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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).

6 votes

Immersed versus embedded surfaces representing a same homology class

In full generality, this is false. I'll construct an example in the blow up $X$ of $\mathbb{CP}^2$ (which is, in fact, an $S^2$-bundle over $S^2$). Kronheimer and Mrowka proved that every symplectic …
Marco Golla's user avatar
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11 votes

Does every embedded 2-sphere in $\mathbb{R}^n$ bound an embedded ball?

Haefliger proved that $S^k$ in $\mathbb{R}^n$ is unknotted if $2n > 3(k+1)$. Therefore, any 2-sphere embedded in $\mathbb{R}^n$ with $n\ge 5$ is unknotted. On the other hand, as Mark Grant points out …
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11 votes

How to show whether a given knot and its mirror image are the same or not?

There are several ways to try and tell apart a knot from its mirror Probably the most "classic" way to do this is using the signature; You can compute Jones polynomial and check that it's not symmet …
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9 votes
Accepted

What is the easiest way to compute Ozsváth-Szabó tau invariant of a knot?

In principle, a computer can compute $\tau$ for any knot, using grid diagrams (or variants thereof); Baldwin and Gillam computed $\tau$ for knots up to 11 crossings a while ago. However, this is not p …
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17 votes
Accepted

Can Khovanov homology have arbitrarily large torsion?

This paper from earlier this year (Jan 18, to be precise) proves the existence of $\mathbb{Z}/n\mathbb{Z}$-torsion for $n\le 8$ and $\mathbb{Z}/2^s\mathbb{Z}$-torsion for $s\le23$. It also states at t …
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5 votes
Accepted

Why is the dividing set nonempty when a convex surface has Legendrian boundary?

When I first read the question, I found it really odd, but now I'm convinced that this is really true. I'll use the setup of Etnyre's lecture notes on convex surfaces, around pages 5-6. Just to recap …
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2 votes

On the proof of Robert Lipshitz's formula on Maslov index.

The coefficients of the regions that meet at $p$ reflect the fact that the domain $\mathcal{D}$ has (or doesn't have) a corner at $p$. If $p$ is not a corner (i.e. if either $p\not\in\mathbf{x}\cup\m …
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12 votes

Behavior of genus function on a 4-manifold for sums

Sometimes the function $G$ can be constantly 0: consider the class $x = [S^2\times\{p\}]$ in $H_2(S^2\times F)$, where $F$ is a surface. Then $G(nx)$ can be realised by an embedded sphere for all $n$: …
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8 votes

Heegaard splittings of Brieskorn spheres

All Brieskorn spheres are small Seifert fibred spaces (small SFS, in brief), i.e. they admit a fibration $S^1 \to \Sigma(p,q,r) \to S^2$ with three multiple fibres. This is easier to see when $p,q,r$ …
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5 votes
Accepted

Upper bounds on the genus of the surface produced by Seifert's algorithm

Just a note about terminology: the minimal genus of a Seifert surface arising from Seifert's algorithm to a diagram of $K$ is called the canonical genus of $K$. This paper by Brittenham and Jensen pr …
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3 votes
Accepted

Minimal genus of characteristic surfaces?

If you want something specific to characteristic classes, the only thing I know you can leverage on is the fact that the complement of any surface representing a characteristic class is spin. Let's fi …
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5 votes
Accepted

Lickorish-Wallace theorem for torsion spin$^c$ 3-manifolds?

First, let me remark that $S^3$ has a unique spin$^c$ structure $\mathfrak{t}_0$, which is also torsion. (I decided to call it $\mathfrak{t}_0$, because I prefer to use $\mathfrak{t}$ for spin$^c$ str …
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2 votes
Accepted

Glueing two 2-tangles

No, $L$ and $T_1$ together do not determine $T_2$. Take a non-trivial knot $K$ and suppose that $L$ is $K\#K$, and $T_1$ is the trivial tangle. Then $T_2$ can certainly be: a 2-tangle with one bounda …
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2 votes

Simple examples of equivariant cobordism

For the concrete case of the cyclic group $C_p$ acting linearly on $S^3$, there's a very explicit construction. Call $\omega = e^{2\pi i/p}$. Fix an integer $q$ coprime with $p$, and let us look at th …
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22 votes

Manifold embedded in $R^{n+1}$ with a submanifold that doesn't embed in $R^n$

If you are interested in an example in codimension 2 examples (which also happens to involve two orientable manifolds), according to the table on page 2 in the survey Davis, Donald M. Embeddings o …
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