A counter-example for the reversed direction of Casson-Gordon's theorem
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For a knot $K$, let $Sigma_K$ be the double cyclic branched cover of a knot.
By the classical work of Casson and Gordon, we know that if $K$ is smoothly slice, then $Sigma_K$ bounds a rational homology ball.
Is there any well-known counter-example for the reversed direction?
EDIT More general statement is true due to the work of Casson & Gordon, just seen in ACP18.
For any prime $p$ and positive integer $r$, the $p^r$-fold cyclic branched cover of a knot $ K$, denoted by $Sigma_{p^r}(K)$, is a $mathbb Z_p$-homology sphere.
Theorem: If $K$ is smoothly slice, then $Sigma_{p^r}(K)$ bounds $mathbb Z_p$-homology ball.
gt.geometric-topology
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up vote
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For a knot $K$, let $Sigma_K$ be the double cyclic branched cover of a knot.
By the classical work of Casson and Gordon, we know that if $K$ is smoothly slice, then $Sigma_K$ bounds a rational homology ball.
Is there any well-known counter-example for the reversed direction?
EDIT More general statement is true due to the work of Casson & Gordon, just seen in ACP18.
For any prime $p$ and positive integer $r$, the $p^r$-fold cyclic branched cover of a knot $ K$, denoted by $Sigma_{p^r}(K)$, is a $mathbb Z_p$-homology sphere.
Theorem: If $K$ is smoothly slice, then $Sigma_{p^r}(K)$ bounds $mathbb Z_p$-homology ball.
gt.geometric-topology
New contributor
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
For a knot $K$, let $Sigma_K$ be the double cyclic branched cover of a knot.
By the classical work of Casson and Gordon, we know that if $K$ is smoothly slice, then $Sigma_K$ bounds a rational homology ball.
Is there any well-known counter-example for the reversed direction?
EDIT More general statement is true due to the work of Casson & Gordon, just seen in ACP18.
For any prime $p$ and positive integer $r$, the $p^r$-fold cyclic branched cover of a knot $ K$, denoted by $Sigma_{p^r}(K)$, is a $mathbb Z_p$-homology sphere.
Theorem: If $K$ is smoothly slice, then $Sigma_{p^r}(K)$ bounds $mathbb Z_p$-homology ball.
gt.geometric-topology
New contributor
For a knot $K$, let $Sigma_K$ be the double cyclic branched cover of a knot.
By the classical work of Casson and Gordon, we know that if $K$ is smoothly slice, then $Sigma_K$ bounds a rational homology ball.
Is there any well-known counter-example for the reversed direction?
EDIT More general statement is true due to the work of Casson & Gordon, just seen in ACP18.
For any prime $p$ and positive integer $r$, the $p^r$-fold cyclic branched cover of a knot $ K$, denoted by $Sigma_{p^r}(K)$, is a $mathbb Z_p$-homology sphere.
Theorem: If $K$ is smoothly slice, then $Sigma_{p^r}(K)$ bounds $mathbb Z_p$-homology ball.
gt.geometric-topology
gt.geometric-topology
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edited Nov 10 at 15:09
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asked Nov 10 at 10:04
ToposLogos
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Here's a particularly subtle counterexample, from the work of Kirk and Livingston (Topology Vol. 38, No. 3, pp. 663--671, 1999). They show that the pretzel knots $J = P(-3,5,7,2)$ and $K = P(5,-3,7,2)$ are not concordant (even locally flat). These two knots are related by mutation (switch the first two pairs of twists in the standard picture of a pretzel knot). On the other hand, they have the same double branched cover, say $Sigma$, a certain Brieskorn sphere. (This is a general fact about mutations, but is clear in this setting.)
Then the knot $J # -K$ is not slice, but its double branched cover $Sigma # -Sigma$ bounds an integer homology ball (in fact $(Sigma - text{int} B^3) times I$).
There are many earlier examples in the literature for pairs of knots with the same double branched covers. Presumably it's not hard to check that for some of these, the knots are not concordant, using eg knot signatures. So one could construct many more examples. Likewise there are pairs of knots with the same n-fold branched covers (for a fixed $n$) and I'm sure you could construct counterexamples for those as well.
Thanks a lot for your explanatory response, Professor Ruberman.
– ToposLogos
Nov 10 at 16:56
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1
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Probably, I found some more explicit examples:
Akbulut and Larson recently showed in AL18 that family of Brieskorn spheres $Sigma(2,4n+1,12n+5)$ and $Sigma(3,3n+1,12n+5)$ bound rational homology ball.
These Brieskorn spheres are respectively $2$- and $3$-fold cyclic branched cover of torus knots $T_{4n+1,12n+5}$ and $T_{3n+1,12n+5}$, which are not smoothly slice, for example due to Milnor slice genus proven by Kronheimer and Mrowka in KM93.
New contributor
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Here's a particularly subtle counterexample, from the work of Kirk and Livingston (Topology Vol. 38, No. 3, pp. 663--671, 1999). They show that the pretzel knots $J = P(-3,5,7,2)$ and $K = P(5,-3,7,2)$ are not concordant (even locally flat). These two knots are related by mutation (switch the first two pairs of twists in the standard picture of a pretzel knot). On the other hand, they have the same double branched cover, say $Sigma$, a certain Brieskorn sphere. (This is a general fact about mutations, but is clear in this setting.)
Then the knot $J # -K$ is not slice, but its double branched cover $Sigma # -Sigma$ bounds an integer homology ball (in fact $(Sigma - text{int} B^3) times I$).
There are many earlier examples in the literature for pairs of knots with the same double branched covers. Presumably it's not hard to check that for some of these, the knots are not concordant, using eg knot signatures. So one could construct many more examples. Likewise there are pairs of knots with the same n-fold branched covers (for a fixed $n$) and I'm sure you could construct counterexamples for those as well.
Thanks a lot for your explanatory response, Professor Ruberman.
– ToposLogos
Nov 10 at 16:56
add a comment |
up vote
8
down vote
accepted
Here's a particularly subtle counterexample, from the work of Kirk and Livingston (Topology Vol. 38, No. 3, pp. 663--671, 1999). They show that the pretzel knots $J = P(-3,5,7,2)$ and $K = P(5,-3,7,2)$ are not concordant (even locally flat). These two knots are related by mutation (switch the first two pairs of twists in the standard picture of a pretzel knot). On the other hand, they have the same double branched cover, say $Sigma$, a certain Brieskorn sphere. (This is a general fact about mutations, but is clear in this setting.)
Then the knot $J # -K$ is not slice, but its double branched cover $Sigma # -Sigma$ bounds an integer homology ball (in fact $(Sigma - text{int} B^3) times I$).
There are many earlier examples in the literature for pairs of knots with the same double branched covers. Presumably it's not hard to check that for some of these, the knots are not concordant, using eg knot signatures. So one could construct many more examples. Likewise there are pairs of knots with the same n-fold branched covers (for a fixed $n$) and I'm sure you could construct counterexamples for those as well.
Thanks a lot for your explanatory response, Professor Ruberman.
– ToposLogos
Nov 10 at 16:56
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Here's a particularly subtle counterexample, from the work of Kirk and Livingston (Topology Vol. 38, No. 3, pp. 663--671, 1999). They show that the pretzel knots $J = P(-3,5,7,2)$ and $K = P(5,-3,7,2)$ are not concordant (even locally flat). These two knots are related by mutation (switch the first two pairs of twists in the standard picture of a pretzel knot). On the other hand, they have the same double branched cover, say $Sigma$, a certain Brieskorn sphere. (This is a general fact about mutations, but is clear in this setting.)
Then the knot $J # -K$ is not slice, but its double branched cover $Sigma # -Sigma$ bounds an integer homology ball (in fact $(Sigma - text{int} B^3) times I$).
There are many earlier examples in the literature for pairs of knots with the same double branched covers. Presumably it's not hard to check that for some of these, the knots are not concordant, using eg knot signatures. So one could construct many more examples. Likewise there are pairs of knots with the same n-fold branched covers (for a fixed $n$) and I'm sure you could construct counterexamples for those as well.
Here's a particularly subtle counterexample, from the work of Kirk and Livingston (Topology Vol. 38, No. 3, pp. 663--671, 1999). They show that the pretzel knots $J = P(-3,5,7,2)$ and $K = P(5,-3,7,2)$ are not concordant (even locally flat). These two knots are related by mutation (switch the first two pairs of twists in the standard picture of a pretzel knot). On the other hand, they have the same double branched cover, say $Sigma$, a certain Brieskorn sphere. (This is a general fact about mutations, but is clear in this setting.)
Then the knot $J # -K$ is not slice, but its double branched cover $Sigma # -Sigma$ bounds an integer homology ball (in fact $(Sigma - text{int} B^3) times I$).
There are many earlier examples in the literature for pairs of knots with the same double branched covers. Presumably it's not hard to check that for some of these, the knots are not concordant, using eg knot signatures. So one could construct many more examples. Likewise there are pairs of knots with the same n-fold branched covers (for a fixed $n$) and I'm sure you could construct counterexamples for those as well.
answered Nov 10 at 14:45
Danny Ruberman
10.6k3557
10.6k3557
Thanks a lot for your explanatory response, Professor Ruberman.
– ToposLogos
Nov 10 at 16:56
add a comment |
Thanks a lot for your explanatory response, Professor Ruberman.
– ToposLogos
Nov 10 at 16:56
Thanks a lot for your explanatory response, Professor Ruberman.
– ToposLogos
Nov 10 at 16:56
Thanks a lot for your explanatory response, Professor Ruberman.
– ToposLogos
Nov 10 at 16:56
add a comment |
up vote
1
down vote
Probably, I found some more explicit examples:
Akbulut and Larson recently showed in AL18 that family of Brieskorn spheres $Sigma(2,4n+1,12n+5)$ and $Sigma(3,3n+1,12n+5)$ bound rational homology ball.
These Brieskorn spheres are respectively $2$- and $3$-fold cyclic branched cover of torus knots $T_{4n+1,12n+5}$ and $T_{3n+1,12n+5}$, which are not smoothly slice, for example due to Milnor slice genus proven by Kronheimer and Mrowka in KM93.
New contributor
add a comment |
up vote
1
down vote
Probably, I found some more explicit examples:
Akbulut and Larson recently showed in AL18 that family of Brieskorn spheres $Sigma(2,4n+1,12n+5)$ and $Sigma(3,3n+1,12n+5)$ bound rational homology ball.
These Brieskorn spheres are respectively $2$- and $3$-fold cyclic branched cover of torus knots $T_{4n+1,12n+5}$ and $T_{3n+1,12n+5}$, which are not smoothly slice, for example due to Milnor slice genus proven by Kronheimer and Mrowka in KM93.
New contributor
add a comment |
up vote
1
down vote
up vote
1
down vote
Probably, I found some more explicit examples:
Akbulut and Larson recently showed in AL18 that family of Brieskorn spheres $Sigma(2,4n+1,12n+5)$ and $Sigma(3,3n+1,12n+5)$ bound rational homology ball.
These Brieskorn spheres are respectively $2$- and $3$-fold cyclic branched cover of torus knots $T_{4n+1,12n+5}$ and $T_{3n+1,12n+5}$, which are not smoothly slice, for example due to Milnor slice genus proven by Kronheimer and Mrowka in KM93.
New contributor
Probably, I found some more explicit examples:
Akbulut and Larson recently showed in AL18 that family of Brieskorn spheres $Sigma(2,4n+1,12n+5)$ and $Sigma(3,3n+1,12n+5)$ bound rational homology ball.
These Brieskorn spheres are respectively $2$- and $3$-fold cyclic branched cover of torus knots $T_{4n+1,12n+5}$ and $T_{3n+1,12n+5}$, which are not smoothly slice, for example due to Milnor slice genus proven by Kronheimer and Mrowka in KM93.
New contributor
edited Nov 10 at 16:56
New contributor
answered Nov 10 at 15:05
ToposLogos
1736
1736
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