A figure from the proof with no explanation.
Hello! I am Will Hobkirk, a PhD candidate at The University of Sydney.
My supervisor is Jonathan Spreer and my auxiliary supervisor is Stephan Tillman.
My research focus is in Algebraic and Geometric Topology. My favourite things are $\mathbb{Z}_k$-folded surfaces. Here is a little introduction. If you would like more information please send me an email! You can also check out my University Page. This is unpublished research but I will update here when the paper this all comes from is up on Arxiv.
Lens spaces are a family of 3-manifolds obtained by identifying the boundaries of two solid tori. The Lens space $L(p,q)$ where $\gcd(p,q)= 1$ is given by identifying the solid tori such that the meridional curve of one solid torus is identified with a $(p,q)$-torus knot on the boundary of the other solid torus.
$S^3 = L(1,0)$
$S^1 \times S^2 = L(0,1)$
$\mathbb{R}\text{P}^3 = L(2,1)$
$L(7,2)$
The Lens spaces $L(2n,q)$ have a non-trivial $H_2(\mathbb{Z}_2)$ which is realised geometrically by a non-orientable surface $X$ embedded in $L(2n,1)$ which does not separate the embedding manifold (equivalently $X$ is connected). The maximum Euler characteristic of such a non-orientable surface embedded in $L(2n,1)$ is $2-n$ by a result of Bredon and Wood. When $n=1$ this essentially says that $\mathbb{R}\text{P}^2 \subset \mathbb{R}\text{P}^3$. This embedding looks like:
The embedding of $\mathbb{R}\text{P}^2 \subset \mathbb{R}\text{P}^3$.
The maximum euler characteristic connected non-orientable surface that embeds in $L(4,1)$ is the Klein bottle. $\mathbb{R}\text{P}^2$ does NOT embed in $L(4,1)$.
The maximum euler characteristic connected non-orientable surface that embeds in $L(6,1)$ has non-orientable genus of $3$. $\mathbb{R}\text{P}^2$ and the Klein bottle do NOT embed in $L(6,1)$.
What is the correct object to realise the non-trivial $H_2(\mathbb{Z}_k)$ classes of the Lens spaces $L(kn,q)$? The answer: $\mathbb{Z}_k$-folded surfaces. A $\mathbb{Z}_k$-folded surface is an oriented folded surface in the sense of Turaev where the orientation of the interior induces a consistent orientation along each component of the singular locus and the index of every component of the singular locus is k.
A singular point in a $\mathbb{Z}_3$-folded surface with a neighbourhood homeomorphic to $K_{3,1} \times \mathbb{R}$ satifying orientation constraints.
A point in a $\mathbb{Z}_5$-folded surface with a neighbourhood homeomorphic to $K_{5,1} \times \mathbb{R}$ satifying orientation constraints.
The maximum Euler characteristic of a non-separating $\mathbb{Z}_3$-folded surface in the Lens space $L(3n,1)$ is $2-n$. The following is this $\mathbb{Z}_3$-folded surface $X\subset L(3,1)$ of Euler characteristic $1$.
This is the non-separating $\mathbb{Z}_3$-folded surface $X\subset L(3,1)$ of Euler characteristic $1$.
This is the non-separating $\mathbb{Z}_3$-folded surface $X\subset L(6,1)$ of Euler characteristic $0$.
This is the $\mathbb{Z}_3$-folded surface $X\subset L(9,1)$ of Euler characteristic $-1$.
One part of the proof involved a generalisation the notion of a compression surgery to define the following surgeries of $\mathbb{Z}_3$-folded surfaces.
The usual notion of a disk compression of a surface.
A $2$-compression of a $\mathbb{Z}_3$-folded surface (before and after).
A $4$-compression of a $\mathbb{Z}_3$-folded surface (before and after).
The rest of the proof involves the combinatorics of intersection patterns of $\mathbb{Z}_3$-folded surfaces with the splitting torus and a meridional disk of one of the two solid tori in the genus $1$ Heegaard decomposition of $L(3n,1)$:
A figure from the proof with no explanation.
These $\mathbb{Z}_k$-folded surfaces are a new object but they contain a lot of information about the triangulations of $3$-manifolds due to the existence of canonical $\mathbb{Z}_k$-folded surfaces
The elementary pieces of a canonical $\mathbb{Z}_3$-folded surface (up to symmetries of the tetrahedron).
With very little work this can be used to show that the complexity of $L(3n,1)$ is at least $n$. We can surely do better...
The elementary pieces of a canonical $\mathbb{Z}_5$-folded surface (up to symmetries of the tetrahedron).