I’m sometimes bored while flying, and I like looking out the window (though if I can, I usually pick aisle seats so I can exit more quickly).

I realized something rather amusing. I closed one eye, and held two fingers about an inch apart and a foot away from my open eye. Then, I timed how long it took an object on the ground to move from the one finger to the other finger an inch away; it took about ten seconds.

Let $h$ be the distance in feet from my eye to that point on the ground. By similar triangles, moving an inch when one foot away from my eye means moving $h$ inches on the ground. The distance from my eye to the ground is (wild guess!) 60,000 feet, so the point on the ground actually moved 60,000 inches, or 5000 feet, about a mile. Moving a mile in ten seconds is moving six miles per minute, or 360 miles per hour.

I seem to recall that 450 mph is actually how fast a commercial jet might go, so at least I’m within an order of magnitude. Now 450 miles per hour would have been 39,600 feet per minute, or 6600 feet in ten seconds, or 79,200 inches in ten seconds, so maybe I should’ve estimated 80,000 feet to the ground. But there are so many other sources of error in this technique…

Are there other fun things to estimate when trapped on a plane?

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!

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).

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!

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.