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Topology of cell complexes and manifolds, classification of manifolds (e.g. smoothing, surgery), low dimensional topology (e.g. knot theory, invariants of 4-manifolds), embedding theory, combinatorial and PL topology, geometric group theory, infinite dimensional topology (e.g. Hilbert cube manifolds, theory of retracts).

7 votes
Accepted

Linking number a complete invariant of link homotopy

The linking number is the same as the homology class that one component represents in the complement of the other. You can reduce any $2$-component link to a normal form by first homotoping one compon …
Jim Conant's user avatar
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7 votes
Accepted

rational cohomology of symmetric groups

Take the $n$-skeleton. It has trivial rational homology except possibly in degree $n$. Now add enough $n+1$-cells from the $n+1$-skeleton to kill this top homology. You won't have created any $n+1$-di …
Jim Conant's user avatar
  • 4,898
3 votes

Fibered example of topologically slice knots

A common source of topologically slice knots are those with Alexander polynomial $1$. However these are not fibered. This follows from a classical result, that $2\mathrm{genus}(K) = \mathrm{deg}(\Delt …
Jim Conant's user avatar
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6 votes
3 answers
381 views

Surgery along an arc connecting the components of a $2$-component link gives the unknot

Math Overflow seems to have a dearth of low dimensional topology, but this seems like an interesting question. Let $L$ be a $2$-component link in $S^3$. Suppose that there is a framed arc joining the …
Jim Conant's user avatar
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2 votes

Utility of virtual knot theory?

This paper by Chrisman and Manturov is as close as possible to an answer to my original question. From their introduction: By classical knot theory we mean the study of knots and links in the $3$- …
Jim Conant's user avatar
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9 votes
Accepted

Vassilliev invariants of knots and their cables

As I mentioned in a comment, for the degree $2$ invariant $v_2$ which is the coefficient of $z^2$ in the Conway Polynomial, we have that $v_2(K_{p,q})=av_2(K)+b$. If $K$ is the unknot, this implies th …
Jim Conant's user avatar
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10 votes
Accepted

Knot theory without planar diagrams?

Here is an example close to my heart. We show that the second coefficient of the Conway polynomial can be defined by counting quadrisecants. (Collinearities of $4$ points on the knot.) (source: rybu …
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0 votes

generators of Out(F_n) and homology

You can cook up lots of normal subgroups by looking at any characteristic subgroup of the free group. For example, if $F^{(k)}$ is the $k$th term of the lower central series, there is a surjection $$O …
Jim Conant's user avatar
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2 votes
Accepted

Where can one find reference proving that Braid group induces isomorphism between punctured ...

The isomorphism of fundamental groups comes from a diffeomorphism of spaces: $D\times[0,1]\setminus B$ is diffeomorphic to the product of an n-punctured disk with $[0,1]$. To see this, note that you c …
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3 votes
Accepted

Handle slides homeomorphism

I'm too lazy to draw this right now, but I think I can describe it anyway. Consider a twisted band attached right next to a dual pair of untwisted bands. I'll write this $A\bar{A}BCBC$. Here the overb …
Jim Conant's user avatar
  • 4,898
2 votes
Accepted

Quasi-Lie algebras in nature?

Tom Goodwillie: The homotopy groups of a (say, simply connected) space $X$ form a graded Lie algebra under Whitehead product, in which the even-dimensional part (which is actually the $\pi_n$ for …
38 votes
3 answers
3k views

Why are there no wild arcs in the plane?

On math.stackexchange it was asked whether all arcs in the plane are ambient-isotopic. I suggested that one could prove this by appealing to the Schönflies theorem, which you can do as long as you can …
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13 votes
3 answers
739 views

Algorithm for detecting ribbon or slice links?

A link in $S^3$ is said to be slice if it bounds a collection of flat disks into the $4$-ball. Here "flat," means that there is a (locally) trivial normal bundle. This condition can be strengthened to …
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13 votes

Whitehead doubles of any knots

Ian's answer is very elegant, but in case you're looking for a more computational approach, you could use the Seifert form. Namely, if you take a Seifert surface $\Sigma$ for a knot, look at the form …
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6 votes
Accepted

Boundary links and ribbon links.

The answer to your first question is no. There are non-ribbon boundary links whose components are unknotted! Indeed the Bing double of a knot is a boundary link with unknotted components, but it has r …
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