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.










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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.










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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.










      share|cite|improve this question









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      ToposLogos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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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      edited Nov 10 at 15:09





















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      asked Nov 10 at 10:04









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          2 Answers
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          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.






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          • Thanks a lot for your explanatory response, Professor Ruberman.
            – ToposLogos
            Nov 10 at 16:56


















          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.






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            2 Answers
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            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.






            share|cite|improve this answer





















            • Thanks a lot for your explanatory response, Professor Ruberman.
              – ToposLogos
              Nov 10 at 16:56















            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.






            share|cite|improve this answer





















            • Thanks a lot for your explanatory response, Professor Ruberman.
              – ToposLogos
              Nov 10 at 16:56













            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.






            share|cite|improve this answer












            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 10 at 14:45









            Danny Ruberman

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            • 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




            Thanks a lot for your explanatory response, Professor Ruberman.
            – ToposLogos
            Nov 10 at 16:56










            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.






            share|cite|improve this answer










            New contributor




            ToposLogos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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              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.






              share|cite|improve this answer










              New contributor




              ToposLogos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.




















                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.






                share|cite|improve this answer










                New contributor




                ToposLogos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                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.







                share|cite|improve this answer










                New contributor




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                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 10 at 16:56





















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                answered Nov 10 at 15:05









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