ approx mathbbH^2$ ($n = 2$) is missing. It cannot be reduced to $n = 1$.
– Paul Frost
Jul 19 at 12:15
@PaulFrost You're right. I didn't see that before. Thank you. I'll amend my answer.
– Sou
Jul 19 at 12:24
I like your proof for $n = 2$. (+1)
– Paul Frost
Jul 19 at 17:38
@PaulFrost : Thanks. I like your proof too.
– Sou
Jul 19 at 17:40
The proof of $barmathbbR_+ times barmathbbR_+ approx mathbbH^2$ ($n = 2$) is missing. It cannot be reduced to $n = 1$.
– Paul Frost
Jul 19 at 12:15
The proof of $barmathbbR_+ times barmathbbR_+ approx mathbbH^2$ ($n = 2$) is missing. It cannot be reduced to $n = 1$.
– Paul Frost
Jul 19 at 12:15
@PaulFrost You're right. I didn't see that before. Thank you. I'll amend my answer.
– Sou
Jul 19 at 12:24
@PaulFrost You're right. I didn't see that before. Thank you. I'll amend my answer.
– Sou
Jul 19 at 12:24
I like your proof for $n = 2$. (+1)
– Paul Frost
Jul 19 at 17:38
I like your proof for $n = 2$. (+1)
– Paul Frost
Jul 19 at 17:38
@PaulFrost : Thanks. I like your proof too.
– Sou
Jul 19 at 17:40
@PaulFrost : Thanks. I like your proof too.
– Sou
Jul 19 at 17:40
add a comment |Â
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2
I think the problem is the line $$f(overlinemathbbR^n_+ setminus (0,dots,0)) =f( (0,infty) times dotstimes (0,infty)) $$ as the two sets whose f-images you are taking are NOT equal. For example, $(1,0,dots,0) in overlinemathbbR^n_+ setminus (0,dots,0)$ but $(1,0,dots,0) notin (0,infty) times dotstimes (0,infty)$
– itinerantleopard
Jul 18 at 3:12
2
To see that they are indeed homeomorphic, it might help to think about some small-dimensional cases first. When $n=1$, the space $ overlinemathbbR_+^1$ is actually equal to $mathbbH^1$. When $n=2$, the space $overlinemathbbR^2_+ $ is the closed upper right quadrant of the plane and $mathbbH^2$ is the closed upper half plane. The former can be homeomorphically deformed into the latter by "unfolding" the boundary until it lies flat against the $x$-axis. The geometric picture in higher dimensions is similar. Hope that helps!
– itinerantleopard
Jul 18 at 3:18