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?

Let $f(n)$ be the number of Google hits for the integer $n$. Then $f(578)$ is about 100 million, and $f(1156)$, that is, the number of hits for a number twice as big, is about 40 million, a bit less than half as big. Doubling the input continues to halve the output: $f(2312)$ is about 20 million (half again!), and $f(4624)$ is about 8 million, and $f(9248)$ is about 4 million.

There are about half as many pages talking about numbers that are twice as big. This is an example of a power law, and indeed, a log-log plot of $f$ looks linear to my blurry vision:

Doing a linear regression in R gives the red line, or in symbols, $$f(x) \approx 5,800,000,000 / x^{1.029}.$$ Rather humorously, this means that $f(a)/f(b) \approx b/a$. In the end, this is not so surprising: Zipf’s law says that, in a corpus of naturally occuring text, the frequency of a word is inversely proportional to its rank; here, we have a similar phenomenon at work: roughly, the popularity of a number is inversely proportional to its size.

In other words, while the number of integers expressible with fewer than $n$ bits grows exponentially in $n$, the number of pages discussing integers expressible with fewer than $n$ bits grows linearly in $n$; being silly, I’d say that this is an asymptotic version of the claim that most large numbers are uninteresting. After all, popular numbers have a lot of fan sites.

I searched for each number between 1 and 500 on Google, and recorded the (estimated) number of hits. I’m not aware of anyone having done this before; in any case, I made a chart:

Click on the above chart to see a bigger version. You can also look more closely at the first hundred numbers, or look at the above data with a log scale on the y-axis.

I have some observations and questions:

• There’s some periodicity in the above data (every 5, every 10, every 100).
• Can you explain how quickly the distribution falls off (is it exponentially decaying, for instance)?
• The most popular numbers are, in decreasing order of popularity: 2, 3, 10, 4, 5, 11, 6, 7, 8, 20, 15, 30, 14, 18, 1, 24, 21, 19, 25, 22, 28, 29, 50, and so on.
• The most popular numbers ending in 0 are, in decreasing order of popularity and having been divided by ten: 1, 2, 3, 5, 10, 4, 8, 9, 7, 20, 6, 50, 15, 12, 30, 25, 11, 40, 13, 18, 16, 14, and so on. Is the distribution of numbers ending in 0 related to the distribution of all numbers?
• Are certain families of numbers more popular? Are prime numbers or square numbers particularly popular?

You can download my comma-separated data file if you would like to play with the data yourself. Note, however, that I got this data from Google’s SOAP interface, which, for reasons I don’t understand, doesn’t give the same number of “estimated hits” as the web page interface.

In seminar today, Okun pointed out the following interesting observation; for any finitely generated group $G$, you can define its growth series $G(t) = \sum_{g \in G} t^{\ell(g)}$, where $\ell(g)$ is the length of the shortest word for $g$. The first observation is that $G(t)$ is often a rational function, in which case $G(1)$ makes sense. The second observation is that $G(1)$ is “often” equal to $\chi(G)$. This is an example of weighted $L^2$ cohomology.

Grigorchuk’s group (and generally any group with intermediate (i.e., subexponential but not polynomial) growth) does not have a rational growth function; the coefficients in a power series for a rational function grow either polynomially or exponentially. This observation appears in

Stoll, Michael. Rational and transcendental growth series for the higher Heisenberg groups. Invent. Math. 1996. 85–109. MR.

More significantly, this paper constructs groups which, being nilpotent, have polynomial growth, but nonetheless have generating sets for which that the corresponding growth series is not rational.

Cartan proved that every finite-dimensional real Lie algebra $\germ g$ comes from a connected, simply-connected Lie group $G$. I hadn’t known the proof of this result (and apparently it is rather uglier than one might hope), but

Gorbatsevich, V. V.. Construction of a simply connected group with a given Lie algebra. Uspekhi Mat. Nauk 1986. 177–178. MR.

gives a short proof of it, which I presented to the undergraduates in my Lie group seminar. I’ll sketch the proof now.

Theorem. For every Lie algebra $\mathfrak{g}$, there is a simply-connected, connected Lie group $G$ having $\mathfrak{g}$ as its Lie algebra.

First, if $\mathfrak{g} \subset \mathfrak{gl}(V)$, then the exponential map gives $U = \exp \mathfrak{g}$, and we define $G = \bigcup_{k=1}^\infty U^k \subset GL(V)$. It turns out $G$ is a Lie group, and $\mathfrak{g}$ is its Lie algebra.

If $\mathfrak{g}$ has no center, then $\rm{ad} : \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ is injective, so we have realized $\mathfrak{g}$ as a Lie subalgebra of endomorphisms of a vector space, and by the above, there is a Lie group $G \subset GL(\mathfrak{g})$ with $\mathfrak{g}$ as its Lie algebra. Taking its universal cover $\tilde{G}$ proves the theorem in this case.

Now we induct on the dimension of the center $Z(\mathfrak{g})$. Let $Z \subset Z(\mathfrak{g})$ be a one-dimensional central subspace of $\mathfrak{g}$, and construct a short exact sequence $0 \to Z \to \mathfrak{g} \to \mathfrak{g}’ \to 0$. But this central extension of $\mathfrak{g}’$ by $Z = \R$ corresponds to a 2-cocycle $\omega \in H^2(\mathfrak{g}; \R)$.

Lemma. Let $D : H^2(G;\R) \to H^2(\mathfrak{g}; \R)$ be the map which differentiates a (smooth!) $2$-cocycle of the group cohomology of $G$. The map $D$ is injective.

Consequently, we can find $f \in H^2(G;\R)$ with $Df = \omega$. Since $\dim Z(\mathfrak{g}’) < \dim Z(\mathfrak{g})$, by induction there is a Lie group $G’$ having $\mathfrak{g}’$ as its Lie algebra. We build the central extension of $G’$ by $\R$ using the cocycle $f$, namely, $0 \to \R \to G \to G’ \to 0$, where $G \cong G’ \times \R$ and the operation is $(g_1, t_1) \cdot (g_2, t_2) = (g_1 g_2, t_1 + t_2 + f(g_1,g_2))$. Since $Df = \omega$, it turns out that the Lie algebra corresponding to $G$ is $\mathfrak{g}$. We finish the proof by taking the universal cover $\tilde{G}$.