# 2015-09-22

I want to start writing about "Heegaard splittings" and "trisections" in many different low dimensions, so as to think about many different relative cases.

So let's start with dimension one. (Why not dimension zero?) Recall, of course, that a closed $1$-manifold is a disjoint union of circles, and a compact $1$-manifold with boundary is a disjoint union of circles and arcs.

Well, let me just start straight in with a strange definition:

A genus $g$ Heegaard splitting of a closed $1$--manifold $M$ is a decomposition $M = M_1 \cup M_2$ such that (1) each $M_i \cong \natural^g S^0 \times B^1$ and (2) $M_1 \cap M_2 \cong \#^g S^0 \times S^0$.

Hmmm... what does this mean? Remember that $S^0$ is a pair of points, specifically $\{-,1,1\}$, and that $B^1$ is an interval, specifically $[-1,1]$, with $\partial B^1 = S^0$ of course. So $S^0 \times B^1$ is a pair of arcs. What about $\natural$? That means "boundary connected sum" which means "glue the two $n$-manifolds together by identifying an $(n-1)$-ball in the boundary of one with an $(n-1)$-ball in the boundary of the other". The symbol $\natural^g$ means glue $g$ of them together this way. So $\natural^2 S^0 \times B^1$ means glue two copies of a pair of arcs together by identifying a $0$-ball (a.k.a. point) in the boundary of one with a $0$-ball (point) in the boundary of the other; this just produces $3$ arcs. Thus $\natural^g S^0 \times B^1$ is a fancy name for $g+1$ arcs. Oh, and, when $g=0$, $\natural^0 X$ is defined to be $B^n$, where $n$ is the dimension of $X$, just because $B^n$ is the identity element in the world of boundary connected sums. So $\natural^0 S^0 \times B^1$ is $1$ arc.

Even weirder, perhaps, is $\#^g S^0 \times S^0$. The symbol $\#$ means "connected sum", which means glue two $n$-manifolds together by removing a $n$-ball from the interior of each and identifying the resulting $(n-1)$-sphere boundary components. If $n=0$ we remove a $B^0$, i.e. a point, from each $0$-manifold, i.e. we just remove $2$ points all together, and then we glue them together along a $S^{-1}$, where $S^{-1} = \partial B^0 = \emptyset$. So we don't glue them together at all! In the world of $0$-manifolds, "connected sum" means "take the disjoint union and then remove $2$ points". So, noting that $S^0 \times S^0$ is $4$ points, a little counting shows that $\#^g S^0 \times S^0$ is $2(g+1)$ points. For the $g=0$ case we use that $\#^0 X = S^n$ where $n$ is the dimension of $X$.

So, finally, a genus $g$ Heegaard splitting of a closed $1$-manifold $M$ is a decomposition $M=M_1 \cup M_2$ such that (1) each $M_i$ is a disjoint union of $g+1$ arcs and (2) $M_1 \cap M_2$ is $2(g+1)$ points. So, just break $M$ up into alternating red ($M_1$) and blue ($M_2$) arcs. Lots of words for not much yet.

Tomorrow, we'll look at $(g,k)$ trisections of closed $1$-manifolds.

# 2014-02-06

Here is a question coming from 3D printing: Suppose you have two non-overlapping regions $A$ and $B$ in $\mathbb{R}^3$. Let's assume they have smooth, or piecewise smooth, or piecewise linear, boundaries.

First question: can you move them apart? In other words, to be more precise, is there a rigid Euclidean isotopy of $B$ such that, at the end, there is a plane separating $\mathbb{R}^3$ into two halves, with $A$ in one half and $B$ in the other? We could simplify the question further, for example, by asking that the isotopy be actually a linear translation, i.e. $B$ moves rigidly without rotating along a straight line.

In general, of course, one expects to this question to be, "No," because $A$ and $B$ may be badly interlocked, either for topological or geometric reasons. But even here I don't know how easy it is to solve the question for a given $A$ and $B$. It's also interesting to ask, "What are the simplest examples where the answer is negative?"

For the next question, suppose you have decided that the two cannot be separated this way. Can you subdivide either $A$ or $B$ or both so that, moving each smaller piece independently, you can separate everything? If so, what is the most efficient way to do this?

Now the motivation: An undergraduate student, Fred Hohman, and I have a 3-dimensional design that we would like to print on our 3D printer. We would like part of it to be red and part to be blue. Our printer only prints one color per object. You can change the color and print a new object, but each print by itself has to be one color. So we want to print this thing in pieces, and then assemble the pieces afterwards. In fact, we'd really like to make a 3D jigsaw puzzle out of it, so you can take them apart and put them back together and get a good feeling for the real 3D layout, or anatomy, of everything.

For those in the know, by the way, the two parts are: a trefoil knot and a thickened up Seifert surface. Later, we want multiple Seifert surfaces that are in fact fibers of the standard fibration of the complement of the trefoil.

# 2014-01-26

Well, it has been a year and a couple of months since I last wrote here, but now I'm turning over a new leaf and will try to write very regularly. I will bounce back and forth between quite technical posts and posts targeted towards people with less background. I thought I'd start back with something nice and simple - this picture:

What it is it? This illustrates stereographic projection, and I made the image using Blender, a 3D modelling program that I am using quite a bit in a second-semester geometry course that I am teaching at the University of Georgia. Stereographic projection is a projection of the sphere onto a plane. Here, a point light source is located at the top of a transparent sphere. On that sphere are a random collection of great circles, one of which is the equator. (A great circle is just a circle like the equator, which cuts the sphere in half. I got these circles by taking copies of the equator and just rotating them around various axes passing through the origin.) At the bottom of the sphere is a plane tangent to the sphere, colored this funny blue-green. The circles on the sphere (they are actually tubes surrounding the circles) cast shadows on the plan. That is mostly what you see here: the shadows of the circles. In the middle you can also see the actual sphere. The perspective is from high above the sphere. Here is a screenshot of the circular tubes being built in Blender, the light source is the orange thing at the top:

# 2012-11-13

So Rob Kirby and I have come up with the notion of a trisection of a $4$-manifold, as a completely natural generalization of a Heegaard splitting  of a $3$-manifold. I want to think right now about generalizing this to arbitrary dimensions. There should be a good notion of a Heegaard $(n-1)$-splitting on an $n$-manifold, and maybe more generally a Heegaard $m$-splitting of an $n$-manifold. (This latter will probably only make sense for certain $k$'s and $n$'s, in particular I'm hoping it works for $m=n-1$ and $m=n$. This is because we also have observed that it makes sense to talk about trisections of $3$-manifolds, which actually correspond to nothing more than open book decompositions.)

So, a Heegaard $m$-splitting of an $n$-manifold $X$ should, first off, be a decomposition of $X$ into $m$ codimension-$0$ pieces, $X_1, \ldots, X_m$, each diffeomorphic to $\natural^{k_1} S^1 \times B^{n-1}$ , for some fixed $k_1$. Then the pairwise intersections $X_i \cap X_j$, for $i \neq j$, should be diffeomorphic to $\natural^{k_2} S^1 \times B^{n-2}$, for some fixed $k_2$. Triple intersections $X_i \cap X_j \cap X_k$, for $i \neq j \neq k$, should be diffeomorphic to $\natural^{k_3} S^1 \times B^{n-3}$, and so on. All the way down to the complete intersection $X_1 \cap X_2 \cap \ldots \cap X_m$, which should be the common boundary of all the $(m-1)$-fold intersections, i.e. it should be diffeomorphic to $\sharp^{k_{m-1}} S^1 \times S^{n-m}$.

I guess this is a perfectly well-defined concept. The immediate questions are:

1. Does $\sharp^k S^1 \times S^{n-1}$ have a natural Heegard $(n-1)$-splitting?
2. Does $S^n$ have a simplest nontrivial Heegaard $(n-1)$-splitting which we can use for a stabilization operation?

It's also fairly clear from this that the cases m=n-1 and m=n should be the most tractable, because there the central fiber $X_1 \cap \ldots \cap X_m$ is either a surface or a collection of circles. Let's focus on $m=n-1$.

OK, I think the answer to question (1) is easy but I'm not seeing how to write it right now. More later.