Can the surface of a 3-dimension cube be diffeomorphic to a 3-dimension sphere?












0














We all know the surface of a cube homeomorphic to a sphere $S^3$ by retraction, but I'm confused that whether the surface of a cube can be diffeomorphic to $S^3$.






smooth-manifolds







share|cite|improve this question
























  • Do you mean the surface of the unit hypercube in $Bbb R^4$, with the differential structure inherited from $Bbb R^4$? I don't really know what happens at the edges and vertices there, but it looks messy.
    – Arthur
    Dec 2 '18 at 14:51












  • Can you define what you mean by "3-dimensional cube" and by "3-dimensional sphere"? I would think that the former is the three-fold product of the unit interval, and the latter is the unit sphere in $Bbb R^4$. But those two spaces are not homeomorphic (e.g. their homology distinguishes them).
    – Danu
    Dec 2 '18 at 15:34










  • I think you mean a $3$ dimensional ball, which is a solid object like the earth. We use sphere to represent the surface of a ball.
    – Ross Millikan
    Dec 2 '18 at 16:07












  • In addition, in order to ask if two things are diffeomorphic, both things must be smooth manifolds. So, you must first choose some kind of atlas on the cube.
    – Jason DeVito
    Dec 4 '18 at 14:24










  • So the question is, if there exist a kind of atlas on cube, such that the cube with the atlas can be a smooth manifolds.
    – Lau
    Dec 4 '18 at 14:26
















0














We all know the surface of a cube homeomorphic to a sphere $S^3$ by retraction, but I'm confused that whether the surface of a cube can be diffeomorphic to $S^3$.






smooth-manifolds







share|cite|improve this question
























  • Do you mean the surface of the unit hypercube in $Bbb R^4$, with the differential structure inherited from $Bbb R^4$? I don't really know what happens at the edges and vertices there, but it looks messy.
    – Arthur
    Dec 2 '18 at 14:51












  • Can you define what you mean by "3-dimensional cube" and by "3-dimensional sphere"? I would think that the former is the three-fold product of the unit interval, and the latter is the unit sphere in $Bbb R^4$. But those two spaces are not homeomorphic (e.g. their homology distinguishes them).
    – Danu
    Dec 2 '18 at 15:34










  • I think you mean a $3$ dimensional ball, which is a solid object like the earth. We use sphere to represent the surface of a ball.
    – Ross Millikan
    Dec 2 '18 at 16:07












  • In addition, in order to ask if two things are diffeomorphic, both things must be smooth manifolds. So, you must first choose some kind of atlas on the cube.
    – Jason DeVito
    Dec 4 '18 at 14:24










  • So the question is, if there exist a kind of atlas on cube, such that the cube with the atlas can be a smooth manifolds.
    – Lau
    Dec 4 '18 at 14:26














0












0








0







We all know the surface of a cube homeomorphic to a sphere $S^3$ by retraction, but I'm confused that whether the surface of a cube can be diffeomorphic to $S^3$.






smooth-manifolds







share|cite|improve this question















We all know the surface of a cube homeomorphic to a sphere $S^3$ by retraction, but I'm confused that whether the surface of a cube can be diffeomorphic to $S^3$.






smooth-manifolds




smooth-manifolds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 14:27

























asked Dec 2 '18 at 14:35









Lau

541315




541315












  • Do you mean the surface of the unit hypercube in $Bbb R^4$, with the differential structure inherited from $Bbb R^4$? I don't really know what happens at the edges and vertices there, but it looks messy.
    – Arthur
    Dec 2 '18 at 14:51












  • Can you define what you mean by "3-dimensional cube" and by "3-dimensional sphere"? I would think that the former is the three-fold product of the unit interval, and the latter is the unit sphere in $Bbb R^4$. But those two spaces are not homeomorphic (e.g. their homology distinguishes them).
    – Danu
    Dec 2 '18 at 15:34










  • I think you mean a $3$ dimensional ball, which is a solid object like the earth. We use sphere to represent the surface of a ball.
    – Ross Millikan
    Dec 2 '18 at 16:07












  • In addition, in order to ask if two things are diffeomorphic, both things must be smooth manifolds. So, you must first choose some kind of atlas on the cube.
    – Jason DeVito
    Dec 4 '18 at 14:24










  • So the question is, if there exist a kind of atlas on cube, such that the cube with the atlas can be a smooth manifolds.
    – Lau
    Dec 4 '18 at 14:26


















  • Do you mean the surface of the unit hypercube in $Bbb R^4$, with the differential structure inherited from $Bbb R^4$? I don't really know what happens at the edges and vertices there, but it looks messy.
    – Arthur
    Dec 2 '18 at 14:51












  • Can you define what you mean by "3-dimensional cube" and by "3-dimensional sphere"? I would think that the former is the three-fold product of the unit interval, and the latter is the unit sphere in $Bbb R^4$. But those two spaces are not homeomorphic (e.g. their homology distinguishes them).
    – Danu
    Dec 2 '18 at 15:34










  • I think you mean a $3$ dimensional ball, which is a solid object like the earth. We use sphere to represent the surface of a ball.
    – Ross Millikan
    Dec 2 '18 at 16:07












  • In addition, in order to ask if two things are diffeomorphic, both things must be smooth manifolds. So, you must first choose some kind of atlas on the cube.
    – Jason DeVito
    Dec 4 '18 at 14:24










  • So the question is, if there exist a kind of atlas on cube, such that the cube with the atlas can be a smooth manifolds.
    – Lau
    Dec 4 '18 at 14:26
















Do you mean the surface of the unit hypercube in $Bbb R^4$, with the differential structure inherited from $Bbb R^4$? I don't really know what happens at the edges and vertices there, but it looks messy.
– Arthur
Dec 2 '18 at 14:51






Do you mean the surface of the unit hypercube in $Bbb R^4$, with the differential structure inherited from $Bbb R^4$? I don't really know what happens at the edges and vertices there, but it looks messy.
– Arthur
Dec 2 '18 at 14:51














Can you define what you mean by "3-dimensional cube" and by "3-dimensional sphere"? I would think that the former is the three-fold product of the unit interval, and the latter is the unit sphere in $Bbb R^4$. But those two spaces are not homeomorphic (e.g. their homology distinguishes them).
– Danu
Dec 2 '18 at 15:34




Can you define what you mean by "3-dimensional cube" and by "3-dimensional sphere"? I would think that the former is the three-fold product of the unit interval, and the latter is the unit sphere in $Bbb R^4$. But those two spaces are not homeomorphic (e.g. their homology distinguishes them).
– Danu
Dec 2 '18 at 15:34












I think you mean a $3$ dimensional ball, which is a solid object like the earth. We use sphere to represent the surface of a ball.
– Ross Millikan
Dec 2 '18 at 16:07






I think you mean a $3$ dimensional ball, which is a solid object like the earth. We use sphere to represent the surface of a ball.
– Ross Millikan
Dec 2 '18 at 16:07














In addition, in order to ask if two things are diffeomorphic, both things must be smooth manifolds. So, you must first choose some kind of atlas on the cube.
– Jason DeVito
Dec 4 '18 at 14:24




In addition, in order to ask if two things are diffeomorphic, both things must be smooth manifolds. So, you must first choose some kind of atlas on the cube.
– Jason DeVito
Dec 4 '18 at 14:24












So the question is, if there exist a kind of atlas on cube, such that the cube with the atlas can be a smooth manifolds.
– Lau
Dec 4 '18 at 14:26




So the question is, if there exist a kind of atlas on cube, such that the cube with the atlas can be a smooth manifolds.
– Lau
Dec 4 '18 at 14:26










1 Answer
1






active

oldest

votes


















2














Let me answer this in lower dimensions:



First, consider the bi-unit square $S$ (the boundary of $[-1, 1] times [-1, 1]$ in $Bbb R^2$) and the unit circle $C$ in $Bbb R^2$.



There's a nice map ("radial projection") from the first to the second defined as follows:



$$
f(x, y) = begin{cases}
(frac{x}{sqrt{y^2+1}}, frac{y}{sqrt{y^2 + 1}})& x = pm 1 \
(frac{x}{sqrt{x^2+1}}, frac{y}{sqrt{x^2 + 1}})& y = pm 1 \
end{cases}
$$
which cane be written, more generally, as $f(x, y) = frac{1}{sqrt{x^2 + y^2}} (x, y)$, but I wanted to emphasize that it's defined on the square, and that at the four corners, the expressions involved shift from a dependence on $x$ to a dependence on $y$ or vice versa.



The inverse of $f$ is
$$
g(x, y) = begin{cases}
(frac{x}{|x|}, frac{y}{|x|}) & |x| ge |y|\
(frac{x}{|y|}, frac{y}{|y|}) & |x| le |y|
end{cases}
$$



Now one smoothness structure on $S$ can be described as follows: we say that a function
$$
u: S to Bbb R
$$ is smooth if and only iff
$$
u circ g : C to Bbb R
$$
is smooth.



With this definition, the square $S$ is a smooth manifold, and it's diffeomorphic to the circle $C$, with $g$ being the diffeomorphism.



There's a small caveat here, though: the inclusion map
$$
i : S to Bbb R^2: (x, y) mapsto (x, y)
$$
that embeds the square in the plane is not a smooth embedding.



What about higher dimensions? Well, an exact analog to $f$ can be written using more variables (and the "general" formula that I gave after the "cases" formula); the analog of $g$ involves 3 cases for a cube in 3-space (look at $|x|, |y|, |z|$: each case handles the situations where one of these is the largest).



When we get to 4-space, it's the same deal, but with another variable, and the formula for $g$ has four cases.



But everything else I said (including the fact that with the resulting smoothness structure, the standard embedding of the square/cube/hypercube is not a smooth embedding) remains true.






share|cite|improve this answer























  • what's $s$ in $scirc g$?
    – Lau
    Dec 5 '18 at 3:16










  • Fixed. It's "u" -- just a typo. Sorry about that.
    – John Hughes
    Dec 5 '18 at 12:44










  • I think I have got your point, thanks a lot.
    – Lau
    Dec 5 '18 at 12:47











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Let me answer this in lower dimensions:



First, consider the bi-unit square $S$ (the boundary of $[-1, 1] times [-1, 1]$ in $Bbb R^2$) and the unit circle $C$ in $Bbb R^2$.



There's a nice map ("radial projection") from the first to the second defined as follows:



$$
f(x, y) = begin{cases}
(frac{x}{sqrt{y^2+1}}, frac{y}{sqrt{y^2 + 1}})& x = pm 1 \
(frac{x}{sqrt{x^2+1}}, frac{y}{sqrt{x^2 + 1}})& y = pm 1 \
end{cases}
$$
which cane be written, more generally, as $f(x, y) = frac{1}{sqrt{x^2 + y^2}} (x, y)$, but I wanted to emphasize that it's defined on the square, and that at the four corners, the expressions involved shift from a dependence on $x$ to a dependence on $y$ or vice versa.



The inverse of $f$ is
$$
g(x, y) = begin{cases}
(frac{x}{|x|}, frac{y}{|x|}) & |x| ge |y|\
(frac{x}{|y|}, frac{y}{|y|}) & |x| le |y|
end{cases}
$$



Now one smoothness structure on $S$ can be described as follows: we say that a function
$$
u: S to Bbb R
$$ is smooth if and only iff
$$
u circ g : C to Bbb R
$$
is smooth.



With this definition, the square $S$ is a smooth manifold, and it's diffeomorphic to the circle $C$, with $g$ being the diffeomorphism.



There's a small caveat here, though: the inclusion map
$$
i : S to Bbb R^2: (x, y) mapsto (x, y)
$$
that embeds the square in the plane is not a smooth embedding.



What about higher dimensions? Well, an exact analog to $f$ can be written using more variables (and the "general" formula that I gave after the "cases" formula); the analog of $g$ involves 3 cases for a cube in 3-space (look at $|x|, |y|, |z|$: each case handles the situations where one of these is the largest).



When we get to 4-space, it's the same deal, but with another variable, and the formula for $g$ has four cases.



But everything else I said (including the fact that with the resulting smoothness structure, the standard embedding of the square/cube/hypercube is not a smooth embedding) remains true.






share|cite|improve this answer























  • what's $s$ in $scirc g$?
    – Lau
    Dec 5 '18 at 3:16










  • Fixed. It's "u" -- just a typo. Sorry about that.
    – John Hughes
    Dec 5 '18 at 12:44










  • I think I have got your point, thanks a lot.
    – Lau
    Dec 5 '18 at 12:47
















2














Let me answer this in lower dimensions:



First, consider the bi-unit square $S$ (the boundary of $[-1, 1] times [-1, 1]$ in $Bbb R^2$) and the unit circle $C$ in $Bbb R^2$.



There's a nice map ("radial projection") from the first to the second defined as follows:



$$
f(x, y) = begin{cases}
(frac{x}{sqrt{y^2+1}}, frac{y}{sqrt{y^2 + 1}})& x = pm 1 \
(frac{x}{sqrt{x^2+1}}, frac{y}{sqrt{x^2 + 1}})& y = pm 1 \
end{cases}
$$
which cane be written, more generally, as $f(x, y) = frac{1}{sqrt{x^2 + y^2}} (x, y)$, but I wanted to emphasize that it's defined on the square, and that at the four corners, the expressions involved shift from a dependence on $x$ to a dependence on $y$ or vice versa.



The inverse of $f$ is
$$
g(x, y) = begin{cases}
(frac{x}{|x|}, frac{y}{|x|}) & |x| ge |y|\
(frac{x}{|y|}, frac{y}{|y|}) & |x| le |y|
end{cases}
$$



Now one smoothness structure on $S$ can be described as follows: we say that a function
$$
u: S to Bbb R
$$ is smooth if and only iff
$$
u circ g : C to Bbb R
$$
is smooth.



With this definition, the square $S$ is a smooth manifold, and it's diffeomorphic to the circle $C$, with $g$ being the diffeomorphism.



There's a small caveat here, though: the inclusion map
$$
i : S to Bbb R^2: (x, y) mapsto (x, y)
$$
that embeds the square in the plane is not a smooth embedding.



What about higher dimensions? Well, an exact analog to $f$ can be written using more variables (and the "general" formula that I gave after the "cases" formula); the analog of $g$ involves 3 cases for a cube in 3-space (look at $|x|, |y|, |z|$: each case handles the situations where one of these is the largest).



When we get to 4-space, it's the same deal, but with another variable, and the formula for $g$ has four cases.



But everything else I said (including the fact that with the resulting smoothness structure, the standard embedding of the square/cube/hypercube is not a smooth embedding) remains true.






share|cite|improve this answer























  • what's $s$ in $scirc g$?
    – Lau
    Dec 5 '18 at 3:16










  • Fixed. It's "u" -- just a typo. Sorry about that.
    – John Hughes
    Dec 5 '18 at 12:44










  • I think I have got your point, thanks a lot.
    – Lau
    Dec 5 '18 at 12:47














2












2








2






Let me answer this in lower dimensions:



First, consider the bi-unit square $S$ (the boundary of $[-1, 1] times [-1, 1]$ in $Bbb R^2$) and the unit circle $C$ in $Bbb R^2$.



There's a nice map ("radial projection") from the first to the second defined as follows:



$$
f(x, y) = begin{cases}
(frac{x}{sqrt{y^2+1}}, frac{y}{sqrt{y^2 + 1}})& x = pm 1 \
(frac{x}{sqrt{x^2+1}}, frac{y}{sqrt{x^2 + 1}})& y = pm 1 \
end{cases}
$$
which cane be written, more generally, as $f(x, y) = frac{1}{sqrt{x^2 + y^2}} (x, y)$, but I wanted to emphasize that it's defined on the square, and that at the four corners, the expressions involved shift from a dependence on $x$ to a dependence on $y$ or vice versa.



The inverse of $f$ is
$$
g(x, y) = begin{cases}
(frac{x}{|x|}, frac{y}{|x|}) & |x| ge |y|\
(frac{x}{|y|}, frac{y}{|y|}) & |x| le |y|
end{cases}
$$



Now one smoothness structure on $S$ can be described as follows: we say that a function
$$
u: S to Bbb R
$$ is smooth if and only iff
$$
u circ g : C to Bbb R
$$
is smooth.



With this definition, the square $S$ is a smooth manifold, and it's diffeomorphic to the circle $C$, with $g$ being the diffeomorphism.



There's a small caveat here, though: the inclusion map
$$
i : S to Bbb R^2: (x, y) mapsto (x, y)
$$
that embeds the square in the plane is not a smooth embedding.



What about higher dimensions? Well, an exact analog to $f$ can be written using more variables (and the "general" formula that I gave after the "cases" formula); the analog of $g$ involves 3 cases for a cube in 3-space (look at $|x|, |y|, |z|$: each case handles the situations where one of these is the largest).



When we get to 4-space, it's the same deal, but with another variable, and the formula for $g$ has four cases.



But everything else I said (including the fact that with the resulting smoothness structure, the standard embedding of the square/cube/hypercube is not a smooth embedding) remains true.






share|cite|improve this answer














Let me answer this in lower dimensions:



First, consider the bi-unit square $S$ (the boundary of $[-1, 1] times [-1, 1]$ in $Bbb R^2$) and the unit circle $C$ in $Bbb R^2$.



There's a nice map ("radial projection") from the first to the second defined as follows:



$$
f(x, y) = begin{cases}
(frac{x}{sqrt{y^2+1}}, frac{y}{sqrt{y^2 + 1}})& x = pm 1 \
(frac{x}{sqrt{x^2+1}}, frac{y}{sqrt{x^2 + 1}})& y = pm 1 \
end{cases}
$$
which cane be written, more generally, as $f(x, y) = frac{1}{sqrt{x^2 + y^2}} (x, y)$, but I wanted to emphasize that it's defined on the square, and that at the four corners, the expressions involved shift from a dependence on $x$ to a dependence on $y$ or vice versa.



The inverse of $f$ is
$$
g(x, y) = begin{cases}
(frac{x}{|x|}, frac{y}{|x|}) & |x| ge |y|\
(frac{x}{|y|}, frac{y}{|y|}) & |x| le |y|
end{cases}
$$



Now one smoothness structure on $S$ can be described as follows: we say that a function
$$
u: S to Bbb R
$$ is smooth if and only iff
$$
u circ g : C to Bbb R
$$
is smooth.



With this definition, the square $S$ is a smooth manifold, and it's diffeomorphic to the circle $C$, with $g$ being the diffeomorphism.



There's a small caveat here, though: the inclusion map
$$
i : S to Bbb R^2: (x, y) mapsto (x, y)
$$
that embeds the square in the plane is not a smooth embedding.



What about higher dimensions? Well, an exact analog to $f$ can be written using more variables (and the "general" formula that I gave after the "cases" formula); the analog of $g$ involves 3 cases for a cube in 3-space (look at $|x|, |y|, |z|$: each case handles the situations where one of these is the largest).



When we get to 4-space, it's the same deal, but with another variable, and the formula for $g$ has four cases.



But everything else I said (including the fact that with the resulting smoothness structure, the standard embedding of the square/cube/hypercube is not a smooth embedding) remains true.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 12:44

























answered Dec 4 '18 at 14:43









John Hughes

62.4k24090




62.4k24090












  • what's $s$ in $scirc g$?
    – Lau
    Dec 5 '18 at 3:16










  • Fixed. It's "u" -- just a typo. Sorry about that.
    – John Hughes
    Dec 5 '18 at 12:44










  • I think I have got your point, thanks a lot.
    – Lau
    Dec 5 '18 at 12:47


















  • what's $s$ in $scirc g$?
    – Lau
    Dec 5 '18 at 3:16










  • Fixed. It's "u" -- just a typo. Sorry about that.
    – John Hughes
    Dec 5 '18 at 12:44










  • I think I have got your point, thanks a lot.
    – Lau
    Dec 5 '18 at 12:47
















what's $s$ in $scirc g$?
– Lau
Dec 5 '18 at 3:16




what's $s$ in $scirc g$?
– Lau
Dec 5 '18 at 3:16












Fixed. It's "u" -- just a typo. Sorry about that.
– John Hughes
Dec 5 '18 at 12:44




Fixed. It's "u" -- just a typo. Sorry about that.
– John Hughes
Dec 5 '18 at 12:44












I think I have got your point, thanks a lot.
– Lau
Dec 5 '18 at 12:47




I think I have got your point, thanks a lot.
– Lau
Dec 5 '18 at 12:47


















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Marschland

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Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si...
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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7 ...
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