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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

Centre of orbifold fundamental group of torus (Klein bottle) with one cone point

The groups have trivial center, as pointed out by Sam Nead. Another, more combinatorial, way to show this is to apply the algorithm from [1], which decides whether any given one-relator group has a no …
Carl-Fredrik Nyberg Brodda's user avatar
5 votes

Prove these are not surface groups

Here's an argument that uses only basic combinatorial group theory (Reidemeister-Schreier). Let $n \geq 2$, and let $G = \langle a_1, b_1, \dots, a_g, b_g | [a_1, b_1]^n [a_2, b_2] \cdots [a_g, b_g] \ …
Carl-Fredrik Nyberg Brodda's user avatar
1 vote

Surveys on unknotting number

Lackenby discussed the unknotting number (and mentions some explicit knots for which we do not know the unknotting number!) in Lackenby, Marc, Elementary knot theory, Woodhouse, N. M. J. (ed.), Lectur …
Carl-Fredrik Nyberg Brodda's user avatar
2 votes

General properties of free-by-cyclic groups

They are all residually finite. They are not all subgroup separable/LERF. They do not all have decidable submonoid membership problem. Residual finiteness is a result which can be found in the (ap …
2 votes

Looking through a bunch of links for unlinks?

I have never used this particular piece of software, but I have heard good things about the Book Knot Simplifier, which is capable of doing much of what you are describing that you'd want it to -- tho …
Carl-Fredrik Nyberg Brodda's user avatar
8 votes
1 answer
348 views

Finite two-relator groups and quotients of knot groups

Let $G$ be a one-relator group $\langle A \mid R = 1 \rangle$. Then clearly $G$ is finite if and only if it is cyclic of finite order, i.e. can be given by a presentation $\langle a \mid a^n = 1 \rang …
Carl-Fredrik Nyberg Brodda's user avatar