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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
Kirby diagram of the complement of a subhandlebody of a smooth closed 4-manifold
When the decomposition of $X$ has no 3-handles, this is often feasible. The trick is to turn the handle decomposition of $X$ upside down (see Example 5.5.5 in Gompf and Stipsicz's 4-manifolds and Kirb …
5
votes
Accepted
Why is this Brieskorn manifold a rational homology sphere?
The key is pointed out by HJRW in his comment: there's a missing piece in your explanation, which is the genus of the base $B$ of the Seifert fibration. Némethi writes:
$$
2g-2 = (n−2)A/a−\sum q_i,
$$ …
7
votes
Accepted
If a lattice can be embedded into $\mathbb Q^n,\langle-1\rangle^n$, then can it be embedded ...
As Ian Agol points out in his comment, $-E_8$ is an example of a lattice that embeds in $\mathbb{Q}^8$ but not in $\mathbb{Z}^n$ for any $n$.
The embedding in $\mathbb{Q}^8$ is given explicitly in Con …
1
vote
Necessary condition for invertible knot concordance from both ends
No, this is not true: any doubly-slice knots gives a counterexample to your statement.
A knot $K$ is doubly-slice if there is an unknotted 2-sphere $F$ in $S^4$ such that $(S^3,K)$ is diffeomorphic to …
7
votes
$3$-manifold that is a surgery on a knot
About your third question:
Are there any partial results for branched double covers of knots?
The Montesinos trick tells you that if $K$ has unknotting number 1, then its double cover is $\pm\frac{\ …
5
votes
Euler number of a Seifert bundle as a generalization of an Euler number of a circle bundle o...
Why is the Euler number of a Seifert bundle a "natural" generalization of a circle bundle over a surface?
I can see at least three reasons:
A circle bundle is a Seifert manifold with no singular fi …
7
votes
Generalization of the sphere theorem in dimension at least 4
Aru Ray and Danny Ruberman wrote a paper (here the arxiv version) about Dehn's lemma in dimension 4. From the abstract:
We investigate certain 4-dimensional analogues of the classical 3-dimensional D …
7
votes
Detecting a "bad map" in Fintushel-Stern knot surgery
There is nothing inherently "bad" with other choices. My guess is that Fintushel and Stern chose this identification for three reasons: first, they can give a nice formula for how the Seiberg–Witten i …
7
votes
Accepted
Picture of the isotopy class of a degree $d$ smooth complex curve
This might be a bit annoying to justify (see below), but I'm quite sure that the surface you're looking for is constructed as follows. (And in any case this is too long for a comment.)
Start with the …
9
votes
Accepted
Are two slice surfaces with minimal genus isotopic?
The question is very loaded and the question would almost require a survey...
Anyway, the answer to your questions is mostly no. Let $S \subset S^4$ be a 2-knot (i.e. an embedded 2-sphere), $p \in S$ …
6
votes
Accepted
A Mazur manifold bounded by $\Sigma(2,3,13)$
I don't know where you took the picture from, but that link is not symmetric (and it likely does not describe the correct manifold).
Indeed, the 0-framed component of the link you drew is a trefoil: i …
10
votes
Accepted
Two surfaces in a 4-manifold whose algebraic intersection number is zero
Yes, this can be done by tubing one surface along the other.
Suppose that you have two intersection points $p_+, p_- \in \Sigma_1 \cap \Sigma_2$ of opposite signs. Suppose also that $\Sigma_1$ and $\S …
6
votes
Accepted
Homology classes in connected sum of $\Bbb CP^2$'s that can be represented by smoothly embed...
As Mike says in his comment, the answer is not known in full generality, not even for $n=2$, so for classes in $\mathbb{CP}^2\#\mathbb{CP}^2$. I think the paper that he links (https://arxiv.org/abs/22 …
5
votes
Accepted
Examples of homology sphere that bound a nonsmoothable contractible 4-manifold
There are certainly lots of obstructions coming from gauge theory, for instance Frøyshov's $h$-invariant or Ozsváth and Szabó's $d$-invariant. If an integer homology sphere $P$ has $d(P) \neq 0$, then …
2
votes
Accepted
Are all exact Lagrangian spheres, vanishing cycles?
No, this is not necessarily true. In fact, this is never the case if $E = B^4$, and $F$ has connected boundary and genus $g \ge 2$.
Since $H_1(E) = 0$, the vanishing cycles span $H_1(F)$.
An Euler cha …