A very long while ago I posted some solutions to Lights Out; back then, I solved the $n$-by-$n$ board by row-reducing an $n^2$-by-$n^2$ matrix.

Since then, both Boris Okun and Brent Werness pointed out to me that I should’ve solved Lights Out by using a scanning algorithm: propagating the button presses down one row at a time, and exponentiating the propagation matrix to make sure that I don’t get stuck at the last row.

This is much faster.

With this method, here is a (scaled down, auto-leveled) 2000-by-2000 solution:

And here is a (very much scaled-down, auto-leveled) 5000-by-5000 solution:

My advisor, Shmuel Weinberger, was teaching Math 113, and asked for some pictures of the following procedure:

• Start with a triangle in the plane.
• Reflect that triangle across its three sides.
• And repeat, reflecting the resulting triangles through their sides, and so forth.

I made a couple movies of this, illustrating this procedure as you move through the space of triangles. Observe how, for only four shapes of triangles, the resulting set of triangle vertices is discrete.

Movie with only a few triangles

Download the original as a QuickTime movie.

Movie with more triangles

Download the original as a QuickTime movie.

I recently gave a beamer talk, which gave me the chance to point the beamer at my blackboard.

Here’s a little movie I made:

I’m grading for the first year topology course at Chicago, and one of their homework problems asked them to show that pairs of (indistinguishable!) points on a circle correspond to points on the Möbius strip; in other words, the quotient of the torus $T^2 = S^1 \times S^1$ by the $\Z/2$-action which exchanges the two $S^1$ factors is a Möbius strip.

In the above animation, you can see the identification in action: the two red points on the green circle correspond to the red dot on the Möbius strip.

I gave a talk in the Farb and Friends Student Seminar (back in March!) on:

Davis, Michael W. and Januszkiewicz, Tadeusz. Hyperbolization of polyhedra. J. Differential Geom. 1991. 347–388. MR.

This is an awesome paper—well-worth a few words on every blog!

The construction is way easier than you might think. The ingredients:

• A model space $X$ with a map $f : X \to \Delta^n$
• Any simplicial complex $K$ with a nondegenerate (edge-non-collapsing) map $K \to \Delta^n$ (if having a map to $\Delta^n$ seems like a bother, note that the barycentric subdivision $K’$ comes with a map to $\Delta^n$ for free).

Let $X_J = f^{-1}(J)$ for $J$ a subcomplex of $\Delta^n$; we think of this as decomposing $X$ into pieces resembling a simplex.

Now the construction is easy: replace each simplex in $K$ with a corresponding piece of $X$. Or more formally, build the fiber product of $X$ and $|K|$ over $\Delta^n$; this fiber product is denoted by $X \tilde{\Delta} K$ in the paper. From this, we get a natural map $f_K : X \tilde{\Delta} K \to K$.

The vague upshot is this: features of $X$ translate into features of $X \tilde{\Delta} K$, while nonetheless preserving features of $K$. Here are a couple of examples of how assumptions on $X$ lead to consequence for $X \tilde{\Delta} K$.

• If $X$ is path-connected, and for each codimension 1 face $\alpha$ of $\Delta^n$, we have $X_{\alpha} \neq \varnothing$, then $\pi_1(f_K) : \pi_1(X \tilde{\Delta} K) \to \pi_1(K)$ is a surjection.
• If $X$ and $K$ are PL-manifolds, and $\dim X_J = \dim J$, and $\partial X_J = X_{\partial J}$, then $X \tilde{\Delta} K$ is a PL-manifold.

I’ll briefly introduce the Lights Out puzzle: the game is played on an n-by-n grid of buttons which, when pressed, toggle between a lit and unlit state. The twist is that toggling a light also toggles the state of its neighbors (above, below, right, left—although, on the boundary, lights have fewer neighbors). All the buttons are lit when the game begins, and the goal is to turn all the lights off.

There are two key observations:

• toggling a light twice amounts to doing nothing,
• toggling light $A$ and then light $B$ has the same effect as toggling $B$ and then toggling $A$.

As a result, the order in which we press the buttons is irrelevant. So to solve the n-by-n puzzle, we just need to know whether a button needs to be pressed. My old website had some pictures I made showing solutions for boards of various sizes—pictures where a white pixel meant “press” and a black pixel meant “don’t press.” I assembled these pictures into a video, showing solutions to the Lights Out puzzle for $n \leq 200$:

For as cool as that looks, there’s not much to be discovered (as far as I can tell) from watching these frames flash by. But it does look like about half the buttons have to be pressed to solve the puzzle: why is that?

The still frames of the movie are available here as PNGs in a zipped archive. Here is a solution to the 400-by-400 board:

Finding that solution involved row-reducing a $(400 \cdot 400 + 1)$-by-$400 \cdot 400$ matrix—that’s a matrix with over 25 billion entries. On the other hand, each entry is one bit, so that matrix fits (not coincidentally) in 3 gigabytes of memory. One could surely do better, considering how sparse the matrix is: perhaps we could have a little contest for solving very large Lights Out games.

Besides the fact that all these pictures look awesome, Lights Out is a neat example to motivate some linear algebra over a finite field. It illustrates how satisfying an “easy” local condition (each light must be turned off) might require a globally complicated solution—a lesson for mathematics and for life!

I made a movie recently for my advisor. The movie is so pretty, that I thought I’d share it here: may I present to you randomly drawn dots, where two dots are the same color when they touch!

I’ll be a bit more explicit: a dot is drawn at a random location; if it does not overlap any previous dots, it gets a new color. Otherwise, the dot takes the color of the component it touches. Sometimes a new dot connects many components, and in this case, the new component takes on the color of the largest among the old components.

There’s a lot of neat questions to be asked about such a process: for instance, after drawing n dots, how many components should we expect to see? As you can see in the movie, when you draw only a few dots, most of those dots are isolated and have their own color; but after drawing a ridiculously large number of dots, they are all connected and the same color. And inbetween, something more interesting happens.

Here’s an example of “something more interesting” taken from a larger picture than the above movie:

In this fun paper,

Kreck, Matthias. An inverse to the Poincaré conjecture. Arch. Math. (Basel) 2001. 98–106. MR.

it is pointed out that

• homology is a very basic invariant, and
• closed manifolds are very basic objects

and so a very basic question is: what sequences of abelian groups are the homology groups of a closed simply connected manifold?

It isn’t very hard to realize any sequence of abelian groups up to the middle dimension, but that middle dimension is tricky (e.g., classify $(n-1)$-connected $2n$-manifolds).

Anyway, I was wondering: is this realization question solvable for homology with coefficients in $\Z/2\Z$ or $\Q$?

Granger causality is a technique for determining whether one time series can be used to forecast another; since the Intrade market provides time series data for political questions, we can look at whether political outcomes can be used to forecast other political outcomes.

There’s a library for the statistical package R to do the Granger test, and Intrade produces CSV market data. I fed the market data for various contracts since January 1, 2008 into R, and the output of that into GraphViz to make a nice-looking visualization; in particular, I connect $a$ to $b$ if $a$ Granger-causes $b$ with $p$-value less than 0.05. Darker arrows have smaller $p$-values. This is all an embarassing misuse of statistics and $p$-values, but it is quick and easy to do, and the results are fun to see.

Here is the graph for a lag of one day (i.e., does yesterday’s value of $a$ predict today’s value of $b$):

Here is the graph for a lag of two days (i.e., can the two previous days of data for $a$ be used to forecast the next day of data for $b$):

And here is the graph for a lag of three days:

Don’t take this too seriously. And one word of warning: an arrow from $a$ to $b$ does not mean that if $a$ is more likely, then $b$ is more likely—rather, it ought to mean that past knowledge of $a$ can be used to forecast $b$. I suppose it would be interesting to add some color for the direction of the relationship, and maybe I’ll do that when I have another free hour.

I made some movies of some of my favorite complexes: let $I^n$ be the $n$-dimensional cube, and let $e_1, \ldots, e_n$ be the $n$ edges around the origin, and let $e_i e_j$ be the square face containing the edges $e_i$ and $e_j$. Define a subcomplex $\Sigma^2_n \subset I^n$ consisting of the squares $$e_1 e_2, e_2 e_3, \ldots, e_{n-1} e_n, e_n e_1$$ and all the squares in $I^n$ parallel to these. It turns out that $\Sigma^2_n$ is a surface with a lot of symmetries.

In particular $\Sigma^2_4$ is a torus in $\R^4$, and here is a movie of it spinning:

I’m particularly fond of this, as you can really see that four squares are coming together at each vertex (hence, it has zero curvature), and you can see the hole in the torus as it spins.

The complex $\Sigma^2_5$ is a genus five surface in $\R^5$, and here is a movie of it spinning:

I represented the extra dimensions with color—not that it helps much!

For $X$ a metric space, and $S \subset X$, define the length spectrum of S to be $D_S := { d(x,y) : x, y \in S }$. It might be better to call this the “distance spectrum” or “distance set.”

Ian, during his Pizza seminar, gave the following definition: a set $S \subset \R^n$ is a $k$-distance set if $D_S$ has cardinality no greater than $k$. In words, the distances between points in a $k$-distance set take on no more than $k$ possible values.

The question that Ian answered is the following: how big can a $k$-distance set in $\R^n$ be? Clever linear algebra shows that the size grows polynomially in $n$ with degree $k$. A related exercise is the following: suppose $S \subset \R^n$ and $D_S$ is countable; prove that $S$ is countable.

Now here is my question: suppose $S \subset \R^n$ and $D_S$ is measurable with measure $m$. Can one then bound the measure of $S$? Ian asked this for the counting measure, but presumably one can get results for Lebesgue measure. Likewise, one can ask this for spaces other than $\R^n$.

All this talk of spectra has gotten me thinking very vaguely about a bunch of stuff–some random ideas! One context in which I have seen spectra is for lattices in Lie groups; I don’t know, but definitely ought to know how much control the spectrum exerts on the lattice. As a baby example, it is true that one can recover a lattice $\Lambda \subset \R^2$ from its length spectrum? Similarly, a Riemannian manifold has a length spectrum, and the “marked length rigidity conjecture” asks how much of the Riemannian structure is related to this. For information:

Furman, Alex. Coarse-geometric perspective on negatively curved manifolds and groups. 2002. 149–166. MR.

Finally, it is possible to define a “spectral distance” (I’m mis-using so many word here!) between two lattices in a Lie group. Namely, given $\Lambda_1, \Lambda_2 \subset G$, define $d(\Lambda_1, \Lambda_2) = d_{GH}(D_{\Lambda_1}, D_{\Lambda_2})$, i.e., the Hausdorff distance between their spectra. Though you’d probably want something slightly more refined (to count multiplicities). You could likewise say that two manifolds are “nearly isospectral” if their spectra are not so far apart in Gromov-Hausdorff distance. I have no idea whether this is a good idea; it probably isn’t. In any case,

Sunada, Toshikazu. Riemannian coverings and isospectral manifolds. Ann. of Math. (2) 1985. 169–186. MR.

constructs isospectral manifolds, and it would be interesting to know how easy it is to construct nearly isospectral manifolds. A braver person than I might conjecture that two manifolds are isospectral if they are $\epsilon$-nearly isospectral for small enough $\epsilon$.

At last, can one detect arithmeticity of a lattice from its spectrum? I suppose if I were really hip, I would ask: can a geometer hear arithmeticity? I think Sunada’s examples are all arithmetic?

There are usually courses at Mathcamp about surfaces; there should be courses about orbifolds! For instance, knowing that the smallest hyperbolic orbifold is the (2,3,7)-orbifold, having orbifold Euler characteristic $-1/84$, immediately gives that a closed hyperbolic surface of genus $g$ has no more than $84(g-1)$ isometries (preserving orientation); this is “Hurwitz’ $84(g-1)$ theorem.”

Just to show off this theorem, here is a cubical complex which is a surface with lots of symmetries (and the clever reader will recognize this as coming right out of Davis’ construction of aspherical manifolds): consider the $n$-dimensional cube $I^n$, and let $e_1, \ldots, e_n$ be the $n$ edges around the origin, and $e_i e_j$ be the square face containing the edges $e_i$ and $e_j$. Define a subcomplex $\Sigma^2_n \subset I^n$ consisting of the squares $e_1 e_2, e_2 e_3, \ldots, e_{n-1} e_n, e_n e_1$ and all squares in $I^n$ parallel to these. Now $\Sigma^2_n$ is a orientable surface (the link of any vertex is an $n$-cycle, i.e., topologically an $S^1$). Don’t be fooled by the notation: $\Sigma^2_n$ has genus much larger than $n$.

In fact, let’s calculate the genus. Every vertex of $I^n$ is contained in $\Sigma^2_n$, and there are $2^n$ vertices in $I^n$. Likewise, every edge in $I^n$ is contained in $\Sigma^2_n$, and there are $n 2^{n-1}$ edges in $I^n$. Finally, there are $2^{n-2}$ squares parallel to each of $e_i e_j$, so there are $n 2^{n-2}$ square faces in $\Sigma^2_n$. Thus, $$\chi(\Sigma^2_n) = 2^n - n 2^{n-1} + n 2^{n-2} = 2^{n} - n \cdot 2^{n-2},$$ and so $\Sigma^2_n$ is a surface of genus $g = 1 + 2^{n-3} (n - 4)$. This is a maybe a good exercise for someone first learning about Euler characteristic, but not especially interesting…

So here’s the punchline–or rather the punch-question–why is the genus growing exponentially in $n$? Because $\Sigma^2_n$ is very symmetric! And Hurwitz says to get so much symmetry, we need (linearly) as much genus. And we can find exponentially many symmetries of $\Sigma^2_n$ without any work. For starters, the group $(\Z/2\Z)^n$ acts on $\Sigma^2_n$ by reflecting through hyperplanes, as does the group $\Z/n\Z$ cyclically permuting the basis $e_1, \ldots, e_n$. If we want to be precise, let $G_n$ be the resulting group of order $n 2^{n}$. Quotienting $\Sigma^2_n$ by these symmetries gives an orbifold $\Sigma^2_n / G_n$, which one observes to be a square with cone points on each vertex (three with cone angle $\pi/2$ and one with cone angle $2 \pi/n$) and reflections in each of the four sides. Thus, the orbifold Euler characteristic of $\Sigma^2_n / G_n$ is $3/4 + 1/n - 4/2 + 1 = 1/n - 1/4$, so the Euler characteristic of $\Sigma^2_n$ must be $(1/n - 1/4) \cdot n \cdot 2^{n} = 2^{n} - n \cdot 2^{n-2}$, just like we got before. One might argue that this method was “easier” than the previous method for calculating $\chi(\Sigma^2_n)$, but that misses the point—I (and probably everyone else) calculated the number of edges of $I^n$ by using a group action, if only implicitly.

The point is, even without doing any calculations or thinking very hard, the number of symmetries of $\Sigma^2_n$ is growing exponentially in $n$, and therefore the genus must be growing exponentially in $n$ as well—the orbifold makes this reasoning precise.

There’s a lot of stuff left to be discovered about the number of automorphisms of genus $g$ surfaces. For instance, it’s known that the $84(g-1)$ bound is attained for infinitely many genera, but there are also infinitely many genera for which it is not attained. Let $N(g)$ be the maximal order of the automorphism group of a genus $g$ surface; Maclachlan and Accola proved (in 1968) that $N(g) \geq 8(g+1)$. This bound is sharp, too. There’s a beautiful paper

Belolipetsky, Mikhail and Jones, Gareth A.. A bound for the number of automorphisms of an arithmetic Riemann surface. Math. Proc. Cambridge Philos. Soc. 2005. 289–299. MR.

working out what happens in the arithmetic and non-arithmetic case. Anyway, what is known about the set of $g$ for which $84(g-1)$ is attained? What is the asymptotic density of this set?

At a recent Pizza Seminar, Matt Day gave a lovely talk explaining why it isn’t possible to classify 4-manifolds.

An algorithm for deciding whether two closed 4-manifolds are homeomorphic gives an algorithm for deciding whether a closed 4-manifold is simply connected, and therefore (since every finitely presented group is the fundamental group of a 4-manifold), and algorithm for deciding when a group is trivial. Here’s the reduction: we are given a 4-manifold $M$, and we compute its signature $\sigma(M)$. By Freedman, there are no more than two closed simply connected 4-manifolds, $M_1$ and $M_2$, having the same signature as $M$; we construct $M_1$ and $M_2$, and we use the homeomorphism decision procedure to test if $M \cong M_1$ or $M \cong M_2$.

Since there is no algorithm for deciding when a group is trivial, there can not be an algorithm for deciding when two closed 4-manifolds are homeomorphic.

Here is a paper discussing some of these issues:

Chernavsky, A. V. and Leksine, V. P.. Unrecognizability of manifolds. Ann. Pure Appl. Logic 2006. 325–335. MR.

In particular, that paper discusses Novikov’s proof that $S^n$ cannot be recognized for $n \geq 5$.

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.

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 $\tilde{G}$.

Jenn is a fabulous program for visualizing the Cayley graphs of finite Coxeter groups. The pictures are absolutely beautiful (oh, symmetry!).

Today in Geometry/Topology seminar, quasihomomorphisms $\Z \to \Z$ were discussed, i.e., the set of maps $f : \Z \to \Z$ such that $| f(a+b) - f(a) - f(b) |$ is uniformly bounded, modulo the relation of being a bounded distance apart. These come up when defining rotation and translation numbers, for instance.

Anyway, Uri Bader mentioned that these quasihomomorphisms form a field, isomorphic to $\R$, under pointwise addition and composition. I hadn’t realized that this is a general construction. Given a finitely generated group (with fixed generating set, so we have the word metric $d$ on the group), I can define a quasihomomorphism $f : G \to G$ by demanding $d(f(ab),f(a)f(b))$ be uniformly bounded, and where two quasihomomorphisms $f, g$ are equivalent if $d(f(a),g(a))$ is uniformly bounded. Let’s call the resulting object $\hat{G}$ for now.

What can be said about $\hat{G}$? For instance, what is $\hat{F_2}$?

There are some questions about outer space that I would like to be able to answer. Some nice survey articles look to be:

Bestvina, Mladen. The topology of $\rm Out(F_n)$. 2002. 373–384. MR.

and also:

Vogtmann, Karen. Automorphisms of free groups and outer space. Geom. Dedicata 2002. 1–31. MR.

Here is a ridiculously simple question I have wondered about: given $A, B \subset F_n$, say with $[F_n : A] = [F_n : B]$, how can I tell if $A$ and $B$ are conjugate? I suspect I’m being stupid here.

In light of my recent comments on LINCOS and communiating with extraterristrials, I found an article:

Ruelle, David. Conversations on mathematics with a visitor from outer space. 2000. 251–259. MR.

Putnam also makes use of the idea of mathematicians from other planets, to more philosophical ends.

By the Gauss-Bonnet theorem, the volume of a hyperbolic 4-manifold is proportional to its Euler characteristic. There are examples, constructed explicitly in

Ratcliffe, John G. and Tschantz, Steven T.. The volume spectrum of hyperbolic 4-manifolds. Experiment. Math. 2000. 101–125. MR.

of hyperbolic 4-manifolds with every positive integer as their Euler characteristic. These examples are non-compact (with five or six cusps, I believe). But

Ratcliffe, John G.. The geometry of hyperbolic manifolds of dimension at least 4. 2006. 269–286. MR.

observes that there are restrictions on the Euler characteristic that a closed hyperbolic 4-manifold may possess. In particular, it is shown in

Chern, Shiing-shen. On curvature and characteristic classes of a Riemann manifold. Abh. Math. Sem. Univ. Hamburg 1955. 117–126. MR.

that the Pontrjagin numbers of a hyperbolic manifold $M$ vanish. But the signature $\sigma(M)$ is a rational linear combination of those Pontrjagin numbers, so $\sigma(M) = 0$. And by Poincare duality, $\chi(M) \equiv \sigma(M) \pmod 2$, so $\chi(M)$ is even. A natural question to ask is: does there exist a hyperbolic 4-manifold $M$ with $\chi(M) = 2$? Now if such an $M$ also had $H_1(M) \neq 0$, we would know the volume spectrum of closed hyperbolic 4-manifolds.

This certainly seems to parallel the case for 2-manifolds: all negative integers are the Euler characteristic of a hyperbolic 2-manifold, and all even negative integers are the Euler characteristic of a closed hyperbolic 2-manifold.

The vanishing of Pontrjagin numbers for hyperbolic manifolds also holds for pinched negative curvature under some conditions:

Ratcliffe, John G. and Tschantz, Steven T.. The volume spectrum of hyperbolic 4-manifolds. Experiment. Math. 2000. 101–125. MR.

It is also a fact that the Stiefel-Whitney numbers vanish for a closed hyperbolic manifold (and the vanishing of the top Stiefel-Whitney class is the same thing as having even Euler characteristic).

Gromov, M. and Piatetski-Shapiro, I.. Nonarithmetic groups in Lobachevsky spaces. Inst. Hautes Études Sci. Publ. Math. 1988. 93–103. MR.

Vinberg, …

Margulis’ amazing arithmeticity theorem says that irreducible lattices in Lie groups of high ($>2$) rank are arithmetic. But ${\rm SO}(n,1)$ has rank 1, so a question is how to produce non-arithmetic lattices. For ${\rm SO}(3,1)$, there are non-arithmetic lattices coming from hyperbolic knot complements.

G–P-S produces higher dimensional examples by taking two hyperbolic (arithmetic) manifolds, cutting along totally geodesic hypersurfaces, and gluing. Are there are examples of non-arithmetic hyperbolic manifolds without any totally geodesics hypersurfaces?

There are complements of $T^2$’s in $S^4$ which are hyperbolic, and maybe these would provide some examples.

From Recent Advances in the Langlands Program, quoted in This Week’s Finds:

First of all, it should be remarked that according to the Tannakian phylosophy, one can reconstruct a group from the category of its finite-dimensional representations, equipped with the structure of the tensor product.

I suppose one should think of this as the categorification of Pontrjagin duality?

For a long while, I had wondered how this goes; this Introduction to Tannaka Duality and Quantum Groups will probably answer my questions.

Here is a question that I haven’t been able to find very much about:

What are the finite subgroups of the rotation groups $SO(n)$?

For examples, I can take a Coxeter group, and choose elements corresponding to rotations (e.g., the subgroup generated by products of generators), but that’s not going to produce very many examples.

Here’s a very easy theorem.

Theorem. All closed orientable 3-manifolds are parallelizable. All closed orientable 3-manifolds are the boundary of a 4-manifold.

Proof: Let $M$ be an orientable $3$-manifold. Recall that the Wu class $v$ is the unique cohomology class such that $\langle v \cup x, [M] \rangle = \langle Sq(x), [M] \rangle$, and Wu’s theorem says that $w(M) = Sq(v)$. The up-shot is that Stiefel-Whitney classes are homotopy invariants, even though they are defined using the tangent bundle.

Since $M$ is orientable, we have $w_1(M) = 0$. Since $\dim M = 3$, the Steenrod squares $Sq^2$ and $Sq^3$ kill everything, so $v_2 = 0$ and $v_3 = 0$. By Wu’s theorem, $w_2(M) = Sq^1(v_1) + v_2 = 0$, and $w_3(M) = Sq^1(v_2) + v_3 = 0$. In other words, all the Stiefel-Whitney classes vanish. ∎

Orientability matters; after all, being orientable is the same thing as $w_1$ vanishing. For example, $RP^2 \times S^1$ is not parallelizable, since $w_1(RP^2 \times S^1) = w_1(RP^2) + w_1(S^1) \neq 0$.