2019
DOI: 10.1090/tran/7389
|View full text |Cite
|
Sign up to set email alerts
|

Links with non-trivial Alexander polynomial which are topologically concordant to the Hopf link

Abstract: We give infinitely many 2-component links with unknotted components which are topologically concordant to the Hopf link, but not smoothly concordant to any 2-component link with trivial Alexander polynomial. Our examples are pairwise non-concordant.

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
1
1
1
1

Citation Types

0
6
0

Year Published

2020
2020
2024
2024

Publication Types

Select...
5
1

Relationship

1
5

Authors

Journals

citations
Cited by 6 publications
(6 citation statements)
references
References 40 publications
0
6
0
Order By: Relevance
“…Similar procedures show that CFK ∞ (T (7,9)) is not stably equivalent to CFK ∞ (T (2, 7) # T (7, 8)), and, in fact, the following general theorem holds. Theorem 1.2.…”
Section: Introductionmentioning
confidence: 77%
See 2 more Smart Citations
“…Similar procedures show that CFK ∞ (T (7,9)) is not stably equivalent to CFK ∞ (T (2, 7) # T (7, 8)), and, in fact, the following general theorem holds. Theorem 1.2.…”
Section: Introductionmentioning
confidence: 77%
“…We have that ( 4 5 ), we compute how far the line of slope − 3 2 needs to be moved so that the elements represented by (1,8) and (3,5) are homologous in F 4 5 ,r . In the diagram we see that we need F 4 5 ,r to contain the elements represented by (3,7) and (2,8). The minimal r which accomplishes this is r = 23 5 , as shown in ( 4 5 ) < − 8 5 .…”
Section: Resultsmentioning
confidence: 99%
See 1 more Smart Citation
“…In fact, a careful check of the Alexander and the algebraic filtrations shows that CFKfalse(T3q+2,3false)=Spandouble-struckZ2false[U,U1false]false⟨z0,,z2q+2false⟩. Summing up we get that CFKfalse(T3q+1,3false)CFKfalse(T3,2false)=CFKfalse(T3q+2,3false)A with A acyclic. Note that this can also be seen as a consequence of [, Lemma 3.18]. To prove the corresponding statement for CFKI we need to check that (1)the involution of CFKIfalse(T3q+1,3false)CFKIfalse(T3,2false) restricts to ιT3q+2,3 on the sub complex spanned by z0,,z2q+2, (2)ιT3q+1,3×ιT3,2leaves A invariant. Here CFKIfalse(T3q+1,3false)CFKIfalse(T<...>…”
Section: Obstructions From the Kim‐livingston Secondary Invariantmentioning
confidence: 97%
“…For knots K and J, if for some s ∈ (0, 2) and m ∈ R we have Remark 4.11. As noted in Allen's paper [1], in [6] Kim, Krcatovich and Park gave a condition for the knot complex of the connected sum of two L-space knots to be stably equivalent to a staircase complex. In particular, Lemma 3.18 in [6] implies that CFK ∞ (T (p, 2p − 1)) is stably equivalent to CFK ∞ (T (p − 1, p)#T (p, p + 1)).…”
Section: Calculations Of Secondary Upsilon Invariantmentioning
confidence: 99%