3 ms·

Link to paper: https://arxiv.org/abs/1808.02923 https://arxiv.org/abs/1808.02923 I have not read the paper, but I have skimmed it briefly, and it looks pretty

Link to paper: https://arxiv.org/abs/1808.02923 https://arxiv.org/abs/1808.02923 I have not read the paper, but I have skimmed it briefly, and it looks pretty exciting. This isn't a case of "here's a new invariant, and, oh, BTW, it works to show the Conway knot isn't slice." It's an actual new technique. And, at first glance, it looks like a pretty simple technique. I didn't immediately see anything here that wasn't just a neat combination of low dimensional topology and basic knot theory techniques. I'd be interested to see what this technique could do with knots having more than 12 crossings.

kmill 6y ago
Yeah, it's pretty cool! On one hand you have knot traces, which are spaces that completely determine whether a knot is slice, and on the other you have the Rassmussen s-invariant, which can sometimes tell whether a knot isn't slice. It turns out that knot traces contain different information from the s-invariant. One way sliceness is studied is through knot concordance, which is a certain restricted kind of deformation of a knot through 4-d space. All slice knots are concordant to each other. The Rasmussen s-invariant is invariant under concordance. So, the particular pair of knots C and K' cannot be concordant since they have different Rassmussen s-invariants. One consequence is that it's not true that knots with diffeomorphic knot traces are concordant in general. Another interesting thing is that C and K' are related by positive mutation, which makes them topologically concordant (weaker than concordance). Since K' is slice, this implies C is topologically slice (weaker than sliceness), though this was already known by Freedman's work because the Alexander polynomial of C is 1. Altogether, C is topologically slice but not slice (not the first example) while also being a positive mutant of a slice knot (which is what made the problem so difficult for so long). The knot trace X(K) is a 4-dimensional manifold whose boundary is a 3-manifold called the 0-surgery of K. There was a conjecture that knots with diffeomorphic 0-surgeries are concordant (perhaps up to mirror images). While it was already disproved, this pair of knots gives another counterexample. In fact, she gives infinitely many pairs of counterexamples in an earlier paper: https://arxiv.org/abs/1702.03974 https://arxiv.org/abs/1702.03974

kmill 6y ago

Yeah, it's pretty cool! On one hand you have knot traces, which are spaces that completely determine whether a knot is slice, and on the other you have the Rassmussen s-invariant, which can sometimes tell whether a knot isn't slice. It turns out that knot traces contain different information from the s-invariant. One way sliceness is studied is through knot concordance, which is a certain restricted kind of deformation of a knot through 4-d space. All slice knots are concordant to each other. The Rasmussen s-invariant is invariant under concordance. So, the particular pair of knots C and K' cannot be concordant since they have different Rassmussen s-invariants. One consequence is that it's not true that knots with diffeomorphic knot traces are concordant in general. Another interesting thing is that C and K' are related by positive mutation, which makes them topologically concordant (weaker than concordance). Since K' is slice, this implies C is topologically slice (weaker than sliceness), though this was already known by Freedman's work because the Alexander polynomial of C is 1. Altogether, C is topologically slice but not slice (not the first example) while also being a positive mutant of a slice knot (which is what made the problem so difficult for so long). The knot trace X(K) is a 4-dimensional manifold whose boundary is a 3-manifold called the 0-surgery of K. There was a conjecture that knots with diffeomorphic 0-surgeries are concordant (perhaps up to mirror images). While it was already disproved, this pair of knots gives another counterexample. In fact, she gives infinitely many pairs of counterexamples in an earlier paper: https://arxiv.org/abs/1702.03974 https://arxiv.org/abs/1702.03974