Mixing
What is the type of this surface? (square with two bridges)
$begingroup$
Take a square (without border) and build two bridges on it. You can go under the bridge or across the bridge (as my attempt at drawing it poorly attempts to describe).
Since you can freely move the bridges around, on one hand I would say this surface is the same as a torus: just one bridge would turn it into a cylinder (connecting two sides) and then putting another bridge on it turns it into a mug, which is a torus. (also orientation is ^ ^, > >)
On the other hand the square with just one bridge would seem to have two different "generators" (i.e. not homotopic to a point) already, the two white ones I drew (imagine them on the same bridge). But then the square with two bridges would have four... which means it is not a torus.
I am confused: what is the type of this surface?
homotopy-theory surfaces
asked Jan 16 at 11:42
AnalysisStudent0414AnalysisStudent0414
4,376928
$endgroup$
add a comment |
$begingroup$
Take a square (without border) and build two bridges on it. You can go under the bridge or across the bridge (as my attempt at drawing it poorly attempts to describe).
Since you can freely move the bridges around, on one hand I would say this surface is the same as a torus: just one bridge would turn it into a cylinder (connecting two sides) and then putting another bridge on it turns it into a mug, which is a torus. (also orientation is ^ ^, > >)
On the other hand the square with just one bridge would seem to have two different "generators" (i.e. not homotopic to a point) already, the two white ones I drew (imagine them on the same bridge). But then the square with two bridges would have four... which means it is not a torus.
I am confused: what is the type of this surface?
homotopy-theory surfaces
asked Jan 16 at 11:42
AnalysisStudent0414AnalysisStudent0414
4,376928
$endgroup$
add a comment |
$begingroup$
Take a square (without border) and build two bridges on it. You can go under the bridge or across the bridge (as my attempt at drawing it poorly attempts to describe).
Since you can freely move the bridges around, on one hand I would say this surface is the same as a torus: just one bridge would turn it into a cylinder (connecting two sides) and then putting another bridge on it turns it into a mug, which is a torus. (also orientation is ^ ^, > >)
On the other hand the square with just one bridge would seem to have two different "generators" (i.e. not homotopic to a point) already, the two white ones I drew (imagine them on the same bridge). But then the square with two bridges would have four... which means it is not a torus.
I am confused: what is the type of this surface?
homotopy-theory surfaces
asked Jan 16 at 11:42
AnalysisStudent0414AnalysisStudent0414
4,376928
$endgroup$
Take a square (without border) and build two bridges on it. You can go under the bridge or across the bridge (as my attempt at drawing it poorly attempts to describe).
Since you can freely move the bridges around, on one hand I would say this surface is the same as a torus: just one bridge would turn it into a cylinder (connecting two sides) and then putting another bridge on it turns it into a mug, which is a torus. (also orientation is ^ ^, > >)
On the other hand the square with just one bridge would seem to have two different "generators" (i.e. not homotopic to a point) already, the two white ones I drew (imagine them on the same bridge). But then the square with two bridges would have four... which means it is not a torus.
I am confused: what is the type of this surface?
homotopy-theory surfaces
homotopy-theory surfaces
asked Jan 16 at 11:42
AnalysisStudent0414AnalysisStudent0414
4,376928
asked Jan 16 at 11:42
AnalysisStudent0414AnalysisStudent0414
4,376928
asked Jan 16 at 11:42
AnalysisStudent0414AnalysisStudent0414
4,376928
asked Jan 16 at 11:42
AnalysisStudent0414AnalysisStudent0414
4,376928
asked Jan 16 at 11:42
AnalysisStudent0414AnalysisStudent0414
4,376928
4,376928
add a comment |
add a comment |
1 Answer
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$begingroup$
The right white path you drew is homotopically trivial. It lives exclusively on the square. Your entire surface is homotopic to the number symbol '8' if you shrink the square down to 2 paths connecting the bridge ends.
So the homotopy group is a free product of two copies of $mathbb{Z}$, for a torus it is the direct sum of two copies of $mathbb{Z}$.
answered Jan 16 at 11:51
quaraguequarague
398110
$endgroup$
$begingroup$
This was my main point of confusion: so... here building the bridges is just connecting/identifying two separate points. What would be a better geometric equivalent for the "real-life situation"? (in which you couldn't shrink that path to a point).
$endgroup$
– AnalysisStudent0414
Jan 16 at 11:55
$begingroup$
(I realised this is a stupid question: if you cannot shrink the right white path, then the bridges are disconnected from the surface right?)
$endgroup$
– AnalysisStudent0414
Jan 16 at 13:18
1
$begingroup$
If the bridges are strips/lines then yes. I guess if your bridges are cylinders instead of just strips/lines you have a choice how to attach them. You could either leave the square intact or cut out the disk that is bounded by the circle where the bridge attaches to the squares. These spaces differ homotopically.
$endgroup$
– quarague
Jan 16 at 14:14
$begingroup$
Oh right! my bridge is not a cylinder so the "mug" thing in the question is completely wrong. Thank you, you made me see what the mistake was.
$endgroup$
– AnalysisStudent0414
Jan 16 at 14:23
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
The right white path you drew is homotopically trivial. It lives exclusively on the square. Your entire surface is homotopic to the number symbol '8' if you shrink the square down to 2 paths connecting the bridge ends.
So the homotopy group is a free product of two copies of $mathbb{Z}$, for a torus it is the direct sum of two copies of $mathbb{Z}$.
answered Jan 16 at 11:51
quaraguequarague
398110
$endgroup$
$begingroup$
This was my main point of confusion: so... here building the bridges is just connecting/identifying two separate points. What would be a better geometric equivalent for the "real-life situation"? (in which you couldn't shrink that path to a point).
$endgroup$
– AnalysisStudent0414
Jan 16 at 11:55
$begingroup$
(I realised this is a stupid question: if you cannot shrink the right white path, then the bridges are disconnected from the surface right?)
$endgroup$
– AnalysisStudent0414
Jan 16 at 13:18
1
$begingroup$
If the bridges are strips/lines then yes. I guess if your bridges are cylinders instead of just strips/lines you have a choice how to attach them. You could either leave the square intact or cut out the disk that is bounded by the circle where the bridge attaches to the squares. These spaces differ homotopically.
$endgroup$
– quarague
Jan 16 at 14:14
$begingroup$
Oh right! my bridge is not a cylinder so the "mug" thing in the question is completely wrong. Thank you, you made me see what the mistake was.
$endgroup$
– AnalysisStudent0414
Jan 16 at 14:23
add a comment |
$begingroup$
The right white path you drew is homotopically trivial. It lives exclusively on the square. Your entire surface is homotopic to the number symbol '8' if you shrink the square down to 2 paths connecting the bridge ends.
So the homotopy group is a free product of two copies of $mathbb{Z}$, for a torus it is the direct sum of two copies of $mathbb{Z}$.
answered Jan 16 at 11:51
quaraguequarague
398110
$endgroup$
$begingroup$
This was my main point of confusion: so... here building the bridges is just connecting/identifying two separate points. What would be a better geometric equivalent for the "real-life situation"? (in which you couldn't shrink that path to a point).
$endgroup$
– AnalysisStudent0414
Jan 16 at 11:55
$begingroup$
(I realised this is a stupid question: if you cannot shrink the right white path, then the bridges are disconnected from the surface right?)
$endgroup$
– AnalysisStudent0414
Jan 16 at 13:18
1
$begingroup$
If the bridges are strips/lines then yes. I guess if your bridges are cylinders instead of just strips/lines you have a choice how to attach them. You could either leave the square intact or cut out the disk that is bounded by the circle where the bridge attaches to the squares. These spaces differ homotopically.
$endgroup$
– quarague
Jan 16 at 14:14
$begingroup$
Oh right! my bridge is not a cylinder so the "mug" thing in the question is completely wrong. Thank you, you made me see what the mistake was.
$endgroup$
– AnalysisStudent0414
Jan 16 at 14:23
add a comment |
$begingroup$
The right white path you drew is homotopically trivial. It lives exclusively on the square. Your entire surface is homotopic to the number symbol '8' if you shrink the square down to 2 paths connecting the bridge ends.
So the homotopy group is a free product of two copies of $mathbb{Z}$, for a torus it is the direct sum of two copies of $mathbb{Z}$.
answered Jan 16 at 11:51
quaraguequarague
398110
$endgroup$
The right white path you drew is homotopically trivial. It lives exclusively on the square. Your entire surface is homotopic to the number symbol '8' if you shrink the square down to 2 paths connecting the bridge ends.
So the homotopy group is a free product of two copies of $mathbb{Z}$, for a torus it is the direct sum of two copies of $mathbb{Z}$.
answered Jan 16 at 11:51
quaraguequarague
398110
answered Jan 16 at 11:51
quaraguequarague
398110
answered Jan 16 at 11:51
quaraguequarague
398110
answered Jan 16 at 11:51
quaraguequarague
398110
398110
$begingroup$
This was my main point of confusion: so... here building the bridges is just connecting/identifying two separate points. What would be a better geometric equivalent for the "real-life situation"? (in which you couldn't shrink that path to a point).
$endgroup$
– AnalysisStudent0414
Jan 16 at 11:55
$begingroup$
(I realised this is a stupid question: if you cannot shrink the right white path, then the bridges are disconnected from the surface right?)
$endgroup$
– AnalysisStudent0414
Jan 16 at 13:18
1
$begingroup$
If the bridges are strips/lines then yes. I guess if your bridges are cylinders instead of just strips/lines you have a choice how to attach them. You could either leave the square intact or cut out the disk that is bounded by the circle where the bridge attaches to the squares. These spaces differ homotopically.
$endgroup$
– quarague
Jan 16 at 14:14
$begingroup$
Oh right! my bridge is not a cylinder so the "mug" thing in the question is completely wrong. Thank you, you made me see what the mistake was.
$endgroup$
– AnalysisStudent0414
Jan 16 at 14:23
add a comment |
$begingroup$
This was my main point of confusion: so... here building the bridges is just connecting/identifying two separate points. What would be a better geometric equivalent for the "real-life situation"? (in which you couldn't shrink that path to a point).
$endgroup$
– AnalysisStudent0414
Jan 16 at 11:55
$begingroup$
(I realised this is a stupid question: if you cannot shrink the right white path, then the bridges are disconnected from the surface right?)
$endgroup$
– AnalysisStudent0414
Jan 16 at 13:18
1
$begingroup$
If the bridges are strips/lines then yes. I guess if your bridges are cylinders instead of just strips/lines you have a choice how to attach them. You could either leave the square intact or cut out the disk that is bounded by the circle where the bridge attaches to the squares. These spaces differ homotopically.
$endgroup$
– quarague
Jan 16 at 14:14
$begingroup$
Oh right! my bridge is not a cylinder so the "mug" thing in the question is completely wrong. Thank you, you made me see what the mistake was.
$endgroup$
– AnalysisStudent0414
Jan 16 at 14:23
$begingroup$
This was my main point of confusion: so... here building the bridges is just connecting/identifying two separate points. What would be a better geometric equivalent for the "real-life situation"? (in which you couldn't shrink that path to a point).
$endgroup$
– AnalysisStudent0414
Jan 16 at 11:55
$begingroup$
This was my main point of confusion: so... here building the bridges is just connecting/identifying two separate points. What would be a better geometric equivalent for the "real-life situation"? (in which you couldn't shrink that path to a point).
$endgroup$
– AnalysisStudent0414
Jan 16 at 11:55
$begingroup$
(I realised this is a stupid question: if you cannot shrink the right white path, then the bridges are disconnected from the surface right?)
$endgroup$
– AnalysisStudent0414
Jan 16 at 13:18
$begingroup$
(I realised this is a stupid question: if you cannot shrink the right white path, then the bridges are disconnected from the surface right?)
$endgroup$
– AnalysisStudent0414
Jan 16 at 13:18
1
1
$begingroup$
If the bridges are strips/lines then yes. I guess if your bridges are cylinders instead of just strips/lines you have a choice how to attach them. You could either leave the square intact or cut out the disk that is bounded by the circle where the bridge attaches to the squares. These spaces differ homotopically.
$endgroup$
– quarague
Jan 16 at 14:14
$begingroup$
If the bridges are strips/lines then yes. I guess if your bridges are cylinders instead of just strips/lines you have a choice how to attach them. You could either leave the square intact or cut out the disk that is bounded by the circle where the bridge attaches to the squares. These spaces differ homotopically.
$endgroup$
– quarague
Jan 16 at 14:14
$begingroup$
Oh right! my bridge is not a cylinder so the "mug" thing in the question is completely wrong. Thank you, you made me see what the mistake was.
$endgroup$
– AnalysisStudent0414
Jan 16 at 14:23
$begingroup$
Oh right! my bridge is not a cylinder so the "mug" thing in the question is completely wrong. Thank you, you made me see what the mistake was.
$endgroup$
– AnalysisStudent0414
Jan 16 at 14:23
add a comment |
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