Is there a $4$-component link such that upon removing any one of them you get the Borromean link?
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1
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Is there a $4$-component link such that upon removing any one of them you get the Borromean link?
I've managed to get close but not quite. What I have gets me something similar to the Borromean link but two of the components actually form what I think is the Whitehead link.
knot-theory
edited Nov 2 at 13:31
amWhy
191k27223437
asked Nov 2 at 13:03
Darth Geek
10.8k12341
add a comment |
up vote
1
down vote
favorite
Is there a $4$-component link such that upon removing any one of them you get the Borromean link?
I've managed to get close but not quite. What I have gets me something similar to the Borromean link but two of the components actually form what I think is the Whitehead link.
knot-theory
edited Nov 2 at 13:31
amWhy
191k27223437
asked Nov 2 at 13:03
Darth Geek
10.8k12341
1
There is a numberphile video on this topic, where Professor John Hunton from the University of Leicester states that there is a generalization of Borromean links only for odd numbers.
– Wauzl
Nov 2 at 13:07
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a $4$-component link such that upon removing any one of them you get the Borromean link?
I've managed to get close but not quite. What I have gets me something similar to the Borromean link but two of the components actually form what I think is the Whitehead link.
knot-theory
edited Nov 2 at 13:31
amWhy
191k27223437
asked Nov 2 at 13:03
Darth Geek
10.8k12341
Is there a $4$-component link such that upon removing any one of them you get the Borromean link?
I've managed to get close but not quite. What I have gets me something similar to the Borromean link but two of the components actually form what I think is the Whitehead link.
knot-theory
knot-theory
edited Nov 2 at 13:31
amWhy
191k27223437
asked Nov 2 at 13:03
Darth Geek
10.8k12341
edited Nov 2 at 13:31
amWhy
191k27223437
asked Nov 2 at 13:03
Darth Geek
10.8k12341
edited Nov 2 at 13:31
amWhy
191k27223437
edited Nov 2 at 13:31
amWhy
191k27223437
edited Nov 2 at 13:31
amWhy
191k27223437
191k27223437
asked Nov 2 at 13:03
Darth Geek
10.8k12341
asked Nov 2 at 13:03
Darth Geek
10.8k12341
asked Nov 2 at 13:03
Darth Geek
10.8k12341
10.8k12341
1
There is a numberphile video on this topic, where Professor John Hunton from the University of Leicester states that there is a generalization of Borromean links only for odd numbers.
– Wauzl
Nov 2 at 13:07
add a comment |
1
There is a numberphile video on this topic, where Professor John Hunton from the University of Leicester states that there is a generalization of Borromean links only for odd numbers.
– Wauzl
Nov 2 at 13:07
1
1
There is a numberphile video on this topic, where Professor John Hunton from the University of Leicester states that there is a generalization of Borromean links only for odd numbers.
– Wauzl
Nov 2 at 13:07
There is a numberphile video on this topic, where Professor John Hunton from the University of Leicester states that there is a generalization of Borromean links only for odd numbers.
– Wauzl
Nov 2 at 13:07
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
You can sort of think of the Borromean rings as lying on three faces of a tetrahedron, with the center triangle of the usual presentation at a vertex. By adding a component corresponding to the fourth face in a way so that the link has tetrahedral symmetry, you get the following:
While the outer component looks funny, it is the same as any of the other three, in the sense that there is an isotopy of the link sending this diagram to itself, but moving an inner component to the outer component.
answered Nov 10 at 7:09
Kyle Miller
7,762827
This is exactly what I was looking for! Clever way to distort the fourth component in a way that becomes a normal circle when one of the other ones is removed. I'm guessing that since the construction is done with tetrahedral symmetry, an $n$-component equivalent might not be so easy to construct. Or do you think one can do the same thing starting with an $n$-simplex and then project onto $mathbb{R}^3$?
– Darth Geek
Nov 10 at 10:14
@DarthGeek Looking at this carefully again this morning, I think I need to retract my claim that it has tetrahedral symmetry. It might be possible to do such a construction for certain polyhedra with three faces per vertex, with one component per face, but I'm not sure what condition allows for the property that removing all but three around a particular vertex leaves the Borromean rings.
– Kyle Miller
Nov 10 at 18:57
add a comment |
up vote
1
down vote
This is essentially the same answer as that of Kyle, but with a more artistic aproach. If you vote this anwser, please vote Kyle's too since he's the one that inspired this design in the first place.
answered Nov 13 at 13:02
Darth Geek
10.8k12341
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You can sort of think of the Borromean rings as lying on three faces of a tetrahedron, with the center triangle of the usual presentation at a vertex. By adding a component corresponding to the fourth face in a way so that the link has tetrahedral symmetry, you get the following:
While the outer component looks funny, it is the same as any of the other three, in the sense that there is an isotopy of the link sending this diagram to itself, but moving an inner component to the outer component.
answered Nov 10 at 7:09
Kyle Miller
7,762827
This is exactly what I was looking for! Clever way to distort the fourth component in a way that becomes a normal circle when one of the other ones is removed. I'm guessing that since the construction is done with tetrahedral symmetry, an $n$-component equivalent might not be so easy to construct. Or do you think one can do the same thing starting with an $n$-simplex and then project onto $mathbb{R}^3$?
– Darth Geek
Nov 10 at 10:14
@DarthGeek Looking at this carefully again this morning, I think I need to retract my claim that it has tetrahedral symmetry. It might be possible to do such a construction for certain polyhedra with three faces per vertex, with one component per face, but I'm not sure what condition allows for the property that removing all but three around a particular vertex leaves the Borromean rings.
– Kyle Miller
Nov 10 at 18:57
add a comment |
up vote
2
down vote
accepted
You can sort of think of the Borromean rings as lying on three faces of a tetrahedron, with the center triangle of the usual presentation at a vertex. By adding a component corresponding to the fourth face in a way so that the link has tetrahedral symmetry, you get the following:
While the outer component looks funny, it is the same as any of the other three, in the sense that there is an isotopy of the link sending this diagram to itself, but moving an inner component to the outer component.
answered Nov 10 at 7:09
Kyle Miller
7,762827
This is exactly what I was looking for! Clever way to distort the fourth component in a way that becomes a normal circle when one of the other ones is removed. I'm guessing that since the construction is done with tetrahedral symmetry, an $n$-component equivalent might not be so easy to construct. Or do you think one can do the same thing starting with an $n$-simplex and then project onto $mathbb{R}^3$?
– Darth Geek
Nov 10 at 10:14
@DarthGeek Looking at this carefully again this morning, I think I need to retract my claim that it has tetrahedral symmetry. It might be possible to do such a construction for certain polyhedra with three faces per vertex, with one component per face, but I'm not sure what condition allows for the property that removing all but three around a particular vertex leaves the Borromean rings.
– Kyle Miller
Nov 10 at 18:57
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You can sort of think of the Borromean rings as lying on three faces of a tetrahedron, with the center triangle of the usual presentation at a vertex. By adding a component corresponding to the fourth face in a way so that the link has tetrahedral symmetry, you get the following:
While the outer component looks funny, it is the same as any of the other three, in the sense that there is an isotopy of the link sending this diagram to itself, but moving an inner component to the outer component.
answered Nov 10 at 7:09
Kyle Miller
7,762827
You can sort of think of the Borromean rings as lying on three faces of a tetrahedron, with the center triangle of the usual presentation at a vertex. By adding a component corresponding to the fourth face in a way so that the link has tetrahedral symmetry, you get the following:
While the outer component looks funny, it is the same as any of the other three, in the sense that there is an isotopy of the link sending this diagram to itself, but moving an inner component to the outer component.
answered Nov 10 at 7:09
Kyle Miller
7,762827
answered Nov 10 at 7:09
Kyle Miller
7,762827
answered Nov 10 at 7:09
Kyle Miller
7,762827
answered Nov 10 at 7:09
Kyle Miller
7,762827
7,762827
This is exactly what I was looking for! Clever way to distort the fourth component in a way that becomes a normal circle when one of the other ones is removed. I'm guessing that since the construction is done with tetrahedral symmetry, an $n$-component equivalent might not be so easy to construct. Or do you think one can do the same thing starting with an $n$-simplex and then project onto $mathbb{R}^3$?
– Darth Geek
Nov 10 at 10:14
@DarthGeek Looking at this carefully again this morning, I think I need to retract my claim that it has tetrahedral symmetry. It might be possible to do such a construction for certain polyhedra with three faces per vertex, with one component per face, but I'm not sure what condition allows for the property that removing all but three around a particular vertex leaves the Borromean rings.
– Kyle Miller
Nov 10 at 18:57
add a comment |
This is exactly what I was looking for! Clever way to distort the fourth component in a way that becomes a normal circle when one of the other ones is removed. I'm guessing that since the construction is done with tetrahedral symmetry, an $n$-component equivalent might not be so easy to construct. Or do you think one can do the same thing starting with an $n$-simplex and then project onto $mathbb{R}^3$?
– Darth Geek
Nov 10 at 10:14
@DarthGeek Looking at this carefully again this morning, I think I need to retract my claim that it has tetrahedral symmetry. It might be possible to do such a construction for certain polyhedra with three faces per vertex, with one component per face, but I'm not sure what condition allows for the property that removing all but three around a particular vertex leaves the Borromean rings.
– Kyle Miller
Nov 10 at 18:57
This is exactly what I was looking for! Clever way to distort the fourth component in a way that becomes a normal circle when one of the other ones is removed. I'm guessing that since the construction is done with tetrahedral symmetry, an $n$-component equivalent might not be so easy to construct. Or do you think one can do the same thing starting with an $n$-simplex and then project onto $mathbb{R}^3$?
– Darth Geek
Nov 10 at 10:14
This is exactly what I was looking for! Clever way to distort the fourth component in a way that becomes a normal circle when one of the other ones is removed. I'm guessing that since the construction is done with tetrahedral symmetry, an $n$-component equivalent might not be so easy to construct. Or do you think one can do the same thing starting with an $n$-simplex and then project onto $mathbb{R}^3$?
– Darth Geek
Nov 10 at 10:14
@DarthGeek Looking at this carefully again this morning, I think I need to retract my claim that it has tetrahedral symmetry. It might be possible to do such a construction for certain polyhedra with three faces per vertex, with one component per face, but I'm not sure what condition allows for the property that removing all but three around a particular vertex leaves the Borromean rings.
– Kyle Miller
Nov 10 at 18:57
@DarthGeek Looking at this carefully again this morning, I think I need to retract my claim that it has tetrahedral symmetry. It might be possible to do such a construction for certain polyhedra with three faces per vertex, with one component per face, but I'm not sure what condition allows for the property that removing all but three around a particular vertex leaves the Borromean rings.
– Kyle Miller
Nov 10 at 18:57
add a comment |
up vote
1
down vote
This is essentially the same answer as that of Kyle, but with a more artistic aproach. If you vote this anwser, please vote Kyle's too since he's the one that inspired this design in the first place.
answered Nov 13 at 13:02
Darth Geek
10.8k12341
add a comment |
up vote
1
down vote
This is essentially the same answer as that of Kyle, but with a more artistic aproach. If you vote this anwser, please vote Kyle's too since he's the one that inspired this design in the first place.
answered Nov 13 at 13:02
Darth Geek
10.8k12341
add a comment |
up vote
1
down vote
up vote
1
down vote
This is essentially the same answer as that of Kyle, but with a more artistic aproach. If you vote this anwser, please vote Kyle's too since he's the one that inspired this design in the first place.
answered Nov 13 at 13:02
Darth Geek
10.8k12341
This is essentially the same answer as that of Kyle, but with a more artistic aproach. If you vote this anwser, please vote Kyle's too since he's the one that inspired this design in the first place.
answered Nov 13 at 13:02
Darth Geek
10.8k12341
answered Nov 13 at 13:02
Darth Geek
10.8k12341
answered Nov 13 at 13:02
Darth Geek
10.8k12341
answered Nov 13 at 13:02
Darth Geek
10.8k12341
10.8k12341
add a comment |
add a comment |
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There is a numberphile video on this topic, where Professor John Hunton from the University of Leicester states that there is a generalization of Borromean links only for odd numbers.
– Wauzl
Nov 2 at 13:07