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Is $mathbb S^2$ just a notation or $mathbb S^2cong mathbb Stimes mathbb S$ (in a possible way)?
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I was wondering : $mathbb Stimes mathbb S$ is a torus and $mathbb S^2$ is a sphere. Is $mathbb S^2$ just a notation or it's justified by the fact that $$mathbb S^2cong mathbb Stimes mathbb S ?$$
(I wouldn't see how a torus would be a sphere since the fundamental group of the sphere is trivial whereas, the fundamental group of the torus is $mathbb Z^2$, but maybe it could have a link between sphere and torus that would justify $mathbb S^2$ and $mathbb Stimes mathbb S$.
general-topology manifolds
asked Jul 22 at 22:34
user330587
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up vote
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I was wondering : $mathbb Stimes mathbb S$ is a torus and $mathbb S^2$ is a sphere. Is $mathbb S^2$ just a notation or it's justified by the fact that $$mathbb S^2cong mathbb Stimes mathbb S ?$$
(I wouldn't see how a torus would be a sphere since the fundamental group of the sphere is trivial whereas, the fundamental group of the torus is $mathbb Z^2$, but maybe it could have a link between sphere and torus that would justify $mathbb S^2$ and $mathbb Stimes mathbb S$.
general-topology manifolds
asked Jul 22 at 22:34
user330587
821310
2
Nope. $S^n$ denotes the $n$-dimensional sphere, sitting inside $Bbb R^n+1$. The notation denotes dimension, not direct product.
– Ted Shifrin
Jul 22 at 22:35
@TedShifrin: It's what I thought (for the reason I said). It wouldn't make sense that $mathbb S^2$ and $mathbb Stimes mathbb S$ where isomorphic. Thank you.
– user330587
Jul 22 at 22:38
The usual notation $Bbb T^n$, however, does denote the $n$-dimensional torus, which happens to be $(S^1)^n$, i.e., the cartesian product of $n$ circles. :P Just to confuse matters :P
– Ted Shifrin
Jul 22 at 22:40
I don't think the blackboard bold font $BbbS$ here is customary even for those of us who were brought up on $BbbN$, $BbbZ$, $BbbQ$etc. for the various number systems: "$S^1$" not "$BbbS^1$".
– Rob Arthan
Jul 22 at 22:41
1
@RobArthan, yes, agreed, although it's commonly used for the torus. Maybe more by analysts than by topologists. shrug
– Ted Shifrin
Jul 22 at 22:42
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was wondering : $mathbb Stimes mathbb S$ is a torus and $mathbb S^2$ is a sphere. Is $mathbb S^2$ just a notation or it's justified by the fact that $$mathbb S^2cong mathbb Stimes mathbb S ?$$
(I wouldn't see how a torus would be a sphere since the fundamental group of the sphere is trivial whereas, the fundamental group of the torus is $mathbb Z^2$, but maybe it could have a link between sphere and torus that would justify $mathbb S^2$ and $mathbb Stimes mathbb S$.
general-topology manifolds
asked Jul 22 at 22:34
user330587
821310
I was wondering : $mathbb Stimes mathbb S$ is a torus and $mathbb S^2$ is a sphere. Is $mathbb S^2$ just a notation or it's justified by the fact that $$mathbb S^2cong mathbb Stimes mathbb S ?$$
(I wouldn't see how a torus would be a sphere since the fundamental group of the sphere is trivial whereas, the fundamental group of the torus is $mathbb Z^2$, but maybe it could have a link between sphere and torus that would justify $mathbb S^2$ and $mathbb Stimes mathbb S$.
general-topology manifolds
asked Jul 22 at 22:34
user330587
821310
asked Jul 22 at 22:34
user330587
821310
asked Jul 22 at 22:34
user330587
821310
asked Jul 22 at 22:34
user330587
821310
821310
2
Nope. $S^n$ denotes the $n$-dimensional sphere, sitting inside $Bbb R^n+1$. The notation denotes dimension, not direct product.
– Ted Shifrin
Jul 22 at 22:35
@TedShifrin: It's what I thought (for the reason I said). It wouldn't make sense that $mathbb S^2$ and $mathbb Stimes mathbb S$ where isomorphic. Thank you.
– user330587
Jul 22 at 22:38
The usual notation $Bbb T^n$, however, does denote the $n$-dimensional torus, which happens to be $(S^1)^n$, i.e., the cartesian product of $n$ circles. :P Just to confuse matters :P
– Ted Shifrin
Jul 22 at 22:40
I don't think the blackboard bold font $BbbS$ here is customary even for those of us who were brought up on $BbbN$, $BbbZ$, $BbbQ$etc. for the various number systems: "$S^1$" not "$BbbS^1$".
– Rob Arthan
Jul 22 at 22:41
1
@RobArthan, yes, agreed, although it's commonly used for the torus. Maybe more by analysts than by topologists. shrug
– Ted Shifrin
Jul 22 at 22:42
 |Â
show 1 more comment
2
Nope. $S^n$ denotes the $n$-dimensional sphere, sitting inside $Bbb R^n+1$. The notation denotes dimension, not direct product.
– Ted Shifrin
Jul 22 at 22:35
@TedShifrin: It's what I thought (for the reason I said). It wouldn't make sense that $mathbb S^2$ and $mathbb Stimes mathbb S$ where isomorphic. Thank you.
– user330587
Jul 22 at 22:38
The usual notation $Bbb T^n$, however, does denote the $n$-dimensional torus, which happens to be $(S^1)^n$, i.e., the cartesian product of $n$ circles. :P Just to confuse matters :P
– Ted Shifrin
Jul 22 at 22:40
I don't think the blackboard bold font $BbbS$ here is customary even for those of us who were brought up on $BbbN$, $BbbZ$, $BbbQ$etc. for the various number systems: "$S^1$" not "$BbbS^1$".
– Rob Arthan
Jul 22 at 22:41
1
@RobArthan, yes, agreed, although it's commonly used for the torus. Maybe more by analysts than by topologists. shrug
– Ted Shifrin
Jul 22 at 22:42
2
2
Nope. $S^n$ denotes the $n$-dimensional sphere, sitting inside $Bbb R^n+1$. The notation denotes dimension, not direct product.
– Ted Shifrin
Jul 22 at 22:35
Nope. $S^n$ denotes the $n$-dimensional sphere, sitting inside $Bbb R^n+1$. The notation denotes dimension, not direct product.
– Ted Shifrin
Jul 22 at 22:35
@TedShifrin: It's what I thought (for the reason I said). It wouldn't make sense that $mathbb S^2$ and $mathbb Stimes mathbb S$ where isomorphic. Thank you.
– user330587
Jul 22 at 22:38
@TedShifrin: It's what I thought (for the reason I said). It wouldn't make sense that $mathbb S^2$ and $mathbb Stimes mathbb S$ where isomorphic. Thank you.
– user330587
Jul 22 at 22:38
The usual notation $Bbb T^n$, however, does denote the $n$-dimensional torus, which happens to be $(S^1)^n$, i.e., the cartesian product of $n$ circles. :P Just to confuse matters :P
– Ted Shifrin
Jul 22 at 22:40
The usual notation $Bbb T^n$, however, does denote the $n$-dimensional torus, which happens to be $(S^1)^n$, i.e., the cartesian product of $n$ circles. :P Just to confuse matters :P
– Ted Shifrin
Jul 22 at 22:40
I don't think the blackboard bold font $BbbS$ here is customary even for those of us who were brought up on $BbbN$, $BbbZ$, $BbbQ$etc. for the various number systems: "$S^1$" not "$BbbS^1$".
– Rob Arthan
Jul 22 at 22:41
I don't think the blackboard bold font $BbbS$ here is customary even for those of us who were brought up on $BbbN$, $BbbZ$, $BbbQ$etc. for the various number systems: "$S^1$" not "$BbbS^1$".
– Rob Arthan
Jul 22 at 22:41
1
1
@RobArthan, yes, agreed, although it's commonly used for the torus. Maybe more by analysts than by topologists. shrug
– Ted Shifrin
Jul 22 at 22:42
@RobArthan, yes, agreed, although it's commonly used for the torus. Maybe more by analysts than by topologists. shrug
– Ted Shifrin
Jul 22 at 22:42
 |Â
show 1 more comment
2 Answers
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up vote
6
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Using the smash product, we do have $S^n=S^1wedge S^1wedgecdotswedge S^1$ (with $n$ copies of $S^1$). However, I think that that is only coincidental. Note that this way you do get $S^2$ as a quotient space of $S^1times S^1$.
The main reason, to most people, that $S^n$ is denoted by $S^n$ is probably just that it is the $n$-dimensional sphere, and that's that. The $n$ denotes the dimension of the space, and doesn't explicitly refer to some kind of $n$-times repeated operation.
answered Jul 22 at 22:41
Arthur
98.5k793174
add a comment |Â
up vote
4
down vote
As Ted Shifrin pointed out in comments, $S^n$ is not a product. But there is another way to interpret the notation $S^n$ that makes it seem a little systematic. For any topological space $X$, let $S(X)$ denote the suspension of $X$: This is the topological space obtained from the "cylinder" $Xtimes [0,1]$ by identifying $Xtimes 0$ to a point and $Xtimes 1$ to another point. Then if $S^0$ denotes the zero-sphere (i.e., the two-point discrete space), then it's not hard to show that $S(S^0)$ is homeomorphic to $S^1$, and in general $S^n(S^0) approx S(S^n-1)$ is homeomorphic to $S^n$.
answered Jul 24 at 0:31
Jack Lee
25.2k44262
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
Using the smash product, we do have $S^n=S^1wedge S^1wedgecdotswedge S^1$ (with $n$ copies of $S^1$). However, I think that that is only coincidental. Note that this way you do get $S^2$ as a quotient space of $S^1times S^1$.
The main reason, to most people, that $S^n$ is denoted by $S^n$ is probably just that it is the $n$-dimensional sphere, and that's that. The $n$ denotes the dimension of the space, and doesn't explicitly refer to some kind of $n$-times repeated operation.
answered Jul 22 at 22:41
Arthur
98.5k793174
add a comment |Â
up vote
6
down vote
Using the smash product, we do have $S^n=S^1wedge S^1wedgecdotswedge S^1$ (with $n$ copies of $S^1$). However, I think that that is only coincidental. Note that this way you do get $S^2$ as a quotient space of $S^1times S^1$.
The main reason, to most people, that $S^n$ is denoted by $S^n$ is probably just that it is the $n$-dimensional sphere, and that's that. The $n$ denotes the dimension of the space, and doesn't explicitly refer to some kind of $n$-times repeated operation.
answered Jul 22 at 22:41
Arthur
98.5k793174
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Using the smash product, we do have $S^n=S^1wedge S^1wedgecdotswedge S^1$ (with $n$ copies of $S^1$). However, I think that that is only coincidental. Note that this way you do get $S^2$ as a quotient space of $S^1times S^1$.
The main reason, to most people, that $S^n$ is denoted by $S^n$ is probably just that it is the $n$-dimensional sphere, and that's that. The $n$ denotes the dimension of the space, and doesn't explicitly refer to some kind of $n$-times repeated operation.
answered Jul 22 at 22:41
Arthur
98.5k793174
Using the smash product, we do have $S^n=S^1wedge S^1wedgecdotswedge S^1$ (with $n$ copies of $S^1$). However, I think that that is only coincidental. Note that this way you do get $S^2$ as a quotient space of $S^1times S^1$.
The main reason, to most people, that $S^n$ is denoted by $S^n$ is probably just that it is the $n$-dimensional sphere, and that's that. The $n$ denotes the dimension of the space, and doesn't explicitly refer to some kind of $n$-times repeated operation.
answered Jul 22 at 22:41
Arthur
98.5k793174
edited Jul 22 at 23:19
answered Jul 22 at 22:41
Arthur
98.5k793174
answered Jul 22 at 22:41
Arthur
98.5k793174
answered Jul 22 at 22:41
Arthur
98.5k793174
98.5k793174
add a comment |Â
add a comment |Â
up vote
4
down vote
As Ted Shifrin pointed out in comments, $S^n$ is not a product. But there is another way to interpret the notation $S^n$ that makes it seem a little systematic. For any topological space $X$, let $S(X)$ denote the suspension of $X$: This is the topological space obtained from the "cylinder" $Xtimes [0,1]$ by identifying $Xtimes 0$ to a point and $Xtimes 1$ to another point. Then if $S^0$ denotes the zero-sphere (i.e., the two-point discrete space), then it's not hard to show that $S(S^0)$ is homeomorphic to $S^1$, and in general $S^n(S^0) approx S(S^n-1)$ is homeomorphic to $S^n$.
answered Jul 24 at 0:31
Jack Lee
25.2k44262
add a comment |Â
up vote
4
down vote
As Ted Shifrin pointed out in comments, $S^n$ is not a product. But there is another way to interpret the notation $S^n$ that makes it seem a little systematic. For any topological space $X$, let $S(X)$ denote the suspension of $X$: This is the topological space obtained from the "cylinder" $Xtimes [0,1]$ by identifying $Xtimes 0$ to a point and $Xtimes 1$ to another point. Then if $S^0$ denotes the zero-sphere (i.e., the two-point discrete space), then it's not hard to show that $S(S^0)$ is homeomorphic to $S^1$, and in general $S^n(S^0) approx S(S^n-1)$ is homeomorphic to $S^n$.
answered Jul 24 at 0:31
Jack Lee
25.2k44262
add a comment |Â
up vote
4
down vote
up vote
4
down vote
As Ted Shifrin pointed out in comments, $S^n$ is not a product. But there is another way to interpret the notation $S^n$ that makes it seem a little systematic. For any topological space $X$, let $S(X)$ denote the suspension of $X$: This is the topological space obtained from the "cylinder" $Xtimes [0,1]$ by identifying $Xtimes 0$ to a point and $Xtimes 1$ to another point. Then if $S^0$ denotes the zero-sphere (i.e., the two-point discrete space), then it's not hard to show that $S(S^0)$ is homeomorphic to $S^1$, and in general $S^n(S^0) approx S(S^n-1)$ is homeomorphic to $S^n$.
answered Jul 24 at 0:31
Jack Lee
25.2k44262
As Ted Shifrin pointed out in comments, $S^n$ is not a product. But there is another way to interpret the notation $S^n$ that makes it seem a little systematic. For any topological space $X$, let $S(X)$ denote the suspension of $X$: This is the topological space obtained from the "cylinder" $Xtimes [0,1]$ by identifying $Xtimes 0$ to a point and $Xtimes 1$ to another point. Then if $S^0$ denotes the zero-sphere (i.e., the two-point discrete space), then it's not hard to show that $S(S^0)$ is homeomorphic to $S^1$, and in general $S^n(S^0) approx S(S^n-1)$ is homeomorphic to $S^n$.
answered Jul 24 at 0:31
Jack Lee
25.2k44262
edited Jul 24 at 10:25
answered Jul 24 at 0:31
Jack Lee
25.2k44262
answered Jul 24 at 0:31
Jack Lee
25.2k44262
answered Jul 24 at 0:31
Jack Lee
25.2k44262
25.2k44262
add a comment |Â
add a comment |Â
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2
Nope. $S^n$ denotes the $n$-dimensional sphere, sitting inside $Bbb R^n+1$. The notation denotes dimension, not direct product.
– Ted Shifrin
Jul 22 at 22:35
@TedShifrin: It's what I thought (for the reason I said). It wouldn't make sense that $mathbb S^2$ and $mathbb Stimes mathbb S$ where isomorphic. Thank you.
– user330587
Jul 22 at 22:38
The usual notation $Bbb T^n$, however, does denote the $n$-dimensional torus, which happens to be $(S^1)^n$, i.e., the cartesian product of $n$ circles. :P Just to confuse matters :P
– Ted Shifrin
Jul 22 at 22:40
I don't think the blackboard bold font $BbbS$ here is customary even for those of us who were brought up on $BbbN$, $BbbZ$, $BbbQ$etc. for the various number systems: "$S^1$" not "$BbbS^1$".
– Rob Arthan
Jul 22 at 22:41
1
@RobArthan, yes, agreed, although it's commonly used for the torus. Maybe more by analysts than by topologists. shrug
– Ted Shifrin
Jul 22 at 22:42