Can two spheres be thought of as a single surface?
$begingroup$
Let's stick to $mathbb{R}^3$ for the sake of simplicity. Say I've got a sphere of radius $1$ centered at the origin and another with the same radius centered at $(5,6,7)$. Can one think about them as a single surface?
Some colleagues asked me this question a couple days ago and my answer was no, but then one of them mentioned the example of the two sheet hyperboloid (at one point one of them even said: "so is the set of all surfaces a surface as well?") and I was confused. I looked up the definition of $2$-dimensional manifolds again to make sure, but the answer is still not clear to me.
differential-geometry surfaces
asked Dec 7 '18 at 23:35
Matheus AndradeMatheus Andrade
1,364418
$endgroup$
add a comment |
$begingroup$
Let's stick to $mathbb{R}^3$ for the sake of simplicity. Say I've got a sphere of radius $1$ centered at the origin and another with the same radius centered at $(5,6,7)$. Can one think about them as a single surface?
Some colleagues asked me this question a couple days ago and my answer was no, but then one of them mentioned the example of the two sheet hyperboloid (at one point one of them even said: "so is the set of all surfaces a surface as well?") and I was confused. I looked up the definition of $2$-dimensional manifolds again to make sure, but the answer is still not clear to me.
differential-geometry surfaces
asked Dec 7 '18 at 23:35
Matheus AndradeMatheus Andrade
1,364418
$endgroup$
1
$begingroup$
It is disconnected but... I don't see any problem. You can consider both as your entire space. In fact, I would say that it is a submanifold.
$endgroup$
– Dog_69
Dec 7 '18 at 23:39
$begingroup$
I would say that the two sheet hyperboloid is only "a" surface if we're in projective space but I wouldn't call myself any sort of authority
$endgroup$
– Dan Uznanski
Dec 7 '18 at 23:41
add a comment |
$begingroup$
Let's stick to $mathbb{R}^3$ for the sake of simplicity. Say I've got a sphere of radius $1$ centered at the origin and another with the same radius centered at $(5,6,7)$. Can one think about them as a single surface?
Some colleagues asked me this question a couple days ago and my answer was no, but then one of them mentioned the example of the two sheet hyperboloid (at one point one of them even said: "so is the set of all surfaces a surface as well?") and I was confused. I looked up the definition of $2$-dimensional manifolds again to make sure, but the answer is still not clear to me.
differential-geometry surfaces
asked Dec 7 '18 at 23:35
Matheus AndradeMatheus Andrade
1,364418
$endgroup$
Let's stick to $mathbb{R}^3$ for the sake of simplicity. Say I've got a sphere of radius $1$ centered at the origin and another with the same radius centered at $(5,6,7)$. Can one think about them as a single surface?
Some colleagues asked me this question a couple days ago and my answer was no, but then one of them mentioned the example of the two sheet hyperboloid (at one point one of them even said: "so is the set of all surfaces a surface as well?") and I was confused. I looked up the definition of $2$-dimensional manifolds again to make sure, but the answer is still not clear to me.
differential-geometry surfaces
differential-geometry surfaces
asked Dec 7 '18 at 23:35
Matheus AndradeMatheus Andrade
1,364418
asked Dec 7 '18 at 23:35
Matheus AndradeMatheus Andrade
1,364418
asked Dec 7 '18 at 23:35
Matheus AndradeMatheus Andrade
1,364418
asked Dec 7 '18 at 23:35
Matheus AndradeMatheus Andrade
1,364418
asked Dec 7 '18 at 23:35
Matheus AndradeMatheus Andrade
1,364418
1,364418
1
$begingroup$
It is disconnected but... I don't see any problem. You can consider both as your entire space. In fact, I would say that it is a submanifold.
$endgroup$
– Dog_69
Dec 7 '18 at 23:39
$begingroup$
I would say that the two sheet hyperboloid is only "a" surface if we're in projective space but I wouldn't call myself any sort of authority
$endgroup$
– Dan Uznanski
Dec 7 '18 at 23:41
add a comment |
1
$begingroup$
It is disconnected but... I don't see any problem. You can consider both as your entire space. In fact, I would say that it is a submanifold.
$endgroup$
– Dog_69
Dec 7 '18 at 23:39
$begingroup$
I would say that the two sheet hyperboloid is only "a" surface if we're in projective space but I wouldn't call myself any sort of authority
$endgroup$
– Dan Uznanski
Dec 7 '18 at 23:41
1
1
$begingroup$
It is disconnected but... I don't see any problem. You can consider both as your entire space. In fact, I would say that it is a submanifold.
$endgroup$
– Dog_69
Dec 7 '18 at 23:39
$begingroup$
It is disconnected but... I don't see any problem. You can consider both as your entire space. In fact, I would say that it is a submanifold.
$endgroup$
– Dog_69
Dec 7 '18 at 23:39
$begingroup$
I would say that the two sheet hyperboloid is only "a" surface if we're in projective space but I wouldn't call myself any sort of authority
$endgroup$
– Dan Uznanski
Dec 7 '18 at 23:41
$begingroup$
I would say that the two sheet hyperboloid is only "a" surface if we're in projective space but I wouldn't call myself any sort of authority
$endgroup$
– Dan Uznanski
Dec 7 '18 at 23:41
add a comment |
1 Answer
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oldest
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$begingroup$
Sure, there's no need for a manifold to be connected. Your surface might naturally occur as a preimage of a regular value of a mapping $mathbb R^3to mathbb R$, or as the boundary of a $3$-dimensional connected manifold-with-boundary in $mathbb R^4$, or as the transversal intersection of two connected $3$-dimensional manifolds in $mathbb R^4$. It's also homeomorphic to the product space $S^2times S^0$, where $S^2$ is the familiar $2$-sphere and $S^0$ is the $0$-dimensional sphere consisting of two points. So if you wanted to change the definition of "surface" and "manifold" to rule out your surface, then you'd have to change all that surrounding theory, too.
answered Dec 8 '18 at 0:02
Chris CulterChris Culter
21.1k43886
$endgroup$
$begingroup$
Got it. Thanks!
$endgroup$
– Matheus Andrade
Dec 9 '18 at 17:30
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
Sure, there's no need for a manifold to be connected. Your surface might naturally occur as a preimage of a regular value of a mapping $mathbb R^3to mathbb R$, or as the boundary of a $3$-dimensional connected manifold-with-boundary in $mathbb R^4$, or as the transversal intersection of two connected $3$-dimensional manifolds in $mathbb R^4$. It's also homeomorphic to the product space $S^2times S^0$, where $S^2$ is the familiar $2$-sphere and $S^0$ is the $0$-dimensional sphere consisting of two points. So if you wanted to change the definition of "surface" and "manifold" to rule out your surface, then you'd have to change all that surrounding theory, too.
answered Dec 8 '18 at 0:02
Chris CulterChris Culter
21.1k43886
$endgroup$
$begingroup$
Got it. Thanks!
$endgroup$
– Matheus Andrade
Dec 9 '18 at 17:30
add a comment |
$begingroup$
Sure, there's no need for a manifold to be connected. Your surface might naturally occur as a preimage of a regular value of a mapping $mathbb R^3to mathbb R$, or as the boundary of a $3$-dimensional connected manifold-with-boundary in $mathbb R^4$, or as the transversal intersection of two connected $3$-dimensional manifolds in $mathbb R^4$. It's also homeomorphic to the product space $S^2times S^0$, where $S^2$ is the familiar $2$-sphere and $S^0$ is the $0$-dimensional sphere consisting of two points. So if you wanted to change the definition of "surface" and "manifold" to rule out your surface, then you'd have to change all that surrounding theory, too.
answered Dec 8 '18 at 0:02
Chris CulterChris Culter
21.1k43886
$endgroup$
$begingroup$
Got it. Thanks!
$endgroup$
– Matheus Andrade
Dec 9 '18 at 17:30
add a comment |
$begingroup$
Sure, there's no need for a manifold to be connected. Your surface might naturally occur as a preimage of a regular value of a mapping $mathbb R^3to mathbb R$, or as the boundary of a $3$-dimensional connected manifold-with-boundary in $mathbb R^4$, or as the transversal intersection of two connected $3$-dimensional manifolds in $mathbb R^4$. It's also homeomorphic to the product space $S^2times S^0$, where $S^2$ is the familiar $2$-sphere and $S^0$ is the $0$-dimensional sphere consisting of two points. So if you wanted to change the definition of "surface" and "manifold" to rule out your surface, then you'd have to change all that surrounding theory, too.
answered Dec 8 '18 at 0:02
Chris CulterChris Culter
21.1k43886
$endgroup$
Sure, there's no need for a manifold to be connected. Your surface might naturally occur as a preimage of a regular value of a mapping $mathbb R^3to mathbb R$, or as the boundary of a $3$-dimensional connected manifold-with-boundary in $mathbb R^4$, or as the transversal intersection of two connected $3$-dimensional manifolds in $mathbb R^4$. It's also homeomorphic to the product space $S^2times S^0$, where $S^2$ is the familiar $2$-sphere and $S^0$ is the $0$-dimensional sphere consisting of two points. So if you wanted to change the definition of "surface" and "manifold" to rule out your surface, then you'd have to change all that surrounding theory, too.
answered Dec 8 '18 at 0:02
Chris CulterChris Culter
21.1k43886
answered Dec 8 '18 at 0:02
Chris CulterChris Culter
21.1k43886
answered Dec 8 '18 at 0:02
Chris CulterChris Culter
21.1k43886
answered Dec 8 '18 at 0:02
Chris CulterChris Culter
21.1k43886
21.1k43886
$begingroup$
Got it. Thanks!
$endgroup$
– Matheus Andrade
Dec 9 '18 at 17:30
add a comment |
$begingroup$
Got it. Thanks!
$endgroup$
– Matheus Andrade
Dec 9 '18 at 17:30
$begingroup$
Got it. Thanks!
$endgroup$
– Matheus Andrade
Dec 9 '18 at 17:30
$begingroup$
Got it. Thanks!
$endgroup$
– Matheus Andrade
Dec 9 '18 at 17:30
add a comment |
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$begingroup$
It is disconnected but... I don't see any problem. You can consider both as your entire space. In fact, I would say that it is a submanifold.
$endgroup$
– Dog_69
Dec 7 '18 at 23:39
$begingroup$
I would say that the two sheet hyperboloid is only "a" surface if we're in projective space but I wouldn't call myself any sort of authority
$endgroup$
– Dan Uznanski
Dec 7 '18 at 23:41