2015-10-01

First, unrelated to trisections, I was just last night made aware of (thanks to Tanner Strunk) a video on youtube of Milnor talking about differential topology in 1965: Definitely a classic.

We're still thinking about trisections of surfaces, so I want to spend today's entry on a picture I often attempt to draw in talks. This is a sequence of pictures of a torus in $\mathbb{R}^3$, but since they are pictures on your computer screen of course they are projected onto $\mathbb{R}^2$. I would love to see this example animated, drawn professionally, somehow illustrated by a serious mathematical illustrator. I'll do blackboard pictures again, and today just post the pictures without explanation:

T2inR3.1  T2inR3.2  T2inR3.3  T2inR3.4  T2inR3.5  T2inR3.6  T2inR3.7

 

2015-09-28

On to trisections of closed $2$-manifolds:

A $(g,k)$ trisection of a closed $2$-manifold $\Sigma$ is a decomposition $\Sigma=\Sigma_1 \cup \Sigma_2 \cup \Sigma_3$ such that: (1) each $\Sigma_i \cong \natural^k S^0 \times B^2$, (2) each $\Sigma_i \cap \Sigma_j \cong \natural^g S^0 \times B^1$ and (3) $\Sigma_1 \cap \Sigma_2 \cap \Sigma_3 \cong \#^g S^0 \times S^0$.

As we've seen earlier, these connected sums and boundary connected sums in low dimensions behave strangely, but the upshot is that (1) each $\Sigma_i$ is a disjoint union of $k+1$ disks, (2) each $\Sigma_i \cap \Sigma_j$ is a disjoint union of $g+1$ arcs and (3) $\Sigma_1 \cap \Sigma_2 \cap \Sigma_3$ is $2g+2$ points. The Euler characteristic $\chi(\Sigma)$ is then $(2g+2)-3(g+1)+3(k+1) = 2-g+3k$. Note that a more sensible indexing might be to call it a $(b,c)$ trisection, where $b$ is the number of arcs (bridges) and $c$ is the number of disks (components); this is what Meier and Zupan do, and the relation of course is $g=b-1$, $k=c-1$.

Here is a quick picture prepared for a grant proposal, of a $(0,0)$ trisection of $S^2$ and a $(2,0)$ trisection of $T^2$ (I've used the $(g,k)$ notation not the $(b,c)$ notation here):

S2T2trisections

Another good one for $S^2$ is the cube, where opposite faces form the $\Sigma_i$'s; this is a $(3,1)$ trisection.

2015-09-24

Summarizing, it seems that Heegaard splittings of $0$-manifolds are problematic, or need a different definition, and similarly trisections of $1$-manifolds are problematic; in both cases, the problem is having part of the dimension be $(-1)$-dimensional. But Heegaard splittings of (closed) $1$-manifolds were fine.

Now let's do Heegaard splittings of closed surfaces. Here are two options:

Option 1: A "genus $g$" Heegaard splitting of a closed surface $\Sigma$ is a decomposition $\Sigma=\Sigma_1 \cup \Sigma_2$ such that (1) each $\Sigma_i \cong \natural^g S^1 \times B^1$ and (2) $\Sigma_1 \cap \Sigma_2 \cong \#^g S^1 \times S^0$. 

On other words, each $\Sigma_i$ is a planar surface with $g+1$ boundary componets, a.k.a. a $g$-punctured disk (well, disk with $g$ open disks removed). So just put your surface flat on the table and slice parallel to the table, like this:

SurfaceHS

Note that this doesn't give us much choice. There is only one way to do this for a fixed surface of (actual) genus $h$. And maybe we really want to think of $\Sigma$ as just the double of some surface with boundary, i.e. a $2$-dimensional $1$-handlebody. So...

Option 2: A Heegaard splitting of a closed surface $\Sigma$ is a decomposition $\Sigma=\Sigma_1 \cup \Sigma_2$ such that (1) each $\Sigma_i$ is a $2$-dimensional $1$-handlebody and (2) $\Sigma_1 \cap \Sigma_2$ is a disjoint union of circles.

Now the formalism is breaking down, but this again is something special in low dimensions, that $2$-dimensional $1$-handlebodies are not completely determined by the number of $1$-handles. This suggests also that breaking the formalism a little might help with the $0$- and $1$-dimensional examples that failed. I'll return to this tomorrow, I think. Let's just end with a picture of this kind of Heegaard splitting:

SurfaceHS2

Here each half is a once-punctured torus. All I actually did was cut along the central "neck" first, but then modify that cut by a Dehn twist to make it look fancy.

2015-09-23

OK, yesterday I defined a "genus $g$ Heegaard splitting of a $1$-manifold", which is sort of absurd, but I'm trying to organize my low-dimensional definitions carefully enough to say something nontrivial as we go up in dimension. There is the brief question of whether we should think about Heegaard splittings of $0$-manifolds. Let's try:

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

Condition (2) means that $M_1 \cap M_2 = \emptyset$. For condition (1), recall that $\natural$ means "glue by identifying balls in the boundary", and since the boundary here is empty, $\natural$ should mean disjoint union. Also, the ball is the identity element for boundary connected sum, so $\natural^g X$ is $B^n$, where $n$ is the dimension of $X$. So $\natural^g S^0 \times B^0$ would seem to mean one point if $g=0$, and otherwise $2g$ points. This doesn't sound good, e.g. $B^0$ doesn't have a Heegaard splitting. Well, maybe this only works for an even number of points ($B^0$ is somehow not "closed"). But it also looks like a collection of $6$ points doesn't have a Heegaard splitting. Maybe the problem is the definition of $\natural$ for $0$-manifolds, but this is enough. Maybe the real problem is just that we hit $(-1)$-dimensional stuff in our definition.

What I really want to think about (which may run into the same problem I now realize) is:

A $(g,k)$ trisection of a $1$-manifold $M$ is a decomposition $M=M_1 \cup M_2 \cup M_3$ where (1) each $M_i \cong \natural^k S^0 \times B^1$, (2) each $M_i \cap M_j \cong \natural^g S^0 \times B^0$ and (3) $M_1 \cap M_2 \cap M_3 \cong \#^g S^0 \times S^{-1} = \emptyset$.

Remember that $\natural^k S^0 \times B^1$ is $(k+1)$ arcs and $\natural^g S^0 \times B^0$ is one point if $g=0$ otherwise $2g$ points. E.g. this seems fine ($k=0$, $g=0$):CircleTrisectionAnd so does this ($k=1$, $g=1$): CircleTrisection2But one more step and it falls apart, I think. $k=3$ will give pairwise intersections being $3$ points. Unless we change the way we alternate? No that won't fix it either.

Conclusion: there is no reasonable definition of a trisection in dimension less than $2$ because the triple intersection has to be codimension $2$ and shouldn't be empty, just as there is no reasonable definition of a Heegaard splitting in dimension $0$ because the pairwise intersection needs to be codimension $1$ and nonempty. So, tomorrow, we'll look at Heegaard splittings of surfaces.