This is a question I wandered into accidentally years ago now, which I think other people might be amused to think about (or more likely, put on an abstract algebra exam).

Let $G$ be a group, and $H$ a subgroup of $G \times G$. Is $H$ always isomorphic to $G_1 \times G_2$, for some subgroups $G_1, G_2 < G$? But beware!–I am not requiring (or expecting) any canonicity or naturality for the isomorphism: for instance $G$ sits diagonally in $G \times G$, and it just so happens that $G = { 1 } \times G = G \times { 1 }$, so this is not a counterexample, in spite of the fact that the “horizontal” or “vertical” subgroup is not a canonical choice for the diagonal subgroup.

What is a good name for groups with this property? It’s not completely trivial: cyclic groups, for instance, have this property–not that I think this property is important, but names can be amusing…

I have examples of groups $G$ and $H < G \times G$ with $H$ not (abstractly!) a product of subgroups of $G$. My challenge to you is to find some explicit examples of $H < G \times G$ and prove that $H$ doesn’t decompose.

In the end, I think this is a fun problem for a group theory final exam; I think it nicely highlights the difference between “being isomorphic” and “being equal,” though if one completes the challenge as stated, one probably already understands that distinction… So maybe the best reason for blogging about this is that chiastic title.

I gave a Farb student seminar talk on a lovely paper,

Davis, Michael W.. Groups generated by reflections and aspherical manifolds not covered by Euclidean space. Ann. of Math. (2) 1983. 293–324. MR.

I also used some of the material in

Davis, Michael W.. Exotic aspherical manifolds. 2002. 371–404. MR.

which summarizes other the many applications of the “reflection group trick,” and works through some examples with cubical complexes.

The main result is

Theorem. Suppose $B\pi = K(\pi,1)$ is a finite complex. Then there is a closed aspherical manifold $M^n$ and a retraction $\pi_1(M) \to \pi$.

This manifold $M$ can be explictly constructed by gluing together copies of the regular neighorhood of $B\pi$ embedded in some Euclidean space. The application of this theorem is to “promote” a finite complex to a closed aspherical manifold. For instance, we have a finite complex with non-residually-finite fundamental group: define the group $\pi = \langle a, b : a b^2 a^{-1} = b^3 \rangle$, which is not residually finite, and observe that the presentation 2-complex is aspherical, so we have a finite $B\pi$. Then using the theorem to “promote” this to a closed aspherical manifold, we get a manifold $M^n$ with fundamental group retracting onto $\pi$. But a group retracting onto a non-residually-finite group is also non-residually finite, so we have found a closed aspherical manifold $M^n$ with non-residually-finite fundamental group.

Just to whet your appetite, let me introduce a few of the main players, so as to give a sense of how to glue together copies of the regular neighborhood of $B\pi$.

Let $L$ be a simplicial complex, and $V = L^{(0)}$, the vertices of $L$.

From $L$ we construct two things: some complexes to glue together, and some groups with which to do the gluing. First, we construct the groups. Define $J$ to be the group $(\Z/2\Z)^V$, i.e., the abelian group generated by $v \in V$ with $v^2 = 1$. Next define $W_L$ to be the right-angled Coxeter group having $L^{(1)}$ as its Coxeter diagram; specifically, $W_L$ is the group with generators $v \in V$ and relations $v^2 = 1$ for $v \in V$ and also the relations $v_i v_j = v_j v_i$ if the edge $(v_i,v_j)$ is in $L$. Note that $J$ is the abelianization of $W_L$.

Next we will build the complexes to be glued together with the above groups. Let $K$ be the cone on the barycentric subdivision of $L$, and define closed subspaces ${ K_v }_{v \in V}$ by setting $K_v$ to be the closed star of the vertex $v$ in the subdivision of $L$. Note that $K_v$ are subcomplexes of the boundary of $K$, and that a picture would be worth a thousand words right now.

Having the complexes and the groups, we will glue together copies of $K$ along the $K_v$’s, thinking of the latter as the mirrors. Specifically, define $P_L = (J \times K)/\sim$ with $(g,x) \sim (h,y)$ provided that $x = y$ and $g^{-1} h \in J_{\sigma(x)}$, where $\sigma(x) = { v \in V : x \in K_v }$, and $J_{\sigma(x)}$ is the subgroup of $J$ generated by $\sigma(x)$. That is a mouthful, but it really is just carefully taking a copy $K$ for each group element of $J$ and gluing along the $K_v$’s in the appropriate manner. The resulting compplex $P_L$ has a $J$ action with fundamental domain $K$. Similarly, we use $W_L$ to define a complex $\Sigma_L = (W_L \times K)/\sim$.

The topology of $\Sigma_L$ is related to the complex $L$ that we started with. For example, if $L$ is the triangulation of $S^{n-1}$, then $\Sigma_L$ is a manifold. Similarly, if $L$ is a flag complex, then $\Sigma_L$ is contractible.

The idea, now, is to take some finite complex $B\pi$, embed it in $\R^N$, and take a regular neighborhood; the result is a manifold $X$ with boundary $\partial X$, and with $\pi_1 X = \pi$. Triangulate $\partial X$ as a flag complex, and call the resulting complex $L$. Instead of gluing together copies of $K$, glue together copies of $X$ along the subdivision of $L$ to get $P_L(X) = (J \times X)/\sim$ and $\Sigma_L(X) = (W_L \times X)/\sim$. With some work, we check that $\Sigma_L(X)$ is contractible because $L$ is flag, and that the contractible space $\Sigma_L(X)$ covers the closed manifold $P_L(X)$, which is therefore aspherical. Since $P_L(X) \to X \to P_L(X)$ is a retraction of spaces, we have found our desired aspherical manifold $M = P_L(X)$ with a retraction of fundamental groups.

While on public transportation, my mind wanders… And one might assume the following about me and my buses,

• The bus travels for one unit of time,
• I will get on the bus at a random time (uniformly distributed),
• I will leave the bus at a random time (independent, unformly distributed).

Then the probability that I am on the bus at time $t$ is $p(t) = 2 \cdot t \cdot (1-t)$. So one might expect that the total number of people on the bus at time $t$ to look like $C \cdot t \cdot (1-t)$ for some $C$.

I would enjoy riding a bus from the start to the end, and seeing how accurate this is, though tragically, I rather doubt it is very accurate at all. For starters, the entrance and exit times are correlated (who gets off the bus one stop after they get on?), and there are places where people are more likely to enter, and where people are more likely to exit. In fact, upon further reflection, this is a horrible model of bus ridership.

But, if you, say, averaged all the bus routes to make the entrance and exit distributions more uniform…–is there anywhere I can get this data? Wait, wait, this seems like an awful idea: I’d better stop now.

Thanks to Bryce Johnson for pointing out a mistake in my calculation of the probability $p(t)$ above–I had forgotten to include a factor of two!

When I walk down the street, I create patterns in how I walk, often by controlling my stride length so I will step on cracks every third sidewalk square, or whatnot. If I were a true master, my stride length would be incommensurable with respect to the sidewalk length–surely this was the problem that forced irrationalities upon the Greeks…

Anyway, I was also happy to realize (at a recent retreat) that clothing is nicely categorized by how many disks must be removed from a sphere to produce the particular clothing item. For some examples, consider:

• A sock or a hat is a sphere minus a disk.
• A headband (or tube top) is a sphere minus two disks.
• Jeans are a sphere minus three disks (the fabled “pair of pants”).
• A shirt is a sphere minus four disks (the “lantern”).
• A bathing suit is a sphere minus five disks.
• A fingerless glove might be a sphere minus six disks.
• Two fingerless gloves connected by a band is a sphere minus 11 disks.

Another lovely example is that of some scarves, which are a projective plane minus a disk (i.e., a Mobius strip), and therefore sit flat against one’s neck. I would be very interested in owning more non-orientable clothing (someone, somewhere, must own a non-orientable tank top–though perhaps that mythical object would be too annoying to be allowed to exist).

I’ve been (not surprisingly) drinking quite a bit of coffee lately, and I’ve noticed that many corregated coffee cup holders include a bit of loose glue. At first, I thought this was a mistake, an oversight in the perfection of the coffee cup holder design.

On the contrary, that bit of excess glue melts when the hot coffee is poured into the cup, adhering the corregated holder to the cup–brilliant!