Paradromic rings and Mobius strip Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Drawing a thickened Möbius strip in MathematicaIs there a Möbius torus?Sphere with three Möbius strips glued and sphere with a handle and a Möbius strip gluedA question about Möbius stripTorus/Möbius Band homeomorphismTopologically distinguishing Mobius Strips based on the number of half-twistsKnots from the boundary of Möbius stripsOrbit space of torus homeomorphic to mobius stripHow does a Möbius strip not have an area?Cutting a multiple twisted Möbius strip in half
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Paradromic rings and Mobius strip
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Drawing a thickened Möbius strip in MathematicaIs there a Möbius torus?Sphere with three Möbius strips glued and sphere with a handle and a Möbius strip gluedA question about Möbius stripTorus/Möbius Band homeomorphismTopologically distinguishing Mobius Strips based on the number of half-twistsKnots from the boundary of Möbius stripsOrbit space of torus homeomorphic to mobius stripHow does a Möbius strip not have an area?Cutting a multiple twisted Möbius strip in half
$begingroup$
I'm working on a project about the differences between the original Möbius strip, a strip with an additional even number of half-twists, and a strip with an additional odd number of half-twists. This is a small project and I don't have much knowledge of topology.
Here are some statements about these objects that I would like to prove.
- In every case, the Euler characteristic is 0.
- In the case of the Möbius strip and the odd number of half twists, neither is orientable. But in the even case, since it has 2 sides, it is orientable.
- This one is the most confusing: which surfaces are homeomorphic to the Möbius strip? My teacher said the Möbius strip is a unique surface and it's only homeomorphic to itself, but I read on the internet that it is homeomorphic to a square.
If it really is only homeomorphic to itself, then there is no doubt that a ring with an additional odd number of twists is also homeomorphic to a Möbius strip.
I also think that a ring with an even number of twists is equaled to a torus since they both have 2 edges and 2 surfaces, on the same dimensional.
I would be happy to hear about more interesting mathematical elements that I could use...
Thanks.
general-topology mobius-band non-orientable-surfaces
edited Apr 2 at 16:25
j0equ1nn
1,5741024
asked Jun 23 '18 at 1:33
Ofek PintokOfek Pintok
1267
$endgroup$
add a comment |
$begingroup$
I'm working on a project about the differences between the original Möbius strip, a strip with an additional even number of half-twists, and a strip with an additional odd number of half-twists. This is a small project and I don't have much knowledge of topology.
Here are some statements about these objects that I would like to prove.
- In every case, the Euler characteristic is 0.
- In the case of the Möbius strip and the odd number of half twists, neither is orientable. But in the even case, since it has 2 sides, it is orientable.
- This one is the most confusing: which surfaces are homeomorphic to the Möbius strip? My teacher said the Möbius strip is a unique surface and it's only homeomorphic to itself, but I read on the internet that it is homeomorphic to a square.
If it really is only homeomorphic to itself, then there is no doubt that a ring with an additional odd number of twists is also homeomorphic to a Möbius strip.
I also think that a ring with an even number of twists is equaled to a torus since they both have 2 edges and 2 surfaces, on the same dimensional.
I would be happy to hear about more interesting mathematical elements that I could use...
Thanks.
general-topology mobius-band non-orientable-surfaces
edited Apr 2 at 16:25
j0equ1nn
1,5741024
asked Jun 23 '18 at 1:33
Ofek PintokOfek Pintok
1267
$endgroup$
1
$begingroup$
For $1$, euler characteristic only depends on the homology of $M$, and all three cases have a clear homotopy to their midline circle, and hence euler characteristic $0$.
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:12
1
$begingroup$
for $2.$, glue a $2$-cell to the mobius band and deduce that it is $mathbb RP^2$ which is not orientable for a few reasons (for example, its homology as well.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:13
1
$begingroup$
for $3.$ I think that there are some misconceptions about homeomorphism, since all are suitable quotients of the square, but certainly not homeomorphic to it (which can again be deduced by homology.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:14
$begingroup$
@AndresMejia " euler characteristic only depends on the homology of M", could you please explain this?
$endgroup$
– Ofek Pintok
Jun 23 '18 at 15:14
$begingroup$
I like your question but felt it needed to be stated more clearly and with proper mathematical language. I run some lessons on this concept and find many students have these same questions, so great post!
$endgroup$
– j0equ1nn
Apr 2 at 15:46
add a comment |
$begingroup$
I'm working on a project about the differences between the original Möbius strip, a strip with an additional even number of half-twists, and a strip with an additional odd number of half-twists. This is a small project and I don't have much knowledge of topology.
Here are some statements about these objects that I would like to prove.
- In every case, the Euler characteristic is 0.
- In the case of the Möbius strip and the odd number of half twists, neither is orientable. But in the even case, since it has 2 sides, it is orientable.
- This one is the most confusing: which surfaces are homeomorphic to the Möbius strip? My teacher said the Möbius strip is a unique surface and it's only homeomorphic to itself, but I read on the internet that it is homeomorphic to a square.
If it really is only homeomorphic to itself, then there is no doubt that a ring with an additional odd number of twists is also homeomorphic to a Möbius strip.
I also think that a ring with an even number of twists is equaled to a torus since they both have 2 edges and 2 surfaces, on the same dimensional.
I would be happy to hear about more interesting mathematical elements that I could use...
Thanks.
general-topology mobius-band non-orientable-surfaces
edited Apr 2 at 16:25
j0equ1nn
1,5741024
asked Jun 23 '18 at 1:33
Ofek PintokOfek Pintok
1267
$endgroup$
I'm working on a project about the differences between the original Möbius strip, a strip with an additional even number of half-twists, and a strip with an additional odd number of half-twists. This is a small project and I don't have much knowledge of topology.
Here are some statements about these objects that I would like to prove.
- In every case, the Euler characteristic is 0.
- In the case of the Möbius strip and the odd number of half twists, neither is orientable. But in the even case, since it has 2 sides, it is orientable.
- This one is the most confusing: which surfaces are homeomorphic to the Möbius strip? My teacher said the Möbius strip is a unique surface and it's only homeomorphic to itself, but I read on the internet that it is homeomorphic to a square.
If it really is only homeomorphic to itself, then there is no doubt that a ring with an additional odd number of twists is also homeomorphic to a Möbius strip.
I also think that a ring with an even number of twists is equaled to a torus since they both have 2 edges and 2 surfaces, on the same dimensional.
I would be happy to hear about more interesting mathematical elements that I could use...
Thanks.
general-topology mobius-band non-orientable-surfaces
general-topology mobius-band non-orientable-surfaces
edited Apr 2 at 16:25
j0equ1nn
1,5741024
asked Jun 23 '18 at 1:33
Ofek PintokOfek Pintok
1267
edited Apr 2 at 16:25
j0equ1nn
1,5741024
asked Jun 23 '18 at 1:33
Ofek PintokOfek Pintok
1267
edited Apr 2 at 16:25
j0equ1nn
1,5741024
edited Apr 2 at 16:25
j0equ1nn
1,5741024
edited Apr 2 at 16:25
j0equ1nn
1,5741024
1,5741024
asked Jun 23 '18 at 1:33
Ofek PintokOfek Pintok
1267
asked Jun 23 '18 at 1:33
Ofek PintokOfek Pintok
1267
asked Jun 23 '18 at 1:33
Ofek PintokOfek Pintok
1267
1267
1
$begingroup$
For $1$, euler characteristic only depends on the homology of $M$, and all three cases have a clear homotopy to their midline circle, and hence euler characteristic $0$.
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:12
1
$begingroup$
for $2.$, glue a $2$-cell to the mobius band and deduce that it is $mathbb RP^2$ which is not orientable for a few reasons (for example, its homology as well.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:13
1
$begingroup$
for $3.$ I think that there are some misconceptions about homeomorphism, since all are suitable quotients of the square, but certainly not homeomorphic to it (which can again be deduced by homology.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:14
$begingroup$
@AndresMejia " euler characteristic only depends on the homology of M", could you please explain this?
$endgroup$
– Ofek Pintok
Jun 23 '18 at 15:14
$begingroup$
I like your question but felt it needed to be stated more clearly and with proper mathematical language. I run some lessons on this concept and find many students have these same questions, so great post!
$endgroup$
– j0equ1nn
Apr 2 at 15:46
add a comment |
1
$begingroup$
For $1$, euler characteristic only depends on the homology of $M$, and all three cases have a clear homotopy to their midline circle, and hence euler characteristic $0$.
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:12
1
$begingroup$
for $2.$, glue a $2$-cell to the mobius band and deduce that it is $mathbb RP^2$ which is not orientable for a few reasons (for example, its homology as well.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:13
1
$begingroup$
for $3.$ I think that there are some misconceptions about homeomorphism, since all are suitable quotients of the square, but certainly not homeomorphic to it (which can again be deduced by homology.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:14
$begingroup$
@AndresMejia " euler characteristic only depends on the homology of M", could you please explain this?
$endgroup$
– Ofek Pintok
Jun 23 '18 at 15:14
$begingroup$
I like your question but felt it needed to be stated more clearly and with proper mathematical language. I run some lessons on this concept and find many students have these same questions, so great post!
$endgroup$
– j0equ1nn
Apr 2 at 15:46
1
1
$begingroup$
For $1$, euler characteristic only depends on the homology of $M$, and all three cases have a clear homotopy to their midline circle, and hence euler characteristic $0$.
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:12
$begingroup$
For $1$, euler characteristic only depends on the homology of $M$, and all three cases have a clear homotopy to their midline circle, and hence euler characteristic $0$.
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:12
1
1
$begingroup$
for $2.$, glue a $2$-cell to the mobius band and deduce that it is $mathbb RP^2$ which is not orientable for a few reasons (for example, its homology as well.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:13
$begingroup$
for $2.$, glue a $2$-cell to the mobius band and deduce that it is $mathbb RP^2$ which is not orientable for a few reasons (for example, its homology as well.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:13
1
1
$begingroup$
for $3.$ I think that there are some misconceptions about homeomorphism, since all are suitable quotients of the square, but certainly not homeomorphic to it (which can again be deduced by homology.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:14
$begingroup$
for $3.$ I think that there are some misconceptions about homeomorphism, since all are suitable quotients of the square, but certainly not homeomorphic to it (which can again be deduced by homology.)
$endgroup$
– Andres Mejia
Jun 23 '18 at 2:14
$begingroup$
@AndresMejia " euler characteristic only depends on the homology of M", could you please explain this?
$endgroup$
– Ofek Pintok
Jun 23 '18 at 15:14
$begingroup$
@AndresMejia " euler characteristic only depends on the homology of M", could you please explain this?
$endgroup$
– Ofek Pintok
Jun 23 '18 at 15:14
$begingroup$
I like your question but felt it needed to be stated more clearly and with proper mathematical language. I run some lessons on this concept and find many students have these same questions, so great post!
$endgroup$
– j0equ1nn
Apr 2 at 15:46
$begingroup$
I like your question but felt it needed to be stated more clearly and with proper mathematical language. I run some lessons on this concept and find many students have these same questions, so great post!
$endgroup$
– j0equ1nn
Apr 2 at 15:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To set some notation,
let $n$
be the number of half-twists in the band.
So $n=0$
gives the usual cylinder and $n=1$
gives the usual Möbius strip.
Note that we haven't distinguished which way we're twisting the band,
but let's say we did,
then $n<0$
would correspond to twisting in the opposite direction.
- Yes the Euler characteristic is zero regardless of what $n$
is. One way to see that is to model the surface using a square with edge identifications, as shown below. In either case, up to the identifications, there would be 2 vertices, 3 edges, and 1 face in the cellular decomposition. So the Euler characteristic is $2-3+1=0$.
$hskip 1.6 in$
To prove what you said about orientability would require going back to the definition, in terms of existence of a consistent choice of normal vectors on the surface. In this case, you can think of this like whether or not you can slide something around and end up with it back where it started but flipped (reflected). When we start talking about "sides," this is effectively capturing that idea because "flipping" is like looking from the other "side."
Yes, this part is the most subtle. Topologically, the only objects you are getting here are the cylinder and the Möbius strip, and all that matters is the parity of $n$. This is because (topology definitions) a Möbius strip is a non-orientable surface whose boundary is a simple closed curve, and a cylinder is an orientable surface whose boundary is two simple closed curves. Notice how additional twists wouldn't be detected in the above square diagrams either. But geometrically, each $n$ gives a distinct object. More precisely, each odd value for $n$ gives a different embedding of the Möbius strip into space, and each even value for $n$ gives a different embedding of the cylinder into space. There are even two distinct embeddings of $1$-twist Möbius strips in this sense,
depending which way we twisted
(or, in our notation, choice of $n=pm1$).
Your remark at the end about the torus needs more thought. To get a torus, you would have to also identify the other pair of edges on the square (both identifications preserving orientation). Then you need to be careful about what you call a "surface" (since the whole object is a surface) and what you mean by "dimensional." But there is plenty to think about here just with Möbius strips and cylinders.
answered Apr 2 at 16:53
j0equ1nnj0equ1nn
1,5741024
$endgroup$
add a comment |
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$begingroup$
To set some notation,
let $n$
be the number of half-twists in the band.
So $n=0$
gives the usual cylinder and $n=1$
gives the usual Möbius strip.
Note that we haven't distinguished which way we're twisting the band,
but let's say we did,
then $n<0$
would correspond to twisting in the opposite direction.
- Yes the Euler characteristic is zero regardless of what $n$
is. One way to see that is to model the surface using a square with edge identifications, as shown below. In either case, up to the identifications, there would be 2 vertices, 3 edges, and 1 face in the cellular decomposition. So the Euler characteristic is $2-3+1=0$.
$hskip 1.6 in$
To prove what you said about orientability would require going back to the definition, in terms of existence of a consistent choice of normal vectors on the surface. In this case, you can think of this like whether or not you can slide something around and end up with it back where it started but flipped (reflected). When we start talking about "sides," this is effectively capturing that idea because "flipping" is like looking from the other "side."
Yes, this part is the most subtle. Topologically, the only objects you are getting here are the cylinder and the Möbius strip, and all that matters is the parity of $n$. This is because (topology definitions) a Möbius strip is a non-orientable surface whose boundary is a simple closed curve, and a cylinder is an orientable surface whose boundary is two simple closed curves. Notice how additional twists wouldn't be detected in the above square diagrams either. But geometrically, each $n$ gives a distinct object. More precisely, each odd value for $n$ gives a different embedding of the Möbius strip into space, and each even value for $n$ gives a different embedding of the cylinder into space. There are even two distinct embeddings of $1$-twist Möbius strips in this sense,
depending which way we twisted
(or, in our notation, choice of $n=pm1$).
Your remark at the end about the torus needs more thought. To get a torus, you would have to also identify the other pair of edges on the square (both identifications preserving orientation). Then you need to be careful about what you call a "surface" (since the whole object is a surface) and what you mean by "dimensional." But there is plenty to think about here just with Möbius strips and cylinders.
answered Apr 2 at 16:53
j0equ1nnj0equ1nn
1,5741024
$endgroup$
add a comment |
$begingroup$
To set some notation,
let $n$
be the number of half-twists in the band.
So $n=0$
gives the usual cylinder and $n=1$
gives the usual Möbius strip.
Note that we haven't distinguished which way we're twisting the band,
but let's say we did,
then $n<0$
would correspond to twisting in the opposite direction.
- Yes the Euler characteristic is zero regardless of what $n$
is. One way to see that is to model the surface using a square with edge identifications, as shown below. In either case, up to the identifications, there would be 2 vertices, 3 edges, and 1 face in the cellular decomposition. So the Euler characteristic is $2-3+1=0$.
$hskip 1.6 in$
To prove what you said about orientability would require going back to the definition, in terms of existence of a consistent choice of normal vectors on the surface. In this case, you can think of this like whether or not you can slide something around and end up with it back where it started but flipped (reflected). When we start talking about "sides," this is effectively capturing that idea because "flipping" is like looking from the other "side."
Yes, this part is the most subtle. Topologically, the only objects you are getting here are the cylinder and the Möbius strip, and all that matters is the parity of $n$. This is because (topology definitions) a Möbius strip is a non-orientable surface whose boundary is a simple closed curve, and a cylinder is an orientable surface whose boundary is two simple closed curves. Notice how additional twists wouldn't be detected in the above square diagrams either. But geometrically, each $n$ gives a distinct object. More precisely, each odd value for $n$ gives a different embedding of the Möbius strip into space, and each even value for $n$ gives a different embedding of the cylinder into space. There are even two distinct embeddings of $1$-twist Möbius strips in this sense,
depending which way we twisted
(or, in our notation, choice of $n=pm1$).
Your remark at the end about the torus needs more thought. To get a torus, you would have to also identify the other pair of edges on the square (both identifications preserving orientation). Then you need to be careful about what you call a "surface" (since the whole object is a surface) and what you mean by "dimensional." But there is plenty to think about here just with Möbius strips and cylinders.
answered Apr 2 at 16:53
j0equ1nnj0equ1nn
1,5741024
$endgroup$
add a comment |
$begingroup$
To set some notation,
let $n$
be the number of half-twists in the band.
So $n=0$
gives the usual cylinder and $n=1$
gives the usual Möbius strip.
Note that we haven't distinguished which way we're twisting the band,
but let's say we did,
then $n<0$
would correspond to twisting in the opposite direction.
- Yes the Euler characteristic is zero regardless of what $n$
is. One way to see that is to model the surface using a square with edge identifications, as shown below. In either case, up to the identifications, there would be 2 vertices, 3 edges, and 1 face in the cellular decomposition. So the Euler characteristic is $2-3+1=0$.
$hskip 1.6 in$
To prove what you said about orientability would require going back to the definition, in terms of existence of a consistent choice of normal vectors on the surface. In this case, you can think of this like whether or not you can slide something around and end up with it back where it started but flipped (reflected). When we start talking about "sides," this is effectively capturing that idea because "flipping" is like looking from the other "side."
Yes, this part is the most subtle. Topologically, the only objects you are getting here are the cylinder and the Möbius strip, and all that matters is the parity of $n$. This is because (topology definitions) a Möbius strip is a non-orientable surface whose boundary is a simple closed curve, and a cylinder is an orientable surface whose boundary is two simple closed curves. Notice how additional twists wouldn't be detected in the above square diagrams either. But geometrically, each $n$ gives a distinct object. More precisely, each odd value for $n$ gives a different embedding of the Möbius strip into space, and each even value for $n$ gives a different embedding of the cylinder into space. There are even two distinct embeddings of $1$-twist Möbius strips in this sense,
depending which way we twisted
(or, in our notation, choice of $n=pm1$).
Your remark at the end about the torus needs more thought. To get a torus, you would have to also identify the other pair of edges on the square (both identifications preserving orientation). Then you need to be careful about what you call a "surface" (since the whole object is a surface) and what you mean by "dimensional." But there is plenty to think about here just with Möbius strips and cylinders.
answered Apr 2 at 16:53
j0equ1nnj0equ1nn
1,5741024
$endgroup$
To set some notation,
let $n$
be the number of half-twists in the band.
So $n=0$
gives the usual cylinder and $n=1$
gives the usual Möbius strip.
Note that we haven't distinguished which way we're twisting the band,
but let's say we did,
then $n<0$
would correspond to twisting in the opposite direction.
- Yes the Euler characteristic is zero regardless of what $n$
is. One way to see that is to model the surface using a square with edge identifications, as shown below. In either case, up to the identifications, there would be 2 vertices, 3 edges, and 1 face in the cellular decomposition. So the Euler characteristic is $2-3+1=0$.
$hskip 1.6 in$
To prove what you said about orientability would require going back to the definition, in terms of existence of a consistent choice of normal vectors on the surface. In this case, you can think of this like whether or not you can slide something around and end up with it back where it started but flipped (reflected). When we start talking about "sides," this is effectively capturing that idea because "flipping" is like looking from the other "side."
Yes, this part is the most subtle. Topologically, the only objects you are getting here are the cylinder and the Möbius strip, and all that matters is the parity of $n$. This is because (topology definitions) a Möbius strip is a non-orientable surface whose boundary is a simple closed curve, and a cylinder is an orientable surface whose boundary is two simple closed curves. Notice how additional twists wouldn't be detected in the above square diagrams either. But geometrically, each $n$ gives a distinct object. More precisely, each odd value for $n$ gives a different embedding of the Möbius strip into space, and each even value for $n$ gives a different embedding of the cylinder into space. There are even two distinct embeddings of $1$-twist Möbius strips in this sense,
depending which way we twisted
(or, in our notation, choice of $n=pm1$).
Your remark at the end about the torus needs more thought. To get a torus, you would have to also identify the other pair of edges on the square (both identifications preserving orientation). Then you need to be careful about what you call a "surface" (since the whole object is a surface) and what you mean by "dimensional." But there is plenty to think about here just with Möbius strips and cylinders.
answered Apr 2 at 16:53
j0equ1nnj0equ1nn
1,5741024
answered Apr 2 at 16:53
j0equ1nnj0equ1nn
1,5741024
answered Apr 2 at 16:53
j0equ1nnj0equ1nn
1,5741024
answered Apr 2 at 16:53
j0equ1nnj0equ1nn
1,5741024
1,5741024
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For $1$, euler characteristic only depends on the homology of $M$, and all three cases have a clear homotopy to their midline circle, and hence euler characteristic $0$.
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– Andres Mejia
Jun 23 '18 at 2:12
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for $2.$, glue a $2$-cell to the mobius band and deduce that it is $mathbb RP^2$ which is not orientable for a few reasons (for example, its homology as well.)
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– Andres Mejia
Jun 23 '18 at 2:13
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for $3.$ I think that there are some misconceptions about homeomorphism, since all are suitable quotients of the square, but certainly not homeomorphic to it (which can again be deduced by homology.)
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– Andres Mejia
Jun 23 '18 at 2:14
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@AndresMejia " euler characteristic only depends on the homology of M", could you please explain this?
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– Ofek Pintok
Jun 23 '18 at 15:14
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I like your question but felt it needed to be stated more clearly and with proper mathematical language. I run some lessons on this concept and find many students have these same questions, so great post!
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– j0equ1nn
Apr 2 at 15:46